Rd Sharma XII Vol 2 2020 Solutions for Class 12 Science Maths Chapter 3 Differential Equations are provided here with simple step-by-step explanations. These solutions for Differential Equations are extremely popular among Class 12 Science students for Maths Differential Equations Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma XII Vol 2 2020 Book of Class 12 Science Maths Chapter 3 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma XII Vol 2 2020 Solutions. All Rd Sharma XII Vol 2 2020 Solutions for class Class 12 Science Maths are prepared by experts and are 100% accurate.

Page No 21.106:

Question 1:

dydx+2y=e3x

Answer:

We have,dydx+2y=e3x           .....(1)Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=2Q=e3x I.F.=eP dx          =e2 dx         = e2xMultiplying both sides of (1) by e2x, we gete2x dydx+2y=e2xe3xe2xdydx+2e2xy=e5xIntegrating both sides with respect to x, we gety e2x=e5xdx+Cy e2x=e5x5+Cy=15e3x+Ce-2xHence, y=15e3x+Ce-2x is the required solution.

Page No 21.106:

Question 2:

4dydx+8y=5 e-3x

Answer:

We have,
4dydx+8y=5 e-3x

dydx+2y=54e-3x           .....(1)Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=2Q=54e-3x I.F.=eP dx          =e2 dx         = e2xMultiplying both sides of (1) by e2x, we gete2x dydx+2y=54e2xe-3xe2xdydx+2e2xy=54e-xIntegrating both sides with respect to x, we gety e2x=54e-x dx+Cy e2x=-54e-x+Cy=54e-3x+Ce-2xHence, y=54e-3x+Ce-2x is the required solution.

Page No 21.106:

Question 3:

dydx+2y=6 ex

Answer:

We have, dydx+2y=6ex           .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=2Q=6ex I.F.=eP dx          =e2 dx         = e2xMultiplying both sides of 1 by e2x, we gete2x dydx+2y=6e2xexe2xdydx+2e2xy=6e3xIntegrating both sides with respect to x, we gety e2x=6e3xdx+Cy e2x=6e3x3+Cy e2x=2e3x+CHence, y e2x=2e3x+C is the required solution.

Page No 21.106:

Question 4:

dydx+y=e-2x

Answer:

We have, dydx+y=e-2x           .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=1Q=e-2x  I.F.=eP dx          =e1 dx         = exMultiplying both sides of 1 by ex, we getex dydx+y=exe-2xexdydx+exy=e-xIntegrating both sides with respect to x, we gety ex=e-xdx+Cy ex=-e-x+Cy=-e-2x+Ce-xHence, y=-e-2x+Ce-x is the required solution.

Page No 21.106:

Question 5:

xdydx=x+y

Answer:

We have,xdydx=x+ydydx=1+1xy dydx-1xy=1          .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=-1x Q=1 I.F.=eP dx          =e-1x dx          =e-log x         =elog 1x         =1xMultiplying both sides of 1 by 1x, we get1x dydx-1xy=1x×11xdydx-1x2y=1xIntegrating both sides with respect to x, we gety1x=1x dx+Cyx=log x+CHence, yx=log x+C is the required solution.

Page No 21.106:

Question 6:

dydx+2y=4x

Answer:

We have,dydx+2y=4x           .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=2 Q=4x  I.F.=eP dx          =e2 dx         = e2xMultiplying both sides of 1 by e2x, we gete2x dydx+2y=e2x4x e2xdydx+2e2xy=e2x4x Integrating both sides with respect to x, we gety e2x=4x e2x dx+Cy e2x=4xI e2xII dx+Cy e2x=4xe2x dx-4ddxxe2x dxdx+Cy e2x=4xe2x2-4×12e2x dx+Cy e2x=2x e2x-4×14e2x+Cy e2x=2x e2x-e2x+Cy e2x=2x-1e2x+Cy=2x-1+Ce-2xHence, y=2x-1+Ce-2x is the required solution.

Page No 21.106:

Question 7:

xdydx+y=x ex

Answer:

We have,xdydx+y=x exdydx+1xy=ex        .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=1x Q=ex I.F.=eP dx          =e1x dx          =elog x         =x         Multiplying both sides of 1 by x, we getxdydx+1xy=x exxdydx+y=xexIntegrating both sides with respect to x, we getxy=x exdx+Cxy=xI exII dx+Cxy=xex dx-ddxxex dxdx+Cxy=x ex-ex+Cxy=x-1ex+Cy=x-1xex+CxHence, y=x-1xex+Cx is the required solution.

Page No 21.106:

Question 8:

dydx+4xx2+1y+1x2+12=0

Answer:

We have, dydx+4xx2+1y+1x2+12=0    dydx+4xx2+1y=-1x2+12       .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=4xx2+1 Q=-1x2+12 I.F.=eP dx          =e22xx2+1 dx          =e2log x2+1         =x2+12Multiplying both sides of 1 by x2+12, we getx2+12dydx+4xx2+1y=x2+12-1x2+12 x2+12dydx+4xx2+1y=-1Integrating both sides with respect to x, we getx2+12y=-dx+Cx2+12y=-x+CHence, x2+12y=-x+C  is the required solution.

Page No 21.106:

Question 9:

xdydx+y=x log x

Answer:

We have,
xdydx+y=x log x
Dividing both sides by x, we get
dydx+yx=log xComparing with dydx+Py=Q, we getP=1xQ=log xNow, I.F.=ePdx=e1xdx                         =elogx                         =xSo, the solution is given byy×I.F.=Q×I.F. dx+Cxy=x IIlog xI dx+Cxy=log xxdx-ddxlog xx dxdx+Cxy=x2 log x2-x2dx+Cxy=x2 log x2-x24+C4xy=2 x2log x-x2+K    where, K=2C

Page No 21.106:

Question 10:

xdydx-y=x-1 ex

Answer:

We have, xdydx-y=x-1exdydx-1xy=x-1xex         .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=-1x Q=x-1xex I.F.=eP dx          =e-1x dx          =e-log x         =1xMultiplying both sides of 1 by I.F.=1x, we get1x dydx-1xy=1xx-1xex 1xdydx-1x2y=x-1x2exIntegrating both sides with respect to x, we get1xy=1x-1x2ex  dx+C1xy=exx+Cy=ex+CxHence, y=ex+Cx is the required solution.

Page No 21.106:

Question 11:

dydx+yx=x3

Answer:

We have, dydx+yx=x3         .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=1xQ=x3 I.F.=eP dx          =e1x dx          =elog x         =x         Multiplying both sides of 1 by x, we getx dydx+1xy=x x3 xdydx+y=x4Integrating both sides with respect to x, we getxy=x4 dx+Cxy=x55+C5xy=x5+5C5xy=x5+K            where, K=5CHence, 5xy=x5+K is the required solution.

Page No 21.106:

Question 12:

dydx+y=sin x

Answer:

We have,dydx+y=sin x           .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=1Q=sin x  I.F.=eP dx          =e dx         = exMultiplying both sides of 1 by ex, we getex dydx+y=exsin xexdydx+exy=exsin x Integrating both sides with respect to x, we gety ex=exsin x dx+Cy ex=ex2sin x-cos x+Cy=Ce-x+12sin x-cos xHence, y=Ce-x+12sin x-cos x is the required solution.

Page No 21.106:

Question 13:

dydx+y=cos x

Answer:

We have,dydx+y=cos x           .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=1Q=cos x   I.F.=eP dx          =e dx         = exMultiplying both sides of (1) by ex, we getex dydx+y=excos x exdydx+exy=excos xIntegrating both sides with respect to x, we gety ex=excos x dx+Cy ex=12excos x+sin x+-sin x+cos x dx+Cyex=ex2cos x+sin x+Cy=12cos x+sin x+Ce-xHence, y=12cos x+sin x+Ce-x is the required solution.

Page No 21.106:

Question 14:

dydx+2y=sin x

Answer:

We have,dydx+2y=sin x           .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=2andQ=sin x   I.F.=eP dx          =e2 dx         = e2xMultiplying both sides of 1 by I.F.=e2x, we gete2x dydx+2y=e2xsin x e2xdydx+2e2xy=e2xsin xIntegrating both sides with respect to x, we gety e2x=e2xsin x  dx+Cy e2x=152e2x2sin x-cos x+e2x2 cos x+sin x dx+CPutting e2x2 sin x-cos x=t2e2x2sin x-cos x+e2x2 cos x+sin x dx=dty e2x=15dt+Cy e2x=t5+Cy e2x=e2x52sin x-cos x+Cy=152sin x-cos x+Ce-2xHence, y=152sin x-cos x+Ce-2x is the required solution.

Page No 21.106:

Question 15:

dydx = y tan x − 2 sin x

Answer:

We have, dydx=y tan x-2sin xdydx-y tan x=-2sin x           .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=-tan xQ=-2sin x   I.F.=eP dx          =e-tan x dx         = e-logsec x=cos xMultiplying both sides of 1 by cos x, we getcos x dydx-y tan x=-2sin x×cos xcos xdydx-ysin x=-sin 2x Integrating both sides with respect to x, we gety cos x=-sin 2x dx+Cycos x=cos 2x2+C2y cos x=cos 2x+2C2y cos x=cos 2x+K,      where k=2CHence, 2y cos x=cos 2x+K is the required solution.

Page No 21.106:

Question 16:

1+x2dydx+y=tan-1 x

Answer:

We have, 1+x2dydx+y=tan-1xdydx+y1+x2=tan-1x1+x2        .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=11+x2 Q=tan-1x1+x2 I.F.=eP dx          =e11+x2 dx          =etan-1xMultiplying both sides of 1 by etan-1x, we getetan-1x dydx+y1+x2=etan-1x tan-1x1+x2etan-1xdydx+etan-1xy1+x2=etan-1xtan-1x1+x2Integrating both sides with respect to x, we getetan-1xy=tan-1x×etan-1x1+x2 dx+Cetan-1xy=I+C        .....2Here, I=tan-1x×etan-1x1+x2 dxPutting tan-1 x=t, we get11+x2dx=dt I=t et dt     =tetdt-ddttetdtdt     =t et-et     =t-1et     =tan-1x-1etan-1xSubstituting the value of I in 2, we getetan-1xy=tan-1x-1etan-1x+Cy=tan-1x-1+Ce-tan-1xHence, y=tan-1x-1+Ce-tan-1x is the required solution.

Page No 21.106:

Question 17:

dydx + y tan x = cos x

Answer:

We have,dydx+y tan x=cos x           .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=tan xQ=cos x   I.F.=eP dx          =etan x dx         = elogsec x=sec xMultiplying both sides of 1 by sec x, we getsec xdydx+y tan x=cos x ×sec xsec xdydx+y sec x tan x=1Integrating both sides with respect to x, we gety sec x=dx+Cy sec x=x+CHence, y sec x=x+C  is the required solution.

Page No 21.106:

Question 18:

dydx + y cot x = x2 cot x + 2x

Answer:

We have,dydx+y cot x=x2cot x+2x           .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=cot xQ=x2cot x+2x  I.F.=eP dx          =ecot x dx         = elogsin x=sin xMultiplying both sides of 1 by sin x, we getsin xdydx+ycot x=sin xx2cot x+2xsin xdydx+ycos x=x2cos x+2x sin x Integrating both sides with respect to x, we gety sin x=x2Icos xIIdx+2x sin x dx+Cy sin x=x2cos xdx-ddxx2cos x dxdx+2xsin x  dx+Cy sin x=x2sin x-2xsin x dx+2xsin x dx+Cy sin x=x2sin x+CHence, y sin x=x2sin x+C is the required solution.

Page No 21.106:

Question 19:

dydx+y tan x=x2 cos2 x

Answer:

We have,
dydx+y tan x=x2 cos2 x
Comparing with dydx+Py=Q, we getP=tan x Q=x2 cos2 xNow,I.F.=etan x dx =elog sec x=sec xTherefore, solution is given byy×I.F.=x2 cos2 x×I.F. dx+Cy sec x=x2 cos x dx+Cy sec x=I+CWhere, I=x2IIcos x Idx+C I=x2cos x dx-ddxx2cos x dxdx I=x2sin x-2x sin x dx I=x2sin x-2xI sin xII dx I=x2sin x-2xsin x dx+2ddxxsin x dxdx I=x2sin x+2x cos x -2cos x dx I=x2sin x+2x cos x -2sin x I=x2sin x+2x cos x -2sin x y sec x=x2sin x+2x cos x-2sin x+Cy sec x=x2sin x+2x cos x-2sin x+C

Page No 21.106:

Question 20:

1+x2dydx+y=etan-1 x

Answer:

We have, 1+x2dydx+y=etan-1xdydx+y1+x2=etan-1 x1+x2        .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=11+x2Q=etan-1 x1+x2 I.F.=eP dx          =e11+x2 dx          =etan-1xMultiplying both sides of 1 by etan-1x, we getetan-1x dydx+y1+x2=etan-1xetan-1 x1+x2etan-1xdydx+y etan-1x1+x2=etan-1xetan-1x1+x2Integrating both sides with respect to x, we gety etan-1x=e2tan-1x1+x2 dx+Cy etan-1x=I+C        .....2Here,I=e2tan-1x1+x2 dxPutting tan-1 x=t, we get11+x2dx=dt I=e2t dt     =e2t2     =e2tan-1x2Putting the value of I in 2, we gety etan-1x=e2tan-1x2+C2y etan-1x=e2tan-1x+2C2y etan-1x=e2tan-1x+k, where k=2CHence, 2y etan-1x=e2tan-1x+k is the required solution.

Page No 21.106:

Question 21:

x dy = (2y + 2x4 + x2) dx

Answer:

We have, x dy=2y+2x4+x2dxdydx=2xy+2x3+xdydx-2xy=2x3+x        .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=-2xQ=2x3+x I.F.=eP dx          =e-2x dx          =e-2log x         =1x2Multiplying both sides of 1 by 1x2, we get1x2 dydx-2xy=1x2 2x3+x1x2dydx-2x3y=2x+1xIntegrating both sides with respect to x, we get1x2y=2x+1xdx+C1x2y=x2+log x+Cy=x4+x2log x+Cx2Hence, y=x4+x2log x+Cx2 is the required solution.

Page No 21.106:

Question 22:

1+y2+x-etan-1ydydx=0

Answer:

We have,1+y2+x-etan-1ydydx=0x-etan-1ydydx=-1+y2dydx=-1+y2x-etan-1ydxdy=-x-etan-1y1+y2dxdy+x1+y2=etan-1 y1+y2        .....1Clearly, it is a linear differential equation of the form dxdy+Px=QwhereP=11+y2Q=etan-1 y1+y2  I.F.=eP dy          =e11+y2 dy          =etan-1yMultiplying both sides of 1 by etan-1y, we getetan-1y dxdy+x1+y2=etan-1y etan-1 y1+y2etan-1ydxdy+x etan-1x1+y2=e2tan-1 y1+y2Integrating both sides with respect to y, we getx etan-1y=e2tan-1 y1+y2 dy+Cx etan-1y=I+C        .....2Here,I=e2tan-1 y1+y2 dyPutting tan-1 y=t, we get11+y2dy=dt I=e2t dt     =e2t2     =e2tan-1y2Putting the value of I in 2, we getx etan-1y=e2tan-1y2+C2x etan-1y=e2tan-1y+2C2x etan-1y=e2tan-1y+k     where k=2CHence, 2x etan-1y=e2tan-1y+k is the required solution.

Page No 21.106:

Question 23:

y2dxdy+x-1y=0

Answer:

We have,y2dxdy+x-1y=0y2dxdy+x=1y dxdy+1y2x=1y3         .....1Clearly, it is a linear differential equation of the form dxdy+Px=QwhereP=1y2Q=1y3 I.F.=eP dy          =e1y2dy         = e-1yMultiplying both sides of 1 by e-1y, we get e-1ydxdy+x1y2= e-1y1y3 e-1ydxdy+x1y2 e-1y= e-1y 1y3Integrating both sides with respect to y, we getx e-1y= e-1y1y3dy+Cx e-1y=I+C         .....2whereI= e-1y1y3dyPutting t=1y, we getdt=-1y2dy I=- t Ie-tIIdt    =-te-tdt+ddtte-tdtdt    =te-t+e-t    =t+1e-t    =1y+1e-1yPutting the value of I in 2, we getx e-1y=1y+1e-1y+C x=y+1y+Ce1yHence, x=y+1y+Ce1y is the required solution.

Page No 21.106:

Question 24:

2x-10y3dydx+y=0

Answer:

We have, 2x-10y3dydx+y=02x-10y3dydx=-y dxdy=-1y2x-10y3  dxdy+2yx=10y2        .....1Clearly, it is a linear differential equation of the form dxdy+Px=QwhereP=2yQ=10y2 I.F.=eP dy          =e2ydy         = e2log y=y2Multiplying both sides of 1 by y2, we get  y2dxdy+2yx=  y2×10y2 y2dxdy+2yxy2=10y4Integrating both sides with respect to y, we getx y2= 10y4 dy+Cxy2=2y5+Cx=2y3+Cy-2Hence, x=2y3+Cy-2 is the required solution.

Page No 21.106:

Question 25:

(x + tan y) dy = sin 2y dx

Answer:

We have,x+tan ydy=sin 2y dxdxdy=x cosec 2y+12sec2y      dxdy-x cosec 2y=12sec2y         .....1Clearly, it is a linear differential equation of the form dxdy+Px=QwhereP=-cosec 2yQ=12sec2y I.F.=eP dy          =e-cosec 2y dy         = e-12logtan y=1tan yMultiplying both sides of 1 by 1tan y, we get 1tan ydxdy-x cosec 2y=12 1tan y×sec2y 1tan ydxdy-x cosec 2y1tan y=12 1tan y×sec2y Integrating both sides with respect to y, we get1tan yx= 12 1tan y×sec2y dy+Cxtan y=I+C         .....2where I=12 1tan y×sec2y dyPutting t=tan y, we getdt=sec2 y dy I=121t×dt    =t    =tan yPutting the value of I in 2, we getxtan y=tan y+C x=tan y+Ctan yHence, x=tan y+Ctan y is the required solution.

Page No 21.106:

Question 26:

dx + xdy = ey sec2 y dy

Answer:

We have, dx+x dy=e-ysec2y dy dx=e-ysec2y dy-x dy dxdy=e-ysec2y-x dxdy+x=e-ysec2y     .....1Clearly, it is a linear differential equation of the form dxdy+Px=QwhereP=1Q=e-ysec2y I.F.=eP dy          =edy          =eyMultiplying both sides of 1 by ey, we getey dxdy+x=ey e-ysec2yeydxdy+x ey=sec2yIntegrating both sides with respect to y, we getx ey=sec2y dy+Cx ey=tan y+CHence, x ey=tan y+C is the required solution.

Page No 21.106:

Question 27:

dydx = y tan x − 2 sin x

Answer:

We have,dydx=y tan x-2sin xdydx-y tan x=-2sin x           .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=-tan xQ=-2sin x I.F.=eP dx          =e-tan x dx         = e-logsec x=cos xMultiplying both sides of 1 by cos x, we getcos x dydx-ytan x=-2sin x ×cos xcos xdydx+ysin x=-sin 2xIntegrating both sides with respect to x, we gety cos x=-sin 2x dx+Cy cos x=cos 2x2+Cy cos x=1-2sin2x2+Cy cos x=-sin2x+12+Cy cos x=-sin2x+K     where k=12+Cy=sec x-sin2x+KHence, y=sec x-sin2x+K is the required solution.

Page No 21.106:

Question 28:

dydx + y cos x = sin x cos x

Answer:

We have, dydx+y cos x=sin x cos x           .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=cos xQ=sin x cos x I.F.=eP dx          =ecosx dx         = esin xMultiplying both sides of 1 by esin x, we get esin xdydx+y cos x= esin xsin x cos x esin xdydx+esin xy cos x= esin x sin xcos x Integrating both sides with respect to x, we gety esin x= esin x sin xcos x dx+Cy esin x=I+C           .....2whereI=esin x sin x cos x dxPutting t=sin x, we getdt=cos x dx I=etII tI dt     =tetdt-ddttetdtdt     =t et-et     =ett-1     = esin xsin x-1Putting the value of I in 2, we gety esin x= esin xsin x-1+Cy=sin x-1+Ce-sin x Hence, y=sin x-1+Ce-sin x is the required solution.

Page No 21.106:

Question 29:

Solve the following differential equations:

1+x2dydx-2xy=x2+2x2+1        [CBSE 2005]

Answer:

Given, 1+x2dydx-2xy=x2+2x2+1   
dydx-2x1+x2y=x2+2
This is a linear differential equation.
I.F.=e-2x1+x2dx=11+x2
y11+x2=x2+2x2+1dxy11+x2=1+11+x2dxy11+x2=x+tan-1x+Cy=x+tan-1x+C1+x2

Page No 21.106:

Question 30:

sin xdydx+y cos x=2 sin2 x cos x

Answer:

We have,sin xdydx+y cos x=2 sin2 x cos xdydx+y cot x=2sin x cos x           .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=cot xQ=2sin x cos x I.F.=eP dx          =ecot x dx         = elogsin x=sin xMultiplying both sides of 1 by sin x, we getsin xdydx+y cot x=sin x×2sin xcos xsin xdydx+y cosx=2sin2 xcos xIntegrating both sides with respect to x, we gety sin x=2sin2 x cos x dx+C           .....2Putting sin x=tcos x dx=dtTherefore, 2 becomesy sin x=2t2 dt+Cy sin x=23t3+Cy sin x=23sin3x+CHence, y sin x=23sin3x+C is the required solution.

Page No 21.106:

Question 31:

x2-1dydx+2x+2y=2x+1

Answer:

We have,x2-1dydx+2x+2y=2x+1dydx+2x+2x2-1y=2x+1x2-1  dydx+2x+2x2-1y=2x-1      .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=2x+2x2-1Q=2x-1 I.F.=eP dx          =e2x+2x2-1 dx          =e2xx2-1+4 1x2-1dx          =elogx2-1+4×12logx-1x+1          =elogx2-1×x-12x+12          =elogx-13x+1          =x-13x+1Multiplying both sides of 1 by x-13x+1, we getx-13x+1 dydx+2x+2x2-1y=x-13x+1×2x-1x-13x+1dydx-2x+2x-12x+12y=2x-12x+1Integrating both sides with respect to x, we getx-13x+1y=2x-12x+1 dx+Cx-13x+1y=2x+12-4xx+1 dx+Cx-13x+1y=2x+1-4xx+1 dx+Cx-13x+1y=2x+1-4x+1-1x+1 dx+Cx-13x+1y=2x+1-4+4x+1 dx+Cx-13x+1y=2x-6+8x+1 dx+Cx-13x+1y=x2-6x+8log x+1+Cy=x+1x-13x2-6x+8log x+1+CHence, y=x+1x-13x2-6x+8logx+1 +C is the required solution.

Page No 21.106:

Question 32:

xdydx+2y=x cos x

Answer:

We have, xdydx+2y=x cos xdydx+2xy=cos xComparing with dydx+Py=Q, we getP=2xQ=cos xNow, I.F.=e2xdx =e2log x=x2Solution is given by,y×I.F.=cos x×I.F. dx+Cyx2=x2 cos x dx+Cx2y=I+C          .....1Where,I=x2IIcos x Idx+C I=x2cos x dx-ddxx2cos x dxdx I=x2sin x-2x sin x dx I=x2sin x-2xI sin xII dx I=x2sin x-2xsin x dx+2ddxxsin x dxdx I=x2sin x+2x cos x -2cos x dx I=x2sin x+2x cos x -2sin x I=x2sin x+2x cos x -2sin xTherefore 1 becomesx2y=x2sin x+2x cos x-2sin x+C

Page No 21.106:

Question 33:

dydx-y=xex

Answer:

We have, dydx-y=x ex        .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=-1 Q=ex I.F.=eP dx          =e-dx          =e-xMultiplying both sides of 1 by e-x, we gete-xdydx-y=x exe-xe-xdydx-e-xy=xIntegrating both sides with respect to x, we gete-xy=x dx+Ce-xy=x22+Cy=x22+Ce xHence, y=x22+Ce x is the required solution.

Page No 21.106:

Question 34:

dydx+2y=xe4x

Answer:

We have, dydx+2y=xe4x           .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=2Q=xe4x I.F.=eP dx          =e2dx         = e2xMultiplying both sides of 1 by e2x, we get e2xdydx+2y= e2x×xe4x e2xdydx+ 2e2xy= xe6xIntegrating both sides with respect to x, we get e2xy=e6xII xI dx+Ce2xy=xe6xdx-ddxxe6xdxdx+Ce2xy=xe6x6-e6x36+Cy=xe4x6-e4x36+Ce-2xHence, y=xe4x6-e4x36+Ce-2x is the required solution.

Page No 21.106:

Question 35:

Solve the differential equation x+2y2dydx=y, given that when x = 2, y = 1.

Answer:

 We have, x+2y2dydx=ydxdy=1yx+2y2  dxdy-1yx=2y        .....1Clearly, it is a linear differential equation of the form dxdy+Px=QwhereP=-1yQ=2y I.F.=eP dy          =e-1ydy         = e-log y=1yMultiplying both sides of 1 by 1y, we get1ydxdy-1yx= 1y×2y1ydxdy-1y2x=2Integrating both sides with respect to y, we getx1y= 2dy+Cx1y=2y+Cx=2y2+Cy        .....2Now,y=1 at x=2 2=2+CC=0Putting the value of C in 2, we getx=2y2Hence, x=2y2 is the required solution.



Page No 21.107:

Question 36:

Find one-parameter families of solution curves of the following differential equations:
(or Solve the following differential equations)
(i) dydx+3y=emx, m is a given real number

(ii) dydx-y=cos 2x

(iii) xdydx-y=x+1e-x

(iv) xdydx+y=x4

(v) x log xdydx+y=log x

(vi) dydx-2xy1+x2=x2+2

(vii) dydx+y cos x=esin x cos x

(viii) x+ydydx=1

(ix) dydxcos2 x=tan x-y

(x) e-y sec2 y dy=dx+x dy

(xi) x log xdydx+y=2 log x

(xii) xdydx+2y=x2 log x

Answer:

i We have,dydx+3y=emx           .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=3 Q=emx I.F.=eP dx          =e3 dx         = e3xMultiplying both sides of (1) by e3x, we get e3x dydx+3y=e3xemx  e3xdydx+3 e3xy=em+3xIntegrating both sides with respect to x, we getye3x=em+3x dx+C    when m+30  ye3x=em+3xm+3+Cy=emxm+3+Ce-3x ye3x=e0×x dx+C    when m+3=0 ye3x=dx+Cye3x=x+Cy=x+Ce-3xHence, y=emxm+3+Ce-3x, where m+30 and y=x+Ce-3x, where m+3=0  are required solutions.


ii We have,dydx-y=cos 2x           .....(1)Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=-1Q=cos 2x I.F.=eP dx          =e- dx         = e-xMultiplying both sides of (1) by e-x, we gete-x dydx-y=e-xcos 2x e-xdydx-e-xy=e-xcos 2xIntegrating both sides with respect to x, we gety e-x=e-xcos 2x  dx+C y e-x=I+C           .....(2)Where,I=e-xcos 2x  dx           .....(3)I=12e-xsin 2x-12-e-xsin 2x dxI=12e-xsin 2x+12e-xsin 2x dxI=12e-xsin 2x-14e-xcos 2x-12×12-e-x×-cos 2x dxI=12e-xsin 2x-14e-xcos 2x-14e-xcos 2x dxI=12e-xsin 2x-14e-xcos 2x-14I       From 354I=12e-xsin 2x-14e-xcos 2x5I=2e-xsin 2x-e-xcos 2xI=e-x52sin 2x-cos 2x           .....(4)From (2) and (4) we getye-x=e-x52sin 2x-cos 2x+Cy=152sin 2x-cos 2x+CexHence, y=152sin 2x-cos 2x+Cex is the required solution.


iii We have, xdydx-y=x+1e-xdydx-1xy=x+1xe-x         .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=-1xQ=x+1xe-x I.F.=eP dx          =e-1x dx          =e-log x         =1xMultiplying both sides of 1 by 1x, we get1x dydx-1xy=1xx+1xe-x 1xdydx-1x2y=x+1x2e-xIntegrating both sides with respect to x, we get1xy=1x+1x2e-xdx+C        .....2Putting 1xe-x=t-1xe-x-1x2e-xdx=dt1x+1x2e-xdx=-dtTherefore 2 becomes1xy=-dt+C1xy=-t+C1xy=-1xe-x+Cy=-e-x+CxHence, y=-e-x+Cx is the required solution.


iv We have, xdydx+y=x4dydx+1xy=x3         .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=1x Q=x3 I.F.=eP dx          =e1x dx          =elog x         =xMultiplying both sides of 1 by x, we getx dydx+1xy=x.x3 xdydx+y=x4 Integrating both sides with respect to x, we getxy=x4 dx+Cxy=x55+Cy=x45+CxHence, y=x45+Cx is the required solution.


v We have,x log xdydx+y=log xDividing both sides by x log x, we getdydx+yx log x= log xx log xdydx+yx log x=1 xdydx+1x log xy=1 xComparing with dydx+Py=Q, we getP=1x log x Q=1 xNow,I.F.=ePdx=e1x log xdx                         =eloglog x                         =log xSo, the solution is given byy×I.F.=Q×I.F. dx+Cy log x=1 x×log x dx+Cy log x=log x22+Cy=12log x+Clog x


vi We have, dydx-2xy1+x2=x2+2     .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=-2x1+x2 Q=x2+2 I.F.=eP dx          =e-2x1+x2 dx          =e-log1+x2          =11+x2Multiplying both sides of 1 by 11+x2, we get11+x2 dydx-2xy1+x2=11+x2x2+211+x2dydx-2xy1+x22=x2+2x2+1Integrating both sides with respect to x, we get11+x2y=x2+2x2+1 dx+C11+x2y=x2+1+1x2+1 dx+C11+x2y=dx+1x2+1 dx+C11+x2y=x+tan-1x +Cy=1+x2x+tan-1x +CHence, y=1+x2x+tan-1x +C is the required solution.


vii We have,dydx+y cos x=esin x cos x        .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=cos x Q=esin x cos x I.F.=eP dx          =ecos x dx          =esin xMultiplying both sides of 1 by esin x, we getesin x dydx+y cos x=esin x×esin x cos xesin xdydx+yesin xcos x=e2sin x cos xIntegrating both sides with respect to x, we getesin xy=e2sin x cos x dx+Cesin xy=I+C       .....2Where,I=e2sin x cos x dxPutting t=sinx, we getdt=cos x dx I=e2t dt     =e2t2     =e2sin x2Putting the value of I in 2, we getesin xy=e2sin x2+Cy=esin x2+Ce-sin xHence, y=esin x2+Ce-sin x is the required solution.


viii We have,x+ydydx=1dydx=1x+ydxdy=x+ydxdy-x=y       .....1Clearly, it is a linear differential equation of the form dxdy+Px=QwhereP=-1Q=y I.F.=eP dy          =e-1 dy          =e-yMultiplying both sides of (1) by e-y, we gete-y dxdy-x=e-yye-ydxdy-e-yx=e-yyIntegrating both sides with respect to y, we gete-yx=y Ie-yII dy+Ce-yx=ye-y dy-ddyye-y dydy+Ce-yx=-ye-y -e-y +Ce-yx+ye-y +e-y =Cx+y+1e-y =Cx+y+1=CeyHence, x+y+1=Cey is the required solution.


ix We have,dydxcos2x=tan x-ydydx+1cos2 xy=tan x sec2 xdydx+y sec2 x=tan x sec2 x        .....1Clearly, it is a linear differential equation of the form dydx+Py=QwhereP=sec2 xQ=tan x sec2 x I.F.=eP dx          =esec2 x dx          =etan xMultiplying both sides of 1 by etan x, we getetan x dydx+y sec2 x=etan xtan x sec2 xetan xdydx+yetan xsec2 x=etan xtan x sec2 xIntegrating both sides with respect to x, we getetan xy=etan xtan x sec2 x dx+Cetan xy=I+C        .....2Where,I=etan xtan x sec2 x dxPutting t= tan x, we getdt=sec2 x dx I=tI etII dt    =tet dt-ddttet dtdt    =t et-et    =t-1et    =tan x-1etan xPutting the value of I in 2, we getetan xy=tan x-1etan x+Cy=tan x-1+Ce-tan xHence, y=tan x-1+Ce-tan x is the required solution.


x We have,e-ysec2y dy=dx+x dy dx=e-ysec2y dy-x dy dxdy=e-ysec2y-x dxdy+x=e-ysec2y     .....1Clearly, it is a linear differential equation of the form dxdy+Px=QwhereP=1Q=e-ysec2y I.F.=eP dy          =edy          =eyMultiplying both sides of 1 by ey, we geteydxdy+x=eye-ysec2yeydxdy+eyx=sec2yIntegrating both sides with respect to y, we geteyx=sec2y dy+Ceyx=tan y+Cx=tan y+Ce-yHence, x=tan y+Ce-y is the required solution.


xi We have,x log xdydx+y=2log xDividing both sides by x log x, we getdydx+yx log x=2 log xx log xdydx+yx log x=2 xdydx+1x log xy=2 xComparing with dydx+Py=Q, we getP=1x log xQ=2 xNow, I.F.=ePdx=e1x log xdx                         =eloglog x                         =log xSo, the solution is given by y×I.F.=Q×I.F. dx+Cy log x=21 x×log x dx+CPutting log x=t1xdx=dty log x=2t dt+Cy log x=2t22+Cy log x=t2+Cy log x=log x2+C    log x=ty=log x+Clog x


xii We have,xdydx+2y=x2log xDividing both sides by x, we getdydx+2yx=x log xComparing with dydx+Py=Q, we getP=2xQ=x log xNow, I.F.=ePdx=e2xdx                         =e2logx                         =x2So, the solution is given byy×I.F.=Q×I.F. dx+Cx2y=x3IIlog xI dx+Cx2y=log xx3 dx-ddxlog xx3 dxdx+Cx2y=x4log x4-x34dx+Cx2y=x4log x4-x416+Cy=x2log x4-x216+Cx2y=x2164log x-1+Cx2

Page No 21.107:

Question 37:

Solve each of the following initial value problems:
(i) y'+y=ex, y0=12

(ii) xdydx-y=log x, y1=0

(iii) dydx+2y=e-2x sin x, y0=0

(iv) xdydx-y=x+1e-x, y1=0

(v) 1+y2 dx+x-e-tan-1y dx=0, y0=0

(vi) dydx+y tan x=2x+x2 tan x, y0=1

(vii) xdydx+y=x cos x+sin x, yπ2=1

(viii) dydx+y cot x=4x cosec x, yπ2=0

(ix) dydx+2y tan x=sin x; y=0 when x=π3

(x) dydx-3y cot x=sin 2x; y=2 when x=π2
(xi) dydx+ycotx=2cosx, yπ2=0     
(xii) dy=cosx2-ycosecxdx
(xiii) tanxdydx=2xtanx+x2-y;tanx0 given that y = 0 when x=π2.

Answer:

i We have,y'+y=exdydx+y=ex           .....1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=1 and Q=ex I.F.=eP dx          =e1 dx         = exMultiplying both sides of 1 by I.F.=ex, we getex dydx+y=exexexdydx+exy=e2xIntegrating both sides with respect to x, we gety ex=e2x dx+Cy ex=e2x2+C           .....2Now, y0=12 12 e0=e02+CC=0Putting the value of C in 2, we getyex=e2x2ex=ex2Hence, y=ex2  is the required solution.


ii We have,xdydx-y=log xdydx-yx=log xx           .....1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=-1x and Q=log xx I.F.=eP dx          =e-1x dx          =e-log x          =1xMultiplying both sides of 1 by I.F.=1x, we get1x dydx-1xy=1x×logxx1xdydx-1x2y=logxx2Integrating both sides with respect to x, we gety1x=1x2II×logxI dx+Cyx=log x1x2dx-ddxlog x1x2dxdx+Cyx=-log xx+1x2dx+Cyx=-log xx-1x+Cy=-log x-1+Cx           .....2Now, y1=0 0=-0-1+C1C=1Putting the value of C in 2, we gety=-log x-1+xy=x-1-log xHence, y=x-1-log x is the required solution.


iii We have, dydx+2y=e-2xsin x           .....1Clearly, it is a linear differential equation of the formdydx+Py=Qwhere P=2 and Q=e-2xsin x I.F.=eP dx          =e2 dx         = e2xMultiplying both sides of 1 by I.F.=e2x, we gete2x dydx+2y=e2xe-2xsin xe2x dydx+2y=sin xIntegrating both sides with respect to x, we getye2x=sin x dx+Cye2x=-cos x+C           .....2Now,y0=0 0×e0=-cos 0+CC=1Putting the value of C in 2, we getye2x=-cos x+1ye2x=1-cos xHence, ye2x=1-cos x is the required solution.


iv We have,xdydx-y=x+1e-xdydx-1xy=x+1xe-x         .....1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=-1x and Q=x+1xe-x I.F.=eP dx          =e-1x dx          =e-log x         =1xMultiplying both sides of 1 by I.F.=1x, we get1x dydx-1xy=1xx+1xe-x 1x dydx-1xy=x+1x2e-xIntegrating both sides with respect to x, we get1xy=1x+1x2e-x dx+CPutting 1xe-x=t-1xe-x-1x2e-xdx=dt1x+1x2e-x dx=-dt1xy=-dt+Cyx=-t+Cyx=-e-xx+Cy=-e-x+Cx         .....2Now, y1=0 0=-e-1+CC=e-1Putting the value of C in 2, we gety=-e-x+xe-1y=xe-1-e-xHence, y=xe-1-e-x is the required solution.


v We have,1+y2dx+x-e-tan-1ydy=0x-e-tan-1ydydx=-1+y21+y2dxdy=-x-e-tan-1ydxdy+x1+y2=e-tan-1 y1+y2        .....1Clearly, it is a linear differential equation of the form dxdy+Px=Qwhere P=11+y2 and Q=e-tan-1 y1+y2 I.F.=eP dy          =e11+y2 dy          =etan-1yMultiplying both sides of 1 by I.F.=etan-1y, we getetan-1y dxdy+x1+y2=etan-1ye-tan-1 y1+y2etan-1y dxdy+x1+y2=11+y2Integrating both sides with respect to y, we getetan-1yx=11+y2 dy+Cxetan-1y=tan-1y+C        .....2Now, y0=0 0×e0=0+CC=0Putting the value of C in 2, we getxetan-1y=tan-1y+0xetan-1y=tan-1yHence, xetan-1y=tan-1y  is the required solution.


vi We have,dydx+y tan x=2x+x2tan x      .....1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=tan x and Q=x2cot x+2x I.F.=eP dx          =etan x dx         = elogsec x=sec xMultiplying both sides of 1 by I.F.=sec x, we getsec xdydx+ytan x=sec xx2tan x+2xsec xdydx+ytan x=x2tan x sec x+2x sec xIntegrating both sides with respect to x, we getysec x=x2tan xsec x dx+2xII sec xI dx  +Cy sec x=x2tan x sec x dx+2sec xx dx-2ddxsec xx dxdx+Cy sec x=x2tan x sec x dx+x2sec x-x2tan x sec x dx+Cy sec x=x2sec x+Cy=x2+Ccos x      .....2Now,y0=1 1=0+Ccos 0C=1Putting the value of C in 2, we gety=x2+cos xHence, y=x2+cos x is the required solution.


vii We have,xdydx+y=x cos x+sin xdydx+1xy=cos x+sin xx          .....1Clearly, it is a linear differential equation of the formdydx+Py=Qwhere P=tan x and Q=x2cot x+2x I.F.=eP dx          =e1xdx         = elog x=xMultiplying both sides of (1) by I.F.=x, we getxdydx+1xy=xcos x+sin xxxdydx+1xy=x cos x+sin xIntegrating both sides with respect to x, we getxy=x cos x dx+sin x dx+Cxy=x sin x-1sin xdx-cos x+Cxy=x sin x+cos x-cos x+Cxy=x sin x+C          .....2Now, yπ2=1 1×π2=π2sinπ2+CC=0Putting the value of C in 2, we getxy=x sin xy=sin xHence, y=sin x is the required solution.


viii We have, dydx+y cot x=4x cosec x           .....1Clearly, it is a linear differential equation of the formdydx+Py=Qwhere P=cot x and Q=4x cosec x  I.F.=eP dx          =ecot x dx         = elogsin x=sin xMultiplying both sides of 1 by I.F.=sin x, we getsin xdydx+y cot x=sin x4x cosec xsin xdydx+y cot x=4xIntegrating both sides with respect to x, we gety sin x=4x dx+Cy sin x=2x2+C           .....2Now,yπ2=0 0×sinπ2=2π22+CC=-π22Putting the value of C in 2, we gety sin x=2x2-π22Hence, y sin x=2x2-π22 is the required solution.


ix We have,dydx+2y tan x=sin x           .....1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=2tan x and Q=sin x I.F.=eP dx          =e2tan x dx         = e2logsec x=sec2xMultiplying both sides of (1) by I.F.=sec2 x, we getsec2 x dydx+2y tan x=sec2 x×sin xsec2 x dydx+2y tan x=tan x sec xIntegrating both sides with respect to x, we gety sec2 x=tan x sec x dx+Cy sec2 x=sec x+C           .....2Now, yπ3=0 0secπ32=secπ3+CC=-2Putting the value of C in 2, we gety sec2 x=sec x-2y=cos x-2cos2 xHence, y=cos x-2cos2 x  is the required solution.


x We have, dydx-3y cot x=sin 2x           .....1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=-3cot x and Q=sin 2x I.F.=eP dx          =e-3cot x dx         = e-3logsin x=cosec3xMultiplying both sides of 1 by I.F.=cosec3x, we getcosec3xdydx-3y cot x=sin 2xcosec3xcosec3xdydx-3y cot x=2cot x cosec xIntegrating both sides with respect to x, we gety cosec3x=2cot x cosec x dx+C ycosec3x=-2cosec x+Cy=-2sin2x+Csin3x           .....2Now, yπ2=2 2=-2sin2π2+Csin3 π2C=4Putting the value of C in 2, we gety=-2sin2x+4sin3xy=4sin3x-2sin2xHence, y=4sin3x-2sin2x is the required solution.

xi dydx+ycotx=2cosx, yπ2=0     dydx+ycotx=2cosx       ....1    Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=cotx and Q=2cosx I.F.=eP dx          =ecotx dx         = elogsinx         =sinxMultiplying both sides of 1 by I.F.=sinx, we getsinxdydx+ycotx=2sinxcosxsinxdydx+ycosx=sin2xIntegrating both sides with respect to x, we getysinx=sin2x dx+Cysinx=-cos2x2+C           .....2Now, yπ2=0  0×sinπ2=-cosπ2+CC=-12Putting the value of C in 2, we getysinx=-cos2x2-122ysinx=-1+cos2x2ysinx=-2cos2xy=-cotxcosxHence, y=-cotxcosx  is the required solution.


xiidy=cosx2-ycosecxdxdydx=2cosx-ycotx    dydx+ycotx=2cosx       ....1    Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=cotx and Q=2cosx I.F.=eP dx          =ecotx dx         = elogsinx         =sinxMultiplying both sides of 1 by I.F.=sinx, we getsinxdydx+ycotx=2sinxcosxsinxdydx+ycosx=sin2xIntegrating both sides with respect to x, we getysinx=sin2x dx+Cysinx=-cos2x2+C     Hence, ysinx=-cos2x2+C  is the required solution.

(xiii) 
tanxdydx=2xtanx+x2-ydydx+1tanxy=2xtanx+x2tanxdydx+cotxy=2x+x2cotx
This is a linear differential equation of the form dydx+Py=Q.

Integrating factor, I.F. = ePdx=ecotxdx=elogsinx=sinx

The solution of the given differential equation is given by

y×I.F.=Q×I.F.dx+Cy×sinx=2x+x2cotxsinxdx+Cysinx=2xsinxdx+x2cosxdx+Cysinx=2xsinxdx+x2cosxdx-ddxx2×cosxdxdx+C
ysinx=2xsinxdx+x2sinx-2xsinxdx+Cysinx=x2sinx+Cy=x2+cosecx×C        .....1 
It is given that, y = 0 when x=π2.

0=π22+cosecπ2×CC=-π24
Putting C=-π24 in (1), we get

y=x2-π24cosecx

Hence, y=x2-π24cosecx is the required solution.

Page No 21.107:

Question 38:

Find the general solution of the differential equation xdydx+2y=x2.

Answer:

We have, xdydx+2y=x2  dydx+2xy=x        .....1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=2x and Q=x.  I.F.=eP dx          =e2x dx          =e2log x         =x2         Multiplying both sides of 1 by I.F.=x2, we getx2dydx+2xy=x2x x2dydx+2xy=x3 Integrating both sides with respect to x, we getx2y=x3 dx+Cx2y=x44+Cy=x24+Cx-2Hence, y=x24+Cx-2  is the required solution.

Page No 21.107:

Question 39:

Find the general solution of the differential equation dydx-y=cos x.

Answer:

We have,dydx-y=cos x           .....1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=-1 and Q=cos x I.F.=eP dx          =e- dx         = e-xMultiplying both sides of 1 by I.F.=e-x, we gete-x dydx-y=e-xcos x e-xdydx-e-xy=e-xcos xIntegrating both sides with respect to x, we getye-x=e-xcos x dx+Cye-x=I+C           .....2Here,I=e-xcos x dx           .....3I=e-xsin x--e-xsin x dxI=e-xsin x+e-xsin x dxI=e-xsin x-e-xcos x--e-x×-cos x dxI=e-xsin x-e-xcos x-e-xcos x dxI=e-xsin x-e-xcos x-I   From  3 2I=e-xsin x-cos xI=e-x2sin x-cos x           .....4From   2 and 4 we getye-x=e-x2sin x-cos x+Cy=12sin x-cos x+CexHence, y=12sin x-cos x+Cex  is the required solution.

Page No 21.107:

Question 40:

Solve the differential equation y+3x2dxdy=x

Answer:

We have, y+3x2dxdy=xdydx=y+3x2xdydx-1xy=3x        .....1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=-1x and Q=3x I.F.=eP dx          =e-1x dx          =e-log x         =1xMultiplying both sides of (1) by I.F.=1x, we get1x dydx-1xy=1x3x 1xdydx-1x2y=3Integrating both sides with respect to x, we get1xy=3dx+Cyx=3x+CHence, yx=3x+C  is the required solution.



Page No 21.108:

Question 41:

Find the particular solution of the differential equation dxdy+x cot y=2y+y2 cot y, y ≠ 0 given that x = 0 when y=π2.

Answer:

We have,dxdy+x cot y=2y+y2cot y          .....1Clearly, it is a linear differential equation of the form dxdy+Px=Qwhere P=cot y and Q=2y+y2cot y I.F.=eP dy          =ecot y dy         = elogsin y=sin yMultiplying both sides of 1 by I.F.=sin y, we getsin ydxdy+x cot y=sin yy2cot y+2ysin ydxdy+x cos y=y2cos y+2y sin yIntegrating both sides with respect to y, we getx sin y=y2Icos yIIdy+2y sin y  dy+Cx sin y=y2cos ydy-ddyy2cos y dydy+2y sin y  dy+Cx sin y=y2sin y-2y sin y dy+2ysin y dy+Cx sin y=y2sin y+CNow, y=π2 at x=0 0×sin π2=π42sin π2+CC=-π42Putting the value of C, we getx sin y=y2sin y-π42Hence, x sin y=y2sin y-π42 is the required solution.

Page No 21.108:

Question 42:

Solve the following differential equation:

cot-1y+x dy=1+y2 dx

Answer:

The given differential equation is cot-1y+x dy=1+y2 dx.
This differential equation can be written as
dxdy=cot-1y+x1+y2 dxdy+-11+y2x=cot-1y1+y2 
This is a linear differential equation with P=-11+y2 and Q=cot-1y1+y2.
I.F. =  e-11+y2dy=ecot-1y
Multiply the differential equation by integration factor (I.F.), we get
dxdyecot-1y-x1+y2ecot-1y=cot-1y1+y2ecot-1y ddyxecot-1y=cot-1y1+y2ecot-1y 
Integrating both sides with respect y, we get
xecot-1y=cot-1y1+y2ecot-1y dy+C
Putting t=cot-1y and dt=-11+y2dy, we get
xecot-1y=-tet dt+Cxecot-1y=-ett-1+Cxecot-1y=ecot-1y1-cot-1y+C



Page No 21.134:

Question 1:

The surface area of a balloon being inflated, changes at a rate proportional to time t. If initially its radius is 1 unit and after 3 seconds it is 2 units, find the radius after time t.

Answer:

Let r be the radius and S be the surface area of the balloon at any time t. Then,
S=4πr2dSdt=8π r drdt         .....1Given:dSdtα tdSdt=kt, where k is any constantPutting dSdt=kt in (1), we getkt=8π r drdtkt dt=8πr drIntegrating both sides, we get kt dt=8πr drkt22=8π ×r22+C        .....(2)At t=0 s, r=1 unit and at t=3 s, r=2 units                   Given 0=8π×12+CC=-4πAnd92k=8π×2+C92k=12 πk=83πSubstituting the values of C and k in (2), we get 8t26π=8π ×r22-4π4t23=4r2-4t23=r2-1r2=1+t23r=1+13t2

Page No 21.134:

Question 2:

A population grows at the rate of 5% per year. How long does it take for the population to double?

Answer:

Let P0 be the initial population and P be the population at any time t. Then,
dPdt=5P100dPdt=0.05P

dPP=0.05dt Integrating both sides with respect to t, we getdPP=0.05dt log P=0.05t +CNow,P=P0 at t=0  log P0=0+CC=log P0Putting the value of C, we getlog P=0.05t+log P0logPP0=0.05tTo find the time when the population will double, we haveP=2P0 log2P0P0=0.05tlog 2=0.05tt=log 20.05=20 log 2 years

Page No 21.134:

Question 3:

The rate of growth of a population is proportional to the number present. If the population of a city doubled in the past 25 years, and the present population is 100000, when will the city have a population of 500000?

Answer:

Let the original population be N and the population at any time t be P.
Given: dPdtαP
dPdt=aPdPP=adtlogP=at+C          .....1Now, P=N at t=0Putting P=N and t=0 in 1, we getlogN=CPutting C=logN in 1, we getlogP=at+logNlogPN=at         .....2According to the question,log2NN=25aa=125log2=125×0.6931=0.0277Putting a=0.0277 in 2, we getlogPN=0.0277t           ...3For P=500000 and N=100000:log500000100000=0.0277tt=log 50.0277=1.6090.0277=58.08 years=Approximately 58 years

Page No 21.134:

Question 4:

In a culture, the bacteria count is 100000. The number is increased by 10% in 2 hours. In how many hours will the count reach 200000, if the rate of growth of bacteria is proportional to the number present?

Answer:

Let the bacteria count at any time t be N.Given:dNdtα NdNdt=λN1NdN=λdtIntegrating both sides, we get1NdN=λdtlog N=λt+log C             .....(1)Initially when t=0, then N=100000 Givenlog 100000=0+log Clog C=log 100000After 2 hours number increased by 10%Therefore, increased number=1000001+10%=110000Given: t=2, N=110000Putting t=2, N=110000 in (1), we getlog 110000=2λ+log 10000012log 1110=λSubstituting the values of log C and λ in (1), we getlog N=t2log 1110+log 100000            .....(2)Now,Let t=T when N=200000Substituting these values in (2), we get log 200000=T2log  1110+log 100000log 2=T2log 1110T=2log 2log1110 The count will reach 200000 in 2log 2log1110 hours.

Page No 21.134:

Question 5:

If the interest is compounded continuously at 6% per annum, how much worth Rs 1000 will be after 10 years? How long will it take to double Rs 1000?

Answer:

Let P0 be the initial amount and P be the amount at any time t. Then,
dPdt=6P100dPdt=0.06P
dPP=0.06dt Integrating both sides with respect to t, we getlog P=0.06t +CNow, P=P0 at t=0  log P0=0+CC=log P0Putting the value of C, we getlog P=0.06t +log P0logPP0=0.06te0.06t=PP0To find the amount after 10 years, we gete0.06×10=PP0e0.6=PP01.822=PP0P=1.822P0P=1.822×1000=Rs 1822To find the time after which the amount will double, we haveP=2P0 log2P0P0=0.06tlog 2=0.06tt=0.69310.06=11.55 years

Page No 21.134:

Question 6:

The rate of increase in the number of bacteria in a certain bacteria culture is proportional to the number present. Given the number triples in 5 hrs, find how many bacteria will be present after 10 hours. Also find the time necessary for the number of bacteria to be 10 times the number of initial present.

Answer:

Let the original count of bacteria be N and the count of bacteria at any time t be P.
Given: dPdtαP
dPdt=aPdPP=adtlogP=at+C          .....1Now,P=N at t=0 Putting P=N and t=0 in 1, we getlogN=C Putting C=logN in 1, we getlogP=at+logNlogPN=at         .....2According to the question,log3NN=5aa=15log3=15×1.0986=0.21972Putting a=0.21972 in 2, we getlogPN=0.21972t           .....3 e0.21972t=PN             .....4Putting t=10 in 4 to find the bacteria after 10 hours, we get e0.21972×10=PNe2.1972=PNPN=9P=9NTo find the time taken when the number of bacteria becomes 10 times of the number of initial population, we haveP=10N log10NN=15tlog 3t=5 log 10log 3

Page No 21.134:

Question 7:

The population of a city increases at a rate proportional to the number of inhabitants present at any time t. If the population of the city was 200000 in 1990 and 250000 in 2000, what will be the population in 2010?

Answer:

Let the population at any time t be P.

Given: dPdt α P

dPdt=βPdPP=βdtlogP=βt+log C          ...1Now, At t=1990, P=200000 and at t=2000, P=250000 log 200000=1990β+log C       ...2      log 250000=2000β+log C        ...3Subtracting 3 from 2, we getlog 200000-log 250000=10ββ=110log54Putting β=110log 54 in 2, we getlog 200000=1990×110log54+log Clog 200000=199log54+log C   log C=log 200000-199log54 Putting β=110log 54, log C=log 200000-199 log54 and t=2010 in 1, we getlogP=110×2010log 54+log 200000-199 log54logP=201 log 54+log 200000-199log54logP=log 54201- log54199+log 200000logP=log 5420145199+log 200000logP=log 542+log 200000logP=log2516×200000logP=log  312500P=312500

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Question 8:

If the marginal cost of manufacturing a certain item is given by C' (x) = dCdx= 2 + 0.15 x. Find the total cost function C (x), given that C (0) = 100.

Answer:

We have,dCdx=2+0.15xdC=2+0.15xdxIntegrating both sides with respect to x, we getC=2x+0.152x2+K          .....1At C0=100, we have100=20+0.15202+KK=100Putting the value of T in 1, we getC=2x+0.152x2+100C=0.075x2+2x+100

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Question 9:

A bank pays interest by continuous compounding, that is, by treating the interest rate as the instantaneous rate of change of principal. Suppose in an account interest accrues at 8% per year, compounded continuously. Calculate the percentage increase in such an account over one year.

Answer:

Let P0 be the initial amount and P be the amount at any time t.
We have,
dPdt=8P100dPdt=2P25

dPP=225dtIntegrating both sides with respect to t, we getlog P=225t +C          .....1Now,P=P0 at t=0  log P0=0+CC=log P0Putting the value of C in 1, we getlog P=225t +log P0logPP0=225te225t=PP0To find the amount after 1 year, we havee225=PP0e0.08=PP01.0833=PP0P=1.0833P0Percentage increase =P-P0P0×100%                           =1.0833P0-P0P0×100%                           =0.0833×100%                           =8.33%

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Question 10:

In a simple circuit of resistance R, self inductance L and voltage E, the current i at any time t is given by L didt+ R i = E. If E is constant and initially no current passes through the circuit, prove that i=ER1-e-R/Lt.

Answer:

We have,Ldidt+Ri=Edidt+RLi=EL            .....1 I.F.=eRL dt          =eRLtMultiplying both sides of (1) by I.F.=eRLt, we get eRLt didt+RLi=eRLt×ELeRLtdidt+eRLtRLi=eRLt×ELIntegrating both sides with respect to t, we geteRLti=ELeRLt dt+CeRLti=EL×LReRLt +CeRLti=EReRLt +C            .....2Now, i=0 at t=0 e0×0=ERe0 +CC=-ERPutting the value of C in 2, we geteRLti=EReRLt -ERi=ER -ERe-RLti=ER 1-e-RLt

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Question 11:

The decay rate of radium at any time t is proportional to its mass at that time. Find the time when the mass will be halved of its initial mass.

Answer:

Let the initial amount of radium be N and the amount of radium present at any time t be P.
Given: dPdtαP
dPdt=-aP, where a>0dPP=-adtIntegrating both sides, we getlogP=-at+C          .....1Now, P=N at t=0Putting P=N and t=0 in 1, we getlogN=CPutting C=logN in 1, we getlogP=-at+logNlogNP=atAccording to the question, logNN2=atlog2=att=1alog2Here, a is the constant of proportionality.

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Question 12:

Experiments show that radium disintegrates at a rate proportional to the amount of radium present at the moment. Its half-life is 1590 years. What percentage will disappear in one year?

Answer:

Let the original amount of the radium be N and the amount of radium at any time t be P.
Given: dPdtαP
dPdt=-aPdPP=-adtIntegrating both sides, we getlogP=-at+C          .....1Now,P=N at t=0Putting P=N and t=0 in 1, we getlogN=CPutting C=logN in 1, we getlogP=-at+logNlogPN=-at         .....2According to the question,P=12N at t=1590logN2N=-1590a-log 2=-1590aa=11590log 2Putting a=11590log 2 in 2, we getlogPN=-11590log 2t PN=e-log 21590t       .....3Putting t=1 in 4 to find the bacteria after 1 year, we getPN=0.9996P=0.9996NPercentage of amount disapeared in 1 year =N-PN×100%=N-0.9996NN×100%=0.04%



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Question 13:

The slope of the tangent at a point P (x, y) on a curve is -xy. If the curve passes through the point (3, −4), find the equation of the curve.

Answer:

According to the question,dydx=-xyy dy=-x dx Integrating both sides with respect to x, we gety dy=-x dxy22=-x22+CSince the curve passes through 3,-4, it satisfies the above equation. -422=-322+C8=-92+CC=252Putting the value of C, we gety22=-x22+252x2+y2=25

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Question 14:

Find the equation of the curve which passes through the point (2, 2) and satisfies the differential equation y-xdydx=y2+dydx.

Answer:

We have,
y-xdydx=y2+dydx

dydxx+1=y1-ydyy1-y=dxx+1

Integrating both sides, we getdyy1-y=dxx+11y+11-ydy=dxx+1log y-log 1-y=log x+1+C          .....1Since the curve passes throught the point 2, 2, it satisfies the equation of the curve.log 2-log 1-2=log 2+1+CC=log 23Putting the value of C in 1, we getlog y-log 1-y=log x+1+log 23log y1-y=log 2x+13y1-y=2x+13y1-y=±2x+13y1-y=2x+13  or  y1-y=-2x+13Here, given point 2, 2 does not satisfy y1-y=2x+13But it satisfy y1-y=-2x+13y1-y=-2x+13yy-1=2x+133y=2x+1y-13y=2xy-2x+2y-22xy-2x-y-2=0

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Question 15:

Find the equation of the curve passing through the point 1,π4 and tangent at any point of which makes an angle tan−1 yx-cos2yx with x-axis.

Answer:

The slope of the curve is given as dydx=tan θ.
Here,
θ=tan-1yx-cos2yx

 dydx=tantan-1yx-cos2yxdydx=yx-cos2yx

Let y=vxdydx=v+xdvdx v+xdvdx=v-cos2vxdvdx=-cos2vsec2 v dv=-1xdxIntegrating both sides with respect to x, we getsec2 v dv=-1xdxtan v=-log x+Ctan yx=-log x+CSince the curve passes through 1, π4, it satisfies the above equation. tan π4=-log 1+CC=1Putting the value of C, we gettan yx=-log x+1tan yx=-log x+log etan yx=logex

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Question 16:

Find the curve for which the intercept cut-off by a tangent on x-axis is equal to four times the ordinate of the point of contact.

Answer:

Let the given curve be y = f(x). Suppose P(x,y) be a point on the curve. Equation of the tangent to the curve at P is
Y - y =dydx(X-x) , where (X, Y) is the arbitrary point on the tangent.
Putting Y=0 we get,
0 - y = dydx(X - x)Therefore, X-x=-ydxdyX=x-ydxdyTherefore, cut off by the tangent on the x-axis = x-ydxdyGiven, x-ydxdy=4yTherefore, -ydxdy=4y - xdxdy=x-4yydydx=yx-4y   ........(1)this is a homogeneous differential equation.
Putting y = vx and dydx=v+xdvdx in (1) we getv + xdvdx = vxx-4vxTherefore, v+xdvdx=v1-4vxdvdx=v1-4v-v = 4v21-4v1-4vv2dv=4dxx
Integrating on both sides we get,
1-4vv2dv=4dxxTherefore, dvv2-4dvv=4dxx-1v-4 log v = 4 logx + log c-1v = 4 logx + log c +4 log v4 log(xv) + log c = -1vputting the value of v we get4 log(x×yx) + log c = -xy4 log(y) + log c = -xylog (y4c) = -xyy4c = e-xy

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Question 17:

Show that the equation of the curve whose slope at any point is equal to y + 2x and which passes through the origin is y + 2 (x + 1) = 2e2x.

Answer:

According to the question,dydx=y+2xdydx-y=2x           .....1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=-1 and Q=2x I.F.=eP dx          =e- dx         = e-xMultiplying both sides of 1 by I.F.=e-x, we gete-x dydx-y=e-x2x e-xdydx-e-xy=e-x2x  Integrating both sides with respect to x, we gety e-x=2e-xIIxI  dx+Cye-x=2xe-xdx-2ddxxe-xdxdx+Cye-x=-2xe-x-2e-x+C           .....2Since the curve passes through origin, we have0×e0=-2×0×e0-2e0+CC=2Putting the value of C in 2, we getye-x=-2xe-x-2e-x+2y=-2x-2+2exy+2x+1=2exDISCLAIMER: In the question it should be ex instead of e2x.

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Question 18:

The tangent at any point (x, y) of a curve makes an angle tan−1(2x + 3y) with x-axis. Find the equation of the curve if it passes through (1, 2).

Answer:

The slope of the curve is given as dydx=tan θ.
Here,

θ=tan-1 2x+3y dydx=tantan-1 2x+3ydydx=2x+3y

dydx-3y=2x           .....1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=-3 and Q=2x I.F.=eP dx          =e-3 dx         = e-3xMultiplying both sides of (1), by I.F.=e-3x, we gete-3x dydx-3y=e-3x.2xe-3x dydx-3y=2xe-3xIntegrating both sides with respect to x, we gety e-3x=2xe-3x dx+Cy e-3x=2xIe-3xII dx+Cy e-3x=2xe-3x dx-2ddxxe-3x dxdx+Cy e-3x=-2xe-3x3+2×13e-3x dx+Cy e-3x=-23xe-3x-2×19e-3x+Cy e-3x=-23xe-3x-29e-3x+CSince the curve passes through 1, 2, it satisfies the above equation. 2e-3=-23e-3-29e-3+CC=2e-3+23e-3+29e-3C=269e-3Putting the value of C, we gety e-3x=-23x-29e-3x+269e-3

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Question 19:

Find the equation of the curve such that the portion of the x-axis cut off between the origin and the tangent at a point is twice the abscissa and which passes through the point (1, 2).

Answer:



Portion of the x-axis cut off between the origin and tangent at a point =xydxdy=OT

It is given, OT = 2x

xydxdy=2xx=ydxdy-dxx=dyyxy=k

Since the curve passes through the point (1, 2)
⇒ at x = 1 ⇒ y = 2
k = 2
xy = 2

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Question 20:

Find the equation to the curve satisfying x (x + 1) dydx-y = x (x + 1) and passing through (1, 0).

Answer:

We have,xx+1dydx-y=xx+1dydx-yxx+1=1Comparing with dydx+Py=Q, we getP=-1xx+1Q=1Now,I.F.=e-1xx+1dx             =e-1x-1x+1dx                 =e-logxx+1            =x+1x        So, the solution is given byy×I.F.=Q×I.F. dx +Cx+1xy=x+1x dx +Cx+1xy=dx+1xdx +Cx+1xy=x+log x+CSince the curve passes throught the point 1, 0, it satisfies the equation of the curve.1+110=1+log 1+CC=-1Putting the value of C in the equation of the curve, we getx+1xy=x+log x-1y=xx+1x+log x-1

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Question 21:

Find the equation of the curve which passes through the point (3, −4) and has the slope 2yx at any point (x, y) on it.

Answer:

According to the question,
dydx=2yx
12ydy=1xdxIntegrating both sides with respect to x, we get12ydy=1xdx12log y=log x+CSince the curve passes through 3,-4, it satisfies the above equation. 1 2log -4=log 3+Clog 2-log 3=CC=log 23Putting the value of C, we getlog y=2log x+2log 23log y=log 49x2y=±49x29y-4x2=0 or 9y+4x2=0 The given point does not satisfy the equation 9y-4x2=0.   9y+4x2=0

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Question 22:

Find the equation of the curve which passes through the origin and has the slope x + 3y − 1 at any point (x, y) on it.

Answer:

According to the question,
dydx=x+3y-1

dydx-3y=x-1
Comparing with dydx+Py=Q, we getP=-3 Q=x-1Now, I.F.=e-3dx =e-3xSo, the solution is given byy×I.F.=Q×I.F. dx +Cye-3x=x-1e-3x dx +Cye-3x =xIe-3xIIdx-e-3x dx+Cye-3x=xe-3x dx-ddxxe-3x dxdx-e-3x dx+Cye-3x=-13xe-3x+13e-3x dx-e-3x dx+Cye-3x=-13xe-3x-19e-3x +13e-3x+Cy=-13x-19 +13+Ce3xy=-13x +29+Ce3xSince the curve passes throught the origin, it satisfies the equation of the curve.0=-0+29+Ce0C=-29Putting the value of C in the equation of the curve, we gety=-13x +291-e3xy+13x=291-e3x33y+x=21-e3x

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Question 23:

At every point on a curve the slope is the sum of the abscissa and the product of the ordinate and the abscissa, and the curve passes through (0, 1). Find the equation of the curve.

Answer:

According to the question,

dydx=x+xydydx=x1+y

11+ydy=x dxIntegrating both sides with respect to x, we get11+ydy=x dxlog 1+y=x22+CSince the curve passes through 0, 1, it satisfies the above equation. log 1+1=02+CC=log 2Putting the value of C, we getlog 1+y=x22+log 2log 1+y2=x221+y2=ex22y+1=2ex22

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Question 24:

A curve is such that the length of the perpendicular from the origin on the tangent at any point P of the curve is equal to the abscissa of P. Prove that the differential equation of the curve is y2-2xydydx-x2=0, and hence find the curve.

Answer:

Tangent  at P(x, y) is given by Y - y = dydx(X-x)
If p be the perpendicular from the origin, then
p = xdydx-y1+dydx2=x               (given)x2dydx2-2xydydx + y2=x2 +x2 dydx2y2-2xydydx-x2 =0          Hence proved.Now, y2-2xydydx-x2 =0    dydx = y2-x22xy2xydydx-y2 =-x2 2ydydx-y2 x=-x Let y2=vdvdx-vx=-x       
Multiplying by the integrating factor e-1xdx=1x
v.1x=-x.1xdx + c= -x+cy2x2=-x+cx2 + y2=cx

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Question 25:

Find the equation of the curve which passes through the point (1, 2) and the distance between the foot of the ordinate of the point of contact and the point of intersection of the tangent with x-axis is twice the abscissa of the point of contact.

Answer:


It is given that the distance between the foot of ordinate of point of contact (A) and point of intersection of tangent with x-axis (T) = 2x

CoordinateofT=xydxdy, 0AT=xxydxdy=2xEquationoftangent,Yy=dydx(Xx)y-0=dydxxxydxdyydxdy=2xdxx=2dyylnx=lny2+lncx=cy2As the circle passes through 1, 2.1=c×22c=144x=y2

 

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Question 26:

The normal to a given curve at each point (x, y) on the curve passes through the point (3, 0). If the curve contains the point (3, 4), find its equation.

Answer:

Let P (x, y) be any point on the curve. The equation of the normal at P (x, y) to the given curve is given as
Y-y=-1dydxX-x
It is given that the curve passes through the point (3, 0). Then,
0-y=-1dydx3-x-y=-1dydx3-xydydx=3-xy dy=3-xdxy22=3x-x22+C          .....1Since the curve passes through the point 3, 4, it satisfies the equation.422=33-322+CC=8-9+92C=92-1=72Putting the value of C in 1, we gety22=3x-x22+72y2=6x-x2+7x2+y2-6x-7=0

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Question 27:

The rate of increase of bacteria in a culture is proportional to the number of bacteria present and it is found that the number doubles in 6 hours. Prove that the bacteria becomes 8 times at the end of 18 hours.

Answer:

Let the original count of bacteria be N and the count of bacteria at any time t be P.
Given: dPdtαP
dPdt=aPdPP=adtlog P=at+C          .....1Now, P=N at t=0Putting P=N and t=0 in 1, we getlog N=CPutting C=log N in 1, we getlog P=at+log Nlog PN=at         .....2According to the question,log 2NN=6aa=16log 2Putting a=16log 2  in 2, we getlog PN=t6log 2      .....3Putting t=18 in 3 to find the bacteria after 18 hours, we getlog PN=186 log 2log PN=3log 2log PN=log 8PN=8P=8N

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Question 28:

Radium decomposes at a rate proportional to the quantity of radium present. It is found that in 25 years, approximately 1.1% of a certain quantity of radium has decomposed. Determine approximately how long it will take for one-half of the original amount of  radium to decompose?

Answer:

Let the original amount of radium be N and the amount of radium at any time t be P.
Given: dPdtαP
dPdt=-aPdPP=-a dtIntegrating both sides, we getlog P=-at+C          .....1Now,P=N when t=0 Putting P=N and t=0 in 1, we getlog N=CPutting C=log N in 1, we getlog P=-at+log Nlog PN=-at         .....2According to the question,P=98.9100N=0.989N at t=25log 0.989NN=-25aa=-125log 0.989Putting a=-125log 0.989 in 2, we getlogPN=125log 0.989tTo find the time when the radium becomes half of its quantity, we haveN=12Plog NN2=125log 0.989tlog 2=125log 0.989t  t=25log 2log 0.989=25×0.69310.01106=1566.681567 approx.

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Question 29:

Show that all curves for which the slope at any point (x, y) on it is x2+y22xy are rectangular hyperbola.

Answer:

We have,
dydx=x2+y22xyLet y=vxdydx=v+xdvdx v+xdvdx=x2+v2x22vx2xdvdx=1+v22v-vx dvdx=1+v2-2v22vxdvdx=1-v22v2v1-v2dv=1xdx

Integrating both sides, we get2v1-v2dy=1xdx-log 1-v2=log x-log Clog 1-v2C=-log x1-v2=Cx1-yx2=Cxx2-y2x2=Cxx2-y2=Cx
Thus, x2-y2=Cx is the equation of the rectangular hyperbola.



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Question 30:

The slope of the tangent at each point of a curve is equal to the sum of the coordinates of the point. Find the curve that passes through the origin.

Answer:

According to the question,
dydx=x+y
dydx-y=x
Comparing with dydx+Py=Q, we getP=-1Q=xNow, I.F.=e-dx =e-xSo, the solution is given byy×I.F.=Q×I.F. dx +Cye-x=xIe-xIIdx+Cye-x=xe-x dx-ddxxe-x dxdx+Cye-x=-xe-x+e-x dx+Cye-x=-xe-x-e-x+CSince the curve passes throught the origin, it satisfies the equation of the curve.0e0=-0e0-e0+CC=1Putting the value of C in the equation of the curve, we getye-x=-xe-x-e-x+1ye-x+xe-x+e-x=1y+x+1e-x=1x+y+1=ex

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Question 31:

Find the equation of the curve passing through the point (0, 1) if the slope of the tangent to the curve at each of its point is equal to the sum of the abscissa and the product of the abscissa and the ordinate of the point.

Answer:

According to the question,
dydx=x+xy
dydx-xy=x
Comparing with dydx+Py=Q, we getP=-xQ=xNow, I.F.=e-xdx =e-x22So, the solution is given byy×I.F.=Q×I.F. dx +Cye-x22=xe-x22dx+Cye-x22=I+CNow,I=xe-x22dxPutting -x22=t, we get-xdx=dt I=-etdtI=-etI=-e-x22 ye-x22=-e-x22+C Since the curve passes throught the point 0, 1, it satisfies the equation of the curve.1e0=-e0+CC=2Putting the value of C in the equation of the curve, we getye-x22=-e-x22+2y=-1+2ex22

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Question 32:

The slope of a curve at each of its points is equal to the square of the abscissa of the point. Find the particular curve through the point (−1, 1).

Answer:

According to the question,
dydx=x2
dy=x2dxIntegrating both sides with respect to x, we getdy=x2dxy=x33+CSince the curve passes through -1, 1, it satisfies the above equation. 1=-13+CC=1+13C=43Putting the value of C, we gety=x33+433y=x3+4

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Question 33:

Find the equation of the curve that passes through the point (0, a) and is such that at any point (x, y) on it, the product of its slope and the ordinate is equal to the abscissa.

Answer:

According to the question,

ydydx=xy dy=x dxIntegrating both sides with respect to x, we gety dy=x dxy22=x22+CSince the curve passes through 0, a, it satisfies the above equation. a22=02+CC=a22Putting the value of C, we gety22=x22+a22x2-y2=-a2

Page No 21.136:

Question 34:

The x-intercept of the tangent line to a curve is equal to the ordinate of the point of contact. Find the particular curve through the point (1, 1).

Answer:

Let P(x, y) be any point on the curve. Then slope of the tangent at P is dydx.
It is given that the slope of the tangent at P(x,y) is equal to the ordinate  i.e y. 
Therefore dydx = y
1ydy = dxlog y = x + log Clog y = log ex + log Cy = Cex
Since, the curve passes through (1,1). Therefore, x=1 and y=1 . 
Putting these values in equation obtained above we get,
1=Ce1C=1eputting these values in the equation we get,y=ex-1



Page No 21.137:

Question 1:

Determine the order and degree (if defined) of the following differential equations:
(i) dsdt4+3sd2sdt2=0

(ii) y"' + 2y" + y' = 0
(iii) (y"')2 + (y")3 + (y')4 + y5 = 0
(iv) y"' + 2y" + y' = 0
(v) y" + (y')2 + 2y = 0
(vi) y" + 2y' + sin y = 0
(vii) y"' + y2 + ey' = 0

Answer:

i  dsdt4+3sd2sdt2=0
The highest order derivative in the given equation is d2sdt2 and its power is 1.
Therefore, the given differential equation is of second order and first degree.
i.e., Order = 2 and degree = 1

(ii) y"' + 2y" + y' = 0
The highest order derivative in the given equation is y''' and its power is 1.
Therefore, the given differential equation is of third order and first degree.
i.e., Order = 3 and degree = 1

(iii) (y"')2 + (y")3 + (y')4 + y5 = 0
The highest order derivative in the given equation is y''' and its power is 2.
Therefore, the given differential equation is of third order and second degree.
i.e., Order = 3 and degree = 2

(iv) y"' + 2y" + y' = 0
The highest order derivative in the given equation is y''' and its power is 1.
Therefore, the given differential equation is of third order and first degree.
i.e., Order = 3 and degree = 1

(v) y" + (y')2 + 2y = 0
The highest order derivative in the given equation is y'' and its power is 1.
Therefore, the given differential equation is of second order and first degree.
i.e., Order = 2 and degree = 1

(vi) y" + 2y' + sin y = 0
The highest order derivative in the given equation is y'' and its power is 1.
Therefore, the given differential equation is of second order and first degree.
i.e., Order = 2 and degree = 1

(vii) y"' + y2 + ey' = 0
The highest order derivative in the given equation is y''' and its power is 1.
Therefore, the given differential equation is of third order. This equation cannot be expressed as a polynomial of derivative.
Thus, the degree is not defined.
i.e., Order = 3 and degree is not defined.



Page No 21.138:

Question 2:

Verify that the function y = e−3x is a solution of the differential equation d2ydx2+dydx-6y = 0.

Answer:

We have,d2ydx2+dydx-6y = 0      .....1Now,y=e-3xdydx=-3e-3xd2ydx2=9e-3x
Putting the values of d2ydx2, dydx and y  in 1, we get

LHS=9e-3x-3e-3x-6e-3x      = 0      =RHS

Thus, y = e−3x is the solution of the given differential equation.

Page No 21.138:

Question 3:

In each of the following verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

(i) y = ex + 1 y'' − y' = 0
(ii) y = x2 + 2x + C y' − 2x − 2 = 0
(iii) y = cos x + C y' + sin x = 0
(iv) y = 1+x2 y' = xy1+x2
(v) y = x sin x xy' = y + x x2-y2
(vi) y=a2-x2 x+ydydx=0

Answer:

(i) We have,
y'' − y' = 0                            .....(1)
Now,
y = ex +1

y'=exy''=ex

Putting the above values in (1), we get
LHS=ex-ex =0=RHS
Thus, y = ex +1 is the solution of the given differential equation.

(ii) We have,
y' − 2x − 2 = 0                    .....(1)
Now,
y = x2 + 2x + C
y'=2x+2
Putting the above value in (1), we get
LHS=2x+2-2x-2=0=RHS
Thus, y = x2 + 2x + C is the solution of the given differential equation.

(iii) We have,
y' + sin x = 0                    .....(1)
Now,
y = cos x + C
y'=-sin x
Putting the above value in (1), we get
LHS=-sin x+sin x=0=RHS
Thus, y = cos x + C is the solution of the given differential equation.

(iv) We have,
y' = xy1+x2                  .....(1)
Now,
y = 1+x2
y'=x1+x2

Putting the above value in (1), we get
LHS=x1+x2=x1+x2×1+x21+x2=xy1+x2=RHS
Thus, y = 1+x2 is the solution of the given differential equation.

(v) We have,
xy' = y + x x2-y2     .....(1)
Now,
y = x sin x
y'=sin x+x cosx
Putting the above value in (1), we get
LHS=xsin x+xcos x =xsin x+x2cosx=xsin x+xx cos x=xsin x+xx1-sin2 x=xsin x+xx2-x2sin2 x=y+xx2-y2=RHS
Thus, y = x sin x is the solution of the given differential equation.


(v) We have,
xy' = y + x x2-y2         .....(1)
Now,
y = x sin x
y'=sin x+x cosx
Putting the above value in (1), we get
LHS=xsin x+xcos x =xsin x+x2cosx=xsin x+xx cos x=xsin x+xx1-sin2 x=xsin x+xx2-x2sin2 x=y+xx2-y2=RHS
Thus, y = x sin x is the solution of the given differential equation.

(vi) We have,
x+ydydx=0        .....(1)

Now,
y=a2-x2
y'=-xa2-x2

Putting the above value in (1), we get

LHS=x+y-xa2-x2 =x+a2-x2-xa2-x2=x+a2-x2-xa2-x2=x-x=0=RHS                         

Thus, y=a2-x2 is the solution of the given differential equation.

Page No 21.138:

Question 4:

Form the differential equation representing the family of curves y = mx, where m is an arbitrary constant.

Answer:

We have,
y = mx          (1)
Differentiating both sides, we get
dydx=mdydx=yx               From 1xdydx=yxdydx-y=0

Page No 21.138:

Question 5:

Form the differential equation representing the family of curves y = a sin (x + b), where a, b are arbitrary constant.

Answer:

We have,
y = a sin (x + b)          .....(2)
Differentiating both sides, we get
dydx=a cosx+bd2ydx2=-a sinx+b    d2ydx2=-a×ya          Using 2d2ydx2=-y d2ydx2+y=0

Page No 21.138:

Question 6:

Form the differential equation representing the family of parabolas having vertex at origin and axis along positive direction of x-axis.

Answer:

The equation of the parabola having vertex at origin and axis along the positive direction of x-axis is given by
y2 =4ax         .....(1)
Since there is only one parameter, so we differentiate it only once.
Differentiating with respect to x, we get
2ydydx=4a
Substituting the value of 4a in (1), we get
y2=2ydydx×xy2=2xydydxy2-2xydydx=0

Page No 21.138:

Question 7:

Form the differential equation of the family of circles having centre on y-axis and radius 3 unit.

Answer:

The equation of the family of circles with radius 3 units, having its centre on y-axis, is given by
x2+y-a2=32                                   .....1
Here, a is any arbitrary constant.
Since this equation has only one arbitrary constant, we get a first order differential equation.
Differentiating (1) with respect to x, we get
2x+2y-adydx=0x+y-adydx=0x=a-ydydxxdydx=a-ya=y+xdydx 
Substituting the value of a in (1), we get

x2+y-y-xdydx2=32x2+x2dydx2=9x2dydx2+x2=9dydx2x2dydx2-9dydx2+x2=0x2-9dydx2+x2=0x2-9y'2+x2=0Hence, x2-9y'2+x2=0  is the required differential equation.

Page No 21.138:

Question 8:

Form the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.

Answer:

The equation of the parabola having vertex at origin and axis along the positive direction of y-axis is given by
x2 =4ay         .....(1)
Since there is only one parameter, so we differentiate it only once.
Differentiating with respect to x, we get
2x=4ay'4a=2xy'
Substituting the value of 4a in (1), we get
x2=2xy'×yxy'=2yxy'-2y=0

Page No 21.138:

Question 9:

Form the differential equation of the family of ellipses having foci on y-axis and centre at the origin.

Answer:

The equation of the ellipses having foci on y-axis and centre at the origin is given by
x2a2+y2b2=1            .....(1)
Here,
b > a
Since these are two parameters, so we differentiate the equation twice.
Differentiating with respect to x, we get
2xa2+2yb2y'=0xa2+yb2y'=0                          .....21a2+1b2y'2+yb2y''=0         .....3Multiplying throughout by x, we getxa2+xb2y'2+xyb2y''=0               .....4Subtracting 2 from 4, we get1b2xy'2+xyy''-yy'=0 xy'2+xyy''-yy'=0 

Page No 21.138:

Question 10:

Form the differential equation of the family of hyperbolas having foci on x-axis and centre at the origin.

Answer:

The equation of the family of hyperbolas having centre at the origin and foci on the X-axis is given by
x2a2-y2b2=1                                        .....1
Here, a and b are parameters.
Since this equation contains two parameters, so we get a second order differential equation.
Differentiating (1) with respect to x, we get
2xa2-2yb2y'=0                                .....2
Differentiating (2) with respect to x, we get
2a2-2b2yy''+y'2=01a2=1b2yy''+y'2b2a2=yy''+y'2                            .....(3)
From (2), we get
2xa2=2yb2y'b2a2=yxy'                                    .....(4)
From (3) and (4), we get
yxy'=yy''+y'2yy'=xyy''+xy'2Hence, xyy''+xy'2-yy'=0 is the required differential equation.

Page No 21.138:

Question 11:

Verify that xy = a ex + b ex + x2 is a solution of the differential equation xd2ydx2+2dydx-xy+x2-2=0.

Answer:

We have,xy=aex+be-x+x2Differentiating with respect to x on both sides, we getxdydx+y=aex-be-x+2xAgain differentiating with respect to x on both sides, we getxd2ydx2+dydx+dydx=aex+be-x+2xd2ydx2+2dydx=xy-x2+2       xy=aex+be-x+x2xd2ydx2+2dydx-xy+x2-2=0
Thus, xy = a ex + b ex + x2 is the solution of the given differential equation.

Page No 21.138:

Question 12:

Show that y = C x + 2C2 is a solution of the differential equation 2dydx2+xdydx-y=0.

Answer:

We have,
2dydx2+xdydx-y=0               .....1
Now,
y = C x + 2C2
dydx=CPutting dydx=C and y=Cx+2C2 in 1, we get
LHS=2C2+xC-Cx+2C2=2C2+xC-xC-2C2 =0=RHS
 Thus, y = C x + 2C2 is the solution of the given differential equation.

Page No 21.138:

Question 13:

Show that y2x2xy = a is a solution of the differential equation x-2ydydx+2x+y=0.

Answer:

We have,
x-2ydydx+2x+y=0
Now,y2x2xy = a
2ydydx-2x-y-xdydx=02y-xdydx-2x-y=02y-xdydx=2x+yx-2ydydx=-2x+yx-2ydydx+2x+y=0
 Thus, y2x2xy = a is the solution of the given differential equation.

Page No 21.138:

Question 14:

Verify that y = A cos x + sin x satisfies the differential equation cos xdydx+sin x y = 1.

Answer:

We have,
 cos xdydx+sin xy=1           .....1
Now,
y = A cos x + sin x
dydx=-A sin x+cos xPutting dydx=-A sin x+cos x and y=A cos x+sin x in 1, we get
LHS=cos x-A sin x+cos x+sin x A cos x+sin x=-A sin x cos x+cos2 x+ A cos x sin x+sin2 x=cos2 x+sin2 x=1=RHS
 Thus, y = A cos x + sin x is the solution of the given differential equation.

Page No 21.138:

Question 15:

Find the differential equation corresponding to y = ae2x + be3x + cex where a, b, c are arbitrary constants.

Answer:

We have,
y
= ae2x + be3x + cex            .....(1)
Differentiating with respect to x, we get
dydx=2ae2x-3be-3x+cex         .....2d2ydx2=4ae2x+9be-3x+cexd3ydx3=8ae2x-27be-3x+cexd3ydx3=72ae2x-3be-3x+cex-6ae2x+be-3x+cexd3ydx3=7dydx-6y      Using 1 and 2d3ydx3-7dydx+6y=0

Page No 21.138:

Question 16:

Show that the differential equation of all parabolas which have their axes parallel to y-axis is d3ydx3=0.

Answer:

The equation of the family of parabolas axis parallel to y-axis is given by
x-β2=4ay-α                                         .....(1)
Here, α and β are two arbitrary constants.

Differentiating (1) with respect to x, we get
2x-β=4adydx1=2ad2ydx20=2ad3ydx3d3ydx3=0                                  

Page No 21.138:

Question 17:

From x2 + y2 + 2ax + 2by + c = 0, derive a differential equation not containing a, b and c.

Answer:

We have,
x2 + y2 + 2ax + 2by + c = 0         .....(i)

Differentiating (i) with respect to x, we get

2x+2yy'+2a+2by'=0Again differentiating with respect to x, we get2+2y'2+2yy"+2by"=01+y'2+yy"+by"=0b=-1+y'2+yy"y"We have,1+y'2+yy"+by"=0Again differentiating with respect to x, we get2y'y"+y'y"+yy'''+by'''=0On substituting the value of b we get,3y'y"+yy'''+-1+y'2+yy"y"y'''=03y'y"2+yy"y'''-y'''-y'2y'''-yy'''y"=03y'y"2=y'''1+y'2
 



Page No 21.139:

Question 18:

dydx=sin3 x cos4 x+xx+1

Answer:

We have,dydx=sin3 x cos4 x+xx+1dy=sin3 x cos4 x+xx+1dxIntegrating both sides, we getdy=sin3 x cos4 x+xx+1dxy=sin3 x cos4 x dx +xx+1dx y=I1 +I2           .....1     Here, I1=sin3 x cos4 x dxI1=xx+1dxNow, I1=sin3 x cos4 x dx   =1-cos2 x cos4x sin x dxPutting t=cos x, we getdt=-sin x dx I1=-t4 1-t2dt       =t6-t4dt       =t77-t55+C1       =cos7 x7-cos5 x5+C1 I2=xx+1dxPutting t2=x+1, we get2t dt=dxI2=2t2-1t2dt     =2t4-t2 dt      =2t55-2t33+C2      =2x+1525-2x+1323+C2 Putting the value of I1 and I2 in 1, we gety=cos7 x7-cos5 x5+C1+2x+1525-2x+1323+C2 y=cos7 x7-cos5 x5+2x+1525-2x+1323+C      C=C1+C2Hence, y=cos7 x7-cos5 x5+2x+1525-2x+1323+C   is the solution of the given differential equation.

Page No 21.139:

Question 19:

dydx=1x2+4x+5

Answer:

We have,dydx=1x2+4x+5dydx=1x2+4x+4+1dydx=1x+22+12dy=1x+22+12dxIntegrating both sides, we gety=1x+22+12dxy=tan-1 x+21+Cy=tan-1 x+2+C

Page No 21.139:

Question 20:

dydx=y2+2y+2

Answer:

We have,dydx=y2+2y+2dydx=y2+2y+1+1dydx=y+12+121y+12+12dy=dxIntegrating both sides, we get1y+12+12dy=dxtan-1 y+11+C=xx=tan-1y+1+C

Page No 21.139:

Question 21:

dydx+4x=ex

Answer:

We have,dydx+4x=exdydx=ex-4xdy=ex-4xdxIntegrating both sides, we getdy=ex-4xdxy=ex-2x2+Cy+2x2=ex+C

Page No 21.139:

Question 22:

dydx=x2 ex

Answer:

We have,dydx=x2exdy=x2ex dxIntegrating both sides, we getdy=x2IexII dxdy=x2ex dx-ddxx2ex dxdxy=x2ex-2xex dxy=x2ex-2xI exII dxy=x2ex-2xex dx+2ddxxex dxdxy=x2ex-2xex+2ex+Cy=x2 -2x+2ex+C

Page No 21.139:

Question 23:

dydx-x sin2 x=1x log x

Answer:

We have, dydx-x sin2 x=1x log xdydx=1x log x+x sin2 xdydx=1x log x+x 21-cos 2xdydx=1x log x+x 2-x 2cos 2xIntegrating both sides, we getdy=1x log x+x 2-x 2cos 2x dxdy=1x log xdx+12x dx-12x cos 2xdxdy=1x log xdx+12x dx-12xI×cos 2xII dx y=log log x+x24-x2cos 2xdx+12ddxxcos 2x dxdxy=log log x+x24-x sin 2x4-cos 2x8+C

Page No 21.139:

Question 24:

(tan2 x + 2 tan x + 5) dydx=2 (1 + tan x) sec2 x

Answer:

We have,tan2 x+2 tan x+5dydx=21+tan xsec2 xdy=21+tan xsec2 xtan2 x+2 tan x+5 dxIntegrating both sides, we get dy=21+tan xsec2 xtan2 x+2 tan x+5 dx      .....1Putting tan2 x+2 tan x+5=t2 tan x sec2x+ 2sec2x dx=dt21+tan xsec2 x dx=dtTherefore 1 becomes, dy=1t dty=log t+Cy=log tan2 x+2 tan x+5+C

Page No 21.139:

Question 25:

dydx=sin3 x cos2 x+xex

Answer:

We have,dydx=sin3 x cos2 x+xexdy=sin3 x cos2 x+xexdxIntegrating both sides, we getdy=sin3 x cos2 x+xexdxy=sin3 x cos2 x dx +xexdx y=I1 +I2           .....1     Here, I1=sin3x cos2x dxI2=xexdxNow, I1=sin3 x cos2 x dx=1-cos2 x cos2x sin x dxPutting t=cos x, we getdt=-sin x dxI1=-t2 1-t2dt       =-t2+t4dt       =-t33+t55+C1       =cos5 x5-cos3 x3+C1 I2=xexdx    =xI exII dx    =xex dx-ddxxex dxdx    =xex-ex+C2    =x-1ex+C2Putting the value of I1 and I2 in 1, we gety=cos5 x5-cos3 x3+C1+x-1ex+C2y=cos5 x5-cos3 x3+x-1ex+C, where C=C1+C2

Page No 21.139:

Question 26:

tan y dx + tan x dy = 0

Answer:

We have,
tan y dx + tan x dy = 0
tan xdydx=-tan y cot y dy=-cot x dxIntegrating both sides, we getcot y dy=-cot x dxlog sin y=- log sin x+log Clog sin y+ log sin x=log Clog sin ysin x=log Csin ysin x=Csin x sin y=C

Page No 21.139:

Question 27:

(1 + x) y dx + (1 + y) x dy = 0

Answer:

We have, 
(1 + x) y dx + (1 + y) x dy = 0

dydx=-y1+xx1+y1+yydy=-1+xxdx1y+ydy=-1x+1dxIntegrating both sides, we get1y+1dy=-1x+1dx1ydy+dy=-1xdx-dxlogy+y=-logx-x+Clogxy+y+x=Cx+y+logxy=C

Page No 21.139:

Question 28:

x cos2 y dx = y cos2 x dy

Answer:

We have,
x cos2 y dx = y cos2 x dy
y sec2y dy=x sec2x dxIntegrating both sides, we getyI sec2yIIdy=xI sec2xII dxysec2ydy-dydy×sec2y dydy=xsec2x dx-dxdx×sec2x dxdxy tan y-tan y dy=x tan x-tan x dx-Cy tan y-log sec y=x tan x-log sec x-Cx tan x-y tan y=logsec x-logsec y+C

Page No 21.139:

Question 29:

cos y log (sec x + tan x) dx = cos x log (sec y + tan y) dy

Answer:

We have,
cos y log (sec x + tan x) dx = cos x log (sec y + tan y) dy
logsec y+tan ycos ydy=logsec x+tan xcos xdxIntegrating both sides, we getlogsec y+tan ycos ydy=logsec x+tan xcos xdx     .....1Putting logsec y+tan y=t and logsec x+tan x=usec2y+sec y tan ysec y+tan ydy=dt and sec2x+sec x tan xsec x+tan xdx=dusec y dy=dt and sec x dx=duTherefore, 1 becomest dt=u dut22=u22+Clogsec y+tan y22=logsec x+tan x22+Clogsec y+tan y2=logsec x+tan x2+2Clogsec y+tan y2=logsec x+tan x2+k, where k=2C

Page No 21.139:

Question 30:

cosec x (log y) dy + x2y dx = 0

Answer:

We have,
cosec x log ydy+x2y dx=0log yydy=-x2 cosec xdxlog yydy=-x2 sin x dxIntegrating both sides, we getlog yydy=-x2 sin x dx      .....1Putting log y=t1ydy=dtTherefore,  1 becomest dt=-x2 sin x dxt22=-x2I sin xII dx12log y2 =-x2 sin x dx-ddxx2sin x dxdx12log y2 =x2 cos x+2x cos x dx12log y2 =x2cos x-2xI cos xII dx12log y2 =x2cos x-2xcos x dx+2ddxxcos x dxdx12log y2 =x2cos x-2x sin x-2cos x+C12log y2 =x2-2cos x-2x sin x+C12log y2+2-x2cos x+2x sin x=C

Page No 21.139:

Question 31:

(1 − x2) dy + xy dx = xy2 dx

Answer:

We have,
1-x2dy+xy dx=xy2dx1-x2dy=xy2dx-xy dx1-x2dy=xy2-ydx1y2-ydy=x1-x2dxIntegrating both sides, we get1y2-ydy=x1-x2dx1y2-y+14-14dy=x1-x2dx1y-122-122dy=-12-2x1-x2dx12×12log y-12-12y-12+12=-12log 1-x2+log Clog y-1y=-12log 1-x2+log C2 log y-1y=-log 1-x2+2 log Clog y-12y2=-log1-x2+2 log Clog y-12y2+log 1-x2=log C2logy-121-x2y2=log C2y-121-x2y2=C2y-121-x2=y2C2

Page No 21.139:

Question 32:

dydx=sin x+x cos xy2 log y+1

Answer:

We have,
dydx=sin x+x cos xy2 log y+1y 2 log y+1dy= sin x+x cos xdxIntegrating both sides, we get2yII log y Idy+y dy=sin x dx+x cos x dx2log yy dy-2ddylog y×y dydy+y dy=-cos x+x cos x dx-dxdx×cos x dxy2log y-y dy+y dy=-cos x+x sin x dx+cos x+Cy2log y=x sin x+C

Page No 21.139:

Question 33:

x (e2y − 1) dy + (x2 − 1) ey dx = 0

Answer:

We have,xe2y-1dy+x2-1eydx=0xe2y-1dy=1-x2eydxe2y-1eydy=1-x2xdxey-e-ydy=1x-xdxIntegrating both sides, we getey-e-ydy=1x-xdxey+e-y=log x-12x2+Cey+e-y-log x+12x2=C

Page No 21.139:

Question 34:

dydx+1=ex+y

Answer:

We have,
dydx+1=ex+y     .....1Let x+y=v1+dydx=dvdxdydx=dvdx-1Then, 1 becomesdvdx-1+1=evdvdx=eve-vdv=dxIntegrating both sides, we gete-vdv=dx-e-v=x+C-1=evx+C-1=x+Cex+y

Page No 21.139:

Question 35:

dydx=x+y2

Answer:

We have,dydx=x+y2          .....1Let x+y=v1+dydx=dvdxdydx=dvdx-1Therefore, 1 becomes dvdx-1=v2dvdx=v2+11v2+1dv=dxIntegrating both sides, we get1v2+1dv=dxtan-1 v=x+Cv=tanx+Cx+y=tanx+C     

Page No 21.139:

Question 36:

cos (x + y) dy = dx

Answer:

We have,cos x+ydy=dxdydx=1cosx+y     .....1Let x+y=v1+dydx=dvdxdydx=dvdx-1Therefore,  1 becomes dvdx-1=1cos vdvdx=cos v+1cos vcos vcos v+1dv=dxIntegrating both sides, we getcos vcos v+1dv=dxcos v1-cos v1-cos2 vdv=dxcos v1-cos vsin2 vdv=dxcot v cosec v-cot2vdv=dxcot v cosec v-cosec2 v+1dv=dx-cosec v+cot v+v=x+C-cosec x+y+cot x+y+x+y=x+C-cosec x+y+cot x+y+y=Ccosec x+y-cot x+y=y-C1-cos x+ysin x+y=y-C2 sin2 x+y22 sin x+y2 cos x+y2=y-Csin x+y2cos x+y=y-Ctanx+y2=y-C

Page No 21.139:

Question 37:

dydx+yx=y2x2

Answer:

We have,dydx+yx=y2x2dydx=yx2-yxPutting y=vx, we getdydx=v+xdvdx v+xdvdx=v2-vxdvdx=v2-2v1v2-2v dv=1xdxIntegrating both sides, we get1v2-2v dv=1xdx1v2-2v+1-1 dv=1xdx1v-12-12 dv=1xdx12log v-1-1v-1+1 =log x+log Clog v-2v12 =log Cxlog yx-2yx12 =log Cxlog y-2xy12 =log Cxy-2xy12=Cxy-2xy =C2x2y-2x=kx2y, where k=C2

Page No 21.139:

Question 38:

dydx=yx-yxx+y

Answer:

We have,dydx=yx-yxx+yPutting y=vx, we getdydx=v+xdvdx v+xdvdx=vxx-vxxx+vxv+xdvdx=v1-v1+vxdvdx=v1-v1+v-vxdvdx=v-v2-v-v21+vxdvdx=-2v21+v-1+v2v2 dv=1xdxIntegrating both sides, we get-1+v2v2 dv=1xdx-121v2dv-121vdv=1xdx12v-12log v =log x+log Cxy-log yx =2log x+2log Cxy-log yx =log C2x2xy =log C2x2+log yxxy =log C2xyexy =C2xyexy =k xy, where k=C2

Page No 21.139:

Question 39:

(x + y − 1) dy = (x + y) dx

Answer:

We have,x+y-1dy=x+ydxdydx=x+yx+y-1Putting x+y=v, we get1+dydx=dvdxdydx=dvdx-1 dvdx-1=vv-1dvdx=vv-1+1dvdx=v+v-1v-1dvdx=2v-1v-1v-12v-1 dv=dxIntegrating both sides, we getv-12v+1 dv=dx122v2v-1dv-12v-1dv=dx122v-1+12v-1dv-12v-1dv=dx12dv+1212v-1dv-12v-1dv=dx12dv-1212v-1dv=dx12v-14log 2v-1 =x+C12x+y-14log 2x+2y-1 =x+C2x+y-log 2x+2y-1 =4x+4C2x+y-4x-log 2x+2y-1 =4C      2y-x-log 2x+2y-1 =k, where k=4C

Page No 21.139:

Question 40:

dydx-y cot x=cosec x

Answer:

We have,dydx-y cot x=cosec xComparing with dydx+Py=Q, we getP=-cot x Q=cosec xNow,I.F.=e-cot x dx=e-log sin x=elog cosec x=cosec xSo, the solution is given byy cosec x=cosec x×cosec x dx+Cy cosec x=cosec2 x dx+Cy cosec x=-cot x +C

Page No 21.139:

Question 41:

dydx-y tan x=-2 sin x

Answer:

We have,dydx-y tan x=-2sin xComparing with dydx+Py=Q, we getP=-tan x Q=-2sin xNow,I.F.=e-tan xdx=e-logsec x=elogcos x=cos xSo, the solution is given byy cosx=-2sin x cos x dx+Cy cosx=-sin 2x dx+C y cosx=cos 2x2+C

Page No 21.139:

Question 42:

dydx-y tan x=ex sec x

Answer:

We have,dydx-y tan x=ex sec xComparing with dydx+Py=Q, we getP=-tan x Q=ex sec xNow,I.F.=e-tan xdx=e-logsec x=elogcos x=cos xSo, the solution is given by y cosx=cos x exsec x dx+Cy cosx=ex dx+C y cosx=ex+C

Page No 21.139:

Question 43:

dydx-y tan x=ex

Answer:

We have,dydx-y tan x=exComparing with dydx+Py=Q, we getP=-tan x Q=exNow,I.F.=e-tan x dx=e-logsec x=elogcos x=cos xSo, the solution is given byy cos x=excos x dx+Cy cosx=I+C           .....1Where,I=exIIcos xI dx            .....2 I=cos xex dx-ddxcos xex dxdx I=cos x ex+sin x ex dx I=cos x ex+sin xI exII dx I=cos x ex+sin xex dx-ddxsin xex dxdx I=cos x ex+sin xex -cos x ex dx I=cos x ex+sin xex -I        From 2 2I=cos x ex+sin xex  I=ex 2cos x+sin x  y cos x=ex 2cos x+sin x +C        From 1

Page No 21.139:

Question 44:

(1 + y + x2 y) dx + (x + x3) dy = 0

Answer:

1+y+x2ydx+x+x3dy=0dx+y1+x2dx+x1+x2dy=0dx+1+x2 ydx+xdy=01+x2 ydx+xdy=-dxydx+xdy=-11+x2dxydx+xdy=-dx1+x2

On integrating both side we get,

xy=-11+x2dxxy=-tan-1x+cxy+tan-1x=c

Page No 21.139:

Question 45:

(x2 + 1) dy + (2y − 1) dx = 0

Answer:

We have,1+x2dy+2y-1dx=01+x2dy=1-2ydxdy1-2y=11+x2dxIntegrating both sides, we get11-2ydy=11+x2dx-12log1-2y=tan-1x-log C-log1-2y=2tan-1x-2log C-2tan-1x=-log C+log1-2y-2tan-1x=log 1-2yCe-2tan-1x=1-2yCCe-2tan-1x=1-2y1-Ce-2tan-1x=2y12-C2e-2tan-1x=yy=12+Ke-2tan-1x, where K=-C2

Page No 21.139:

Question 46:

y sec2 x + (y + 7) tan x dydx = 0

Answer:

We have,y sec2x+y+7tan xdydx=0y sec2x=-y+7tan xdydx-y-7ydy=sec2xtan xdx-1-7ydy=sec2xtan xdxIntegrating both sides, we get-1-7ydy=sec2xtan xdx-y-7log y=log tan x+log C-y=log tan x+logy7+log C-y=logCy7tan xe-y=Cy7tan xy7tan x=e-yCy7tan x=ke-y, where k=1C

Page No 21.139:

Question 47:

(2ax + x2) dydx = a2 + 2ax

Answer:

2ax+x2dydx=a2+2axdydx=a2+2ax2ax+x2=aa+2xx2a+xLetx=2atan2θ dx=4a tanθsec2θ dθdydx=aa+4atan2θ2a tan2θ 2a1+tan2θdy=a 1+4 tan2θ2 tan2θ 2a sec2θdxdy=a1+4 tan2θ2 tan2θ 2a sec2θ4atanθ sec2θdθ=a 1+4tan2θtanθdθ=a1tanθ+4tanθdθy=acotθ+4 tanθ dθy=alog sinθ+4 log cosθ+cy=alogsinθ4log cosθ+cAs,x=2a tan2θtanθ=x2ay=a log sinθcos4θ+c=alogtanθcos3θ+c=alog x2a×x+2a2a3+cy=alogx12(x+2a)324a2+cy+C=a2 log x+3logx+2a          where C=c-alog4a2

Page No 21.139:

Question 48:

(x3 − 2y3) dx + 3x2 y dy = 0

Answer:

Disclaimer: There seems to be error in the given question.
 

Page No 21.139:

Question 49:

x2 dy + (x2xy + y2) dx = 0

Answer:

We have,x2dy+x2-xy+y2dy=0x2dy=xy-x2-y2dydydx=xy-x2-y2x2This is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we getv+xdvdx=x2v-x2-x2v2x2v+xdvdx=v-1-v2xdvdx=-1-v2dv1+v2=-1xdxIntegrating both sides, we getdv1+v2dv=-1xdxtan-1 v=-log x+log Ctan-1 yx=logCxetan-1yx=CxC=x etan-1yx

Page No 21.139:

Question 50:

y-xdydx=b1+x2dydx

Answer:

We have,y-xdydx=b1+x2dydxy-b=bx2+xdydx1y-bdy=1bx2+xdxIntegrating both sides, we get1y-bdy=1bx2+xdx1y-bdy=1b1x2+1bxdx1y-bdy=1b1x2+1bx+14b2-14b2dx1y-bdy=1b1x+12b2-12b2dxlog y-b=12×12bblog x+12b-12bx+12b+12b+log Clog y-b=log bxbx+1+log Cy-b=Cbxbx+1Cbx=y-bbx+1x=ky-bbx+1, where k=1bC

Page No 21.139:

Question 51:

dydx+2y=sin 3x

Answer:

We have,dydx+2y=sin 3x           .....1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=2 and Q=sin 3x.   I.F.=eP dx          =e2 dx         = e2xMultiplying both sides of 1 by I.F.=e2x, we gete2x dydx+2y=e2xsin 3x e2xdydx+2e2xy=e2xsin 3xIntegrating both sides with respect to x, we gety e2x=e2xsin 3x  dx+Cy e2x=I+C           .....1Where, I=e2xsin 3x dx           .....2I=e2xsin 3x dx-de2xdxsin 3x dxdxI=-e2xcos 3x3+23e2xcos 3x dxI=-e2xcos 3x3+23e2xcos 3x dx-de2xdxcos 3x dxdxI=-e2xcos 3x3+23e2xsin 3x3-23e2xsin 3x dxI=-e2xcos 3x3+2 e2xsin 3x9-49e2xsin 3x dxI=-e2xcos 3x3+2 e2xsin 3x9-49I             Using 213I9=-e2xcos 3x3+2 e2xsin 3x9I=9132 e2xsin 3x9-e2xcos 3x3I=e2x132 sin 3x-3 cos 3x           .....3From 1 and 3, we gety e2x=e2x132 sin 3x-3 cos 3x+Cy=31323sin 3x-cos 3x+Ce-2xHence, y=31323sin 3x-cos 3x+Ce-2x  is the required solution.

Page No 21.139:

Question 52:

dydx+y=4x

Answer:

We have,dydx+y=4x           .....1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=1 and Q=4x I.F.=eP dx          =e dx         = exMultiplying both sides of (1) by I.F.=ex, we getex dydx+y=ex4x exdydx+exy=ex4xIntegrating both sides with respect to x, we gety ex=4x ex dx+Cy ex=4xI exII dx+Cy ex=4xex dx-4ddxxex dxdx+Cy ex=4xex-4ex dx+Cy ex=4xex-4ex+Cy ex=4x-1ex+Cy=4x-1+Ce-xHence, y=4x-1+Ce-x  is the required solution.

Page No 21.139:

Question 53:

dydx+5y=cos 4x

Answer:

We have,dydx+5y=cos 4x           .....1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=15 and Q=cos 4x. I.F.=eP dx          =e 5dx         = e5xMultiplying both sides of (1) by I.F.=e5x, we gete5x dydx+5y=e5xcos 4x e5xdydx+5e5xy=e5xcos 4xIntegrating both sides with respect to x, we gety e5x=e5xcos 4x dx+Cy e5x=I+C       .....2Where,I=e5xcos 4x dx       .....3I=e5xcos 4x dx-de5xdxcos 4x dxdxI=e5xsin 4x4-54e5xsin 4x dxI=e5xsin 4x4-54e5xsin 4x dx-de5xdxsin 4x dxdxI=e5xsin 4x4-54-e5xcos 4x4+54e5xcos 4x dxI=e5xsin 4x4+5e5xcos 4x16-2516e5xcos 4x dxI=e5xsin 4x4+5e5xcos 4x16-2516I        From 34116I=e5xsin 4x4+5e5xcos 4x164116I=e5x164sin 4x+5cos 4xI=e5x414sin 4x+5cos 4x       .....4From 2 and 4 we gety e5x=e5x414sin 4x+5cos 4x+Cy=441sin 4x+54cos 4x+Ce-5xHence, y=441sin 4x+54cos 4x+Ce-5x  is the required solution.

Page No 21.139:

Question 54:

xdydx+x cos2yx=y

Answer:

We have,xdydx+x cos2yx=ydydx+cos2yx=yxdydx=yx-cos2yxPutting y=vx, we getdydx=v+xdvdx v+xdvdx=v-cos2vxdvdx=-cos2 vsec2 v dv=-1xdxIntegrating both sides, we getsec2 v dv=-1xdxtan v =-log x+Ctan yx =-log x+C

Page No 21.139:

Question 55:

cos2 xdydx+y=tan x

Answer:

We have,cos2 xdydx+y=tan xdydx+sec2 xy=tan xsec2 xComparing with dydx+Px=Q, we getP=sec2 x Q=tan xsec2 xNow,I.F.=esec2 x dx=etan xSo, the solution is given byy×etan x=tan xsec2 x×etan x dx + Cyetan x=I + C             .....1Now,I=tan xsec2 x×etan x dxPutting t=tan x, we getdt=sec2 x dx I=tI×etII dt     =t×etdt-dtdt×etdtdt     =tet-etdt     =tet-etI=tan x etan x-etan x=etan xtan x-1Putting the value of I in 1, we getyetan x=etan xtan x-1+ C

Page No 21.139:

Question 56:

x cos x dydx + y (x sin x + cos x) = 1

Answer:

We have,x cos xdydx+y x sin x+cos x=1dydx+tan x+1xy=1x cos xComparing with dydx+Px=Q, we getP=tan x+1x Q=1x cos xNow,I.F.=etan x+1x dx=elog x sec x=x sec xSo, the solution is given byxy sec x=sec2 x dx + Cxy sec x=tan x+C

Page No 21.139:

Question 57:

1+y2+x-e-tan-1 ydydx=0

Answer:

We have,
1+y2+x-e-tan-1 ydydx=0
dxdy=e-tan-1 y-x1+y2dxdy+x1+y2=e-tan-1 y1+y2

Comparing with dxdy+Px=Q, we getP=11+y2 Q=e-tan-1 y1+y2Now,I.F.=e11+y2dy=etan-1 ySo, the solution is given byx×etan-1 y=e-tan-1 y1+y2×etan-1 y dy + Cx×etan-1 y=11+y2 dy + Cxetan-1 y=tan-1 y + C

Page No 21.139:

Question 58:

y2+x+1ydydx=0

Answer:

We have,
y2+x+1ydydx=0
dydx=-y3xy+1dxdy=-xy+1y3dxdy=-xy2-1y3dxdy+xy2=-1y3
Comparing with dxdy+Px=Q, we getP=1y2Q=-1y3Now, I.F.=e1y2dy=e-1ySo, the solution is given byx×e-1y=-e-1y1y3 dy + Cxe-1y=I + C      .....1Now,I=-e-1y1y3 dyPutting t=1y, we getdt=-1y2dy I=tI×e-tII dt     =t×e-tdt-dtdt×e-tdtdt     =-tet+e-tdt     =-te-t-e-t I=-1ye-1y-e-1y=-e-1y1+1yPutting the value of I in 1, we getxe-1y=-e-1y1+1y+ Cx=-1+1y+ Ce1y



Page No 21.140:

Question 59:

2 cos xdydx+4y sin x = sin 2x, given that y = 0 when x = π3.

Answer:

We have,
2cos xdydx+4y sin x=sin 2x
dydx+4ysin x2 cos x=2sin x cos x2 cos xdydx+2y tan x=sin x
Comparing with dydx+Py=Q, we getP=2tan xQ=sin xNow,I.F.=e2tan x dx=e2logsec x=sec2 xSo, the solution is given byy×I.F.=Q×I.F. dx + Cy sec2 x=sin x sec2 x dx + Cy sec2 x=tan x sec x dx + Cy sec2 x=sec x  + Cy=cos x+C cos2x       .....1Now, When x=π3, y=0  0=cos π3  + C cos2 π30=12+C14C=-2Putting the value of C in 1, we gety=cos x-2cos2 x

Page No 21.140:

Question 60:

(1 + y2) dx = (tan−1 yx) dy

Answer:

We have,1+y2dx=tan-1y-xdydxdy=tan-1 y-x1+y2dxdy+x1+y2=tan-1 y1+y2
Comparing with dxdy+Px=Q, we getP=11+y2 Q=tan-1 y1+y2Now, I.F.=e11+y2dy=etan-1 ySo, the solution is given byx×etan-1 y=tan-1 y1+y2×etan-1 y dy + Cxetan-1 y=I + C         .....1Now, I=tan-1 y1+y2×etan-1 y dyPutting t=tan-1 y, we getdt=11+y2dy I=tI×etII dt     =t×etdt-dtdt×etdtdt     =tet-etdt     =tet-et I=tan-1y etan-1 y-etan-1 y    =etan-1 ytan-1 y-1Putting the value of I in 1, we getxetan-1 y=etan-1 ytan-1 y-1+ C

Page No 21.140:

Question 61:

dydx + y tan x = xn cos x, n ≠ − 1

Answer:

We have,dydx+y tan x=xn cos x
Comparing with dydx+Py=Q, we getP=tan x Q=xn cos xNow, I.F.=etan x dx=elogsec x=sec xSo, the solution is given byy×I.F.=Q×I.F. dx + Cy sec x=xn cos x sec x dx + Cy sec x=xn dx + Cy sec x=xn+1n+1  + C

Page No 21.140:

Question 62:

Find the general solution of the differential equation dydx=x+12-y, y2.

Answer:

We have,dydx=x+12-y2-ydy=x+1dxIntegrating both sides, we get2-ydy=x+1dx2y-y22=x22+x+C1x22+x+C1-2y+y22=0x2+2x+y2+2C1-4y=0x2+y2+2x-4y+C=0         Where, C=2C1

Page No 21.140:

Question 63:

Find the particular solution of the differential equation dydx=-4xy2 given that y = 1, when x = 0.

Answer:

We have,dydx=-4xy21y2dy=-4x dxIntegrating both sides, we get1y2dy=-4x dx-1y=-2x2+C      .....1Now, When x=0, y=1  -1=0+CC=-1Putting the value of C in 1, we get-1y=-2x2-11y=2x2+1y=12x2+1

Page No 21.140:

Question 64:

For each of the following differential equations, find the general solution:
(i) dydx=1-cos x1+cos x

(ii) dydx=4-y2, -2<y<2

(iii) dydx=1+x21+y2

(iv) y log y dxx dy = 0

(v) dydx=sin-1x

(vi) dydx+y=1

Answer:

i We have,dydx=1-cos x1+cos xdydx=2sin2 x22cos2 x2dydx=tan2 x2dy=tan2 x2dxIntegrating both sides, we getdy=tan2 x2dxdy=sec2 x2-1dxy= 2 tan x2-x+C

ii We have,dydx=4-y214-y2dy=dxIntegrating both sides, we get14-y2dy=dxsin-1 y2=x+Cy2=sin x+Cy=2sin x+C

iii We have,dydx=1+x21+y211+y2dy=1+x2dxIntegrating both sides, we get11+y2dy=1+x2dxtan-1 y=x+x33+C

iv We have,y log y dx-x dy=0y log y dx=x dy1xdx=1y log ydy1y log ydy=1xdxIntegrating both sides, we get1y log ydy=1xdx       .....1Putting log y=t1ydy=dtTherefore 1 becomes1tdt=1xdxlog t=log x + log Clog log y=log x + log Clog log y=log Cxlog y=Cxy=eCx

v We have,dydx=sin-1xdy=sin-1xdxIntegrating both sides, we getdy=sin-1xdxdy=1II×sin-1xI dx dy=sin-1x1 dx-ddxsin-1x1 dxdxy=x sin-1x-x1-x2dxPutting t2=1-x2, we get2t dt=-2x dx-t dt=x dx y=x sin-1x+dty=x sin-1x+t+Cy=x sin-1x+1-x2+C

vi We have,dydx+y=1dydx=1-y11-ydy=dxIntegrating both sides, we get-1y-1dy=dx1y-1dy=-dxlog y-1=-x+log Clog y-1-log C=-xlog y-1C=-xy-1C=e-xy=1+Ce-x

Page No 21.140:

Question 65:

For each of the following differential equations, find a particular solution satisfying the given condition:
(i) xx2-1dydx=1, y=0 when x=2

(ii) cosdydx=a, y=1 when x=0

(iii) dydx=y tan x, y=1 when x=0

Answer:

i We have,xx2-1dydx=1 dydx=1xx2-1dy=1xx2-1dxIntegrating both sides, we getdy=1xx2-1dxy=1xx2-1dx+Cy=1xx+1x-1dx+C     .....1Let 1xx+1x-1=Ax+Bx+1+Cx-11=Ax+1x-1+Bxx-1+Cxx+11=Ax2-1+Bx2-x+Cx2+x1=x2A+B+C+x-B+C-AComparing both sides, we get-A=1          .....2-B+C=0     .....3A+B+C=0     .....4Solving 2, 3 and 4, we getA=-1B=12C=121xx+1x-1=-1x+12x+1+12x-1Now, 1 becomesy=-1x+12x+1+12x-1dx+Cy=-1xdx+121x-1dx+121x-1dxy=-log x+12log x-1+12log x+1+Cy=12log x-1+12log x+1-log x+CGiven:y2=0 0=12log 2-1+12log 2+1-log 2+CC=log 2-12log 3Substituting the value of C, we gety=12log x-1+12log x+1-log x+log 2-12log 32y=log x-1+log x+1-2log x+2log 2-log 32y=log x-1+log x+1-log x2+log 4-log 32y=logx-1x+1x2-log3-log4y=12logx2-1x2-12log 34

ii We have,cos dydx=a  dydx=cos-1 ady=cos-1 a dxIntegrating both sides, we getdy=cos-1 a dxy=x cos-1 a+CNow, When x=0, y=1  1=0+CC=1Putting the value of C in 1, we gety=x cos-1 a+1cosy-1x=a

iii We have,dydx=y tan x1ydy=tan x dxIntegrating both sides, we get1ydy=tan x dxlog y=log sec x+C     ....1Now, When x=0, y=1  log 1=log 1+CC=0Putting the value of C in 1, we getlog y=log sec xy=sec x

Page No 21.140:

Question 66:

Solve the each of the following differential equations:
(i) x-ydydx=x+2y

(ii) x cosyxdydx=y cosyx+x

(iii) y dx + x log yx dy − 2x dy = 0

(iv) dydx-y=cos x

(v) xdydx+2y=x2, x0

(vi) dydx+2y=sin x

(vii) dydx+3y=e-2x

(viii) dydx+yx=x2

(ix) dydx+sec x y=tan x

(x) xdydx+2y=x2 log x

(xi) x log xdydx+y=2xlog x

(xii) (1 + x2) dy + 2xy dx = cot x dx

(xiii) x+ydydx=1

(xiv) y dx + (xy2) dy = 0

(xv) x+3y2dydx=y

Answer:

i We have,x-ydydx=x+2ydydx=x+2yx-y     .....1Clearly this is a homogeneous equation,Putting y=vxdydx=v+xdvdxSubstituting y=vx and dydx=v+xdvdx 1 becomes,v+xdvdx=x+2vxx-vxv+xdvdx=1+2v1-vxdvdx=1+2v1-v-vxdvdx=1+2v-v+v21-vxdvdx=v2+v+11-v1-vv2+v+1dv=1xdx-vv2+v+1+1v2+v+1dv=1xdx-12×2v+1-1v2+v+1+1v2+v+1dv=1xdx-12×2v+1v2+v+1+12×1v2+v+1+1v2+v+1dv=1xdx-12×2v+1v2+v+1+32×1v2+v+1dv=1xdx-12×2v+1v2+v+1+32×1v2+v+14+34dv=1xdx-12×2v+1v2+v+1+32×1v+122+322dv=1xdxIntegrating both sides, we get-12×2v+1v2+v+1+32×1v+122+322dv=1xdx-122v+1v2+v+1dv+321v+122+322dv=1xdx-12log v2+v+1+32×132tan-1v+1232=log x+C-12log yx2+yx+1+32×132tan-1yx+1232=log x+C-12log y2+xy+x2x2+3tan-12y+x3x=log x+C-12log y2+xy+x2+12log x2+3tan-12y+x3x=log x+C-12log y2+xy+x2+log x+3tan-12y+x3x=log x+C-12log y2+xy+x2+3tan-12y+x3x=Clog y2+xy+x2-23tan-12y+x3x=-2Clog y2+xy+x2=23tan-12y+x3x-2Clog y2+xy+x2=23tan-12y+x3x+k         Where, k=-2C

ii We have,x cos yxdydx=y cosyx+xdydx=y cosyx+xx cos yx     .....1Clearly this is a homogeneous equation,Putting y=vxdydx=v+xdvdxSubstituting y=vx and dydx=v+xdvdx in 1 we getdydx=y cosyx+xx cos yxv+xdvdx=vx cos v+xx cos vv+xdvdx=v cos v+1cos vxdvdx=v cos v+1cos v-vxdvdx=v cos v+1- v cos vcos vxdvdx=1cos vcos v dv=1xdxIntegrating both sides, we getcos v dv=1xdxsin v=log x+log Csin yx=log Cx

iii We have,y dx+x log yxdy-2x dy=0x log yxdy-2x dy=-y dxlog yx-2x dy=-y dxdydx=-ylog yx-2xdydx=yx2-log yx     .....1Clearly this is a homogenous equation,Putting y=vxdydx=v+xdvdxSubstituting y=vx and dydx=v+xdvdx in 1 we getv+xdvdx=v2-log vxdvdx=v2-log v-vxdvdx=v-2v+v log v2-log vxdvdx=-v+v log v2-log v2-log v-v+v log vdv=1xdxlog v-2v log v-vdv=-1xdxlog v-1-1v log v-1dv=-1xdxlog v-1v log v-1dv-1v log v-1dv=-1xdx1vdv-1v log v-1dv=-1xdxIntegrating both sides we get1vdv-1v log v-1dv=-1xdxlog v-I =-log x-log C     .....2Where,I=1v log v-1dvPuting log v=t1vdv=dtI=1t-1dtI=log t-1I=log log v-1     .....3From 2 and 3 we getlog v-log log v-1 =-log x-log Clog vlog v-1 =-log Cxvlog v-1=1Cxlog v-1=vCxlog yx-1=Cy


iv We have,dydx-y=cos xComparing with dydx+Py=Q, we getP=-1 Q=cos xNow, I.F.=e-1dx =e-xSolution is given by,y×I.F.=cos x×I.F. dx+Cye-x=e-x cos x dx+Cye-x=I+C     .....1Where,I=e-xIIcos x Idx     .....2 I=cos xe-x dx-ddxcos xe-x dxdx I=-cos x e-x-sin x e-x dx I=-cos x e-x-sin xI e-xII dx I=-cos x e-x-sin xe-x dx+ddxsin xe-x dxdx I=-cos x e-x+sin x e-x -cos x e-x dx I=-cos x e-x+sin xe-x-I      Using 2 2I=-cos x e-x+sin xe-x  I=12-cos x+sin xe-x     .....3From 1 and 3, we get ye-x=sin x-cos xe-x +Cy=12sin x-cos x+Cex


v We have,xdydx+2y=x2dydx+2xy=xComparing with dydx+Py=Q, we getP=2x Q=xNow,I.F.=e21xdx=e2log x=x2So, the solution is given by y×I.F.=Q×I.F. dx +Cyx2=x3 dx+Cyx2=x44+Cy=x24+Cx-2


vi We have,dydx+2y=sin xComparing with dydx+Py=Q, we getP=2 Q=sin xNow,I.F.=e2dx =e2xSolution is given by,y×I.F.=sin x×I.F. dx+Cye2x=I+C     .....1Where,I=e2xIIsin xIdx       .....2 I=sin xe2x dx-ddxsin xe2x dxdx I=sin x e2x2-12cos x e2x dx I=sin x e2x2-12cos xI e2xII dx I=sin x e2x2-12cos xe2x dx+12ddxcos xe2x dxdx I=sin x e2x2-14cos x e2x -14sin x e2x dx I=sin x e2x2-14cos x e2x -14I            Using 2 I+14I=12sin x e2x-14cos x e2x  54I=142sin x e2x-cos xe2x I=152sin x-cos xe2x       .....3Therefore from 1 and 3, we get ye2x=152sin x-cos xe2x +Cy=152 sin x-cos x+Ce-2x


vii We have,dydx+3y=e-2xComparing with dydx+Py=Q, we getP=3Q=e-2xNow,I.F.=eP dx=e3dx =e3xSo, the solution is given byy×I.F.=Q×I.F. dx +Cye3x=e3x×e-2xdx+Cye3x=ex+Cy=e-2x+Ce-3x


viii We have,dydx+yx=x2 dydx+1xy=x2Comparing with dydx+Py=Q, we getP=1x Q=x2Now,I.F.=e1xdx =elogx=xSo, the solution is given byy×I.F.=Q×I.F. dx +Cyx=x3+Cxy=x44+C


ix We have,dydx+ sec xy=tan xComparing with dydx+Py=Q, we getP=sec xQ=tan xNow,I.F.=esec x dx=elogsec x+tan x=sec x+tan xSo, the solution is given byy×I.F=Q×I.F. dx +Cysec x+tan x=sec x+tan xtan x+Cysec x+tan x=sec x×tan x dx+tan2 x dx+Cysec x+tan x=sec x×tan x dx+sec2 x-1 dx+Cysec x+tan x=sec x+tan x-x+C


x We have,xdydx+2y=x2 log xDividing both sides by x, we getdydx+2yx=x log xComparing with dydx+Py=Q, we getP=2x Q=x log xNow,I.F.=ePdx=e2xdx=e2logx=x2So, the solution is given byy×I.F.=Q×I.F. dx+Cx2y=x3IIlog xI dx+Cx2y=log xx3 dx-ddxlog xx3 dxdx+Cx2y=x4log x4-x34dx+Cx2y=x4log x4-x416+Cy=x2log x4-x216+Cx2y=x2164log x-1+Cx-2


xi We have,x log xdydx+y=2xlog xDividing both sides by x log x, we getdydx+yx log x=2x log xx logxdydx+yx log x=2 x2dydx+1x log xy=2 x2Comparing with dydx+Py=Q, we getP=1x log x Q=2 x2Now, I.F.=ePdx=e1x log xdx=eloglog x=log xSo, the solution is given byy×I.F.=Q×I.F. dx+Cylog x=21x2×log x dx+Cylog x=I+C        .....1Where,I=21x2II log x Idx I=2log x1x2 dx-2ddxlog x1x2 dxdx I=-2xlog x+21x2 dx I=-2xlog x-2x        .....2From 1 and 2 we get ylog x=-2xlog x-2x+Cylog x=-2xlog x+1+C

xii We have,1+x2dy+2xy dx=cot x dxdydx+2x1+x2y=cot x1+x2Comparing with dydx+Py=Q, we getP=2x1+x2Q=cot x1+x2Now,I.F.=e2x1+x2dx =elog1+x2=1+x2So, the solution is given byy×I.F.=Q×I.F. dx +Cy1+x2=cot x1+x2×1+x2 dx+Cy1+x2=cot x dx+Cy1+x2=log  sin x+Cy=1+x2-1log sin x+C1+x2-1


xiii We have,x+ydydx=1dydx=1x+yLet x+y=v1+dydx=dvdxdydx=dvdx-1 dvdx-1=1vdvdx=1v+1dvdx=1+vvv1+vdv=dxIntegrating both sides, we getv1+vdv=dxv+1-11+vdv=dxdv-11+vdv=dxv-log v+1=x-log Cx+y-log x+y+1=x-log Cy-log x+y+1=-log Cy=logx+y+1-log Cy=logx+y+1CCey=x+y+1


xiv We have,y dx+x-y2dy=0y dx=-x-y2dy dxdy=-1yx-y2  dxdy+1yx=y        .....1Clearly, it is a linear differential equation of the form dxdy+Px=Qwhere P=1y  and Q=y I.F.=eP dy          =e1ydy         = elog y=yMultiplying both sides of 1 by I.F.=y, we getydxdy+1yx= y×yydxdy+x=y2Integrating both sides with respect to y, we getxy=y2dy+Cxy=y33+Cx=y23+CyHence, x=y23+Cy  is the required solution.


xv We have,x+3y2dydx=ydxdy=1yx+3y2  dxdy-1yx=3y        .....1Clearly, it is a linear differential equation of the formdxdy+Px=Qwhere P=-1y  and Q=3y I.F.=eP dy          =e-1ydy         = e-log y=1yMultiplying both sides of (1) by I.F.=1y, we get1ydxdy-1yx= 1y×3y1ydxdy-1yx=3Integrating both sides with respect to y, we getx1y= 3dy+Cx1y=3y+Cx=3y2+CyHence, x=3y2+Cy  is the required solution.

Page No 21.140:

Question 67:

Find a particular solution of each of the following differential equations:
(i) 1+x2dydx+2xy=11+x2; y=0, when x=1

(ii) (x + y) dy + (xy) dx = 0; y = 1 when x = 1

(iii) x2 dy + (xy + y2) dx = 0; y = 1 when x = 1

Answer:

i We have,1+x2dydx+2xy=11+x2dydx+2x1+x2y=11+x22Comparing with dydx+Py=Q, we getP=2x1+x2 Q=11+x22Now, I.F.=e2x1+x2dx =elog 1+x2=1+x2So, the solution is given byy×I.F.=Q×I.F. dx +Cy1+x2=11+x2 dx+Cy1+x2=tan-1 x+C     .....1Now, When x=1, y=0  01+1=tan-1 1+CC=-1C=-π4Putting the value of C in 1, we gety1+x2=tan-1 x-π4

ii We have,x+ydy+x-ydx=0dydx=y-xx+yLet y=vxdydx=v+xdvdx v+xdvdx=vx-xx+vxxdvdx=v-11+v-vx dvdx=v-1-v-v21+vxdvdx=-v2+11+v1+vv2+1dv=-1xdx

Integrating both sides, we get1+v1+v2dy=-1xdx112+v2dy+122v1+v2=-1xdxtan-1 v+12log1+v2=-log x+C2tan-1 v+log1+v2+2log x =2C2tan-1 v+log1+v2x2=k      where, k=2C2tan-1 yx+log1+y2x2x2=k2tan-1 yx+log x2+y2=k     .....1Now, When x=1, y=1  2tan-1 1+log 2=kk=π2+log 2Putting the value of k in 1, we get2tan-1 yx+log x2+y2=π2+log 2


iii We have,x2dy+xy+y2dx=0dydx=-xy+y2x2Let y=vxdydx=v+xdvdx v+xdvdx=-vx2+v2x2x2xdvdx=-v+v2-vxdvdx=-2v-v21v2+2vdv=-1xdx

Integrating both sides, we get1v2+2vdy=-1xdx1v2+2v+1-1dy=-1xdx1v+12-12dy=-1xdx12×1logv+1-1v+1+1=-log x+log C12logvv+2=-log x+log Clogvv+2=-2log x+2log Clogvv+2+log x2=log C2logvx2v+2=log C2  vx2v+2=C2 vx2v+2=k,  where k=2Cyxx2yx+2=k   x2yy+2x=k      .....1Now, When x=1, y=1  11+2=k  k=13Putting the value of k in 1, we getx2yy+2x=133x2y=±y+2xBut y1=1 does not satisfy the equation 3x2y=-y+2x.  3x2y=y+2xy=2x3x2-1



Page No 21.141:

Question 68:

Find the equation of the curve passing through the point (1, 1) whose differential equation is x dy = (2x2 + 1) dx, x ≠ 0.

Answer:

We have,x dy=2x2+1dxdy=2x2+1xdxdy=2x+1xdxIntegrating both sides, we getdy=2x+1xdxy=x2+log x+C     .....1Now the given curve passes through 1, 1Therefore, when x=1, y=1 1=1+0+CC=0Putting the value of C in 1, we gety=x2+logx

Page No 21.141:

Question 69:

Find the equation of a curve passing through the point (−2, 3), given that the slope of the tangent to the curve at any point (x, y) is 2xy2.

Answer:

We have,dydx=2xy2y2dy=2x dxIntegrating both sides, we gety2dy=2x dxy33=x2+C     .....1Now the given curve passes theough -2, 3Therefore, when x=-2, y=3 Substituting x=-2 and y=3 in 1 we get333=-22+C 9=4+CC=5Putting the value of C in 1, we gety33=x2+5y3=3x2+15y=3x2+1513

Page No 21.141:

Question 70:

Find the equation of a curve passing through the point (0, 0) and whose differential equation is dydx=ex sin x.

Answer:

We have,
dydx=ex sin x

dy=exsin x dxIntegrating both sides, we getdy=exsin x dxy=I+C     .....1Where I=exIIsin x Idx+C     .....2 I=sin xex dx-ddxsin xex dxdx I=sin x ex-cos x ex dx I=sin x ex-cos xI exII dx I=sin x ex-cos xex dx+ddxcos xex dxdx I=sin x ex-cos x ex -sin x ex dx I=sin x ex-cos x ex -I        From 2 2I=sin x ex-cos x ex I=12exsin x-cos x     .....3From 1 and 3 we get y=12exsin x-cos x+C          ...4Now equation of the curve passes through 0, 0Therefore when x=0; y=0Putting x=0 and  y=0 in 4 we get 0=12e0sin 0-cos 0+CC=12Substituting the value of C in 4, we gety=12exsin x-cos x+122y-1=exsin x-cos x

Page No 21.141:

Question 71:

At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (−4, −3). Find the equation of the curve given that it passes through (−2, 1).

Answer:

The slope of the line having points (x, y) and (−4, −3) is given by y+3x+4.
According to the question,

dydx=2y+3x+4

1y+3dy=2x+4dxIntegrating both sides, we get1y+3dy=21x+4dxlog y+3=2log x+4+log Clog y+3=log Cx+42y+3=Cx+42Since the curve passes through -2, 1, it satisfies the equation of the curve. 1+3=C-2+42C=1Putting the value of C in the equation of the curve, we gety+3=x+42

Page No 21.141:

Question 72:

Show that the family of curves for which the slope of the tangent at any point (x, y) on it is x2+y22xy is given by x2y2 = Cx.

Answer:

We have,
dydx=x2+y22xyLet y=vxdydx=v+xdvdx v+xdvdx=x2+v2x22vx2xdvdx=1+v22v-vxdvdx=1+v2-2v22vxdvdx=1-v22v2v1-v2dv=1xdx

Integrating both sides, we get2v1-v2dy=1xdx-log 1-v2=log x-log Clog 1-v2C=-log x1-v2=Cx1-yx2=Cxx2-y2x2=Cxx2-y2=Cx

Page No 21.141:

Question 73:

Find the equation of a curve passing through the point (0, 1). If the slope of the tangent to the curve at any point (x, y) is equal to the sum of the x-coordinate and the product of the x-coordinate and y-coordinate of that point.

Answer:

According to the question,
dydx=x+xydydx=x1+y1y+1dy=x dxIntegrating both sides, we get1y+1dy=x dxlog y+1=x22+log Clog y+1C=x22y+1=Cex22Since, the curve passes through 0, 1.It satisfies the equation of the curve.1+1=Ce0C=2Puting the value of C in the equation of the curve. We get y+1=2ex22 y=-1+2ex22

Page No 21.141:

Question 74:

Find the equation of the curve passing through the origin given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of the coordinates of the point.

Answer:

According to the question,
dydx=x+y
dydx-y=x
Comparing with dydx+Py=Q, we getP=-1 Q=xNow, I.F.=e-dx =e-xSo, the solution is given byy×I.F.=Q×I.F. dx +Cye-x=xIe-xIIdx+Cye-x=xe-x dx-ddxxe-x dxdx+Cye-x=-xe-x+e-x dx+Cye-x=-xe-x-e-x+CSince the curve passes throught the origin, it satisfies the equation of the curve.0e0=-0e0-e0+CC=1Putting the value of C in the equation of the curve, we getye-x=-xe-x-e-x+1ye-x+xe-x+e-x=1y+x+1e-x=1x+y+1=ex

Page No 21.141:

Question 75:

Find the equation of the curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5.

Answer:

According to the question,
dydx+5=x+y
dydx-y=x-5
Comparing with dydx+Py=Q, we getP=-1 Q=x-5Now,I.F.=e-dx =e-x       So, the solution is given byy×I.F.=Q×I.F. dx +Cye-x=x-5e-xdx +Cye-x=xIe-xIIdx-5e-x dx+Cye-x=xe-x dx-ddxxe-x dxdx+5e-x+Cye-x=-xe-x+e-x dx+5e-x+Cye-x=-xe-x-e-x+5e-x+Cye-x=-xe-x+4e-x+CSince the curve passes throught the point 0, 2, it satisfies the equation of the curve.2e0=-0e0+4e0+CC=-2Putting the value of C in the equation of the curve, we getye-x=-xe-x+4e-x-2y=-x+4-2exy=4-x-2ex

Page No 21.141:

Question 76:

The slope of the tangent to the curve at any point is the reciprocal of twice the ordinate at that point. The curve passes through the point (4, 3). Determine its equation.

Answer:

According to the question,
dydx=12y2y dy=dxIntegrating both sides, we get2y dy=dxy2=x+CSince the curve passes throught the point 4, 3, it satisfies the equation of the curve.9=4+CC=5Putting the value of C in the equation of the curve, we gety2=x+5

Page No 21.141:

Question 77:

The decay rate of radium at any time  t is proportional to its mass at that time. Find the time when the mass will be halved of its initial mass.

Answer:

Let N be the initial amount of radium and P be the amount of radium present at any time t.
We have,dPdtαPdPdt=aP, where a<0dPP=adtlog P=at+C          .....1Now, P=N at t=0Putting P=N and t=0 in 1, we getlog N=C  Putting C=logN in 1, we getlog P=at+log Nlog PN=at         .....2According to the question,log2NN=atlog2=att=1alog2, where a is a constant of proportionality

Page No 21.141:

Question 78:

Experiments show that radium disintegrates at a rate proportional to the amount of radium present at the moment. Its half-life is 1590 years. What percentage will disappear in one year?

Answer:

Let the original amount of radium be N and the amount of radium at any time t be P.

We have,dPdtαPdPdt=-aPdPP=-a dtlog P=-at+C          .....1Now, P=N at t=0 Putting P=N and t=0 in 1, we getlog N=CPutting C=log N in 1, we getlog P=-at+log NlogPN=-at         .....2According to the question,P=12N at t=1590logN2N=-1590aa=11590log 2Putting a=11590log2 in 2, we getlogPN=-11590log2tPN=e-log 21590t       .....3Putting t=1 in 3 to find the bacteria after 1 year, we getPN=e-log 21590 PN=0.9996P=0.9996NPercentage of amount disapeared in 1 year=N-PN×100%=N-0.9996NN×100%=0.04%

Page No 21.141:

Question 79:

A wet porous substance in the open air loses its moisture at a rate proportional to the moisture content. If a sheet hung in the wind loses half of its moisture during the first hour, when will it have lost 95% moisture, weather conditions remaining the same.

Answer:

Let the original amount of moisture in the porous substance be N and the amount of moisture in the porous substance at any time t be P.

Given: dPdtαPdPdt=-aPdPP=-a dtlog P=-at+C          .....1Now, P=N at t=0Putting P=N and t=0 in 1, we getlog N=CPutting C=log N in 1, we getlog P=-at+log Nlog PN=-at         .....2According to the question,P=12N at t=1log N2N=-aa=log 2Putting a=log 2 in 2, we getlog PN=-t log2To find the time when it will loss 95% moisture, we haveP=1-95%N=5100N log 5N100N=-t log 2log 20=t log 2t=log 20log 2



Page No 21.144:

Question 1:

The integrating factor of the differential equation (x log x) dydx+y=2 log x, is given by
(a) log (log x)

(b) ex

(c) log x

(d) x

Answer:

(c) log x

We have,
(x log x) dydx+y=2 log x
Dividing both sides by x log x, we get
dydx+yxlogx=2 log xxlogxdydx+yxlogx=2 xdydx+1xlogxy=2 xComparing with dydx+Py=Q, we getP=1xlogxQ=2 xNow, I.F.=ePdx=e1x log xdx                =eloglog x                =log x

Page No 21.144:

Question 2:

The general solution of the differential equation dydx=yx is
(a) log y = kx

(b) y = kx

(c) xy = k

(d) y = k log x

Answer:

(b) y = kx

We have,
dydx=y x1ydy=1 xdxIntegrating both sides, we get1ydy=1 xdxlog y=log x+log klog y-log x=log klogyx=log kyx= ky= kx

Page No 21.144:

Question 3:

Integrating factor of the differential equation cos xdydx+y sin x = 1, is
(a) sin x
(b) sec x
(c) tan x
(d) cos x

Answer:

(b) sec x

We have,
cos xdydx+y sin x=1
Dividing both sides by cos x, we get
dydx+sin xcos xy=1cos xdydx+tan xy=1cos xComparing with dydx+Py=Q, we getP=tan xQ=1 cos xNow, I.F.=etan xdx=elogsec x                      =sec x

Page No 21.144:

Question 4:

The degree of the differential equation d2ydx22-dydx=y3, is
(a) 1/2
(b) 2
(c) 3
(d) 4

Answer:

(b) 2

We have,
d2ydx22-dydx=y3
The highest order derivative is d2yd2x and its power is 2.Hence, the degree is 2.

Page No 21.144:

Question 5:

The degree of the differential equation 5+dydx25/3=x5d2ydx2, is
(a) 4
(b) 2
(c) 5
(d) 10

Answer:

We have,
5+dydx253=x5 d2yd2xTaking Cube power on both sides, we get5+dydx25=x15 d2yd2x3The highest order  derivative is d2yd2x and its power is 3.Hence, the degree is 3.


Disclaimer: The correct option is not given in the question.

Page No 21.144:

Question 6:

The general solution of the differential equation dydx+y cot x = cosec x, is
(a) x + y sin x = C
(b) x + y cos x = C
(c) y + x (sin x + cos x) = C
(d) y sin x = x + C

Answer:

(d) y sin x = x + C


We have,dydx+y cot x=cosec xdydx+ycot x=cosec xComparing with dydx+Py=Q, we getP=cot x Q=cosec xNow, I.F.=ecot x dx=elogsin x                       =sin xSo, the solution is given byysinx=sin x×cosec x dx+Cy sin x=x +C

Page No 21.144:

Question 7:

The differential equation obtained on eliminating A and B from y = A cos ωt + B sin ωt, is
(a) y" + y' = 0
(b) y" − ω2 y = 0
(c) y" = −ω2 y
(d) y" + y = 0

Answer:

(c) y" = −ω2 y

We have,
y = A cos ωt + B sin ωt                                            .....(1)
Differentiating both sides of (1) with respect to x, we get
dydt=-Aω sin ωt+Bω cos ωt                                            .....(2)
Differentiating both sides of (2) again with respect to x, we get
d2ydt2=-Aω2 cos ωt-Bω2 sin ωtd2ydt2=-ω2A cos ωt+B sin ωtd2ydt2=-ω2y        Using 1y''=-ω2y

Page No 21.144:

Question 8:

The equation of the curve whose slope is given by dydx=2yx; x>0, y>0 and which passes through the point (1, 1) is
(a) x2 = y
(b) y2 = x
(c) x2 = 2y
(d) y2 = 2x

Answer:

(a) x2 = y

We have,
dydx=2yx12×1ydy=1 xdxIntegrating both sides, we get121ydy=1 xdx12log y=log x+log Clog y12-log x=log Clogyx=log Cyx=Cy=Cx          .....1As 1 passes through (1, 1), we get1=CPutting the value of C in 1, we gety=xy=x2

Page No 21.144:

Question 9:

The order of the differential equation whose general solution is given by
y = c1 cos (2x + c2) − (c3 + c4) ax + c5 + c6 sin (xc7) is
(a) 3
(b) 4
(c) 5
(d) 2

Answer:

(c) 5

The given equation can be reduced to : y=c1cos(2x+c2)-(c)ax×ac5+c6sin(x-c7)where c=c3+c4 and ac5 will be a constant

There are 5 constants(c1, c2, c, c6, c7) in the given differential equation.
Hence, the order of the differential equation is 5.

Page No 21.144:

Question 10:

The solution of the differential equation dydx=ax+gby+f represents a circle when
(a) a = b
(b) a = −b
(c) a = −2b
(d) a = 2b

Answer:

(b) a = −b

We have,
dydx=ax+gby+fby+fdy=ax+gdxIntegrating both sides, we getby+fdy=ax+gdxby22+fy=ax22+gx+Cby22+fy-ax22-gx=Cby2+2fy-ax2-2gx-2C=0The above equation represents a circle.Therefore, the coffecients of x2 and y2 must be equal. i.e. -a=ba=-b

Page No 21.144:

Question 11:

The solution of the differential equation dydx+2yx=0 with y(1) = 1 is given by
(a) y=1x2

(b) x=1y2

(c) x=1y

(d) y=1x

Answer:

(a) y=1x2

We have,

dydx+2y x=0dydx=-2y x12×1ydy=-1 xdxIntegrating both sides, we get121ydy=-1 xdx12log y=-log x+log Clog y12+log x=log Clogyx=log Cyx=C            .....1As1 satisfies y1=1, we get1=CPutting the value of C in 1, we getyx=1y=1x2



Page No 21.145:

Question 12:

The solution of the differential equation dydx-yx+1x=0 is given by
(a) y = xex + C
(b) x = yex
(c) y = x + C
(d) xy = ex + C

Answer:

(a) y = xex + C

We have,
dydx-yx+1x=0dydx=yx+1xdyy=x+1xdxIntegrating both sides, we getdyy=x+1xdxdyy=dx+1xdxlog y=x+log x+Clog y-log x=x+Clog yx=x+Cyx=ex+Cy=xex+C

Page No 21.145:

Question 13:

The order of the differential equation satisfying 1-x4+1-y4=ax2-y2 is
(a) 1
(b) 2
(c) 3
(d) 4

Answer:

(a) 1

The order of a differential equation depends on the number of arbitrary constants in it.

Since 1-x4+1-y4=ax2-y2 contains only 1 constant, the order of the differential equation is 1.

Page No 21.145:

Question 14:

The solution of the differential equation y1 y3 = y22 is
(a) x = C1 eC2y + C3
(b) y = C1 eC2x + C3
(c) 2x = C1 eC2y + C3
(d) none of these

Answer:

(b) y = C1 eC2x + C3

y1y3=y22y3y2=y2y1d3ydx3d2ydx2=d2ydx2dydxddxd2ydx2d2ydx2=ddxdydxdydxlnd2ydx2=lndydx+ln C4d2ydx2=C4dydxddxdydxdydx=C4 dxlndydx=C4x+C5dydx=eC4x+C5dy= eC4x+C5 dxy=eC4x+C5C4+C6y=eC4x.eC5C4+C6y=C1eC2x+C3where,C1=eC5C4C4=C2C6=C3

Page No 21.145:

Question 15:

The general solution of the differential equation dydx+y g' (x) = g (x) g' (x), where g (x) is a given function of x, is
(a) g (x) + log {1 + y + g (x)} = C
(b) g (x) + log {1 + yg (x)} = C
(c) g (x) − log {1 + yg (x)} = C
(d) none of these

Answer:

(b) g (x) + log {1 + yg (x)} = C


We have,dydx+y g'x=gxg'x           .....1Clearly, it is a linear differential equation of the form dydx+Py=Qwhere P=g'x and Q=gxg'x.  I.F.=eP dx          =eg'x dx         = egxMultiplying both sides of (1) by I.F., we get egx dydx+yg'x= egx gxg'x egxdydx+ egxy g'x= egxgxg'xIntegrating both sides with respect to x, we gety egx=egxgxg'x dx+Ky egx=I+K where I=egxgxg'x dxNow,  I=egxgxg'x dxPutting gx=t, we getg'x dx=dt I=tIetII dt    =tet dt-ddxtet dtdt    =tet-et    =gxegx-egx y egx=gxegx-egx+Ky egx+egx-gxegx=Ky+1-gx=Ke-gxTaking log on both sides, we getlogy+1-gx=-gx+log Kgx+log1+y-gx=C         Where, C=log K

Page No 21.145:

Question 16:

The solution of the differential equation dydx=1+x+y2+xy2, y0=0 is
(a) y2=expx+x22-1

(b) y2=1+C expx+x22

(c) y = tan (C + x + x2)

(d) y=tanx+x22

Answer:

d y=tanx+x22

We have,
dydx=1+x+y2+xy2dydx=x+1+y2x+1dydx=x+11+y2dy1+y2=x+1dxIntegrating both sides, we getdy1+y2=x+1dxtan-1 y=x22+x+C          .....1Now, y0=0 tan-1 0=02+0+CC=0Putting the value of C in 1, we gettan-1 y=x22+xy=tanx22+x

Page No 21.145:

Question 17:

The differential equation of the ellipse x2a2+y2b2=C is
(a) y"y'+y'y-1x=0

(b) y"y'+y'y+1x=0

(c) y"y'-y'y-1x=0

(d) none of these

Answer:

(a) y"y'+y'y-1x=0

We have,
x2a2+y2b2=C                                    .....1
Differentiating with respect to x, we get
2xa2+2yb2y'=0xa2+yb2y'=0                                  .....2Again differentiating with respect to x, we get1a2+1b2y'2+yb2y''=0             .....3Multiplying throughout by x, we getxa2+xb2y'2+xyb2y''=0                  .....4Subtracting 2 from 4, we get1b2xy'2+xyy''-yy'=0 xy'2+xyy''-yy'=0Dividing both sides by xyy', we gety'y+y''y'-1x=0y''y'+y'y-1x=0 

Page No 21.145:

Question 18:

Solution of the differential equation dydx+yx = sin x is
(a) x (y + cos x) = sin x + C
(b) x (y − cos x) = sin x + C
(c) x (y + cos x) = cos x + C
(d) none of these

Answer:

(a) x (y + cos x) = sin x + C


We have,dydx+yx=sin xdydx+1xy=sin x          .....1Comparing with dydx+Py=Q, we getP=1x Q=sin xNow,I.F.=e1xdx =elogx                     =xTherefore, integration of 1 is given byy×I.F.=x2×I.F. dx+C yx=xI sin xIIdx+Cyx=xsin x dx-ddxxsin x dxdx+Cyx=-x cos x+cos x dx+Cyx+x cos x=sin x+Cxy+cos x=sin x+C

Page No 21.145:

Question 19:

The equation of the curve satisfying the differential equation y (x + y3) dx = x (y3x) dy and passing through the point (1, 1) is
(a) y3 − 2x + 3x2 y = 0
(b) y3 + 2x + 3x2 y = 0
(c) y3 + 2x −3x2 y = 0
(d) none of these

Answer:

(c) y3 + 2x −3x2 y = 0

We have,

yx+y3dx=xy3-xdyHere, xy+y4dx=xy3-x2dyxydx+y4dx-xy3dy+x2dy=0xydx+xdy+y3ydx-xdy=0xdxy+x2y3ydx-xdyx2=0                   xdxy-x2y3xdy-ydxx2=0                    dxyx2y2-yxdyx=0                               Dividing the whole equation by x3y2dxyx2y2=yxdyx
Integrating both sides we get,
dxyx2y2=yxdyx-1xy=yx22-c-1xy-12yx2-c=01xy+12y2x2+c=0y3+2x+2cx2y=0

It is given that the curves passes through (1, 1).
Hence,
y3+2x+2cx2y=013+21+2c11=01+2+2c=02c=-3c=-32
∴ The required curve is y3+2x-2×32x2y=0
y3+2x-3x2y=0

Page No 21.145:

Question 20:

The solution of the differential equation 2xdydx-y=3 represents
(a) circles
(b) straight lines
(c) ellipses
(d) parabolas

Answer:

(d) parabolas

We have,
2xdydx-y=32xdydx=3+y13+ydy=12xdxIntegrating both sides, we get13+ydy=121xdxlog 3+y=12log x+log Clog 3+y-log x12=log Clog 3+yx=log C3+yx=C3+y=CxSquaring both sides, we get3+y2=Cx             .....1Thus, 1 represents the equation of parabolas.

Page No 21.145:

Question 21:

The solution of the differential equation xdydx=y+x tanyx, is
(a) sinxy=x+C

(b) sinyx=Cx

(c) sinxy=Cy

(d) sinyx=Cy

Answer:

(b) sinyx=Cx

We have,
xdydx=y+x tanyxdydx=yx+tanyx        .....1Let y=vxdydx=v+xdvdxPutting the above value in 1, we getv+xdvdx=v+tan vxdvdx=tan vdvtan v=dxxIntegrating both sides, we getlog sin v= log x+log Clog sin v- log x=log Clogsin vx=log Csin vx=Csin v=Cxsinyx=Cx           y=vx



Page No 21.146:

Question 22:

The differential equation satisfied by ax2 + by2 = 1 is
(a) xyy2 + y12 + yy1 = 0
(b) xyy2 + xy12yy1 = 0
(c) xyy2xy12 + yy1 = 0
(d) none of these

Answer:

(b) xyy2 + xy12yy1 = 0

We have,
ax2 + by2 = 1                                           .....(1)
Differentiating both sides of (1) with respect to x, we get
2ax+2bydydx=0                                      .....2
Differentiating both sides of (2) with respect to x, we get
2a+2bdydx2+2byd2ydx2=02byd2ydx2+dydx2=-2ayd2ydx2+dydx2=-2a2byd2ydx2+dydx2=--yxdydx                      Using 2xyd2ydx2+dydx2=ydydxxyd2ydx2+xdydx2=ydydxxyd2ydx2+xdydx2-ydydx=0xyy2+xy12-yy1=0

Page No 21.146:

Question 23:

The differential equation which represents the family of curves y = eCx is
(a) y1 = C2 y
(b) xy1 − ln y = 0
(c) x ln y = yy1
(d) y ln y = xy1

Answer:

(d) y ln y = xy1

We have,
y = eCx
Taking ln on both sides, we get
ln y = Cx ln e
ln y=Cx                                    .....1
Differentiating both sides of (1) with respect to x, we get
1yy1=C
Substituting the value of C in (1), we get
ln y=y1yxy ln y=y1x

Page No 21.146:

Question 24:

Which of the following transformations reduce the differential equation dzdx+zxlog z=zx2log z2 into the form dudx+Px u=Qx
(a) u = log x
(b) u = ez
(c) u = (log z)−1
(d) u = (log z)2

Answer:

(c) u = (log z)−1

Given dzdx+zxlog z=zx2log z2          .....1Let u=log z-1dudx=-1log z2×1z×dzdxdzdx=-zlog z2 dudxSubstituting the value of dzdx from equation (1) we get,-z log z2 dudx+zxlog z=zx2log z2dudx-1x1log z=-1x2dudx-1xlog z-1=-1x2dudx-1xu=-1x2It can be written as,dudx+pxu=Qxwhere, px=-1x             qx=-1x2


The correct option is C.

Page No 21.146:

Question 25:

The solution of the differential equation dydx=yx+ϕyxϕ'yx is
(a) ϕyx=kx

(b) xϕyx=k

(c) ϕyx=ky

(d) yϕyx=k

Answer:

a ϕyx=kx


We have,dydx=yx+ϕyxϕ'yxLet y=vxdydx=v+xdvdx v+xdvdx=v+ϕvϕ'vxdvdx=ϕvϕ'vϕvϕ'vdv=1xdxIntegrating both sides, we getϕ'vϕvdv=1xdxlog ϕv=log x+log klog ϕyx-log x=log klog ϕyxx=log k ϕyxx= kϕyx=kx

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Question 26:

If m and n are the order and degree of the differential equation y25+4y23y3+y3=x2-1, then
(a) m = 3, n = 3
(b) m = 3, n = 2
(c) m = 3, n = 5
(d) m = 3, n = 1

Answer:

(b) m = 3, n = 2


We have,y25+4y23y3+y3=x2-1y3y25+4y23+y32=y3x2-1The highest order derivative is y3 and its highest exponent in this equation is 2.Therefore, order is 3 and degree is 2.Hence,  m=3, n=2

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Question 27:

The solution of the differential equation dydx+1=ex + y, is
(a) (x + y) ex + y = 0
(b) (x + C) ex + y = 0
(c) (xC) ex + y = 1
(d) (xC) ex + y + 1 =0

Answer:

(d) (xC) ex + y + 1 = 0


We have,dydx+1=ex+yLet x+y=v1+dydx=dvdxdydx+1=dvdx dvdx=eve-vdv=dxIntegrating both sides, we get-e-v=x-C-1=evx-Cx-Cex+y+1=0

Page No 21.146:

Question 28:

The solution of x2 + y2 dydx = 4, is
(a) x2 + y2 = 12x + C
(b) x2 + y2 = 3x + C
(c) x3 + y3 = 3x + C
(d) x3 + y3 = 12x + C

Answer:

(d) x3 + y3 = 12x + C

We have,x2+y2dydx=4y2dydx=4-x2y2dy=4-x2dxIntegrating both sides, we gety2dy=4-x2dxy33=4x-x33+Dy3=12x-x3+3Dx3+y3=12x+C, where C=3D

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Question 29:

The family of curves in which the sub tangent at any point of a curve is double the abscissae, is given by
(a) x = Cy2
(b) y = Cx2
(c) x2 = Cy2
(d) y = Cx

Answer:

(a) x = Cy2

Subtangent=ydydx

It is given that subtangent at any point of a curve is double of the abscissa.

ydydx=2xy=2xdydxdxx=2dyylnx=2lny+alnx=lny2+lnclnx=lncy2x=cy2

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Question 30:

The solution of the differential equation x dx + y dy = x2 y dyy2 x dx, is
(a) x2 − 1 = C (1 + y2)
(b) x2 + 1 = C (1 − y2)
(c) x3 − 1 = C (1 + y3)
(d) x3 + 1 = C (1 − y3)

Answer:

(a) x2 − 1 = C (1 + y2)


We have,
x dx + y dy = x2y dyy2x dx

x+xy2dx=x2y-ydyxx2-1dx=y1+y2dy2x2x2-1dx=2y21+y2dyIntegrating both sides, we get122y1+y2dy=122xx2-1dx12log1+y2=12logx2-1-12logClog1+y2=logx2-1-logClog1+y2=logx2-1C1+y2=x2-1CC1+y2=x2-1

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Question 31:

The solution of the differential equation (x2 + 1) dydx + (y2 + 1) = 0, is
(a) y = 2 + x2

(b) y=1+x1-x

(c) y = x (x − 1)

(d) y=1-x1+x

Answer:

d  y=1-x1+x


We have,x2+1dydx+(y2+1)=0x2+1dydx=-y2+11y2+1dy=-1 x2+1dxIntegrating both sides, we get1y2+1dy=-1 x2+1dxtan-1 y=-tan-1 x+tan-1 Ctan-1 y+tan-1 x=tan-1 Ctan-1x+y1-xy=tan-1 Cx+y1-xy=CDisclaimer: The initial value conditions are not given, so the final answer will be obatined only ifC=1. So, x+y=1-xyy+xy=1-xy1+x=1-xy=1-x1+x



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Question 32:

The differential equation xdydx-y=x2, has the general solution
(a) yx3 = 2cx
(b) 2yx3 = cx
(c) 2y + x2 = 2cx
(d) y + x2 = 2cx

Answer:

(b) 2yx3 = cx


We have,

xdydx-y=x2

dydx-1xy=x2Comparing with dydx+Py=Q, we getP=-1x Q=x2Now,I.F.=e-1xdx =e-logx                             =elog1x                             =1xy×I.F=x2×I.Fdx+C y1x=x2×1xdx+Cy1x=xdx+Cy1x=x22+C2y-x3=Cx                          

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Question 33:

The solution of the differential equation dydx-ky=0, y0=1 approaches to zero when x → ∞, if
(a) k = 0
(b) k > 0
(c) k < 0
(d) none of these

Answer:

(c) k < 0


We have,dydx-ky=0dydx=ky1ydy=k dxIntegrating both sides, we get1ydy=kdxlogy=kx+C          .....1Now, y0=1 C=0Putting C=0 in 1, we getlogy=kxekx=yAccording to the question,ek=0Since e-=0 k<0.                          

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Question 34:

The solution of the differential equation 1+x2dydx+1+y2=0, is
(a) tan1 x − tan−1 y = tan−1 C
(b) tan−1 y − tan−1 x = tan−1 C
(c) tan−1 y ± tan−1 x = tan C
(d) tan−1 y + tan−1 x = tan−1 C

Answer:

(d) tan−1y + tan−1x = tan−1C

We have,
1+x2dydx+1+y2=01+x2dydx=-1+y211+y2dy=-1 1+x2dxIntegrating both sides we get,11+y2dy=-1 1+x2dxtan-1y=-tan-1x+tan-1Ctan-1y+tan-1x=tan-1C

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Question 35:

The solution of the differential equation dydx=x2+xy+y2x2, is
(a) tan-1xy=log y+C

(b) tan-1yx=log x+C

(c) tan-1xy=log x+C

(d) tan-1yx=log y+C

Answer:

b tan-1yx=log x+C

We have,
dydx=x2+xy+y2x2          1
This is homogenous differential equation.
Let y=vxdydx=v+xdvdxNow, putting dydx=v+xdvdx and y=vx in 1, we getv+xdvdx=x2+x2v+x2v2x2v+xdvdx=1+v+v2xdvdx=1+v211+v2dv=1xdxIntegrating both sides we get,11+v2dv=1xdxtan-1 v=log x+Ctan-1 yx=log x+C

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Question 36:

The differential equation dydx+Py=Qyn, n>2 can be reduced to linear form by substituting
(a) z = yn −1
(b) z = yn
(c) z = yn + 1
(d) z = y1n

Answer:

(d) z = y1n

We have,
dydx+Py=Qyny-ndydx+Py1-n=Q          .....1Put z=y1-nIntegrating both sides with respect to x, we getdzdx=1-ny-ndydxy-ndydx=11-ndzdxNow, 1 becomes11-ndzdx+Pz=Qdzdx+P1-nz=Q1-nWhich is linear form of differential equation.Therefore, the given differential equation can be reduce to linear form by the substitution,z=y1-n

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Question 37:

If p and q are the order and degree of the differential equation ydydx+x3d2ydx2+xy = cos x, then
(a) p < q
(b) p = q
(c) p > q
(d) none of these

Answer:

(c) p > q

We have,
ydydx+x3d2ydx2+xy=cos x
The highest order derivative is d2yd2x and it's degree is 1So, the order is 2 and the degree is 1.p=2 and q=1Clearly, p>q

Page No 21.147:

Question 38:

Which of the following is the integrating factor of (x log x) dydx+y = 2 log x?
(a) x
(b) ex
(c) log x
(d) log (log x)

Answer:

(c) log x

We have,
x log xdydx+y=2 log x
Dividing both sides by (x log x) we get,
dydx+yx logx=2 log xx logxdydx+yx logx=2 xdydx+1x logxy=2 xComparing with dydx+Py=Q we get,P=1x logx and Q=2 xNow, I.F=ePdx=e1xlogxdx                         =eloglog x                         =log x

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Question 39:

What is integrating factor of dydx + y sec x = tan x?
(a) sec x + tan x
(b) log (sec x + tan x)
(c) esec x
(d) sec x

Answer:

(a) sec x + tan x

We have,

dydx+y sec x=tan xComparing with dydx+Py=Q, we getP=sec x Q=tan xNow,I.F.=esec xdx=elogsec x+tan x=sec x+tan x

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Question 40:

Integrating factor of the differential equation cos xdydx+y sin x=1, is
(a) cos x
(b) tan x
(c) sec x
(d) sin x

Answer:

(c) sec x

We have,
cos xdydx+y sin x=1
Dividing both sides by cos x, we get
dydx+sin xcos xy=1cos xdydx+tan xy=1 cos xComparing with dydx+Py=Q, we getP=tan xQ=2 cos xNow,I.F.=etan xdx=elogsec x=sec x

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Question 41:

The degree of the differential equation d2ydx23+dydx2+sindydx+1=0, is
(a) 3
(b) 2
(c) 1
(d) not defined

Answer:

(d) not defined

We have,
d2ydx23+dydx2+sindydx+1=0
The highest order derivative in this equation is d2yd2x.But the equation cannot be expressed as a polynomial in differential coefficient.Hence, the degree is not defined.

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Question 42:

The order of the differential equation 2x2d2ydx2-3dydx+y=0, is
(a) 2
(b) 1
(c) 0
(d) not defined

Answer:

(a) 2

We have,
2x2d2ydx2-3dydx+y=0
Here, the highest order derivative is d2yd2x.Hence, the order is 2.

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Question 43:

The number of arbitrary constants in the general solution of differential equation of fourth order is
(a) 0
(b) 2
(c) 3
(d) 4

Answer:

(d) 4

The number of arbitrary constants in the general solution of a differential equation of order n is n.
Thus, the number of arbitrary constants in the general solution of differential equation of fourth order is 4.



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Question 44:

The number of arbitrary constants in the particular solution of a differential equation of third order is
(a) 3
(b) 2
(c) 1
(d) 0

Answer:

(d) 0

The number of arbitrary constants in the particular solution of a differential equation is always zero.

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Question 45:

Which of the following differential equations has y = C1 ex + C2 ex as the general solution?
(a) d2ydx2+y=0

(b) d2ydx2-y=0

(c) d2ydx2+1=0

(d) d2ydx2-1=0

Answer:

b  d2ydx2-y=0

We have,
y=C1ex+C2e-x                                    .....1
Differentiating both sides of (1) with respect to x, we get
dydx=C1ex-C2e-x                               .....2
Differentiating both sides of (2) with respect to x, we get
d2ydx2=C1ex+C2e-xd2ydx2=y     Using 1 and 2d2ydx2-y =0

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Question 46:

Which of the following differential equations has y = x as one of its particular solution?
(a) d2ydx2-x2dydx+xy=x

(b) d2ydx2+xdydx+xy=x

(c) d2ydx2-x2dydx+xy=0

(d) d2ydx2+xdydx+xy=0

Answer:

c d2ydx2-x2dydx+xy=0

We have,
y = x           .....(1)
Differentiating both sides of (1) with respect to x, we get
dydx=1               .....2Differentiating again with respect to x, we getd2ydx2=0d2ydx2+x2=x2d2ydx2+x×x=x2×1d2ydx2+xy=x2×1          Using 1d2ydx2+xy=x2dydx          Using 2d2ydx2-x2dydx+xy=0                   

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Question 47:

The general solution of the differential equation dydx=ex+y, is
(a) ex + ey = C
(b) ex + ey = C
(c) ex + ey = C
(d) ex + ey = C

Answer:

(a) ex + ey = C

We have,
dydx=ex+ydydx=ex×eye-ydy=exdxIntegrating both sides, we gete-ydy=exdx-e-y=ex+Dex+e-y=-Dex+e-y=C    Where, C=-D

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Question 48:

A homogeneous differential equation of the form dxdy=hxy can be solved by making the substitution
(a) y = vx
(b) v = yx
(c) x = vy
(d) x = v

Answer:

(c) x = vy

A homogeneous differential equation of the form dxdy=hxy can be solved by substituting x = vy.

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Question 49:

Which of the following is a homogeneous differential equation?
(a) (4x + 6y + 5) dy − (3y + 2x + 4) dx = 0
(b) xy dx − (x3 + y3) dy = 0
(c) (x3 + 2y2) dx + 2xy dy = 0
(d) y2 dx + (x2xyy2) dy = 0

Answer:

(d) y2 dx + (x2xyy2) dy = 0

A differential equation is said to be homogenous if all the terms in the equation have equal degree and it can be written in the form dydx=fx, ygx, y.

In (a), (b) and (c), the degree of all the terms is not equal.
But in the equation y2 dx + (x2xyy2) dy = 0, the degree of all the terms is 2.
Thus, (d) contains a homogeneous differential equation.

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Question 50:

The integrating factor of the differential equation xdydx-y=2x2
(a) ex

(b) ey

(c) 1x

(d) x

Answer:

c  1x


We have,

xdydx-y=2x2dydx-1xy=2xComparing with dydx+Py=Q, we getP=-1x Q=2xNow,I.F.=e-1xdy =e-logx=elog1x=1x                          

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Question 51:

The integrating factor of the differential equation 1-y2dxdy+yx=ay-1<y<1 is
(a) 1y2-1

(b) 1y2-1

(c) 11-y2

(d) 11-y2

Answer:

d  11-y2


We have,1-y2dxdy+yx=aydxdy+y1-y2 x=ay1-y2Comparing with dxdy+Px=Q, we getP=y1-y2 Q=ay1-y2Now, I.F.=ey1-y2dy=e-12-2y1-y2dy=e-12log1-y2=elog11-y2=11-y2

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Question 52:

The general solution of the differential equation y dx-x dyy=0, is
(a) xy = C
(b) x = Cy2
(c) y = Cx
(d) y = Cx2

Answer:

(c) y = Cx


We have,y dx-x dyy=0y dx=x dy1ydy=1 xdxIntegrating both sides, we get1ydy=1 xdxlog y=logx+Dlog y-log x=log Clogyx=log Cyx=Cy=Cx

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Question 53:

The general solution of a differential equation of the type dxdy+P1x=Q1 is
(a) yeP1dy=Q1eP1dydy+C

(b) yeP1dx=Q1eP1dxdx+C

(c) xeP1dy=Q1eP1dydy+C

(d) xeP1dx=Q1eP1dxdx+C

Answer:

c  xeP1dy=Q1eP1dydy+C


We have,
dxdy+P1x=Q1
Comparing with the equation dxdy+Px=Q, we get
P = P1
Q = Q1
The general solution of the equation dxdy+Px=Q is given by
xePdy=QePdydy+C       ...(1)
Putting the value of P and Q in (1), we get
xeP1dy=Q1eP1dydy+C

Page No 21.148:

Question 54:

The general solution of the differential equation ex dy + (y ex + 2x) dx = 0 is
(a) x ey + x2 = C
(b) x ey + y2 = C
(c) y ex + x2 = C
(d) y ey + x2 = C

Answer:

(c) y ex + x2 = C


We have,
ex dy + (yex + 2x) dx = 0
Dividing both sides by exdx, we getdydx+y+2xex=0dydx+y=-2xexComparing with dydx+Py=Q, we getP=1Q=-2xexNow, I.F.=edx =exSolution is given by,                            y×I.F.=Q×I.F. dx+C yex=-ex×2xexdx+Cyex=-2x dx+Cyex=-x2+Cyex+x2=C              



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Question 1:

The order of the differential equation representing the family of parabolas y2 - 4ax is ________________.

Answer:

In the given equation y2 − 4ax, a is the only parameter.

Therefore, the order of the differential equation = 1

Hence, the order of the differential equation representing the family of parabolas y2 − 4ax is 1.

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Question 2:

The degree of the differential equation dydx3+d2ydx22 = 0 is ___________________.

Answer:

In the given differential equation dydx3+d2ydx22 = 0, power of highest order derivative is 2.

​Therefore, the degree of the differential equation = 2

Hence, the degree of the differential equation dydx3+d2ydx22 = 0 is 2.

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Question 3:

The number of arbitrary constants in a particular solution of the differential equation tan x dx + tan y dy = 0 is __________________.

Answer:

The number of arbitrary constants in the general solution is equal to the order of the differential equation.

While the number of arbitrary constants in the particular solution is zero.

Hence, the number of arbitrary constants in a particular solution of the differential equation tanx dx + tany dy = 0 is 0.

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Question 4:

An appropriate substitution to solve the differential equation dxdy=x2logxy-x2xy logxy is ________________.

Answer:

Given: dxdy=x2logxy-x2xy logxy

To solve this differential equation, we use the method of homogeneous differential equation.
Thus we put, x = vy.
​
Hence, an appropriate substitution to solve the differential equation dxdy=x2logxy-x2xy logxy is xy=v.

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Question 5:

The integrating factor of the differential equation xdydx-y = sin x is ____________.

Answer:

Given: xdydx-y = sinx

xdydx-y=sinxdydx-yx=sinxxHere P=-1x and Q=sinxxIntegrating Factor=ePdx                         =e-1xdx                         =e-logx                         =elogx-1                         =x-1                         =1x
Hence, the integrating factor of the differential equation xdydx-y = sinx is 1x.

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Question 6:

The general solution of the differential equation dydx=ex-y is _______________.

Answer:

Given: dydx=ex-y

dydx=ex-ydydx=exe-ydydx=exeyeydy=exdxIntegrating both sides, we geteydy=exdxey=ex+C, where C is any constant

Hence, the general solution of the differential equation dydx=ex-y is ey=ex+C.

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Question 7:

The general solution of the differential equation dydx+yx = 1 is _____________.

Answer:

Given: dydx+yx = 1

dydx+yx=1Here, P=1x and Q=1Integrating Factor I.F.=ePdx                                 =e1xdx                                 =elogx                                 =xThus, solution of the differential equation isy×I.F.=Q×I.F.dx+Cyx=xdx+Cyx=x22+C, where C is any constant

Hence, the general solution of the differential equation dydx+yx = 1 is yx=x22+C.

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Question 8:

The differential equation representing the family of curves y = A sin x + B cos x is _____________.

Answer:

Given: y = A sinx + B cosx


y=A sinx+B cosx       ...1Differentiating both sides with respect to xdydx=A cosx-B sinxDifferentiating both sides with respect to xd2ydx2=-A sinx-B cosx          =-A sinx+B cosx          =-y                        from 1d2ydx2+y=0

Hence, the differential equation representing the family of curves y = A sinx + B cosx is d2ydx2+y=0.

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Question 9:

The linear differential equation e-2xx-yxdydx=1, x0  when written in the form dydx+Py=Q, then P = __________________.

Answer:

Given: e-2xx-yxdxdy=1, x0


e-2xx-yxdxdy=1e-2xx-yx=dydxe-2xx=dydx+yxdydx+yx=e-2xxHere, P=1x and Q=e-2xx

Hence, P = 1x.

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Question 10:

The order of the differential equation representing the family of ellipses having centre at origin and foci on x-axis is ________________.

Answer:

Given: Family of ellipses having centre at origin and foci on x-axis

The equation of family of ellipses having centre at origin and foci on x-axis is x2a2+y2b2=1.
We have 2 arbitrary constants a and b.

Hence, the order is 2.

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Question 11:

The degree of the differential equation 1+d2ydx2=dydx+x is ______________.

Answer:

Given: 1+d2ydx2=dydx+x

To find the degree of the differential equation, the differential equation must be free from fractions and radicals.
Thus,
1+d2ydx2=dydx+xSquaring both sides, we get1+d2ydx22=dydx+x21+d2ydx2=dydx+x2

Here, the power of the highest order derivative is 1.

Hence, the degree is 1.

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Question 12:

The integration factor of the differential equation xdydx = x cos x is __________________.

Answer:

ans

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Question 13:

The degree of the differential equation d2ydx2+edydx = 0 is _________________.

Answer:

Given: d2ydx2+edydx = 0

To find the degree of the differential equation, the differential equation must be free from fractions and radicals.
Thus, in the differential equation d2ydx2+edydx = 0, the degree is not defined.

Hence, the degree is not defined.



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Question 14:

The degree of the differential equation 1+dydx2 = x is ______________.

Answer:

Given: 1+dydx2 = x

To find the degree of the differential equation, the differential equation must be free from fractions and radicals.
Thus,
1+dydx2=xSquaring both sides, we get1+dydx22=x21+dydx2=x2

Here, the power of the highest order derivative is 2.

Hence, the degree is 2.

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Question 15:

The number of arbitrary constants in the general solution of the differential equation of order three is __________________.

Answer:

We know,
Number of arbitrary constants in the general solution of the differential equation = Order of the differential equation

Since, order of the differential equation is 3
Thus, Number of arbitrary constants in the general solution of the differential equation = 3

Hence, the number of arbitrary constants in the general solution of the differential equation of order three is 3.

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Question 16:

The general solution of the differential equation of the type dxdy+Rx=S, Where R and S are  function of y, is _________________.

Answer:

Given: dxdy+Rx=S, where R and S are  function of y


dxdy+Rx=S

The given equation is linear differential equation.

Integrating factor (I.F.) = eRdy

The required solution is

xeRdy=SeRdydy+C, where C is arbitrary constant

Hence, the general solution of the differential equation of the type dxdy+Rx=S, where R and S are function of y, is xeRdy=SeRdydy+C.

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Question 17:

The integrating factor of the differential equation dydx+y=1+yx is _____________________.

Answer:

Given:  dydx+y=1+yx


dydx+y=1+yxdydx+y=1x+yxdydx+y-yx=1xdydx+y1-1x=1xdydx+yx-1x=1xHere, P=x-1x and Q=1xIntegrating factor=ePdx                        =ex-1xdx                        =e1-1xdx                        =ex-logx                        =exe-logx                        =exelogx-1                        =exx-1                        =exx


Hence, the integrating factor of the differential equation dydx+y=1+yx is exx.

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Question 18:

The solution of the differential equation cot y dx = x dy is ________________.

Answer:

Given: coty dx = x dy


coty dx=x dydxx=dycotytany dy=dxxIntegrating both sides, we gettany dy=dxxlogsecy=logx+logC, where logC is arbitrary constantlogsecy=logCxsecy=Cxsecy=±Cxsecy=Ax, where A=±C


Hence, the solution of the differential equation coty dx = x dy is secy=Ax.

Page No 21.150:

Question 19:

The general solution of the differential equation xdydx+2y=x2 is_________________.

Answer:

Given: xdydx+2y=x2


xdydx+2y=x2dydx+2yx=x2xdydx+2xy=xHere, P=2x and Q=xIntegrating factor=ePdx                        =e2xdx                        =e2logx                        =elogx2                        =x2Thus, the required solution is:y×x2=x×x2 dx+Cx2y=x3 dx+Cx2y=x44+C, where C is arbitrary constant


Hence, the general solution of the differential equation xdydx+2y=x2 is x2y=x44+C.

Page No 21.150:

Question 20:

The solution of the differential equation y dx + (x + xy) dy = 0 is ________________.

Answer:

Given: y dx + (x + xydy = 0


y dx+x+xy dy=0y dx=-x+xy dyy dx=-x1+y dydx-x=1+yy dy1+yy dy=-dxxIntegrating both sides, we get1+yy dy=-dxx1y+1 dy=-dxxlogy+y=-logx+logC, where logC is arbitrary constanty=logCx-logyy=logCxyCxy=eyeyxy=Ceyxy=±Ceyxy=A, where A=±C


Hence, the solution of the differential equation y dx + (x + xydy = 0 is eyxy=A.

Page No 21.150:

Question 21:

The order of the differential equation representing the family of circles x2 +(y - a)2 = a2 is __________________.

Answer:

In the given equation x2 +(y a)2 = a2a is the only parameter.

Therefore, the order of the differential equation = 1

Hence, the order of the differential equation representing the family of circles x2 +(y a)2 = a2 is 1.

Page No 21.150:

Question 22:

The number of arbitrary constants in the particular solution of a differential equation of order two is ____________________.

Answer:

The number of arbitrary constants in the general solution is equal to the order of the differential equation.

While the number of arbitrary constants in the particular solution is zero.

Hence, the number of arbitrary constants in the particular solution of a differential equation of order two is 0.

Page No 21.150:

Question 23:

The differential equation of all non-horizontal lines in a plane is __________________.

Answer:

Equation of all non-horizontal lines in a plane is ax + by + c = 0, where a ≠ 0. 

ax+by+c=0Differentiating both the sides w.r.t. yadxdy+b=0Again differentiating both the sides w.r.t. yad2xdy2=0d2xdy2=0   a0


Hence, the differential equation of all non-horizontal lines in a plane is d2xdy2=0.

Page No 21.150:

Question 24:

The differential equation of all non-vertical lines in a plane is __________________.

Answer:

Equation of all non-vertical lines in a plane is ax + by + c = 0, where b ≠ 0. 

ax+by+c=0Differentiating both the sides w.r.t. xa+bdydx=0Again differentiating both the sides w.r.t. xbd2ydx2=0d2ydx2=0   b0


Hence, the differential equation of all non-vertical lines in a plane is d2ydx2=0.

Page No 21.150:

Question 25:

The integrating factor of all differential equation (x2 + 1) dydx + 2xy = x2 - 1 is _______________.

Answer:

Given:  (x2 + 1) dydx + 2xy = x2 − 1


x2+1dydx+2xy=x2-1dydx+2xyx2+1=x2-1x2+1dydx+y2xx2+1=x2-1x2+1Here, P=2xx2+1 and Q=x2-1x2+1Integrating factor=ePdx                        =e2xx2+1dx                        =edtt                  Put x2+1=t   2xdx=dt                         =elogt                        =t                        =x2+1


Hence, the integrating factor of all differential equation (x2 + 1) dydx + 2xy = x2 − 1 is x2+1.

Page No 21.150:

Question 26:

The degree of the differential equation y = x dydx2+dxdy2 is _____________________.

Answer:

Given: y = x dydx2+dxdy2 


y=xdydx2+dxdy2y=xdydx2+dydx-2y=xdydx2+1dydx2ydydx2=xdydx2dydx2+1ydydx2=xdydx4+1

Here, power of highest order derivative is 4.

Hence, the degree of the differential equation y = x dydx2+dxdy2 is 4.

Page No 21.150:

Question 27:

The order of the differential equation representing all circles of radius r is __________________.

Answer:

Equation representing all circles of radius r is ( h)2 +( k)2 = r2, where h, k are two parameters and r is fixed.

Therefore, the order of the differential equation = 2

Hence, the order of the differential equation representing all circles of radius r is 2.

Page No 21.150:

Question 28:

The degree of the differential equation representing the family of curves y = Ax + A3 where A is arbitrary constant, is _______________.

Answer:

Given: y = Ax + A3

Here, A is the only parameter.

Therefore, the order of the differential equation = 1

y=Ax+A3    ...1Differentiating both sides w.r.t. xdydx=A      ...2Substituting the value of A in 1y=dydxx+dydx3

Here, the power of the highest order derivative is 3.

Hence, the degree of the differential equation representing the family of curves y = Ax + A3 where A is arbitrary constant, is 3.

Page No 21.150:

Question 29:

The general solution of the differential equation dxy+dyy = 0 is ________________.

Answer:

Given:  dxy+dyy = 0


dxy+dyy=0dxy=-dyydx=-ydyydx=-dydy=-dxIntegrating both sides, we getdy=-dxy=-x+C, where C is arbitrary constant


Hence, the general solution of the differential equation dxy+dyy = 0 is y=-x+C.

Page No 21.150:

Question 30:

The order and degree of the differential equation d3ydx3-3d2ydx2+2dydx4=y4 ____________ and ___________ respectively.

Answer:

Given: d3ydx3-3d2ydx2+2dydx4=y4 

Here, the order of the differential equation is 3.

And the power of the highest order derivative is 1.

Hence, the order and degree of the differential equation d3ydx3-3d2ydx2+2dydx4=y4 is 3 and 1 respectively.

Page No 21.150:

Question 31:

The differential equation for which y = a cos x + b sin x is a solution is _______________.

Answer:

Given: y = a cosx + b sinx


y=a cosx+b sinx       ...1Differentiating both sides with respect to xdydx=-a sinx+b cosxDifferentiating both sides with respect to xd2ydx2=-a cosx-b sinx          =-a cosx+b sinx          =-y                        from 1d2ydx2+y=0

Hence, the differential equation for which y = a cosx + b sinx is a solution is  d2ydx2+y=0.

Page No 21.150:

Question 32:

The curve for which the slope of the tangent at any point is equal to the ratio of the abscissa and ordinate of the point, is ______________.

Answer:

Given: Slope of the tangent at any point is equal to the ratio of the abscissa and ordinate of the point.

According to the question
dydx=xyy dy=x dxIntegrating both sides, we gety dy=x dxy22=x22+Cy2=x2+2Cy2=x2+A, where A=2C is arbitrary constanty2-x2=A

which is the equation of a rectangular hyperbola.

Hence,the curve for which the slope of the tangent at any point is equal to the ratio of the abscissa and ordinate of the point, is y2-x2=A.

Page No 21.150:

Question 33:

Family y = Ax + A3 of curves will correspond to a differential equation of order _______________ and degree _______________.

Answer:

Given: y = Ax + A3

Here, A is the only parameter.

Therefore, the order of the differential equation is 1.

y=Ax+A3    ...1Differentiating both sides w.r.t. xdydx=A      ...2Substituting the value of A in 1y=dydxx+dydx3

Here, the power of the highest order derivative is 3.

Hence, family y = Ax + A3 of curves will correspond to a differential equation of order 1 and degree 3.

Page No 21.150:

Question 34:

The differential x dy + y dy = 0 equation represents a family of _______________.

Answer:

dx is missing

Page No 21.150:

Question 35:

The differential equation of the family of curves x2 + y2 - 2ay = 0, where a is arbitrary constant, is _______________.

Answer:

Given: x2 + y2 − 2ay = 0


x2+y2-2ay=0       ...1Differentiating both sides with respect to xddxx2+ddxy2-2addxy=02x+2ydydx-2adydx=02adydx=2x+2ydydxadydx=x+ydydxay'=x+yy'               , where dydx=y'a=x+yy'y'Substituting the value of a in 1, we getx2+y2-2x+yy'y'y=0x2y'+y2y'-2x+yy'yy'=0x2y'+y2y'-2xy-2y2y'=0x2y'-y2y'-2xy=0x2-y2dydx-2xy=0

Hence, the differential equation representing the family of curves x2 + y2 − 2ay = 0, is x2-y2dydx-2xy=0.



Page No 21.151:

Question 1:

Define a differential equation.

Answer:

Differential equation:

An equation containing an independent variable, a dependent variable and differential coefficients of the dependent variable with respect to the independent variable is called a differential equation.
for example: dydx=ex+y

Page No 21.151:

Question 2:

Define order of a differential equation.

Answer:

Order of differential equation:

The order of a differential equation is the order of its highest order derivative that apears in the equation.

example: d2ydx2-4dydx=2y
order of the differential equation is 2.

Page No 21.151:

Question 3:

Define degree of a differential equation.

Answer:

Degree of differential equation:

The degree of a differential equation is the power of the highest order derivative occurring in a differential equation when it is written as a polynomial in differential coefficients.

example: d2ydx22-4dydx=2y
the degree of the given differential equation is 2

Page No 21.151:

Question 4:

Write the differential equation representing the family of straight lines y = Cx + 5, where C is an arbitrary constant.

Answer:

We have,y= Cx+5     .....1dydx=CSubstituting the value of C in 1, we gety= dydx×x+5xdydx-y+5=0 Hence, xdydx-y+5=0  is the differential equation representing the family of straight lines y=Cx+5, where C is an arbitary constant. 

Page No 21.151:

Question 5:

Write the differential equation obtained by eliminating the arbitrary constant C in the equation x2y2 = C2.

Answer:

We have,x2-y2=C2Differentiating with respect to x, we get2x-2ydydx=02x=2ydydxx dx=y dyx dx-y dy=0Hence, x dx-y dy=0 is the required differential equation.

Page No 21.151:

Question 6:

Write the differential equation obtained eliminating the arbitrary constant C in the equation xy = C2.

Answer:

We have,xy=C2Differentiating with respect to x, we getxdydx+y=0xdydx=-yx dy=-y dxx dy+y dx=0Hence, x dy+y dx=0 is the required differential equation.

Page No 21.151:

Question 7:

Write the degree of the differential equation a2d2ydx2=1+dydx21/4.

Answer:

 We have,a2d2ydx2=1+dydx21/4a2d2ydx24=1+dydx2Degree of the differential equation is the degree of the highest order derivative.Therefore, the degree must be 4.

Page No 21.151:

Question 8:

Write the order of the differential equation 1+dydx2=7 d2ydx23.

Answer:

1+dydx2=7 d2ydx23The order of a differential equation is the order of its highest order derivatives.Here, the required order is 2.

Page No 21.151:

Question 9:

Write the order and degree of the differential equation y=xdydx+a1+dydx2.

Answer:

We have,y=xdydx+a1+dydx2y-xdydx=a1+dydx2Squaring both sides, we gety-xdydx2=a1+dydx22y2+x2dydx2-2xydydx=a21+dydx2y2+dydx2x2-a2-2xydydx=a2From the above equation, we see that the highest order is 1. So, its order is 1 and the power of the highest order derivative is 2.Thus, it is a differential equation of order 1 and degree 2.

Page No 21.151:

Question 10:

Write the degree of the differential equation d2ydx2+xdydx2=2x2 log d2ydx2.

Answer:

We have,d2ydx2+xdydx2=2x2 log d2ydx2d2ydx2+xdydx2-2x2 log d2ydx2=0Here, we observe that LHS of the differential equation cannot be expressed as a polynomial in dydx. So, its degree is not defined. 

Page No 21.151:

Question 11:

Write the order of the differential equation of the family of circles touching X-axis at the origin.

Answer:




The equation of the family of circles touching x-axis at the origin is
x-02+y-a2=a2x2+y2-2ay=0             .....1Here, a is the parameter.Since this equation contains only one arbitary constant, we differentiate it only once. Differentiating with respect to x, we get2x+2ydydx-2adydx=0a=x+ydydxdydx            .....2Putting the value of a from (2) in (1), we get x2+y2=2yx+ydydxdydxx2-y2dydx=2xySo, this is the required differential equation.Here, order of the differential equation is 1.

Page No 21.151:

Question 12:

Write the order of the differential equation of all non-horizontal lines in a plane.

Answer:

The equation of the non-horizontal lines in a plane isy=mx+c, where m is the slope and c is the intercept on y-axis.Differentiating with respect to x, we getdydx=md2ydx2=0This is the required differential equation. Here, we observe that the order of the required differential equation is 2.

Page No 21.151:

Question 13:

If sin x is an integrating factor of the differential equation dydx+Py=Q, then write the value of P.

Answer:

It is given that sin x is the integrating factor of the differential equation dydx+Py=Q. ePdx=sin xP dx=log sin xP dx=cot x dx          cot x dx=log  sin x+CP=cot x  



Page No 21.152:

Question 14:

Write the order of the differential equation of the family of circles of radius r.

Answer:

ans

Page No 21.152:

Question 15:

Write the order of the differential equation whose solution is y = a cos x + b sin x + c ex.

Answer:

y=a cos x+b sinx +c e-xHere, we see that there are three arbitary constants. Therefore, we differentiate it three times to get rid of all three arbitrary constants.Hence, the order of the differential equation is 3.

Page No 21.152:

Question 16:

Write the order of the differential equation associated with the primitive y = C1 + C2 ex + C3 e−2x + C4, where C1, C2, C3, C4 are arbitrary constants.

Answer:

y=C1+C2ex+C3e-2x+C4the given equation can be reduced to:y=C1+C2ex+C3(e-2x×ec4)Here, ec4 will be a constant.We have 3 constants as C1, C2 and C3.and a differential equation of order n always contains exactly n essential arbitrary constants.Hence, the order of the required differntial equation is 3.

Page No 21.152:

Question 17:

What is the degree of the following differential equation?
5xdydx2-d2ydx2-6y=log x

Answer:

5xdydx2-d2ydx2-6y=log x
Here, we see that  the highest order derivative is d2ydx2 and its power is 1.
Therefore, the given differential equation is of first degree.

Page No 21.152:

Question 18:

Write the degree of the differential equation dydx4+3xd2ydx2=0.

Answer:

dydx4+3xd2ydx2=0
The highest order derivative is d2ydx2 and its power is 1.
Therefore, the given differential equation is of first degree.

Page No 21.152:

Question 19:

Write the degree of the differential equation x d2ydx23+ydydx4+x3=0.

Answer:

d2ydx23+ydydx4+x3=0

The highest order derivative is d2ydx2 and its power is 3.

Therefore, the degree of given differential equation is 3.

Page No 21.152:

Question 20:

Write the differential equation representing family of curves y = mx, where m is arbitrary constant.

Answer:

We have,y=mx     .....1 Differentiating with respect to xdydx=mSubstituting the value of m=dydx in eq 1 we get ,y=xdydxHence, y=xdydx is the required differential equation .

Page No 21.152:

Question 21:

Write the degree of the differential equation x3d2ydx22+xdydx4=0.

Answer:

x3d2ydx22+xdydx4=0

Here, the highest order derivative is d2ydx2 and its power is 2.

Therefore, degree of given differential equation is 2.

Page No 21.152:

Question 22:

Write the degree of the differential equation 1+dydx3=d2ydx22

Answer:

The degree is 2 as the highest derivative is of order 2.

Page No 21.152:

Question 23:

Write the degree of the differential equation
d2ydx2+3dydx2=x2logd2ydx2

Answer:

The given differential equation is not a polynomial equation in derivatives.
Hence, the degree for this differential equation is not defined.

Page No 21.152:

Question 24:

Write the degree of the differential equation
d2ydx22+dydx2=xsindydx

Answer:

The given differential equation is not a polynomial equation in derivatives.
Hence, the degree for this differential equation is not defined.

Page No 21.152:

Question 25:

Write the order and degree of the differential equation d2ydx2+dydx14+x15=0

Answer:

The order is 2 as the highest derivative is 2.
The given differential equation is not a polynomial equation in derivatives.
Hence, the degree for this differential equation is not defined.

Page No 21.152:

Question 26:

The degree ofthe differential equation d2ydx2+edydx=0

Answer:

The given differential equation is not a polynomial equation in derivatives.
Hence, the degree for this differential equation is not defined.

Page No 21.152:

Question 27:

How many arbitrary constants are there in the general solution of the differential equation of order 3.

Answer:

The arbitrary constants in the general solution of the differential equation is equal to the order of the differential equation.
Hence, the number of arbitrary constants in the general solution of the differential equation of order 3 are 3.

Page No 21.152:

Question 28:

Write the order of the differential equation representing the family of curves y = ax + a3.

Answer:

The order of the differential equation is equal to the arbitrary constants present in the general solution of the differential equation.
Hence, the order of the differential equation representing the family of curves y = ax + a3 is 1.

Page No 21.152:

Question 29:

Find the sum of the order and degree of the differential equation y=xdydx3+d2ydx2

Answer:

The order is 2 as the highest derivative is 2.
The degree is 1 as the highest derivative is of order 1.
Hence, the sum of the order and degree of the differential equation y=xdydx3+d2ydx2 is 2 + 1 = 3

Page No 21.152:

Question 30:

Find the solution of the differential equation
x1+y2dx+y1+x2dy=0

Answer:

x1+y2dx+y1+x2dy=0y1+x2dy=-x1+y2dxy1+y2dy=-x1+x2dxy1+y2dy=-x1+x2dx
Let 1+y2=t2 and 1+x2=p22ydy=2tdt and 2xdx=2pdpydy=tdt and xdx=pdpSubstituting in above equation, we getdt=-dpt=-p+C1+x2+1+y2=C

Page No 21.152:

Question 31:

Form the differential equation representing the family of curves y = A sin x, by eliminating the arbitrary constant A.

Answer:

Given: y = A sinx


y=A sinx       ...1Differentiating both sides with respect to xdydx=A cosxA=1cosxdydxSubstituting the value of A in 1, we gety=1cosxdydx sinxy=sinxcosxdydxy=tanxdydxtanxdydx-y=0

Hence, the differential equation representing the family of curves y = A sinx is tanxdydx-y=0.



Page No 21.16:

Question 1:

Form the differential equation of the family of curves represented by y2 = (x − c)3.

Answer:

The equation of the family of curves is
y2=x-c3                                                 ...(1)
where cR is a parameter.
This equation contains only one parameter, so we shall obtain a differential equation of first order.
Differentiating equation (1) with respect to x, we get
2ydydx=3x-c2                                          ...(2)
Dividing equation (1) by equation (2), we get
y22ydydx=x-c33x-c2y2dydx=x-c33y2dydx=x-cc=x-3y2dydx
Substituting the value of c in equation (1), we get
y2=x-x+3y2dydx3y2=27y38dydx38y2dydx3=27y38dydx3-27y=0
It is the required differential equation.

Page No 21.16:

Question 2:

Form the differential equation corresponding to y = emx by eliminating m.

Answer:

The equation of the family of curves is
y=emx                                                ...(1)
where m is a parameter.
This equation contains only one parameter, so we shall get a differential equation of first order.
Differentiating equation (1) with respect to x, we get
dydx=memxdydx=my                             [Using equation (1)]
m=1ydydx                                      ...(2)
Now, from equation (1), we get
ln y = ln emxln y= mx ln eln y= mxm = 1xln y                                         ...(3)
Comparing equations (2) and (3), we get
1xln y = 1ydydxxdydx=y ln y
It is the required differential equation.

Page No 21.16:

Question 3:

Form the differential equations from the following primitives where constants are arbitrary:
(i) y2 = 4ax
(ii) y = cx + 2c2 + c3
(iii) xy = a2
(iv) y = ax2 + bx + c

Answer:

(i) The equation of family of curves is
y2=4ax                                            ...(1)
where a is an arbitrary constant.
This equation contains only one arbitrary constant, so we shall get a differential equation of first order.
Differentiating equation (1) with respect to x, we get
2ydydx=4ay2dydx=a                                                 2
Putting the value of a in equation (1), we get
y2=4y2dydxxy=2xdydx, It is the required differential equation.

(ii) The equation of family of curves is
y=cx+2c2+c3                                            ...(1)
where c is an arbitrary constant.
This equation contains only one arbitrary constant, so we shall get a differential equation of first order.
Differentiating equation (1) with respect to x, we get
dydx=c                                                      ...(2)
Putting the value of c in equation (1), we get
y=xdydx+2dydx2+dydx3 
It is the required differential equation.

(iii) The equation of family of curves is
xy=a2                                                       ...(1)
where a is an arbitrary constant.
This equation contains only one arbitrary constant, so we shall get a differential equation of first order.
Differentiating equation (1) with respect to x, we get
y+xdydx=0
It is the required differential equation.

(iv) The equation of family of curves is
y=ax2+bx+c                                          ...(1)
where a, b and c are arbitrary constants. So, we shall get a differential equation of third order.
Differentiating equation (1) with respect to x, we get
dydx=2ax+b                                               ...(2)
Differentiating equation (2) with respect to x, we get
d2ydx2=2a                                                  ...(3)
Differentiating equation (3) with respect to x, we get
d3ydx3=0 
It is the required differential equation.

Page No 21.16:

Question 4:

Find the differential equation of the family of curves y = Ae2x + Be−2x, where A and B are arbitrary constants.

Answer:

The equation of the family of curves is
y=Ae2x+Be-2x                                        ...(1)
where A and B are arbitrary constants.
This equation contains two arbitrary constants, so we shall get a differential equation of second order.
Differentiating equation (1) with respect to x, we get
dydx=2Ae2x-2Be-2x                                 ...(2)
Differentiating equation (2) with respect to x, we get
d2ydx2=4Ae2x+4Be-2xd2ydx2=4Ae2x+Be-2xd2ydx2=4yIt is the required differential equation.

Page No 21.16:

Question 5:

Find the differential equation of the family of curves, x = A cos nt + B sin nt, where A and B are arbitrary constants.

Answer:

The equation of the family of curves is
x=Acos nt+Bsin nt                                  ...(1)
where A and B are arbitrary constants.
This equation contains two arbitrary constants, so we shall get a differential equation of second order.
Differentiating equation (1) with respect to t, we get
dxdt=-Ansin nt+Bncos nt                          ...(2)
Differentiating equation (2) with respect to t, we get
d2xdt2=-An2cos nt-Bn2sin ntd2xdt2=-n2Acos nt+Bsin ntd2xdt2=-n2xd2xdt2+n2x=0 It is the required differential equation.

Page No 21.16:

Question 6:

Form the differential equation corresponding to y2 = a (bx2) by eliminating a and b.

Answer:

The equation of the family of curves is
y2=ab-x2,                                              ...(1)
where a and b are parameters.
This equation contains two arbitrary constants, so we shall get a differential equation of second order.
Differentiating equation (1) with respect to x, we get
2ydydx=-2ax                                            ...(2)
Differentiating equation (2) with respect to x, we get
dydx2+yd2ydx2=-a                                   ...(3)
From (2) and (3), we get
ydydx=xdydx2+yd2ydx2

It is the required differential equation.

Page No 21.16:

Question 7:

Form the differential equation corresponding to y2 − 2ay + x2 = a2 by eliminating a.

Answer:

The equation of the family of curves is
y2-2ay+x2=a2                                         ...(1)
where a is a parameter.
This equation contains only one arbitrary constant, so we shall get a differential equation of first order.
Differentiating equation (1) with respect to x, we get
2ydydx-2adydx+2x=02ydydx+2x=2adydxy+xdydx=a
Substituting the value of a in equation (2), we get
y2-2y+xdydxy+x2=y+xdydx2y2dydx-2ydydx+xy+x2dydxdydx=ydydx+x2dydx2y2dydx2-2y2dydx2-2xydydx+x2dydx2=y2dydx2+2xydydx+x2x2-2y2dydx2-4xydydx-x2=0 It is the required differential equation.

Page No 21.16:

Question 8:

Form the differential equation corresponding to (xa)2 + (yb)2 = r2 by eliminating a and b.

Answer:

The equation of the family of curves is
x-a2+y-b2=r2                                                 ...(1)
where a and b are parameters.
This equation contains two parameters, so we shall get a second order differential equation.
Differentiating equation (1) with respect to x, we get
2x-a+2y-bdydx=0                                               ...(2)
Differentiating (2) with respect to x, we get
2+2dydx2+2y-bd2ydx2=01+dydx2+y-bd2ydx2=0y-b=-1+dydx2d2ydx2                                            ...(3)
From (2) and (3), we get
x-a-1+dydx2d2ydx2dydx=0x-a=dydx+dydx3d2ydx2                                                    ...(4)
From (1), (3) and (4), we get
dydx+dydx32d2ydx22+1+dydx22d2ydx22=r2dydx2+2dydx4+dydx6+1+2dydx2+dydx4d2ydx22=r2dydx2+2dydx4+dydx6+1+2dydx2+dydx4=r2d2ydx221+3dydx2+3dydx4+dydx6=r2d2ydx221+dydx23=r2d2ydx22It is the required differential equation.



Page No 21.17:

Question 9:

Find the differential equation of all the circles which pass through the origin and whose centres lie on y-axis.

Answer:

The equation of the family of circles that pass through the origin (0, 0) and whose centres lie on the y-axis is given by
x2+y-a2=a2                                            ...(1)
where a is any arbitrary constant.
As this equation has only one arbitrary constant, we shall get a first order differential equation.
Differentiating equation (1) with respect to x, we get
2x+2y-adydx=0x+y-adydx=0x=a-ydydxxdydx=a-ya=y+xdydx                                                               ...(2)
Substituting the value of a in equation (2), we get
x2+y-y-xdydx2=y+xdydx2x2+x2dydx2=y2+2xydydx+x2dydx2x2=y2+2xydydxx2-y2dydx=2xy It is the required differential equation.

Page No 21.17:

Question 10:

Find the differential equation of all the circles which pass through the origin and whose centres lie on x-axis.

Answer:

The equation of the family of circles that pass through the origin (0,0) and whose centres lie on the x-axis is given by
x-a2+y2=a2                                                     ...(1)
where a is any arbitrary constant.
As this equation has only one arbitrary constant, we shall get a first order differential equation.
Differentiating equation (1) with respect to x, we get
2x-a+2ydydx=0x-a+ydydx=0x+ydydx=a                                                        ..(2)
Substituting the value of a in equation (1), we get
x-x-ydydx2+y2=x+ydydx2y2dydx2+y2=x2+2xydydx+y2dydx22xydydx+x2=y2 It is the required differential equation.

Page No 21.17:

Question 11:

Assume that a rain drop evaporates at a rate proportional to its surface area. Form a differential equation involving the rate of change of the radius of the rain drop.

Answer:

Let the surface area of the raindrop be A.
Thus, the rate of evaporation will be given by dVdt.
As per the given condition, 
dVdtAdVdt=-kA
Here, k is a constant. Also, the negative sign appears when V decreases and t increases.
Now, 
V=43πr3
Here, r is the radius of the spherical drop.
ddt43πr3=-k×4πr243×3πr2drdt=-k×4πr2drdt=-k It is the required differential equation.

Page No 21.17:

Question 12:

Find the differential equation of all the parabolas with latus rectum '4a' and whose axes are parallel to x-axis.

Answer:

The equation of the family of parabolas with latus rectum 4a and axis parallel to the x-axis is given by
y-β2=4ax-α                                               ...(1)
where α and β are two arbitrary constants.
As this equation has two arbitrary constants, we shall get second order differential equation.
Differentiating equation (1) with respect to x, we get
2y-βdydx=4a                                                  ...(2)
Differentiating equation (2) with respect to x, we get
y-βd2ydx2+dydxdydx=0                                     ...(3)
Now, from equation (2), we get
y-β=4adydx                                                    ...(4)
From (3) and (4), we get
2adydxd2ydx2+dydx2=02ad2ydx2+dydx3=0 It is the required differential equation.

Page No 21.17:

Question 13:

Show that the differential equation of which y = 2(x2 − 1) + ce-x2 is a solution, is dydx+2xy=4x3.

Answer:

The given equation is 
y=2x2-1+ce-x2                                       ...(1)
where c is a parameter.
As this equation has one arbitrary constant, we shall get a differential equation of first order.
Differentiating equation (1) with respect to x, we get
dydx=22x+ce-x2(-2x)dydx=4x-2xce-x2                                     ...2
From (1) and (2), we get
dydx=4x-2xy-2x2+2
dydx=4x-2xy+4x3-4xdydx+2xy=4x3
Hence, y=2x2-1+ce-x2 is the solution to the differential equation dydx+2xy=4x3.

Page No 21.17:

Question 14:

Form the differential equation having y=sin-1x2+Acos-1x+B, where A and B are arbitrary constants, as its general solution.

Answer:

We have two constnts in the solution, so we will differentiate both sides twice and eliminate the constants A and B.

y=sin-1x2+Acos-1x+Bdydx=2sin-1x1-x2-A1-x21-x2dydx=2sin-1x-A
1-x2d2ydx2+-2x21-x2dydx=21-x21-x2d2ydx2-x1-x2dydx=21-x21-x2d2ydx2-xdydx=21-x2d2ydx2-xdydx-2=0

Page No 21.17:

Question 15:

Form the differential equation of the family of curves represented by the equation (a being the parameter):
(i) (2x + a)2 + y2 = a2
(ii) (2xa)2y2 = a2
(iii) (xa)2 + 2y2 = a2

Answer:

(i) The equation of the family of curves is
2x+a2+y2=a2                                            ...(1)
where a is a parameter.
As this equation has only one parameter, we shall get a differential equation of first order.
Differentiating (1) with respect to x, we get
22x+a×2+2ydydx=0                                    ...(2)
Now, from (1), we get
4x2+4ax+a2+y2=a24ax=-y2-4x2a=-4x2+y24x
Putting the value of a in (2), we get
42x-4x2+y24x+2ydydx=048x2-4x2-y24x+2ydydx=04x2-y2+2xydydx=0y2-4x2-2xydydx=0It is the required differential equation.

(ii) The equation of the family of curves is
2x-a2-y2=a24x2-4ax+a2-y2=a24x2-4ax-y2=0                                                  ...1
where a is a parameter.
As this equation has only one parameter, we shall get a differential equation of first order.
Differentiating (1) with respect to x, we get
8x-4a-2ydydx=0-ydydx+4x=2a                                   ...2
Now, from (1), we get
2a=4x2-y22x                                     ...(3)
From (2) and (3), we get
-ydydx+4x=4x2-y22x-2xydydx+8x2=4x2-y2-2xydydx+4x2+y2=02xydydx=4x2+y2It is the required differential equation.

(iii) The equation of the family of curves is
x-a2+2y2=a2x2-2ax+a2+2y2=a2x2-2ax+2y2=0                                                ...1
where a is a parameter.
As this equation has only one arbitrary constant, we shall get a differential equation of first order.
Differentiating (1) with respect to x, we get
2x-2a+4ydydx=0                                       ...(2)
Now, from (1), we get
2a=x2+2y2x                                             ...(3)
From (2) and (3), we get
2x-x2+2y2x+4ydydx=02x2-x2-2y2+4xydydx=04xydydx+x2-2y2=04xydydx=2y2-x2It is the required differential equation.

Page No 21.17:

Question 16:

Represent the following families of curves by forming the corresponding differential equations (a, b being parameters):
(i) x2 + y2 = a2

(ii) x2y2 = a2

(iii) y2 = 4ax

(iv) x2 + (yb)2 = 1

(v) (xa)2y2 = 1

(vi) x2a2-y2b2=1

(vii) y2 = 4a (xb)

(viii) y = ax3

(ix) x2 + y2 = ax3

(x) y = eax

Answer:

(i) The equation of the family of curves is 
x2+y2=a2                                                                ...(1)
where a is a parameter.
As this equation has only one arbitrary constant, we shall get a differential equation of first order.
Differentiating (1) with respect to x, we get
2x+2ydydx=0x+ydydx=0It is the required differential equation.

(ii) The equation of family of curves is 
x2-y2=a2                                                             ...(1)
where a is a parameter.
As this equation has only one arbitrary constant, we shall get a differential equation of first order.
Differentiating (1) with respect to x, we get
2x-2ydydx=0x-ydydx=0It is the required differential equation.

(iii) The equation of family of curves is 
y2=4ax                                                                 ...(1)
where a is a parameter.
As this equation has only one arbitrary constant, we shall get a differential equation of first order.
Differentiating (1) with respect to x, we get
2ydydx=4a2ydydx=y2x     Using 12xdydx=yy-2xdydx=0It is the required differential equation.

(iv) The equation of family of curves is 
x2+y-b2=1                                                        ...(1)             
where b is a parameter.
As this equation contains only one arbitrary constant, we shall get a differential equation of first order.
Differentiating (1) with respect to x, we get
2x+2y-bdydx=02x+21-x2dydx=0      Using 1x=-1-x2dydxx2=1-x2dydx2x2=dydx2-x2dydx2x21+dydx2=dydx2It is the required differential equation.

(v) The equation of family of curves is
x-a2-y2=1                                                       ...(1)
where a is a parameter.
As this equation has only one arbitrary constant, we shall get a differential equation of first order.
Differentiating (1) with respect to x, we get
2x-a-2ydydx=0x-a-ydydx=01+y2=ydydx      Using 11+y2=y2dydx2y2dydx2-y2=1It is the required differential equation.

(vi) The equation of family of curves is 
x2a2-y2b2=1                                                             ...(1)
where a and b are parameters.
As this equation has two arbitrary constants, we shall get a differential equation of second order.
Differentiating (1) with respect to x, we get
2xa2-2yb2dydx=0,                                                         ...(2)
Differentiating (2) with respect to x, we get
2a2-2b2dydx2-2yb2d2ydx2=02a2=2b2yd2ydx2+dydx2b2a2=yd2ydx2+dydx2                                                             ...3
Now, from (2), we get
2xa2=2yb2dydxb2a2=yxdydx                                                                 ...4
From (3) and (4), we get
yxdydx=yd2ydx2+dydx2xyd2ydx2+dydx2=ydydxIt is the required differential equation.

(vii) The equation of family of curves is
y2=4ax-b                                                               ...(1)                
where a and b are parameters.
As this equation has two arbitrary constants, we shall get a differential equation of second order.
Differentiating (1) with respect to x, we get
2ydydx=4aydydx=2a                                        ...2
Differentiating (2) with respect to x, we get
yd2ydx2+dydx2=0
 
It is the required differential equation.

(viii) The equation of family of curves is 
y=ax3                                                               ...(1)
where a is a parameter.
As this equation has only one arbitrary constant, so we shall get a differential equation of first order.
Differentiating (1) with respect to x, we get
dydx=3ax2dydx=3×yx3×x2                            Using 1xdydx=3yIt is the required differential equation.

(ix) The equation of family of curves is
x2+y2=ax3                                                       ...(1)
where a is a parameter.
As this equation has only one arbitrary constant, we shall get a differential equation of first order.
Differentiating (1) with respect to x, we get
2x+2ydydx=3ax22x+2ydydx=3x2+y2x3x2            Using 12x+2ydydx=3x2+y2x2x2+2xydydx=3x2+3y22xydydx=x2+3y2It is the required differential equation.

(x) The equation of family of curves is 
y=eaxlog y=ax                                                     ...1
where a is a parameter.
As this equation has only one arbitrary constant, we shall get a differential equation of first order.
Differentiating (1) with respect to x, we get
1ydydx=a1ydydx=log yx         Using 1xdydx=y log yIt is the required differential equation.

Page No 21.17:

Question 17:

Form the differential equation representing the family of ellipses having centre at the origin and foci on x-axis.

Answer:

The equation of the family of ellipses having centre at the origin and foci on the x-axis is
x2a2+y2b2= 1                                                  ...(1)
where a and b are the parameters.
As this equation contains two parameters, we shall get a second-order differential equation.
Differentiating (1) with respect to x, we get
2xa2+2yb2dydx=0                                              ...(2)
Differentiating (2) with respect to x, we get
2a2+2b2dydx2+yd2ydx2=02a2=-2b2dydx2+yd2ydx2b2a2=-dydx2+yd2ydx2                              ...(3)
Now, from (2), we get
xa2=-yb2dydxb2a2=-yxdydx                                                ...(4)
From (3) and (4), we get
-yxdydx=-dydx2+yd2ydx2yxdydx=dydx2+yd2ydx2ydydx=xdydx2+xyd2ydx2xyd2ydx2+xdydx2-ydydx=0 It is the required differential equation.

Page No 21.17:

Question 18:

Form the differential equation of the family of hyperbolas having foci on x-axis and centre at the origin.

Answer:

The equation of the family of hyperbolas having the centre at the origin and foci on the x-axis is
x2a2-y2b2=1                                                       ...(1)
where a and b are parameters.
As this equation contains two parameters, we shall get a second-order differential equation.
Differentiating equation (1) with respect to x, we get
2xa2-2yb2dydx=0                                                ...(2)
Differentiating equation (2) with respect to x, we get
2a2-2b2yd2ydx2+dydx2=01a2=1b2yd2ydx2+dydx2b2a2=yd2ydx2+dydx2                                                   3
Now, from equation (2), we get
2xa2=2yb2dydxb2a2=yxdydx                                                      ...(4)
From (3) and (4), we get
yxdydx=yd2ydx2+dydx2ydydx=xyd2ydx2+xdydx2xyd2ydx2+xdydx2-ydydx=0 It is the required differential equation.

Page No 21.17:

Question 19:

Form the differential equation of the family of circles in the second quadrant and touching the coordinate axes.

Answer:

The equation of the family of circles in the second quadrant and touching the co-ordinate axes is
x+a2+y-a2=a2x2+2ax+a2+y2-2ay+a2=a2x2+2ax+y2-2ay+a2=0                                           ...(1)
where a is a parameter.
As this equation contains one parameter, we shall get a differential equation of first order.
Differentiating equation (1) with respect to x, we get
2x+2a+2ydydx-2adydx=0x+ydydx+a-adydx=0x+ydydx+a1-dydx=0a=x+ydydxdydx-1                                                       ... (2)
From (1) and (2), we get
x2+2xx+ydydxdydx-1+y2-2yx+ydydxdydx-1+x+ydydxdydx-12=0x2dydx-12+2xx+ydydxdydx-1+y2dydx-12-2yx+ydydxdydx-1+x+ydydx2=0x2dydx2-2x2dydx+x2+2xxdydx-x+ydydx2-ydydx+y2dydx2-2dydx+1-2yxdydx-x+ydydx2-ydydx+x2+2xydydx+y2dydx2=0x2+2xydydx+y2dydx2=x2+2xy+y2+x2+2xy+y2dydx2x+ydydx2=x+y21+dydx2 It is the required differential equation.

Page No 21.17:

Question 20:

Find the differential equation representing the family of curves y=aebx+5, where a and b are arbitrary constants.

Answer:

Given: y=aebx+5
Differentiating y with respect to x
dydx=aebx+5b=abebx+5=by             Since y=aebx+5     .....(1)
Differentiating (1) again with respect to x we get
d2ydx2=bdydx                                                            .....(2)
Dividing (2) by (1) we get
d2ydx2dydx=bdydxbyyd2ydx2=dydx2

Page No 21.17:

Question 21:

Form the differential equation representing the family of curves y = e2x (a + bx), where a and b are arbitrary constants.

Answer:

ans



Page No 21.24:

Question 1:

Show that y = bex + ce2x is a solution of the differential equation, d2ydx2-3dydx+2y=0.

Answer:

We have,
y=bex+ce2x                                                   ...(1)
Differentiating both sides of equation (1) with respect to x, we get
dydx=bex+2ce2x                                              ...(2)
Differentiating both sides of equation (2) with respect to x, we get
d2ydx2=bex+4ce2x= 3bex+6ce2x-2bex-2ce2x= 3bex+2ce2x-2bex+ce2x= 3dydx-2y  Using equations 1 and 2
d2ydx2-3dydx+2y=0
Hence, the given function is the solution to the given differential equation.

Page No 21.24:

Question 2:

Verify that y = 4 sin 3x is a solution of the differential equation d2ydx2+9y=0.

Answer:

We have,
y=4 sin 3x                                              ...(1)
Differentiating both sides of equation (1) with respect to x, we get
dydx=12 cos 3x                                         ...(2)
Differentiating both sides of equation (2) with respect to x, we get
d2ydx2=-36 sin 3xd2ydx2=-94 sin 3xd2ydx2=-9y             Using equation 1
d2ydx2+9y=0
Hence, the given function is the solution to the given differential equation.

Page No 21.24:

Question 3:

Show that y = ae2x + bex is a solution of the differential equation d2ydx2-dydx-2y=0.

Answer:

We have,
y=ae2x+be-x                                    ...(1)
Differentiating both sides of equation (1) with respect to x, we get
dydx=2ae2x-be-x                               ...(2)
Differentiating both sides of equation (2) with respect to x, we get
d2ydx2=4ae2x+be-xd2ydx2=2ae2x-be-x+2ae2x+2be-xd2ydx2=2ae2x-be-x+2ae2x+be-xd2ydx2=dydx+2y          Using equations 1 and 2 
d2ydx2-dydx-2y=0
Hence, the given function is the solution to the given differential equation.

Page No 21.24:

Question 4:

Show that the function y = A cos x + B sin x is a solution of the differential equation d2ydx2+y=0.

Answer:

We have,
y=A cos x+B sin x                                            ...(1)
Differentiating both sides of equation (1) with respect to x, we get
dydx=-A sin x+B cos x                                     ...(2)
Differentiating both sides of equation (2) with respect to x, we get
d2ydx2=-A cos x-B sin xd2ydx2=-A cos x+B sin xd2ydx2=-y        Using equation 1
d2ydx2+y=0
Hence, the given function is the solution to the given differential equation.



Page No 21.25:

Question 5:

Show that the function y = A cos 2xB sin 2x is a solution of the differential equation d2ydx2+4y=0.

Answer:

We have,
y=A cos 2x-B sin 2x                                   ...(1)
Differentiating both sides of (1) with respect to x, we get
dydx=-2A sin 2x-2B cos 2x                         ...(2)
Differentiating both sides of (2) with respect to x, we get
d2ydx2=-4A cos 2x+4B sin 2xd2ydx2=-4A cos 2x-B sin 2xd2ydx2=-4y                  Using 1
d2ydx2+4y=0
Hence, the given function is the solution to the given differential equation.

Page No 21.25:

Question 6:

Show that y = AeBx is a solution of the differential equation d2ydx2=1ydydx2.

Answer:

We have,
y=AeBx                                           ...(1)
Differentiating both sides of (1) with respect to x, we get
dydx=ABeBx                                      ...(2)
Differentiating both sides of (2) with respect to x, we get
d2ydx2=AB2eBxd2ydx2=ABeBx2AeBxd2ydx2=1ydydx2          Using 1 and 2d2ydx2=1ydydx2
Hence, the given function is the solution to the given differential equation.

Page No 21.25:

Question 7:

Verify that y = ax+b is a solution of the differential equation d2ydx2+2xdydx=0.

Answer:

We have,
y=ax+b                                  ...(1)
Differentiating both sides of (1) with respect to x, we get
dydx=-ax2                              ...(2)
Differentiating both sides of (2) with respect to x, we get
d2ydx2=2ax3d2ydx2=-2x-ax2d2ydx2=-2xdydx                  Using 2d2ydx2+2xdydx=0
Hence, the given function is the solution to the given differential equation.

Page No 21.25:

Question 8:

Verify that y2 = 4ax is a solution of the differential equation y = x dydx+adxdy.

Answer:

We have,
y2=4ax                                          ...(1)
Differentiating both sides of (1) with respect to x, we get
2ydydx=4a
dydx=2ay                                    ...(2)
Now, differentiating both sides of (1) with respect to y, we get
2y=4adxdy
dxdy=y2a                                   ...(3)
 xdydx+adxdy=x2ay+ay2a         Using 2 and 3xdydx+adxdy=2axy+y2xdydx+adxdy= y22y+y2                       Using 1xdydx+adxdy= y2+y2xdydx+adxdy=yy=xdydx+adxdy
Hence, the given function is the solution to the given differential equation.

Page No 21.25:

Question 9:

Show that Ax2 + By2 = 1 is a solution of the differential equation x yd2ydx2+dydx2=ydydx.

Answer:

We have,
Ax2+By2=1                                           ...(1)
Differentiating both sides of (1) with respect to x, we get
2Ax+2Bydydx=0                                     ...(2)
Differentiating both sides of (2) with respect to x, we get

2A+2Bdydx2+2Byd2ydx2=02Byd2ydx2+dydx2=-2Aydydx+dydx2=-2A2Bydydx+dydx2=--yxdydx                      Using 2xydydx+dydx2=ydydx
Hence, the given function is the solution to the given differential equation.

Page No 21.25:

Question 10:

Show that y = ax3 + bx2 + c is a solution of the differential equation d3ydx3=6a.

Answer:

We have,
y=ax3+bx2+c                                         ...(1)
Differentiating both sides of (1) with respect to x, we get
dydx=3ax2+2bx                                        ...(2)
Differentiating both sides of (2) with respect to x, we get
d2ydx2=6ax+2b                                         ...(3)
Differentiating both sides of (3) with respect to x, we get
d3ydx3=6a
Hence, the given function is the solution to the given differential equation.

Page No 21.25:

Question 11:

Show that y = c-x1+cx is a solution of the differential equation (1 + x2) dydx + (1 + y2) = 0.

Answer:

We have,

y=c-x1+cx                                                     ...(1)

Differentiating both sides of (1) with respect to x, we get

dydx=1+cx-1-c-xc1+cx2dydx=-1-cx-c2+cx1+cx2dydx=-1+c21+cx2                                                    ...2

Now,

1+x2dydx+1+y2=-1+x21+c21+cx2+1+c-x21+cx2                      Using 1 and 2=-1+x21+c21+cx2+1+cx2+c-x21+cx2=-1+x21+c21+cx2+1+2cx+c2x2+c2-2cx+x21+cx2=-1+x21+c21+cx2+1+x2+c21+x21+cx2=-1+x21+c21+cx2+1+x21+c21+cx2=01+x2dydx+1+y2=0

Hence, the given function is the solution to the given differential equation.

Page No 21.25:

Question 12:

Show that y = ex (A cos x + B sin x) is the solution of the differential equation
d2ydx2-2dydx+2y=0.

Answer:

We have,
y=exA cos x+B sin x                                                      ...(1)
Differentiating both sides of (1) with respect to x, we get
dydx=exA cos x+B sin x+ex-A sin x+B cos x                     ...(2)
Differentiating both sides of (2) with respect to x, we get
d2ydx2=exA cos x+B sin x+ex-A sin x+B cos x+ex-A sin x+B cos x+ex-A cos x-B sin xd2ydx2=2ex-A sin x+B cos xd2ydx2=2ex-A sin x+B cos x+2exA cos x+B sin x-2exA cos x+B sin xd2ydx2=2dydx-2y            Using 1 and 2d2ydx2-2dydx+2y=0
Hence, the given function is the solution to the given differential equation.

Page No 21.25:

Question 13:

Verify that y = cx + 2c2 is a solution of the differential equation 2dydx2+xdydx-y=0.

Answer:

We have,
y=cx+2c2                                            ...(1)
Differentiating both sides of (1) with respect to x, we get
dydx=c                                                 ...(2)
Now,
2dydx2+xdydx-y=2c2+cx-cx-2c2=0             Using 1 and 22dydx2+xdydx-y=0
Hence, the given function is the solution to the given differential equation.

Page No 21.25:

Question 14:

Verify that y = − x − 1 is a solution of the differential equation (yx) dy − (y2x2) dx = 0.

Answer:

We have,
y=-x-1                                       ...(1)
Differentiating both sides of (1) with respect to x, we get
dydx=-1                                        ...(2)
Now,
dydx-y2-x2y-x=dydx-y+x=-1--x-1+x                   Using 1 and 2=-1+1=0dydx=y2-x2y-xy-xdy=y2-x2dxy-xdy-y2-x2dx=0
Hence, the given function is the solution to the given differential equation.

Page No 21.25:

Question 15:

Verify that y2 = 4a (x + a) is a solution of the differential equations y1-dydx2=2xdydx.

Answer:

We have,
y2=4ax+a                                        ...(1)
Differentiating both sides of (1) with respect to x, we get
2ydydx=4aydydx=2adydx=2ay                                                      ...2
Now,
y1-dydx2-2xdydx=y1-4a2y2-2x2ay=yy2-4a2y2-4axy=y2-4a2y-4axy=4ax+4a2-4a2y-4axy                  Using 1=4axy-4axy=0y1-dydx2=2xdydx
Hence, the given function is the solution to the given differential equation.

Page No 21.25:

Question 16:

Verify that y=cetan-1 x is a solution of the differential equation 1+x2d2ydx2+2x-1dydx=0.

Answer:

We have,
y=cetan-1x                                                    ...(1)
Differentiating both sides of (1) with respect to x, we get
dydx=cetan-1x11+x2                                       ...(2)
Differentiating both sides of (2) with respect to x, we get
d2ydx2=c1+x2etan-1x11+x2-etan-1x2x1+x22d2ydx2=cetan-1x-2xetan-1x1+x22d2ydx2=c1-2xetan-1x1+x221+x2d2ydx2=c1-2xetan-1x1+x21+x2d2ydx2=1-2xdydx                     Using 21+x2d2ydx2+2x-1dydx=0
Hence, the given function is the solution to the given differential equation.

Page No 21.25:

Question 17:

Verify that y=em cos-1x satisfies the differential equation 1-x2d2ydx2-xdydx-m2y=0

Answer:

We have,
y=em cos-1x                                                         ...(1)
Differentiating both sides of (1) with respect to x, we get
dydx=mem cos-1x-11-x2dydx=-mem cos-1x1-x2                                                         ...2
Differentiating both sides of (2) with respect to x, we get
d2ydx2=ddx-mem cos-1x1-x2d2ydx2=-m1-x2mem cos-1x-11-x2-em cos-1x12-2x1-x21-x21-x2d2ydx2=-m-mem cos-1x+xem cos-1x1-x21-x2d2ydx2=m2em cos-1x-mxem cos-1x1-x21-x2d2ydx2=m2y+xdydx                      Using 1 and 21-x2d2ydx2-xdydx-m2y=0
Hence, the given function is the solution to the given differential equation.

Page No 21.25:

Question 18:

Verify that y = log x+x2+a22 satisfies the differential equation a2+x2d2ydx2+xdydx=0.

Answer:

We have,
y=log x+x2+a22                                                 ...(1)
Differentiating both sides of (1) with respect to x, we get
dydx=ddxlog x+x2+a22=ddx2 log x+x2+a2=21+122xx2+a2x+x2+a2=2x2+a2+xx2+a2x+x2+a2=2x2+a2                                                                        ...2
Differentiating both sides of (2) with respect to x, we get
d2ydx2=2-122xx2+a2x2+a2a2+x2d2ydx2=-2xx2+a2a2+x2d2ydx2=-xdydx                            Using 2a2+x2d2ydx2+xdydx=0
Hence, the given function is the solution to the given differential equation.

Page No 21.25:

Question 19:

Show that the differential equation of which y=2x2-1+ce-x2 is a solution is dydx+2xy=4x3

Answer:

We have,
y=2x2-1+ce-x2                                                 ...(1)
Differentiating both sides of (1) with respect to x, we get
dydx=4x-ce-x22x=2x2-ce-x2=-2x2x2-2+ce-x2-2x2=-2x2x2-1+ce-x2-2x2=-2xy-2x2                              Using 1dydx=-2xy+4x3dydx+2xy=4x3
Hence, the given function is the solution to the given differential equation.

Page No 21.25:

Question 20:

Show that y = ex + ax + b is solution of the differential equation exd2ydx2=1.

Answer:

We have,
y=e-x+ax+b                                              ... (i)
Differentiating both sides of equation (i) with respect to x, we have
dydx=-e-x+a                                             ... (ii)
Differentiating both sides of equation (ii) with respect to x, we have
d2ydx2=e-xexd2ydx2=1
Hence, the given function is a solution of the given differential equation.

Page No 21.25:

Question 21:

For each of the following differential equations verify that the accompanying function is a solution:

Differential equation Function
(i) xdydx=y y = ax

(ii) x+ydydx=0

y=±a2-x2

(iii) xdydx+y=y2

y=ax+a

(iv) x3d2ydx2=1

y=ax+b+12x

(v) y=dydx2

y=14x±a2

Answer:

(i) We have,
y=ax                .....1
Given differential equation: xdydx=y
Differentiating both sides of (1) with respect to x, we get
dydx=adydx=yx      Using 1xdydx=y
Hence, the given function is the solution to the given differential equation.

(ii) We have,
y=±a2-x2y2=a2-x2                  .....1
Given differential equation: x+ydydx=0
Differentiating both sides of (1) with respect to x, we get
2y dydx=-2xy dydx=-xx+y dydx=0
Hence, the given function is the solution to the given differential equation.

(iii) We have,
y=ax+axy+ay=axy=a1-yxy1-y=a1-yxy=1a          .....1
given differential equation: xdydx+y=y2
Differentiating both sides of (1) with respect to x, we get
xy0-dydx-1-yxdydx+yxy2=0xy-dydx-1-yxdydx+y=0-xydydx-xdydx-y+xydydx+y2=0-xdydx-y+y2=0xdydx+y=y2
Hence, the given function is the solution to the given differential equation.

(iv) We have,
y=ax+b+12x          .....1
Differentiating both sides of (1) with respect to x, we get
dydx=a-12x2          .....2Now differentiating both sides of 2 with respect to x, we getd2ydx2=-12×-2x3d2ydx2=1x3x3 d2ydx2=1
Hence, the given function is the solution to the given differential equation.

(v) We have,
y=14x±a2          .....1
Differentiating both sides of (1) with respect to x, we get
dydx=14×2x±adydx=12x±aSquaring both sides we getdydx2=12x±a2dydx2=14x±a2dydx2=y                   Using 1y=dydx2
Hence, the given function is the solution to the given differential equation.



Page No 21.28:

Question 1:

Differential equation xdydx=1, y1=0

Function y = log x

Answer:

We have,
y=log x                                              ...(1)
Differentiating both sides of (1) with respect to x, we get
dydx=1x
or, 
xdydx=1
It is the given differential equation.
Thus, y=log x satisfies the given differential equation.
Hence, it is a solution.
Also, when x=1, y=log 1=0, i.e.,  y1=0.
Hence, y=log x is the solution to the given initial value problem.

Page No 21.28:

Question 2:

Differential equation dydx=y, y0=1

Function y = ex

Answer:

We have,
y=ex                                                 ...(1)
Differentiating both sides of (1) with respect to x, we get
dydx=ex
dydx=y                    [Using (1)]
It is the given differential equation.
Here, y=ex satisfies the given differential equation; hence, it is a solution.
Also, when x=0, y=e0=1, i.e., y(0)=1.
Hence, y=ex is the solution to the given initial value problem.

Page No 21.28:

Question 3:

Differential equation d2ydx2+y=0, y 0=0, y' 0=1

Function y = sin x

Answer:

We have,
y=sin x                                               ...(1)
Differentiating both sides of (1) with respect to x, we get
dydx=cos x                                           ...(2)
Differentiating both sides of (2) with respect to x, we get
d2ydx2=-sin xd2ydx2=-y   Using 1
d2ydx2+y=0
It is the given differential equation.
Here, y=sin x satisfies the given differential equation; hence, it is a solution.
Also, when x=0, y=sin 0=0, i.e., y0=0.
And, when x=0, y'=cos 0=1, i.e., y'0=1.
Hence, y=sin x is the solution to the given initial value problem.

Page No 21.28:

Question 4:

Differential equation d2ydx2-dydx=0, y 0=2, y'0=1

Function y = ex + 1

Answer:

We have,
y=ex+1                                             ...(1)
Differentiating both sides of (1) with respect to x, we get
dydx=ex                                            ...(2)
Differentiating both sides of (2) with respect to x, we get
d2ydx2=exd2ydx2=dydx                Using 2d2ydx2-dydx=0 It is the given differential equation.
y=ex+1 satisfies the given differential equation; hence, it is a solution.
Also, when x=0, y=e0+1=1+1=2, i.e. y0=2.
And, when x=0, y'=e0=1, i.e. y'0=1.
Hence, y=ex+1 is the solution to the given initial value problem.

Page No 21.28:

Question 5:

Differential equation dydx+y=2, y 0=3

Function y = ex + 2

Answer:

We have,
y=e-x+2                                            ...(1)
Differentiating both sides of (1) with respect to x, we get
dydx=-e-xdydx=-y-2            Using 1dydx+y=2 It is the given differential equation.
y=e-x+2 satisfies the given differential equation; hence, it is a solution.
Also, when x=0, y=e0+2=1+2=3, i.e. y0=3.
Hence, y=e-x+2 is the solution to the given initial value problem.

Page No 21.28:

Question 6:

Differential equation d2ydx2+y=0, y 0=1, y' 0=1

Function y = sin x + cos x

Answer:

We have,
y=sin x+ cos x                                             .....(1)
Differentiating both sides of (1) with respect to x, we get
dydx=cos x-sin x                                           .....(2)
Differentiating both sides of (2) with respect to x, we get
d2ydx2=-sin x-cos xd2ydx2=-sin x+cos xd2ydx2=-y   Using 1

d2ydx2+y=0
It is the given differential equation.
Therefore, y=sin x+ cos x satisfies the given differential equation.
Also, when x=0; y=sin 0+cos 0=1, i.e. y0=1.
And, when x=0; y'=cos 0-sin 0=1, i.e. y'0=1.
Hence, y=sin x+ cos x  is the solution to the given initial value problem.

Page No 21.28:

Question 7:

Differential equation d2ydx2-y=0, y 0=2, y' 0=0

Function y = ex + ex

Answer:

We have,
y = ex + ex                                         .....(1)
Differentiating both sides of (1) with respect to x, we get
dydx=ex-e-x                                            .....(2)
Differentiating both sides of (2) with respect to x, we get

d2ydx2=ex+e-xd2ydx2=y   Using 1

d2ydx2-y=0
It is the given differential equation.
Therefore, y = ex + ex satisfies the given differential equation.
Also, when x=0; y=e0+e0=1+1, i.e. y0=2.
And, when x=0; y1=e0-e0=1-1, i.e. y'0=0.
Hence, y = ex + ex is the solution to the given initial value problem.

Page No 21.28:

Question 8:

Differential equation d2ydx2-3dydx+2y=0, y 0=1, y' 0=3

Function y = ex + e2x

Answer:

We have,
y = ex + e2x                                               .....(1)
Differentiating both sides of (1) with respect to x, we get
dydx=ex+2e2x                                                 .....(2)
Differentiating both sides of (2) with respect to x, we get
d2ydx2=ex+4e2xd2ydx2=3ex+2e2x-2ex+e2xd2ydx2=3dydx-2y   Using 1 and 2d2ydx2-3dydx+2y=0

d2ydx2-3dydx+2y=0
It is the given differential equation.
Therefore, y = ex + e2x satisfies the given differential equation.
Also, when x=0; y=e0+e0=1+1, i.e. y0=2.
And, when x=0; y'=e0+2e0=1+2, i.e. y'0=3.
Hence, y = ex + e2x is the solution to the given initial value problem.

Disclaimer: In the question instead of y(0) = 1, it should have been y(0) = 2.

Page No 21.28:

Question 9:

Differential equation d2ydx2-2dydx+y=0, y 0 =1, y' 0=2

Function y = xex + ex

Answer:

We have,
y = xex + ex                .....(1)
Differentiating both sides of (1) with respect to x, we get
dydx=xex+ex+exdydx=xex+2ex                        ....2 
Differentiating both sides of (2) with respect to x, we get

d2ydx2=xex+ex+2exd2ydx2=xex+3exd2ydx2=2xex+2ex-xex+exd2ydx2=2dydx-y            Using 1 and 2d2ydx2-2dydx+y=0  

d2ydx2-2dydx+y=0  

It is the given differential equation.
Thus, y = xex + ex satisfies the given differential equation.
Also, when x=0, y=0+1=1, i.e. y0=1.
And, when x=0, y'=0+2=2, i.e. y'0=2.
Hence, y = xex + ex is the solution to the given initial value problem.



Page No 21.34:

Question 1:

dydx=x2+x-1x, x0

Answer:

We have, dydx=x2+x-1xdy=x2+x-1xdxIntegrating both sides, we getdy=x2+x-1xdxy=x33+x22-logx+CClearly, y=x33+x22-logx+C is defined for all xR except x=0.Hence, y=x33+x22-logx+C, where xR-0, is the solution to the given differential equation.

Page No 21.34:

Question 2:

dydx=x5+x2-2x, x0

Answer:

We have, dydx=x5+x2-2xdy=x5+x2-2xdxIntegrating both sides, we getdy=x5+x2-2xdxy=x66+x33-2logx+CClearly, y=x66+x33-2logx+C is defined for all xR except x=0.Hence, y=x66+x33-2logx+C, where xR-0, is the solution to the given differential equation.

Page No 21.34:

Question 3:

dydx+2x=e3x

Answer:

We have, dydx+2x=e3xdydx=e3x-2xdy=e3x-2xdxIntegrating both sides, we getdy=e3x-2xdxy=e3x3-2x22+Cy=e3x3-x2+Cy+x2=e3x3+CSo, y+x2=e3x3+C is defined for all xR. Hence, y+x2=e3x3+C, where xR, is the solution to the given differential equation.

Page No 21.34:

Question 4:

x2+1dydx=1

Answer:

We have, x2+1dydx=1dydx=1x2+1dy=1x2+1dxIntegrating both sides, we getdy=1x2+1dxy=tan-1x+CSo, y=tan-1x+C is defined for all xR. Hence, y=tan-1x+C, where xR, is the solution to the given differential equation.

Page No 21.34:

Question 5:

dydx=1-cos x1+cos x

Answer:

We have, dydx=1-cos x1+cos xdydx=2sin2 x22cos2 x2dydx=tan2 x2dy=tan2 x2dxIntegrating both sides, we getdy=tan2 x2dxdy=sec2 x2-1dxy= 2 tan x2-x+CSo, y= 2 tan x2-x+C is defined for all xR. Hence, y= 2 tan x2-x+C, where xR, is the solution to the given differential equation.

Page No 21.34:

Question 6:

x+2dydx=x2+3x+7

Answer:

We have, x+2dydx=x2+3x+7dydx=x2+3x+7x+2dy=x2+3x+7x+2dxIntegrating both sides, we getdy=x2+3x+7x+2dxdy=x2+3x+2+5x+2dxdy=x+2x+1+5x+2dxdy=x+1+5x+2dxy=x22+x+5 logx+2+CSo, y=x22+x+5 logx+2+C is defined for all xR except x=-2.Hence, y=x22+x+5 logx+2+C, where xR-2, is the solution to the given differential equation.

Page No 21.34:

Question 7:

dydx=tan-1 x

Answer:

We have, dydx=tan-1xdy=tan-1xdxIntegrating both sides, we getdy=tan-1xdxy=1II×tan-1xI dx y=tan-1x1 dx-ddxtan-1x1 dxdxy=x tan-1x-x1+x2dxy=x tan-1x-122x1+x2dxy=x tan-1x-12log1+x2+CSo, y=x tan-1x-12log1+x2+C is defined for all xR. Hence, y=x tan-1x-12log1+x2+C is the solution to the given differential equation.

Page No 21.34:

Question 8:

dydx=log x

Answer:

We have, dydx=log xdy=log xdxIntegrating both sides, we getdy=log xdxdy=1II×log xI dx dy=log x1 dx-ddxlog x1 dxdxy=xlog x-xxdxy=xlog x-1dxy=xlog x-xy=xlog x-1+Cy=xlog x-1+CSo, y=xlog x-1+C is defined for all xR except x=0. Hence, y=xlog x-1+C, where xR-0, is the solution to the given differential equation.
 

Page No 21.34:

Question 9:

1xdydx=tan-1 x, x0

Answer:

We have, 1xdydx=tan-1 x
dydx=x tan-1xdy=x tan-1xdxIntegrating both sides, we getdy=x tan-1xdxy=xII×tan-1xI dx dy=tan-1xx dx-ddxtan-1xx dxdxy=x2 tan-1x2-12x21+x2dxy=x2 tan-1x2-12x2+1-11+x2dxy=x2 tan-1x2-121-11+x2dxy=x2 tan-1x2-12x+tan-1x2+Cy=x2+1 tan-1x2-12x+CHence, y=x2+1 tan-1x2-12x+C is the solution to the given differential equation.

Page No 21.34:

Question 10:

dydx=cos3 x sin2 x+x2x+1

Answer:

We have, dydx=cos3 x sin2 x+x2x+1dy=cos3 x sin2 x+x2x+1dxIntegrating both sides, we getdy=cos3 x sin2 x+x2x+1dxy=cos3 x sin2 x dx +x2x+1dx y=I1 +I2           .....1

where I1=cos3 x sin2 x dx I2=x2x+1dxNow, I1=cos3 x sin2 x dx=sin2 x 1-sin2 xcos x dxPutting t= sin x, we getdt=cos x dx I1=t2 1-t2dt=t2-t4dt=t33-t55+C1=sin3 x3-sin5 x5+C1 I2=x2x+1dx

Putting t2=2x+1, we get2t dt=2dxtdt=dxNow,  I2=t2-12t× t dt=12t4-t2 dt=t510-t36+C2=2x+15210-2x+1326+C2 

Putting the values of I1 and I2 in1, we gety=sin3 x3-sin5 x5+C1+2x+15210-2x+1326+C2 y=sin3 x3-sin5 x5+2x+15210-2x+1326+C              Where, C=C1+C2Hence, y=sin3 x3-sin5 x5+2x+15210-2x+1326+C is the solution to the given differential equation.

Page No 21.34:

Question 11:

(sin x + cos x) dy + (cos x − sin x) dx = 0

Answer:

We have, sin x+cos xdy+cos x-sin xdx=0dy=-cos x-sin xsin x+cos xdxIntegrating both sides, we getdy=-cos x-sin xsin x+cos xdxy=-cos x-sin xsin x+cos xdxPutting sin x+cos x=t cos x-sin x dx=dt  y=-dtty=-logt+Cy=-logsin x+cos x+Cy+logsin x+cos x=CHence, y+logsin x+cos x=C is the solution to the given differential equation.

Page No 21.34:

Question 12:

dydx-x sin2 x=1x log x

Answer:

We have, dydx-x sin2 x=1xlog xdydx=1xlog x+x sin2 xdydx=1xlog x+x 21-cos 2xdydx=1xlog x+x 2-x 2cos 2xdy=1xlog x+x 2-x 2cos 2xdxIntegrating both sides, we getdy=1xlog x+x 2-x 2cos 2xdxy=1xlog xdx+12x dx-12x cos 2xdxy=loglog x+12×x22-12xI×cos 2xII dx y=loglog x+x24-x2cos 2xdx+12ddxxcos 2x dxdxy=loglog x+x24-xsin 2x4-cos 2x8+CHence, y=loglog x+x24-xsin 2x4-cos 2x8+C is the solution to the given differential equation.

Page No 21.34:

Question 13:

dydx=x5 tan-1x3

Answer:

We have,dydx=x5 tan-1x3dy=x5 tan-1x3dxIntegrating both sides, we getdy=x5 tan-1x3dxy=x5 tan-1x3dxPutting t=x3, we getdt=3x2dx y=13t tan-1 t dt=13tII×tan-1 tI dx =13tan-1 tt dt-ddttan-1 tt dxdt=13×t2 tan-1 t2-16t21+t2dt=t2 tan-1 t6-16t2+1-11+t2dt=t2 tan-1 t6-16dt+1611+t2dt=t2 tan-1 t6-16t+tan-1 t6 +C=x6 tan-1 x36-16x3+tan-1 x36+C=16x6 tan-1 x3-x3+tan-1 x3+CHence, y=16x6 tan-1 x3-x3+tan-1 x3+C is the solution to the given differential equation.

Page No 21.34:

Question 14:

sin4 xdydx=cos x

Answer:

We have, sin4 xdydx= cos xdy=cos xsin4 xdxIntegrating both sides, we getdy=cos xsin4 xdxy=cos xsin4 xdxPutting sin x=tcos x dx=dt y=1t4dt=t-3-3+C=-sin-3 x3+C=-13cosec3 x+C Hence, y=-13cosec3 x+C is the solution to the given differential equation.

Page No 21.34:

Question 15:

cos xdydx-cos 2x=cos 3x

Answer:

We have, cos xdydx-cos 2x=cos 3xdy=cos 3x+cos 2xcos xdxdy=4cos3 x-3cos x+2cos2 x-1cos xdxdy=4cos2 x-3+2cos x-sec xdxdy=22cos2 x-1-1+2cos x-sec xdxdy=2cos 2x-1+2cos x-sec xdxIntegrating both sides, we getdy=2cos 2x-1+2cos x-sec xdxy=sin 2x-x+2sin x-logsec x+tan x+CHence, y=sin 2x-x+2sin x-logsec x+tan x+C is the solution to the given differential equation.

Page No 21.34:

Question 16:

1-x4 dy=x dx

Answer:

We have, 1-x4dy=x dxdy=x1-x4dxIntegrating both sides, we getdy=x1-x4dxy=x1-x4dxPutting x2=t2x dx=dty=12dt1-t2=sin-1t2+C=12sin-1x2+CHence, y=12sin-1x2+C is the solution to the given differential equation.

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Question 17:

a+x dy+x dx=0

Answer:

We have, a+xdy+xdx=0a+xdy=-xdxdy=-xa+xdxdy=-x+a-aa+xdxdy=-a+x-aa+xdxIntegrating both sides, we getdy=-a+x-aa+xdxdxy=-2a+x323+2aa+x+Cy+23a+x32-2aa+x=CHence, y+23a+x32-2aa+x=C is the solution to the given differential equation.

Page No 21.34:

Question 18:

1+x2dydx-x=2 tan-1 x

Answer:

We have,1+x2dydx-x=2tan-1x1+x2dydx=x+2tan-1xdy=x1+x2+21+x2tan-1xdxdy=12×2x1+x2+21+x2tan-1x dxIntegrating both sides, we getdy=12×2x1+x2+21+x2tan-1x dxy=122x1+x2dx+211+x2tan-1x dxy=12log1+x2+211+x2tan-1x dxPutting tan-1x=t11+x2dx=dty=12log1+x2+2t dt=12log1+x2+t2+C=12log1+x2+tan-1x2+CHence, y=12log1+x2+tan-1x2+C is the solution to the given differential equation.

Page No 21.34:

Question 19:

dydx=x log x

Answer:

We have, dydx=x log xdy=x log xdxIntegrating both sides, we getdy=x log xdxy=xII×log xI dx  y=log xx dx-ddxlog xx dxdxy=log x×x2 2-1x×x22dxy=12x2 log x-x2dxy=12x2 log x-x24+CHence, y=12x2 log x-x24+C is the solution to the given differential equation.

Page No 21.34:

Question 20:

dydx=x ex-52+cos2 x

Answer:

We have, dydx=x ex-52+cos2 xdydx=x ex-52+cos 2x2+12dydx=x ex+cos 2x2-2dy=x ex+cos 2x2-2dxIntegrating both sides, we getdy=x ex+cos 2x2-2dxy=x ex dx+12cos 2x dx-2dxy=xex dx-ddxxex dxdx+12×sin 2x2-2xy=x ex-ex+14sin 2x-2x+CHence, y=x ex-ex+14sin 2x-2x+C is the solution to the given differential equation.

Page No 21.34:

Question 21:

x3+x2+x+1dydx=2x2+x

Answer:

We have, x3+x2+x+1dydx=2x2+x dydx=2x2+xx3+x2+x+1 dy=2x2+xx+1x2+1dxIntegrating both sides, we getdy=2x2+xx+1x2+1dxy=2x2+xx+1x2+1dxLet 2x2+xx+1x2+1=Ax+1+Bx+Cx2+12x2+x=Ax2+A+Bx2+Bx+Cx+C2x2+x=A+Bx2+B+Cx+A+CComparing the coefficients on both sides, we getA+B=2               .....1B+C=1               .....2A+C=0               .....3Solving 1, 2 and 3, we getA=12B=32C=-12y=121x+1dx+32x-12x2+1 dx=121x+1dx+123xx2+1dx-121x2+1dx=121x+1dx+342xx2+1dx-121x2+1dx=12logx+1+34logx2+1-12tan-1 x+CHence, y=12logx+1+34logx2+1-12tan-1 x+C  is the solution to the given differential equation.

Page No 21.34:

Question 22:

sindydx=k ; y0=1

Answer:

We have, sin dydx=kdydx=sin-1 kdy=sin-1 kdxIntegrating both sides, we getdy=sin-1 k dxy=xsin-1 k +C     .....1It is given that y0=1.1=0×sin-1 k+CC=1Substituting the value of C in 1, we gety=x sin-1 k +1y-1=x sin-1 k Hence, y-1=x sin-1 k is the solution to the given differential equation.

Page No 21.34:

Question 23:

edy/dx=x+1 ; y0=3

Answer:

We have, edydx=x+1Taking log on both sides, we getdydx log e=logx+1dydx=logx+1dy=logx+1dxIntegrating both sides, we getdy=logx+1dxy=1II×log x+1I dx y=log x+11 dx-ddxlog x+11 dxdxy=x log x+1-xx+1dxy=x log x+1-1-1x+1dxy=x log x+1-x+log x+1+C     .....1It is given that y0=3.3=0×log 0+1-0+log 0+1+CC=3Substituting the value of C in 1, we gety=x log x+1+log x+1-x+3y=x+1 log x+1-x+3Hence, y=x+1 log x+1-x+3 is the solution to the given differential equation.

Page No 21.34:

Question 24:

C' (x) = 2 + 0.15 x ; C(0) = 100

Answer:

We have, C' x=2+0.15xdCdx=2+0.15xdC=2+0.15xdxIntegrating both sides, we getdC=2+0.15x dxC=2x+0.152x2+D     .....1It is given that C0=100.100=20 +0.1520+DD=100Substituting the value of D in 1, we getC=2x +0.152x2+100Hence, C=2x +0.152x2+100 is the solution to the given differential equation.

Page No 21.34:

Question 25:

xdydx+1=0 ; y -1=0

Answer:

We have, xdydx+1=0 dydx=-1xdy=-1xdxIntegrating both sides, we getdy=-1xdxy=-logx+C     .....1It is given that y-1=0.0=-log-1+CC=0Substituting the value of C in 1, we gety=-logxHence, y=-logx is the solution to the given differential equation.

Page No 21.34:

Question 26:

xx2-1dydx=1, y2=0

Answer:

We have, xx2-1dydx=1 dydx=1xx2-1dy=1xx2-1dxIntegrating both sides, we getdy=1xx2-1dxy=1xx2-1dxy=1xx-1x+1dxLet 1xx-1x+1 =Ax+Bx-1+Cx+11=Ax2-1+Bx2+x+Cx2-x1=A+B+Cx2+B-Cx-AEquating the coefficients on both sides we getA+B+C=0     .....1B-C=0            .....2A=-1               .....3Solving 1, 2 and 3, we getA=-1B=12C=12y=121x-1dx-1xdx+121x+1dx=12logx-1-logx+12logx+1+C=12logx-1+12logx+1-logx+CIt is given that y2=0.0=12log2-1+12log2+1-log2+CC=log2-12log3Substituting the value of C, we gety=12logx-1+12logx+1-logx+log2-12log32y=logx-1+logx+1-2logx+2log2-log32y=logx-1+logx+1-logx2+log 4-log 32y=log4x-1x+13x2y=12log4x2-13x2Hence, y=12log4x2-13x2 is the solution to the given differential equation.



Page No 21.38:

Question 1:

 dydx+1+y2y=0

Answer:

We have,dydx+1+y2y=0
dydx=-1+y2ydxdy=-y1+y2dx=-y1+y2dyIntegrating both sides, we getdx=-y1+y2dyx=-y1+y2dyPutting 1+y2=t, we get2y dy=dtx=-121tdtx=-12logt+Cx=-12log1+y2+Cx+12log1+y2=CHence, x+12log1+y2=C is the required solution.

Page No 21.38:

Question 2:

dydx=1+y2y3

Answer:

We have,dydx=1+y2y3
dxdy=y31+y2dx=y31+y2dyIntegrating both sides, we getdx=y31+y2dyx=y+y3-y1+y2dyx=1+y2y-y1+y2dyx=y dy-y1+y2dyx=y22-y1+y2dyPutting 1+y2=t we get 2y dy=dtx=y22-121tdtx=y22-12logt+Cx=y22-12log1+y2+C      t=1+y2Hence, x=y22-12log1+y2+C  is the required solution.

Page No 21.38:

Question 3:

dydx=sin2 y

Answer:

We have,dydx=sin2 ydxdy=1sin2ydx=cosec2y dyIntegrating both sides, we getdx=cosec2y dyx=-cot y+Cx+ cot y=CHence, x+ cot y=C is the required solution.

Page No 21.38:

Question 4:

dydx=1-cos 2y1+cos 2y

Answer:

We have,dydx=1-cos 2y1+cos 2ydxdy=1+ cos 2y1-cos 2ydx=1+ cos 2y1-cos 2ydydx=2 cos2y2 sin2ydydx=cot2y dyIntegrating both sides, we getdx=cot2y dyx=cosec2y-1 dyx=cosec2y dy-dyx=-cot y-y+Cx+ cot y+y=CHence, x+ cot y+y=C is the required solution.



Page No 21.4:

Question 1:

d3xdt3+d2xdt2+dxdt2=et

Answer:

In this differential equation, the order of the highest order derivative is 3 and its power is 1. So, it is a differential equation of order 3 and degree 1.

It is a non-linear differential equation because the differential coefficient dxdt has exponent 2, which is greater than 1.

Page No 21.4:

Question 2:

d2ydx2+4y=0

Answer:

In this differential equation, the order of the highest order derivative is 2 and its power is 1. So, it is a differential equation of order 2 and degree 1.

It is a linear differential equation.



Page No 21.5:

Question 3:

dydx2+1dy/dx=2

Answer:

dydx2+1dydx=2
dydx3+1dydx=2
dydx3+1=2dydx
dydx3-2dydx+1=0

In this equation, the order of the highest order derivative is 1 and its highest power is 3. So, it is a differential equation of order 1 and degree 3.
It is a non-linear differential equation because the differential coefficient dydx has exponent 3, which is greater than 1.

Page No 21.5:

Question 4:

1+dydx2=cd2ydx21/3

Answer:

1+dydx2=cd2ydx213Squaring both sides, we get1+dydx2=cd2ydx223Taking cubes of both sides, we getcd2ydx22=1+dydx23c2d2ydx22=1+3dydx2+3dydx4+dydx6

In this differential equation, the order of the highest order derivative is 2 and its power is 2. So, it is a differential equation of order 2 and degree 2.

It is a non-linear differential equation, as its degree is more than 1.

Page No 21.5:

Question 5:

d2ydx2+dydx2+xy=0

Answer:

d2ydx2+dydx2+xy=0

In this differential equation, the order of the highest order derivative is 2 and its power is 1. So, it is a differential equation of order 2 and degree 1.
It is a non-linear differential equation, as the differential coefficient dydx has exponent 2, which is greater than 1.

Page No 21.5:

Question 6:

d2ydx23=dydx

Answer:

d2ydx23=dydxd2ydx213=dydx12Taking cubes of both the sides, we getd2ydx2=dydx32Squaring both the sides, we getd2ydx22=dydx3d2ydx22-dydx3=0

In this differential equation, the order of the highest order derivative is 2 and its power is 2. So, it is a differential equation of order 2 and degree 2.

Thus, it is a non-linear differential equation, as its degree is 2, which is greater than 1.

Page No 21.5:

Question 7:

d4ydx4=c+dydx23/2

Answer:

d4ydx4=c+dydx232Squaring both sides, we getd4ydx42=c+dydx23d4ydx42=c3+3c2dydx2+3cdydx4+dydx6

In this differential equation, the order of the highest order derivative is 4 and its power is 2. So, it is a differential equation of order 4 and degree 2.

Thus, it is a non-linear differential equation, as its degree is 2, which is greater than 1.

Page No 21.5:

Question 8:

x+dydx=1+dydx2

Answer:

x+dydx=1+dydx2x+dydx=1+dydx212Squaring both sides, we getx+dydx2=1+dydx2x2+2xdydx+dydx2=1+dydx22xdydx+x2=1

In this differential equation, the order of the highest order derivative is 1 and the power is 1. So, it is a differential equation of order 1 and degree 1.

Hence, it is a linear differential equation.

Disclaimer: The answer given in the book has some error. The solution here is created according to the question given in the book.

Page No 21.5:

Question 9:

yd2xdy2=y2+1

Answer:

yd2xdy2=y2+1

In this differential equation, the order of the highest order derivative is 2 and its power is 1. So, it is a differential equation of order 2 and degree 1.

It is a linear differential equation.

Page No 21.5:

Question 10:

s2d2tds2+stdtds=s

Answer:

s2d2tds2+stdtds=ssd2tds2+tdtds=1

In this differential equation, the order of the highest order derivative is 2 and its power is 1. So, it is a differential equation of order 2 and degree 1.

It is a non-linear differential equation, as it contains the product of the dependent variable t and its differential co-efficient dtds.

Page No 21.5:

Question 11:

x2d2ydx23+ydydx4+y4=0

Answer:

x2d2ydx23+ydydx4+y4=0

In this differential equation, the order of the highest order derivative is 2 and its power is 3. So, it is a differential equation of order 2 and degree 3.

It is a non-linear differential equation, as its degree is more than 1.

Page No 21.5:

Question 12:

d3ydx3+d2ydx23+dydx+4y=sin x

Answer:

d3ydx3+d2ydx23+dydx+4y=sin x

In this differential equation, the order of the highest order derivative is 3 and its power is 1. So, it is a differential equation of degree 3 and order 1.
It is a non-linear differential equation, as its differential co-efficient d2ydx2 has exponent 3, which is greater than 1.

Page No 21.5:

Question 13:

(xy2 + x) dx + (yx2y) dy = 0

Answer:

xy2+xdx+y-x2ydy=0xy2+1dx=yx2-1dyxy2+1yx2-1=dydxxy2+1dydx-yx2-1=0y2+1dydx-yx-1x=0

In this differential equation, the order of the highest order derivative is 1 and its power is 1. So, it is a differential equation of degree 1 and order 1.

It is a non-linear equation, as the product containing dependent variable and its differential co-efficient y2dydx is present in it.

Page No 21.5:

Question 14:

1-y2 dx+1-x2 dx=0

Answer:

1-y2dx+1-x2dy=01-y2dx=-1-x2dy-1-y21-x2=dydx1-x2dydx+1-y2=0

In this differential equation, the order of the highest order derivative is 1 and its power is 1. So, it is a differential equation of order 1 and degree 1.

It is a non-linear equation, as the exponent of dependent variable y is more than 1 (on expanding 1-y2 binomially).

Page No 21.5:

Question 15:

d2ydx2=dydx2/3

Answer:

d2ydx2=dydx23Taking cubes of both sides, we getd2ydx23=dydx2

In this differential equation, the order of the highest order derivative is 2 and its power is 3. So, it is a differential equation of order 2 and degree 3.

It is a non-linear differential equation, as it has degree 3, which is greater than 1.

Page No 21.5:

Question 16:

2d2ydx2+31-dydx2-y=0

Answer:

2d2ydx2+31-dydx2-y=02d2ydx2=-31-dydx2-ySquaring both sides, we get4d2ydx22=91-dydx2-y4d2ydx22+9dydx2+9y-9=0

In this differential equation, the order of the highest order derivative is 2 and its power is 2. So, it is a differential equation of order 2 and degree 2.

It is a non-linear differential equation, as it has degree 2, which is greater than 1.

Page No 21.5:

Question 17:

5d2ydx2=1+dydx23/2

Answer:

5d2ydx2=1+dydx232Squaring both sides, we get25d2ydx22=1+dydx2325d2ydx22=1+3dydx2+3dydx4+dydx625d2ydx22-dydx6-3dydx4-3dydx2-1=0

In this differential equation, the order of the highest order derivative is 2 and its power is 2. So, it is a differential equation of order 2 and degree 2.

It is a non-linear differential equation, as its degree is 2, which is greater than 1.

Page No 21.5:

Question 18:

y=xdydx+a1+dydx2

Answer:

y=xdydx+a1+dydx2y-xdydx=a1+dydx2Squaring both sides, we gety-xdydx2=a21+dydx2y2-2xydydx+x2dydx2=a2+a2dydx2x2-a2dydx2-2xydydx+y2-a2=0

In this differential equation, the order of the highest order derivative is 1 and its highest power is 2. So, it is a differential equation of order 1 and degree 2.

It is a non-linear differential equation, as its degree is 2, which is greater than 1.

Page No 21.5:

Question 19:

y=px+a2p2+b2, where p=dydx

Answer:

y=px+a2p2+b2y-px=a2p2+b2Squaring both sides, we gety-px2=a2p2+b2y2-2pxy+p2x2=a2p2+b2x2-a2p2-2pxy+y2-b2=0x2-a2dydx2-2xydydx+y2-b2=0                  Substituting p=dydx

In this differential equation, the order of the highest order derivative is 1 and its highest power is 2. So, it is a differential equation of order 1 and degree 2.

It is a non-linear differential equation, as its degree is 2, which is greater than 1.

Page No 21.5:

Question 20:

d2ydx2+3dydx2=x2 logd2ydx2

Answer:

d2ydx2+3dydx2=x2 log d2ydx2

In this differential equation, the order of the highest order derivative is 2.
Clearly, the R.H.S. of the differential equation cannot be expressed as a polynomial in d2ydx2.
Thus, its degree is not defined.

The order of the differential equation is 2 and its degree is not defined.
It is a non-linear differential equation, as one of its differential co-efficients, that is, dydx, has exponent 2, which is greater than 1.

Page No 21.5:

Question 21:

x2d2ydx2=1+dydx24

Answer:

ans

Page No 21.5:

Question 22:

d2ydx22+dydx2=x sin d2ydx2

Answer:

d2ydx22+dydx2=x sin d2ydx2

In this differential equation, the order of the highest order derivative is 2.
Clearly, the R.H.S. of the differential equation cannot be expressed as a polynomial in d2ydx2. So, its degree is not defined.

The order of the differential equation is 2 and its degree is not defined.

It is a non-linear differential equation, as one of its differential co-efficients, that is, dydx, has exponent 2, which is more than 1.

Page No 21.5:

Question 23:

(y'')2 + (y')3 + sin y = 0

Answer:

y''2+y'3+sin y=0

In this differential equation, the order of the highest order derivative is 2 and its power is 2. So, the order of the differential equation is 2 and its degree is 2.

It is a non-linear differential equation, as its degree is 2, which is more than 1.

Page No 21.5:

Question 24:

d2ydx2+5xdydx-6y=log x

Answer:

d2ydx2+5xdydx-6y=log x

In this differential equation, the order of the highest order derivative is 2 and its power is 1. So, the order of the differential equation is 2 and its degree is 1.

It is a linear differential equation.

Disclaimer: The answer in the book has some error. The solution here is created according to the question given in the book.

Page No 21.5:

Question 25:

d3ydx3+d2ydx2+dydx+y sin y=0

Answer:

d3ydx3+d2ydx2+dydx+y sin y=0

In this differential equation, the order of the highest order derivative is 3 and its power is 1. So, the order of the differential equation is 3 and its degree is 1.

It is a non-linear differential equation, as the exponent of the dependent variable is not equal to 1 (by expanding y.sin y).

Disclaimer: The answer given in the book has some error. The solution here is created according to the question given in the book.

Page No 21.5:

Question 26:

dydx+ey=0

Answer:

dydx+ey=0

In this differential equation, the order of the highest order derivative is 1 and its power is 1. So, the order of the differential equation is 1 and its degree is 1.

It is a non-linear differential equation, as the exponent of the dependent variable is not equal to 1 (as per expansion series of ey).

Page No 21.5:

Question 27:

dydx3-4dydx2+7y=sin x

Answer:

dydx3-4dydx2+7y=sin x

In this differential equation, the order of the highest order derivative is 1 and its highest power is 3. So, the order of the differential equation is 1 and its degree is 3.

It is a non-linear differential equation, as its degree is 3, which is greater than 1.

Disclaimer: The answer given in the book has some error. The solution here is created according to the question given in the book.



Page No 21.54:

Question 1:

x-1dydx=2 xy

Answer:

We have,x-1dydx=2 xyx-1dy=2xy dx2xx-1dx=1ydyIntegrating both sides, we get2xx-1dx=1ydy2x-1+1x-1dx=1ydy2dx+21x-1dx=1ydy2x+2 logx-1=log y+C

Page No 21.54:

Question 2:

(1 + x2) dy = xy dx

Answer:

We have,1+x2 dy=xy dx1ydy=x1+x2dxIntegrating both sides, we get1ydy=x1+x2dxSubstituting 1+x2=t, we get2x dx=dt1ydy=121tdtlogy=12log t  +log C logy=12log1+x2+log C             ( t=1+x2)logy=logC1+x2y=C1+x2Hence, y=C1+x2 is the required solution.

Page No 21.54:

Question 3:

dydx=ex+1 y

Answer:

We have,dydx=ex+1 y1ydy=ex+1 dxIntegrating both sides, we get 1ydy=ex+1 dxlog y=ex+x+CHence, log y=ex+x+C is the required solution.

Page No 21.54:

Question 4:

x-1dydx=2x3y

Answer:

We have,x-1dydx=2x3y1ydy=2x3x-1dxIntegrating both sides, we get1ydy=2x3x-1dxlog y=2x3-1+1x-1dxlog y=2x-1x2+x+1+1x-1dxlog y=2x2+x+1dx+21x-1dxlog y=23x3+x2+2x+log x-1+CHence, log y=23x3+x2+2x+log x-1+C is the required solution.

Page No 21.54:

Question 5:

xy (y + 1) dy = (x2 + 1) dx

Answer:

We have,xyy+1dy=x2+1dxyy+1dy=x2+1xdxy2+ydy=x+1xdxIntegrating both sides, we get y2+ydy=x+1xdxy2 dy+ y dy=x dx+1xdxy33+y22=x22+log x+CHence, y33+y22=x22+log x+C is the required solution.

Page No 21.54:

Question 6:

5dydx=ex y4

Answer:

We have,5dydx=ex y45y4dy=exdxIntegrating both sides, we get 5y4dy=exdx-53y3=ex+CHence, -53y3=ex+C is the required solution.

Page No 21.54:

Question 7:

x cos y dy = (xex log x + ex) dx

Answer:

We have,x cos y dy=x exlog x+ex dx cos y dy= ex log x+1xexdxIntegrating both sides, we get cos y dy= ex log x+1xexdxsin y=log x  ex dx-1xex dx+1xex dxsin y=ex log x+CHence, sin y=ex log x+C is the required solution.

Page No 21.54:

Question 8:

dydx=ex+y+x2 ey

Answer:

We have,dydx=ex+y+x2 eydydx=exey+x2eydydx=eyex+x21eydy=ex+x2 dxIntegrating both sides, we get 1eydy=ex+x2 dx-e-y=ex+x33+C

Page No 21.54:

Question 9:

xdydx+y=y2

Answer:

We have,xdydx+y=y2xdydx=y2-y1y2-ydy=1xdxIntegrating both sides, we get 1y2-ydy=1xdx1yy-1dy=1xdx               .....1Let 1yy-1=Ay+By-11=Ay-1+ByPutting y=0, we get1=-AA=-1Putting y=1, we get1=B1yy-1=-1y+1y-11yy-1dy=-1y dy+1y-1dy            .....2 From (1) & (2), we get  -1y dy+1y-1dy =1xdx -log y+log y-1 =log x+log Clog y-1y- log x=log Clogy-1xy=log Cy-1xy=Cy-1=CxyHence, y-1=Cxy is the required solution.

Page No 21.54:

Question 10:

(ey + 1) cos x dx + ey sin x dy = 0

Answer:

We have,ey+1 cos x dx +ey sin x dy=0ey sin x dy=-ey+1 cos x dxeyey+1dy=-cos xsin xdxeyey+1dy=-cot x dxIntegrating both sides, we get eyey+1dy=-cot x dxPutting ey+1=t, we getey dy=dtdtt=-cot x dxlog t =-log  sin x +log C    log  ey+1+log sin x=log Clogey+1 sin x=log Cey+1 sin x=Cey+1 sin x=CHence, ey+1 sin x=C is the required solution. 

Page No 21.54:

Question 11:

x cos2 y  dx = y cos2 x dy

Answer:

We have,x cos 2 y dx=y cos 2x dyxcos2xdx=ycos2ydyx sec2x dx=y sec2y dyIntegrating both sides, we get xI sec2xII dx=yI sec2yII dyxsec2x dx-ddxxsec2x dxdx=ysec2y dy-ddyysec2y dydyx tan x-tan x dx=y tan y-tan y dyx tan x-log  sec x=y tan y-log  sec  y+Cx tan x- y tan y=log  sec x-log  sec y+CHence, x tan x- y tan y=log  sec x-log  sec y+C is the required solution.

Page No 21.54:

Question 12:

xy dy = (y − 1) (x + 1) dx

Answer:

We have,xy dy=y-1x+1 dxyy-1dy=x+1xdxIntegrating both sides, we getyy-1dy=x+1xdxy-1+1y-1dy=x+1xdxdy+1y-1dy=dx+1xdxy+log y-1=x+log  x +Cy-x=log x- log y-1+CHence, y-x=log  x - log y-1+C is the required solution.

Page No 21.54:

Question 13:

xdydx+cot y=0

Answer:

We have,xdydx+cot y=0xdydx=- cot y1xdx=-1cot ydy1xdx=-tan y dyIntegrating both sides, we get1xdx=-tan y dyln  x =-ln sec y+ln Cln  x =ln  cos y +ln Cx=C cos y Hence, x=C cos y is the required solution.

Page No 21.54:

Question 14:

dydx=x ex log x+exx cos y

Answer:

We have,dydx=xex logx +exx cos yx cos y dy=x exlog x+ex dx cos y dy= ex log x+1xexdxIntegrating both sides, we get cos y dy= ex log x+1xexdxsin y=log x  ex dx-1xex dx+1xex dxsin y=ex log x+CHence, sin y=ex log x+C is the required solution.

Page No 21.54:

Question 15:

dydx=ex+y+ey x3

Answer:

We have,dydx=ex+y+ey x3dydx=exey+eyx3dydx=eyex+x3ex+x3 dx=1eydyIntegrating both sides, we get ex+x3 dx=1eydyex+x44=-e-y+Cex+e-y+x44=CHence,ex+e-y+x44=C is the required solution.

Page No 21.54:

Question 16:

y1+x2+x1+y2dydx=0

Answer:

We have,y1+x2+x1+y2dydx=0x1+y2dydx=-y1+x2x1+y2dy=-y1+x2 dx1+y2ydy=-1+x2xdx

Integrating both sides, we get1+y2ydy=- 1+x2xdxPutting 1+y2=t2 and 1+x2=u2, we get2y dy=2t dt and 2x dx=2u dudy=tydt and dx=uxdut2y2dt=-u2x2dxt2t2-1dt=-u2u2-1du

t2-1+1t2-1dt=-u2-1+1u2-1dudt+1t2-1dt=-du-1u2-1dut+12logt-1t+1=-u-12logu-1u+1+CSubstituting t by 1+y2 and u by 1+x2

1+y2+12log1+y2-11+y2+1=-1+x2-12log1+x2-11+x2+1+C1+y2+1+x2+12log1+x2-11+x2+1+12log1+y2-11+y2+1=CHence, 1+y2+1+x2+12log1+x2-11+x2+1+12log1+y2-11+y2+1=C is the required solution.

Page No 21.54:

Question 17:

1+x2 dy+1+y2 dx=0

Answer:

We have,1+x2 dy+1+y2 dx=01+x2 dy=-1+y2 dx11+y2 dy=-11+x2 dxIntegrating both sides, we get11+y2 dy=-11+x2 dxlog y+1+y2=-log x+1+x2+log Clog y+1+y2+log x+1+x2=log Clog y+1+y2x+1+x2=log Cy+1+y2x+1+x2=CHence, log y+1+y2x+1+x2=C is the required differential equation.

Page No 21.54:

Question 18:

1+x2+y2+x2y2+xydydx=0

Answer:

We have,1+x2+y2+x2y2+xydydx=01+x21+y2+xydydx=0xydydx=-1+x21+y2xydydx=-1+x21+y2y1+y2dy=-1+x2xdxIntegrating both sides, we gety1+y2dy=-1+x2xdxy1+y2dy=-x1+x2x2dxPutting 1+y2=t  and 1+x2=u22y dy=dt      and 2x dx=2uduy dy=dt2      and xdx=uduIntegral becomes,12dtt=-u×uu2-1dut=-u2u2-1dut=-u2u2-1dut=-1+1u2-1dut=-(1)du-1u2-1dut=-u-12logu-1u+1+C1+y2=-1+x2-12log1+x2-11+x2+1+C1+y2+1+x2+12log1+x2-11+x2+1=C

Page No 21.54:

Question 19:

dydx=exsin2 x+sin 2xy2 log y+1

Answer:

   dydx=exsin2x+sin 2xy2log y+1y2log y+1dy=exsin2x+sin 2xdx2y log y+ydy=exsin2x+exsin 2xdx2y log y dy + y dy=exsin2x dx+exsin 2x dxIntegrating both sides, we get2yII log yI dy+y dy=ex IIsin2xI dx+exsin 2x dx2log yy dy-ddylog yy dydy+y dy=sin2xex dx-ddxsin2xex dxdx+exsin 2x dx2log y y22-1yy22dy+y dy=sin2x ex-2sin xcos x  exdx+exsin 2x dx+Cy2log y-y dy+y dy=exsin2x-exsin 2x dx+exsin 2x dx+Cy2log y=exsin2x+C

Page No 21.54:

Question 20:

dydx=x2 log x+1sin y+y cos y

Answer:

We have,dydx=x2 log x+1sin y+y cos y

sin y+y cos y dy=x2 log x +1 dxIntegrating both sides, we getsin y+y cos y dy=x2 log x +1 dxsin y dy+y cos y dy=2x log x dx+x dx-cos y+ycos y dy-ddyy cos y dydy=2log x x dx-ddxlog xx dxdx+x22-cos y+y sin y-sin y dy=2 log x× x22-1x×x22 +x22 -cos y+y sin y+cos y=x2 log x-x22+x22+C y sin y=x2 log x+CHence, y sin y=x2 log x+C is the required solution.

Page No 21.54:

Question 21:

(1 − x2) dy + xy dx = xy2 dx

Answer:

We have,1-x2 dy+xy dx= xy2 dx 1-x2 dy=xy2 dx- xy dx1-x2 dy=xy y-1 dx1yy-1 dy=x1-x2 dxIntegrating both sides, we get1yy-1 dy=x1-x2 dx      .....(1)Considering LHS of (1),Let 1yy-1=Ay+By-11=Ay-1+By         .....(2)   Substituting y=1 in (2),1=B Substituting y=0 in (2),1=-AA=-1Substituting the values of A and B in 1yy-1=Ay+By-1, we get1yy-1=-1y+1y-11yy-1dy=-1ydy+1y-1dy                             =-log y+ log y-1+C1         Now, considering RHS of (2), we havex1-x2 dxHere, putting 1-x2=t, we get -2x dx=dtx1-x2 dx=-121tdt                      =-12log t+C2                      =-12log 1-x2+C2              t=1-x2Now, substituting the value of 1yy-1dy and x1-x2 dx in (1), we get-log y+log y-1+C1=-12log 1-x2+C2-log y+log y-1=-12log 1-x2+C whereC=C2-C1

Page No 21.54:

Question 22:

tan y dx + sec2 y tan x dy = 0

Answer:

We have,tan y dx+sec2y tanx dy=0sec2y tanx dy=-tan y dxsec2ytan y dy=-1tanxdx1cos2y×cosysinydy=-cot x dx1siny cosydy=-cot x dx2sin 2y dy=-cot x dx2 cosec 2y dy=-cot x dxIntegrating both sides, we get2cosec 2y dy=-cot x dxlog tan x=-log sin x= log Clog tan x+log sin x= log Clog tan x×sin x= log Ctan x×sin x=C

Page No 21.54:

Question 23:

(1 + x) (1 + y2) dx + (1 + y) (1 + x2) dy = 0

Answer:

We have,1+x1+y2 dx+1+y1+x2dy=01+x1+y2 dx=-1+y1+x2dy1+x1+x2dx=-1+y1+y2dyIntegarting both sides, we get 1+x1+x2dx=-1+y1+y2dy11+x2dx+x1+x2dx=-11+y2dy-y1+y2dySubstituting 1+x2=t in the second integral of LHS and 1+y2=u in the second integral of RHS, we get2x dx=dt   and    2ydy=du11+x2dx+121tdt=-11+y2dy-121udu tan-1x+12log t=-tan-1 y-12log u+Ctan-1x+12log 1+x2=-tan-1 y-12log 1+y2+Ctan-1x+tan-1y+12log 1+x2+12log 1+y2=Ctan-1x+tan-1 y+12log 1+x21+y2=CHence, tan-1x+tan-1 y+12log 1+x21+y2=C is the required solution.

Page No 21.54:

Question 24:

tan ydydx = sin (x + y) + sin (xy)

Answer:

We have,tan y dydx=sin x+y+sin x-ytan y dydx=sin x cos y+cos x sin y+sin x cos y-cos x sin ytan y dydx=2 sin xcos y tan ycos ydy=2 sin x dxtany sec y dy=2 sinx dxIntegrating both sides, we get tany sec y dy=2sin x dxsec y=-2 cos x+Csec y+2 cos x=CHence, sec y+2 cos x=C is the required solution.



Page No 21.55:

Question 25:

cos x cos ydydx=-sin x sin y

Answer:

We have,cos x cos y dydx=-sin x siny cos ysin  ydy=-sin xcos xdxcot y dy=-tan x dxIntegrating both sides, we get cot y dy=-tan x dxlog sin y=-log  sec x +log Clog sin y=log  cos x  +log Csin y=C cos xHence, sin y=C cos x is the required solution.

Page No 21.55:

Question 26:

dydx+cos x sin ycos y=0

Answer:

We have,dydx+cos x sin ycos y=0dydx=-cos x sin ycos ycos ysin ydy=-cos x dxcot y dy=-cos x dxIntegrating both sides, we get cot y dy=-cos x dxlog  sin y=-sin x+CHence, log  sin y=-sin x+C is the required solution.

Page No 21.55:

Question 27:

x1-y2 dx+y1-x2 dy=0

Answer:

We have,x1-y2 dx+y1-x2 dy=0y1-x2 dy=-x1-y2 dxy1-y2 dy=-x1-x2 dxIntegrating both sides, we gety1-y2 dy=-x1-x2 dxSubstituting 1-y2=t and 1-x2=u, we get-2y dy=dt and -2x dy=du -121tdt=121udu-t12=u12+K1-x2+1-y2=-K1-x2+1-y2=C     where, C=-KHence, 1-x2+1-y2=C is the required solution. 

Page No 21.55:

Question 28:

y (1 + ex) dy = (y + 1) ex dx

Answer:

We have,y1+ex dy=y+1 ex dxyy+1dy=ex1+exdxIntegrating both sides, we get yy+1dy=ex1+exdxSubstituting 1+ex=t, we getexdx=dtyy+1dy=1tdty+1-1y+1dy=1tdtdy-1y+1dy=1tdty-log y+1=log t+Cy-log y+1=log 1+ex+C

Page No 21.55:

Question 29:

(y + xy) dx + (xxy2) dy = 0

Answer:

We have,y+xy dx+x-xy2 dy=0y1+xdx=xy2-1dy1+xxdx=y2-1ydyIntegrating both sides, we get 1+xxdx=y2-1ydy1xdx+dx=y dy-1ydylog x+x=y22-log y+Clog x+x-y22+log y=CHence, log x+x-y22+log y=C is the required solution.

Page No 21.55:

Question 30:

dydx=1-x+y-xy

Answer:

We have,dydx=1-x+y-xydydx=1+y-x1+ydydx=1+y1-x11+ydy=1-x dxIntegrating both sides, we get11+ydy=1-x dxlog 1+y=x-x22+CHence, log 1+y=x-x22+C is the required solution.

Page No 21.55:

Question 31:

(y2 + 1) dx − (x2 + 1) dy = 0

Answer:

We have,y2+1 dx-x2+1 dy=0y2+1 dx=x2+1 dy1x2+1dx=1y2+1dyIntegrating both sides, we get1x2+1dx=1y2+1dytan-1 x=tan-1 y+Ctan-1 x-tan-1y=CHence, tan-1 x-tan-1y=C is the required solution.

Page No 21.55:

Question 32:

dy + (x + 1) (y + 1) dx = 0

Answer:

We have,dy+x+1y+1 dx=0dy=-x+1y+1 dx1y+1dy=-x+1 dxIntegrating both sides, we get1y+1dy=-x+1 dxlog y+1=-x22-x+Clog y+1+x22+x=CHence, log y+1+x22+x=C is the required solution.

Page No 21.55:

Question 33:

dydx=1+x21+y2

Answer:

We have,dydx=1+x21+y211+y2dy=1+x2 dx Integrating both sides, we get11+y2dy=1+x2 dx tan-1y=x+x33+CHence, tan -1y=x+x33+C is the required solution.

Page No 21.55:

Question 34:

x-1dydx=2x3 y

Answer:

We have,x-1dydx=2x3 y1ydy=2x3x-1dxIntegrating both sides, we get 1ydy=2x3x-1dxlog y=2x3-1+1x-1dxlog y=2x3-1x-1dx+1x-1dxlog y=2x-1x2+x+1x-1dx+1x-1dxlog y=2x2+x+1 dx+1x-1dxlog y=2 x33+x22+x+log x-1+Clog y=23x3+x2+2x+log x-12+Cy=e23x3+x2+2x+log x-12+Cy=eC×elog x-12×e23x3+x2+2xy=C1x-12 e23x3+x2+2x          eln x=x and where, C1=eC   y=C1x-12 e23x3+x2+2x is required solution.

Page No 21.55:

Question 35:

dydx=ex+y+e-x+y

Answer:

We have,dydx=ex+y+e-x+ydydx=eyex+e-xe-ydy=ex+e-x dxIntegrating both sides, we gete-ydy=ex+e-x dx-e-y=ex-e-x+Ce-x-e-y=ex+CHence, e-x-e-y=ex+C is the required solution.

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Question 36:

dydx=cos2 x-sin2 x cos2 y

Answer:

We have,dydx=cos2 x-sin2 x cos2 ydydx=cos 2x cos2y1cos2ydy=cos 2x dxsec2y dy=cos 2x dxIntegrating both sides, we getsec2y dy=cos 2x dxtan y=sin 2x2+CHence, tan y=sin 2x2+C is the required solution.

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Question 37:

Solve the following differential equation: 
(i) (xy2 + 2xdx + (x2 y + 2ydy = 0
(ii) cosecx logy dydx+ x2y2=0

Answer:

(i) (xy2 + 2xdx + (x2 y + 2ydy = 0

We have,xy2+2x dx+x2y+2y dy=0xy2+2 dx+yx2+2 dy=0xy2+2 dx=-yx2+2 dyxx2+2 dx=-yy2+2 dyIntegrating both sides, we getxx2+2 dx=-yy2+2 dy122xx2+2 dx=-122yy2+2 dy12log x2+2=-12log y2+2+log C12log x2+2+12log y2+2=log Clog x2+2+log y2+2=2log Clog x2+2y2+2=log C2x2+2y2+2=C2x2+2y2+2=Ky2+2=Kx2+2

(ii)
cosecx logy dydx+ x2y2=0cosecx logy dydx=- x2y2 1y2logy dy=- x2cosecxdx 1y2logy dy=-x2sinxdx 1y2logy dy=-x2sinxdx
-logyy+1y×1y=--x2cosx+2xcosxdx+C-logyy-1y=--x2cosx+2xsinx-2sinxdx+C-1+logyy=--x2cosx+2xsinx+2cosxdx+C-1+logyy-x2cosx+2xsinx+cosx=C
 

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Question 38:

Solve the following differential equations:
(i) xydydx=1+x+y+xy

(ii) y1-x2dydx=x1+y2
(iii) yexydx=xexy+y2dy, y0
(iv) 1+y2tan-1xdx+2y1+x2dy=0

Answer:

(i) We have, xydydx=1+x+y+xyxydydx=1+x1+yy1+ydy=1+xxdxIntegrating both sides, we get y1+ydy=1+xxdx1+y-11+ydy=1+xxdxdy-11+ydy=1xdx+dxy-log 1+y= log x+x+Cy= log x+log 1+y+x+Cy=log x1+y+x+C Hence, y=log x1+y+x+C is the required solution.

(ii) We have,y1-x2dydx=x1+y2y1+y2dy=x1-x2dxIntegrating both sides ,y1+y2dy=x1-x2dxSubstituting 1+y2=t and 1-x2=u 2ydy=dt and -2x dx=du121tdt=-121udu12log t =-12log u+log C12log 1+y2=-12log 1-x2+log C12log 1+y2+log 1-x2=log Clog 1+y21-x2=2 log C 1+y21-x2=C2   1+y21-x2=C1       , where C1=C2Hence,1+y21-x2=C1 is the required solution. 

(iii)

yexydx=xexy+y2dyyexydx=xexydy+y2dyyexydx-xexydy=y2dyydx-xdyexy=y2dyydx-xdyy2exy=dyexydxy=dyexydxy=dyexy=y+C
(iv)
1+y2tan-1xdx+2y1+x2dy=01+y2tan-1xdx=-2y1+x2dytan-1x1+x2dx=-2y1+y2dytan-1x1+x2dx=-2y1+y2dytan-1x22=-log1+y2+Ctan-1x22+log1+y2=C

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Question 39:

dydx=y tan 2x, y0=2

Answer:

We have,dydx=y tan 2x, y0=21ydy=tan 2x dxIntegrating both sides, we get 1ydy=tan 2x dxlog y=12log  sec 2x+12log Cy2=C sec 2x       .....1It is given that at x=0, y=2. C=4Substituting the value of C in (1), we get y2 =4cos 2xy=2cos 2x Hence, y=2cos 2x is the required solution.

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Question 40:

2xdydx=3y, y1=2

Answer:

We have,2xdydx=3y,   y1=22ydy=3xdxIntegrating both sides, we get 21ydy=31xdx2 log y=3 log x+log Clog y2=log x3+log Cy2=Cx3       ....(1)It is given that at x=1, y=2.Substituting the values of x and y in (1), we getC=4Now, substituting the value of C in (1), we gety2=4x3Hence, y2=4x3 is the required solution.

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Question 41:

xydydx=y+2, y2=0

Answer:

We have,xydydx=y+2, y2=0yy+2dy=1xdxIntegrating both sides, we getyy+2dy=1xdxy+2-2y+2dy=1xdxdy-21y+2dy=log x+Cy-2 log y+2=log x+C      .....(1) It is given that at x=2, y=0.Substituting the values of x and y in (1), we get-2log 2-log 2=C-log 22×2=CC=-log 8Substituting the value of C in (1), we gety-2 log y+2=log x-log 8y-2 log y+2=log x8Hence, y-2log y+2=log x8 is the required solution.

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Question 42:

dydx=2ex y3, y0=12

Answer:

We have,dydx=2ex y3,    y0=121y3dy=2exdxIntegrating both sides, we get 1y3dy=2exdx-12y2=2ex+C          .....(1)Given: at x=0, y=12Substituting the values of x and y in (1), we get-12×14=2e0+CC=-2-2C=-4Substituting the value of C in (1), we get-12y2=2ex-4y28-4ex=1Hence, y28-4ex=1 is the required solution.

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Question 43:

drdt=-rt, r0=r0

Answer:

We have,drdt=-rt, r0=r0

1rdr =-t dtIntegrating both sides, we get1rdr =-t dtlog r=-t22+C          ....(1)Given: t=0, r=r0. Substituting the values of x and y in (1), we getlog  r0 =0+CC=log  r0 Substituting the value of C in (1), we get
log r=-t22+log  r0    log  r -log  r0 =-t22log rr0=-t22r=r0e-t22Hence, r=r0e-t22 is the required solution.

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Question 44:

dydx=y sin 2x, y0=1

Answer:

We have,dydx=y sin2x, y0=11ydy =sin 2x dxIntegrating both sides, we get1ydy =sin 2x dxlog y=-cos 2x2+C          .....(1)Given:  x=0, y=1.Substituting the values of x and y in (1), we getlog  1 =-12+CC=12Substituting the value of C in (1), we get
log y=-cos 2x2+12log y =1-cos 2x2log y=sin 2xy=esin2xHence, y=esin2x is the required solution.

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Question 45:

(i) dydx=y tan x, y0=1

(ii) 2xdydx=5y, y1=1

(iii) dydx=2e2x y2, y0=-1

(iv) cos ydydx=ex, y0=π2

(v) dydx=2xy, y0=1

(vi) dydx=1+x2+y2+x2y2, y0=1

(vii) xydydx=x+2y+2, y1=-1
(viii) dydx=1+x+y2+xy2 when y = 0, x = 0  [NCERT EXEMPLAR]
(ix) 2y+3-xydydx=0, y(1) = −2 [NCERT EXEMPLAR]
(x) extan y dx+2-exsec2y dy=0, y0=π4         [CBSE 2018]

Answer:

(i) dydx=y tanx, y0=11ydy =tan x dxIntegrating both sides, we get1ydy =tan x dxlog y=log sec x+C           .....(1)We know that at x=0 and y=1.Substituting the values of x and y in (1), we getlog  1 =log  1 +CC=0Substituting the value of C in (1), we get

log y=log  sec x+0y= sec xHence, y=sec x, where x-π2,π2, is the required solution. 


(ii) 2xdydx=5y, y1=12ydy=5x dxIntegrating both sides, we get 21ydy =51x dx2log y=5log  x+ C           .....(1)We know that at x=1 and y=1.Substituting the values of x and y in (1), we get2log  1 =5log  1 +CC=0Substituting the value of C in (1), we get 

 2 log y=5 log  x+0y=x52Hence, y=x52 is the required solution.

(iii) dydx=2e2xy2, y0=-11y2dy=2e2x dxIntegrating both sides, we get1y2dy =2e2x dx-1y=e2x+ C           .....(1)We know that at x=0, y=-1.Substituting the values of x and y in (1), we get1=1+CC=0Substituting the value of C in (1), we get

-1y=e2xy =-e-2xHence, y=-e-2x is the required solution.

(iv) cos ydydx=ex, y0=π2cos y dy=ex dxIntegrating both sides, we getcos y dy=ex dxsin y=ex+C           .....(1)We know that at x=0, y=π2.Substituting the values of x and y in (1), we get1=1+CC=0Substituting the value of C in (1), we get

sin y=exy=sin-1exHence, y=sin-1ex is the required solution.

(v) dydx=2xy, y0=11ydy=2x dxIntegrating both sides, we get 1ydy=2x dxlog y=x2+C           .....(1)We know that at x=0, y=1.Substituting the values of x and y in (1), we get0=0+CC=0Substituting the value of C in (1), we get log y=x2y=ex2Hence, y=ex2 is the required solution.

(vi) dydx=1+x2+y2+x2y2,   y0=1dydx=1+x21+y2dy1+y2=1+x2 dxIntegrating both sides, we get dy1+y2=1+x2 dxtan-1y=x+x33+C           .....(1)We know that at x=0, y=1.Substituting the values of x and y in (1), we getπ4=0+0+CC=π4Substituting the value of C in (1), we get tan-1y=x+x33+π4Hence,tan-1y=x+x33+π4 is the required solution.

(vii) xydydx=x+2y+2,   y1=-1yy+2dy=x+2xdxy+2-2y+2dy=x+2xdx1-2y+2dy=1+2xdxIntegrating both sides, we get 1-2y+2dy=1+2xdxy-2log y+2=x+2log x+C           .....(1)We know that at x=1, y=-1.Substituting the values of x and y in (1), we get-1-2log 1=1+2log 1+C-1=1+CC=-2Substituting the value of C in (1), we get y-2log y+2=x+2log x-2Hence, y-2log y+2=x+2log x-2 is the required solution.
(viii) dydx=1+x+y2+xy2
dydx=1+x+y21+xdydx=1+x1+y2dy1+y2=1+xdxdy1+y2=1+xdxtan-1y=x+x22+C         .....1Now, tan-10=0+0+C           y=0, x=0C=0Substituting the value of C in (1), we gettan-1y=x+x22y=tanx+x22

(ix) 2y+3-xydydx=0
2y+3=xydydx2xdx=yy+3dy2xdx=y+3-3y+3dy2xdx=1-3y+3dy2xdx=1-3y+3dy2logx=y-3logy+3+Clogx2+logy+33=y+Clogx2y+33=y+C    .....1
log12-2+33=-2+CC=2Substituting the value of C in (1), we getlogx2y+33=y+2x2y+33=ey+2

(x)
extany dx+2-exsec2y dy=0, y0=π4
extan y dx=ex-2sec2y dyUsing variable separablesec2ytan ydy=exex-2dxIntegrating both sidesec2ytan ydy=exex-2dxLet tan y=p  sec2y dy=dpand ex-2=q exdx=dq
dpp=dqqlnp=lnq+lnc, where c is constant of integrationp=qcReplacing the valuestan y=cex-2When x=0, y=π41=c(-1)c=-1tan y=2 - exy=tan-12-ex

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Question 46:

Solve the differential equation xdydx+cot y=0, given that y=π4, when x = 2.

Answer:

We have,xdydx+cot y=0xdydx=-cot ytan y dy=-1xdxIntegrating both sides, we gettan y dy=-1xdxlog  sec y=- log x+log Clog  x  sec y =log Cx sec y=C            .....(1) Given: x=2 , y= π4.Substituting the values of x and y in (1), we get2 sec π4=CC=2Substituting the value of C in (1), we getx sec y=2x=2 cos yHence, x=2 cos y is the required solution.

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Question 47:

Solve the differential equation 1+x2dydx+1+y2=0, given that y = 1, when x = 0.

Answer:

We have,1+x2dydx+1+y2=0 , y=1 when x=01+x2dydx=-1+y211+y2 dy=-11+x2dxIntegrating both sides, we get11+y2 dy=-11+x2dxtan-1y=-tan-1x+Ctan-1y+tan-1x=C         .....(1) Given: x=0, y= 1.Substituting the values of x and y in (1), we get π4+0=CC=π4Substituting the value of C in (1), we gettan-1y+tan-1x=π4tan-1x+tan-1y=π4tan-1x+y1-xy=π4x+y1-xy=1x+y=1-xyHence, x+y=1-xy is the required solution.

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Question 48:

Solve the differential equation dydx=2xlog x+1sin y+y cos y, given that y = 0, when x = 1.

Answer:

We have,dydx=2xlog x+1sin y+ycos ysin y+ycos y dy=2xlog x+1 dxIntegrating both sides, we getsin y+ycos y dy=2xlog x+1 dxsin y dy+ ycos y dy=2x log x dx+2x dx-cos y+ycos y dy-ddyycos y dydy=2log x x dx-ddxlog xx dxdx+x2-cos y+y sin y+cos y=2 log x×x22-x24+x2+Cy sin y=x2log x-x22+x2+Cy sin y=x2log x+x22+C       .....(1)Given: x=1, y=0.Substituting the values of x and y in (1), we get 0=0+12+CC=-12Substituting the value of C in (1), we gety sin y=x2log x+x22-122y sin y=2x2log x+x2-1Hence, 2y sin y=2x2log x+x2-1 is the required solution.



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Question 49:

Find the particular solution of edy/dx = x + 1, given that y = 3, when x = 0.

Answer:

We have,edydx=x+1dydx=log x+1dy=log x+1 dxIntegrating both sides, we get dy=log x+1 dxy=log x+11 dx-ddxlog x+11 dxdxy= x log x+1-1x+1×x dxy= x log x+1-1-1x+1 dxy= x log x+1-dx+1x+1dxy= x log x+1-x+log x+1+Cy= x+1 log  x+1-x+C       .....(1)It is given that at x=0 and y=3.Substituing the values of x and y in (1), we get C=3Therefore, substituting the value of C in (1), we gety=x+1 log  x+1-x+3Hence, y=x+1 log  x+1-x+3 is the required solution.

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Question 50:

Find the solution of the differential equation cos y dy + cos x sin y dx = 0 given that y = π2, when x = π2.

Answer:

We have,cos y dy+ cos x sin y dx=0cos y dy=- cos x sin y dxcot y dy=-cos x dxIntegrating both sides, we getcot y dy=-cos x dxlog sin y=- sin x+Clog sin y+ sin x=C          ....(1)It is given that at x=π2, y=π2.Substitutuing the values of x and y in 1, we getlog sin π2+ sin π2=CC=1Therefore, substituting the value of C in (1), we get log siny+ sin x=1Hence, log sin y+sin x=1 is the required solution.

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Question 51:

Find the particular solution of the differential equation dydx=-4xy2 given that y = 1, when x = 0.

Answer:

We have,dydx=-4xy21y2dy=-4x dxIntegrating both sides, we get1y2dy=-4x dx -1y=-4×x22+C-1y=-2x2+C       .....(1)It is given that at x=0, y=1.Substituting the values of x and y in (1), we getC=-1Therefore, substituting the value of C in (1), we get -1y=-2x2-1y=12x2+1Hence, y=12x2+1 is the required solution.

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Question 52:

Find the equation of a curve passing through the point (0, 0) and whose differential equation is dydx=ex sin x.

Answer:

We have to find the equation of the curve that passes through the point (0,0)and whose differential equation is dydx=ex sin x.dy=ex sin x dxIntegarting both sides, we get dy=ex sin x dxy=ex sin x dx          .....1y=ex sin x dx-ddxex sin x dx  dxy=-ex cos x+excos x dxy=-ex cos x+ex  cos x dx-ddxex cos x dxdxy=-ex cos x+ex sin x-exsin x dxy=-ex cos x+ex sin x-y+C             Using 12y=ex sin x-cos x+C          .....2The curve passes through the point (0,0)When, x=0; y=0Substituting the value of x and y in 2, we get0=1 0-1+CC=1Substituting the value of C in 2, we get2y=ex sin x-cos x+1Required equation of curve is 2y=ex sin x-cos x+1

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Question 53:

For the differential equation xydydx = (x + 2) (y + 2). Find the solution curve passing through the point (1, −1).

Answer:

We have,xydydx=x+2y+2yy+2dy=x+2xdxIntegrating both sides, we getyy+2dy=x+2xdxdy-21y+2dy=dx+21xdxy-2 log y+2=x+2 log x+C            .....(1)This equation represents the family of solution curves of the given differential equation.         We have to find a particular member of the family, which passes through the point 1,-1.Substituting x=1 and y=-1 in (1), we get-1-2 log 1=1+2 log 1+CC=-2Putting C=-2 in (1), we get  y-2 log y+2=x+2 log x-2  y-x+2=log x2 y+22 Hence, y-x+2=log x2 y+22 is the equation of the required curve.

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Question 54:

The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of the balloon after t seconds.

Answer:

Let r be the radius and V be the volume of the balloon at any time 't'.
Then, we have,
V=43πr3Given:dVdt=-k    where k>0ddt43πr3=-k4πr2 drdt=-k4πr2dr=-kdt Integrating both sides, we get4πr2dr=-kdt 43πr3=-kt+C               ...(1)It is given that at t=0, r=3. Substituting t=0 and r=3 in (1), we get C=36πPutting C=36π in (1), we get 43πr3=-kt+36π              ...(2)It is also given that at t=3, r=6. Putting t=3 and r=6 in (1), we get288 π=-3k+36πk=-84πPutting k=-84 π in (2), we get43πr3=84π t+36 πr3=63 t+27r=63 t+2713

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Question 55:

In a bank principal increases at the rate of r% per year. Find the value of r if â‚ą100 double itself in 10 years (loge 2 = 0.6931).

Answer:

Let P be the principal at any instant t.
Given:
dPdt=r100PdPP=r100dtIntegrating both sides, we getdPP=r100dtlog P=rt100+C      ......(1)Initially, i.e. at t=0, let P= P0. Putting P=P0, we get log P0=C, Putting C=log P0 in (1), we getlog P=rt100+log P0log PP0=rt100Substituting P0=100, P=2P0=200 and t=10 in (2), we get log 2 =r10r=10 log 2     =10× 0.6931     =6.931

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Question 56:

In a bank principal increases at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years (e0.5 = 1.648).

Answer:

Let at any instant t, the principal be P.Here, it is given that the principal increases at the rate of 5% per year.dPdt=5P100dPP=120dtIntegrating both sides, we get ln P=t20+ln C               ....(1) Initially at t=0, it is given that P=Rs 1000.ln 1000=ln CSubstituting the value of ln C in (1), we get ln P=t20+ln 1000Putting t=10, we getln P1000=0.5P1000=e0.5P=1000×1.648        =1648Therefore, Rs 1000 will be worth Rs 1648 after 10 years.

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Question 57:

In a culture the bacteria count is 100000. The number is increased by 10% in 2 hours. In how many hours will the count reach 200000, if the rate of growth of bacteria is proportional to the number present.

Answer:

Let at any time the bacteria count be N.Given: dNdtα NdNdt=λN1NdN=λdtIntegrating both sides, we get1NdN=λdtln N=λt+ln C             ...(1)Given:at t=0, N=100000therefore, ln C=ln 100000Putting the value in (1) we get,ln N=λt+ln 100000Also, at t=2N=110000Putting the values of t and N in (1), we getln 110000=2λ+ ln 10000012ln 1110=λSubstituting the values of ln C and λ in (1), we getln N=12ln 1110t+ ln 100000            ....(2)When N=200000, let t=T.Substituting these values in (2), we get ln 200000=T2ln  1110+ln 100000ln 2=T2ln 1110T=2ln 2ln 1110Therefore, in 2ln 2ln 1110 hours, the count will reach 200000.

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Question 58:

If y(x) is a solution of the different equation 2+sinx1+ydydx=-cosx and y(0) = 1, then find the value of y(π/2). [CBSE 2014, NCERT EXEMPLAR]

Answer:

2+sinx1+ydydx=-cosx11+ydy=-cosx2+sinxdx11+ydy=-cosx2+sinxdxlog1+y=-log2+sinx+logClog1+y2+sinx=logC1+y2+sinx=C         .....1
Now, y(0) = 1
1+12+0=CC=4
Substituting the value of C in (1), we get
(1 + y)(2 + sinx) = 4
1+y=42+sinxy=42+sinx-1yπ2=42+sinπ2-1=43-1=13

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Question 59:

Find the particular solution of the differential equation
(1 – y2) (1 + log x) dx + 2xy dy = 0, given that y = 0 when x = 1.

Answer:

Given:
1-y21+logxdx+2xydy=01-y21+logxdx=-2xydy1+logx2xdx=-y1-y2dy       .....1Let: 1+logx=t     and    1-y2=p1xdx=dt and -2ydy=dpTherefore, 1 becomes 

t2dt=12pdpt24=logp2+C        .....2Substituting the values of t and p in 2, we get1+logx24=log1-y22+C        .....3At x=1 and y=0, 3 becomesC=14Substituting the value of C in 3, we get1+logx24=log1-y22+141+logx2=2log1-y2+1Or logx2+logx2=log1-y22 It is the required particular solution.



Page No 21.66:

Question 1:

dydx=x+y+12

Answer:

We have,dydx=x+y+12Putting x+y+1=v1+dydx=dvdxdydx=dvdx-1dvdx-1=v2dvdx=v2+11v2+1dv=dxIntegrating both sides, we get1v2+1dv=dxtan-1 v=x+Ctan-1x+y+1=x+C   

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Question 2:

dydxcosx-y=1

Answer:

We have,dydxcosx-y=1dydx=1cosx-yPutting x-y=v1-dydx=dvdxdydx=1-dvdx1-dvdx=1cos vdvdx=1-1cos vdvdx=cos v-1cos vcos vcos v-1dv=dxIntegrating both sides, we getcos vcos v-1dv=dx-cos v1+cos v1-cos2 vdv=dx-cos v1+cos vsin2 vdv=dx-cot v cosec v+cot2vdv=dx-cot v cosec v+cosec2 v-1dv=dx--cosec v-cot v-v=x+Ccosec x-y+cot x-y+x-y=x+Ccosec x-y+cot x-y-y=C1+cos x-ysin x-y-y=Ccotx-y2=y+C

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Question 3:

dydx=x-y+32x-y+5

Answer:

We have,dydx=x-y+32x-y+5Putting x-y=v1-dydx=dvdxdydx=1-dvdx1-dvdx=v+32v+5dvdx=1-v+32v+5dvdx=2v+5-v-32v+5dvdx=v+22v+52v+5v+2dv=dxIntegrating both sides, we get2v+5v+2dv=dx2v+4+1v+2dv=dx2v+4v+2+1v+2dv=dx2dv+1v+2dv=dx2v+log v+2=x+C2x-y+logx-y+2=x+C

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Question 4:

dydx=x+y2

Answer:

We have,dydx=x+y2Let x+y=v1+dydx=dvdxdydx=dvdx-1dvdx-1=v2dvdx=v2+11v2+1dv=dxIntegrating both sides, we get1v2+1dv=dxtan-1 v=x+Cv=tanx+Cx+y=tanx+C

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Question 5:

x+y2dydx=1

Answer:

We have,x+y2dydx=1dydx=1x+y2Let x+y=v1+dydx=dvdxdydx=dvdx-1dvdx-1=1v2dvdx=1v2+1v2v2+1dv=dxIntegrating both sides, we getv2v2+1dv=dxv2+1-1v2+1dv=dx1-1v2+1dv=dxv-tan-1 v=x+Cx+y-tan-1 x+y=x+Cy-tan-1 x+y=C

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Question 6:

cos2x-2y=1-2dydx

Answer:

We have,cos2x-2y=1-2dydx2dydx=1-cos2x-2yLet x-2y=v1-2dydx=dvdx2dydx=1-dvdx1-dvdx=1-cos2 vdvdx=cos2 vsec2 v dv=dxIntegrating both sides, we getsec2 v dv=dxtan v=x-Ctanx-2y=x-Cx=tanx-2y+C

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Question 7:

dydx=secx+y

Answer:

We have,dydx=secx+ydydx=1cosx+yLet x+y=v1+dydx=dvdxdydx=dvdx-1dvdx-1=1cos vdvdx=cos v+1cos vcos vcos v+1dv=dxIntegrating both sides, we getcos vcos v+1dv=dxcos v1-cos v1-cos2 vdv=dxcos v1-cos vsin2 vdv=dxcos v-cos2 vsin2 vdv=dxcot v cosec v-cot2 vdv=dxcot v cosec v-cosec2 v+1dv=dx-cosec v+cot v+v=x+C-cosec x+y+cot x+y+x+y=x+C-cosec x+y+cot x+y+y=C-1+cos x+ysin x+y+y=C-tanx+y2+y=Cy=tanx+y2+C

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Question 8:

dydx=tanx+y

Answer:

We have,dydx=tanx+ydydx=sinx+ycosx+yLet x+y=v 1+dydx=dvdxdydx=dvdx-1dvdx-1=sin vcos vdvdx=sin vcos v+1dvdx=sin v+cos vcos vcos vsin v+cos vdv=dxIntegrating both sides, we getcos vsin v+cos vdv=dx12sin v+cos v+cos v-sin vsin v+cos vdv=dx12dv+12cos v-sin vsin v+cos vdv=dx12v+12cos v-sin vsin v+cos vdv=xPutting sin v+cos v=tcos v-sin vdv=dt12v+12dtt=x12v+12log t=x+C12x+y+12log sin x+y+cos x+y=x+C12y-x+12log sin x+y+cos x+y=Cy-x+log sin x+y+cos x+y=2Cy-x+log sin x+y+cos x+y=K      where, K=2C

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Question 9:

(x + y) (dxdy) = dx + dy

Answer:

We have,
(x + y) (dxdy) = dx + dy

x dx +y dx -x dy- y dy= dx+dyx+y-1dx=x+y+1dydydx=x+y-1x+y+1Let x+y=v 1+dydx=dvdxdydx=dvdx-1 dvdx-1=v-1v+1dvdx=v-1v+1+1dvdx=v-1+v+1v+1dvdx=2vv+1v+12vdv=dxIntegrating both sides, we getv+12vdv=dx12dv+121vdv=dx12v+12logv=x+C12x+y+12logx+y=x+C12y-x+12logx+y=C

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Question 10:

x+y+1dydx=1

Answer:

We have,
x+y+1dydx=1dydx=1x+y+1

Let x+y+1=v 1+dydx=dvdxdydx=dvdx-1dvdx-1=1vdvdx=1v+1vv+1dv=dxIntegrating both sides, we getvv+1dv=dxv+1-1v+1dv=dx1-1v+1dv=dxv-logv+1=x+Kx+y+1-logx+y+1+1=x+Ky-logx+y+2=K-1y-logx+y+2=C1     C1=K-1y-C1=logx+y+2ey-C1=x+y+2eyeC1=x+y+2e-C1ey=x+y+2Cey=x+y+2           C=e-C1x=Cey-y-2

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Question 11:

dydx+1=ex+y

Answer:

dydx+1=ex+y                .....(1)
Let x + y = t
1+dydx=dtdx
Substituting the value of x + y = t and 1+dydx=dtdx in (1), we get
dtdx=ete-tdt=dx-e-t=x+C-e-x+y=x+C     t=x+y



Page No 21.83:

Question 1:

x2 dy + y (x + y) dx = 0

Answer:

We have,x2dy+yx+y dx=0x2dy=-yx+y dxdydx=-yx+yx2This is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we get v+xdvdx=-vxx+vxx2v+xdvdx=-v1+vxdvdx=-v-v-v2xdvdx=-v2+2vdvv2+2v=-dxxdvvv+2=-dxxIntegrating both sides, we get  dvvv+2=-dxx121v-1v+2dv=-dxx121vdv-1v+2dv=-dxx12log v-log v+2=-log x +log C12log vv+2=log Cx log vv+2=2log Cxlog vv+2=log C2x2vv+2=C2x2yxyx+2=C2x2yy+2x=C2x2x2y=C2y+2xx2y=Ky+2x        Where, K=C2

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Question 2:

dydx=y-xy+x

Answer:

We have,dydx=y-xy+xThis is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we getv+xdvdx=vx-xvx+xv+xdvdx=xv-1xv+1xdvdx=v-1v+1-vxdvdx=v-1-v2-vv+1xdvdx=-v2+1v+1v+1v2+1dv=-1xdxIntegrating both sides, we getv+1v2+1dv=-1xdxvv2+1dv+1v2+1dv=-1xdx122vv2+1dv+1v2+1dv=-1xdx12log v2+1+tan-1 v=-log x+C12log v2+1+log x+tan-1 v=Clog v2+1+2 log x+2 tan-1 v=2Clog v2+1+log x2+2 tan-1 v=2Clog v2+1 x2+2 tan-1 v=2C Substituting v=yx, we getlog y2x2+1 x2+2 tan-1 yx=2C log y2+x2 +2 tan-1 yx=k              where k=2C

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Question 3:

dydx=y2-x22xy

Answer:

We have,dydx=y2-x22xyThis is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we getv+xdvdx=v2x2-x22vx2xdvdx=v2-12v-vxdvdx=-v2+12v2vv2+1dv=-1xdxIntegrating both sides, we get2vv2+1dv=-1xdxlog v2+1=-log x+log Clog v2+1=log Cxv2+1=CxPutting v=yx, we getyx2+1=Cx   y2+x2=Cx Hence, x2+y2=Cx  is the required solution.

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Question 4:

xdydx=x+y

Answer:

We have,xdydx=x+ydydx=x+yxThis is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we getv+xdvdx=x+vxxv+xdvdx=1+vxdvdx=1+v-vxdvdx=1dv=1xdxIntegrating both sides, we get dv=1xdxv=log x+CPutting v=yx, we getyx=log x+C  y=xlog x+Cx Hence, y=xlog x+Cx is the required solution.

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Question 5:

(x2y2) dx − 2xy dy = 0

Answer:

We have,x2-y2 dx-2xy dy=0dydx=x2-y22xyThis is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we getv+xdvdx=x2-vx22xvxv+xdvdx=x2-v2x22vx2v+xdvdx=1-v22vxdvdx=1-v22v-vxdvdx=1-3v22v2v1-3v2dv=1xdxIntegrating both sides, we get 2v1-3v2dv=1xdx-13-6v1-3v2dv=1xdx-13log 1-3v2=log x+log Clog 1-3v2=-3log Cxlog 1-3v2=log 1Cx31-3v2=1Cx3Putting v=yx, we get1-3yx2=1Cx3x2-3y2x2=1C3x3xx2-3y2=1C3xx2-3y2=K        where, K=1C3

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Question 6:

dydx=x+yx-y

Answer:

We have,dydx=x+y x-yThis is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we getv+xdvdx=x+vx x-vxv+xdvdx=1+v 1-vxdvdx=1+v1-v-vxdvdx=1+v21-v1-v1+v2dv=1xdxIntegrating both sides, we get1-v1+v2dv=1xdx11+v2dv-v1+v2dv=1xdx11+v2dv-122v1+v2dv=1xdxtan-1v-12log 1+v2=log x+CPutting v=yx, we gettan-1yx-12log 1+y2x2=log x+Ctan-1yx=12log 1+y2x2+log x+C tan-1yx=12log x2+y2x2+log x+C tan-1yx=12log x2+y2-12log x2+log x+C tan-1yx=12log x2+y2-log x+log x+C tan-1yx=12log x2+y2+C Hence, tan-1yx=12log x2+y2+C  is the required solution.

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Question 7:

2xydydx=x2+y2

Answer:

We have,2xydydx=x2+y2dydx=x2+y22xyThis is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we get v+xdvdx=x2+v2x22x2vv+xdvdx=1+v22vxdvdx=1+v22v-vxdvdx=1-v22v2v1-v2dv=1xdxIntegrating both sides, we get2v1-v2dv=1xdx- log 1-v2=log x+log C-log 1-v2x=log CPutting v=yx, we get-log x2-y2x2x=log Cxx2-y2=Cx=Cx2-y2 Hence, x=Cx2-y2 is the required solution.

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Question 8:

x2dydx=x2-2y2+xy

Answer:

We have,x2dydx=x2-2y2+xydydx=x2-2y2+xyx2This is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we get v+xdvdx=x2-2v2x2+x2vx2v+xdvdx=1-2v2+vxdvdx=1-2v211-2v2dv=1xdxIntegrating both sides, we get 11-2v2dv=1xdx112-2v2=1xdx122log 1+2v1-2v=log x+log Clog 1+2v1-2v=22log x+22 log Clog 1+2v1-2v=log Cx221+2v1-2v=Cx22Putting v=yx, we getx+2yx-2y=Cx22Hence, x+2yx-2y=Cx22 is the required solution.

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Question 9:

xydydx=x2-y2

Answer:

We have,xydydx=x2-y2dydx=x2-y2xyThis is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we get v+xdvdx=x2-v2x2vx2v+xdvdx=1-v2vxdvdx=1-v2v-vxdvdx=1-2v2vv1-2v2dv=1xdxIntegrating both sides, we getv1-2v2dv=1xdx-14log 1-2v2=log x+log Clog 1-2v2=-4log  x-4 log Clog 1-2v2x4=log 1C4Putting v=yx, we getlog x2x2-2y2=log 1C4x2x2-2y2=C1whereC1=1C4Hence, x2x2-2y2=C1 is the required solution.

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Question 10:

y ex/y dx = (xex/y + y) dy

Answer:

We have,y exydx=x exy+ydydxdy=x exy+yy exydxdy=xy exy+1exydxdy=xy+e-xyThis is a homogeneous differential equation.Putting x=vy and dxdy=v+ydvdy, we getv+ydvdy=v+e-vydvdy=e-vevdv=1ydyIntegrating both sides, we getevdv=1ydyev=log y+CPutting v=yx, we getexy=log y+CHence, exy=log y+C is the required solution.

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Question 11:

x2dydx=x2+xy+y2

Answer:

x2dydx=x2+xy +y2dydx=x2+xy +y2x2This is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we get v+xdvdx=x2+x2v +v2x2x2v+xdvdx=1+v+v2xdvdx=1+v211+v2dv=1xdxIntegrating both sides, we get 11+v2dv=1xdxtan-1 v=log x+CPutting v=yx, we gettan-1 yx=log x+CHence, tan-1 yx=log x+C is the required solution.

Page No 21.83:

Question 12:

(y2 − 2xy) dx = (x2 − 2xy) dy

Answer:

We have,y2-2xy dx=x2-2xy dydydx=y2-2xyx2-2xyThis is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we get v+xdvdx=v2x2-2vx2x2-2vx2v+xdvdx=v2-2v1-2vxdvdx=3v2-3v1-2v1-2v3v2-3vdv=1xdxIntegrating both sides, we get1-2v3v2-3vdv=1xdx-2v-13v2-3vdv=1xdx-136v-33v2-3vdv=1xdxPutting 3v2-3v=t6v-3 dv=dt-131tdt=1xdx-13log t=log x+log CSubstituting the value of t, we get-13log 3v2-3v=log x+log C-13log v2-v-13log3=log x+log C-13log v2-v=log x+log C-13log3-13log v2-v=log x+log C1          where, log C1=log C-13log3Substituting the value of v, we get-13log yx2-yx=log x+log C1-13log y2x2-yx=log C1xlog y2-xyx2=-3log C1xlog y2-xyx2=log 1C13x3y2-xyx2=1C13x3xy2-x2y=1C13x2y-xy2=-1C13x2y-xy2=K       where, log K=-1C13

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Question 13:

2xy dx + (x2 + 2y2) dy = 0

Answer:

2xy dx+x2+2y2 dy=0dydx=-2xyx2+2y2This is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we getv+xdvdx=-2vx2x2+2v2x2v+xdvdx=-2v1+2v2xdvdx=-2v1+2v2-vxdvdx=-3v-2v31+2v21+2v23v+2v3dv=-1xdxIntegrating both sides, we get1+2v23v+2v3dv=-1xdxSubstituting 3v+2v3=t, we get31+2v2 dv=dt13dttdv=-1xdx13log t=-log x+log C13log 3v+2v3=-log x+log Clog 3v+2v3=-3 log x+3 log Clog 3v+2v3×x3 =log C33v+2v3×x3=C3Putting v=yx, we get3×yx+2×y3x3×x3=C33yx2+2y3=C1Hence, 3yx2+2y3=C1 is the required solution.

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Question 14:

3x2 dy = (3xy + y2) dx

Answer:

We have,3x2 dy=3xy+y2 dxdydx=3xy+y23x2This is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we get v+xdvdx=3vx2+v2x23x2v+xdvdx=3v+v23xdvdx=v233v2dv=1xdxIntegrating both sides, we get31v2dv=1xdx-3×1v =log x +C-3v=log x +CPutting v=yx, we get-3xy=log x  +CHence, -3xy=log x +C is the required solution.

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Question 15:

dydx=x2y+x

Answer:

We have,dydx=x2y+xThis is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we get v+xdvdx=x2vx+xv+xdvdx=12v+1xdvdx=12v+1-vxdvdx=1-2v2-v2v+12v+11-2v2-vdv=1xdxIntegrating both sides, we get 2v+11-2v2-vdv=1xdx2v+12v2+v-1dv=-1xdx2v+12vv+1-1v+1dv=-1xdx2v+12v-1v+1dv=-1xdx             .....(1)Solving left hand side integral of (1),  we getUsing partial fraction,Let 2v+12v-1v+1=A2v-1+Bv+1 A+2B=2           .....(2) And A-B=1               .....(3) Solving (2) and (3), we get A=43 and B=132v+12v-1v+1dv=4312v-1dv+131v+1dv                                       =43×2log 2v-1+13log v+1 +log C From (1), we get 23log 2v-1+13v+1 +log C =-log x+log C1log 2v-12v+1=-3log x+log C2      log 2v-12v+1=log C23x32v-12v+1=C23x3Putting v=yx, we get2y-xx2y+xx=C23x3x+y2y-x2=k

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Question 16:

(x + 2y) dx − (2xy) dy = 0

Answer:

x+2ydx-2x-y dy=0dydx=x+2y2x-yThis is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we get v+xdvdx=x+2vx2x-vxxdvdx=1+2v2-v-vxdvdx=1+v22-v2-v1+v2dv=1xdxIntegrating both sides, we get2-v1+v2dv=1xdx           .....(1)21+v2dv-v1+v2dv=1xdx21+v2dv-122v1+v2dv=1xdx2 tan-1v-12log 1+v2=log x+log C2 tan-1v=log x+log C+log 1+v2122 tan-1v=log Cx1+v2Cx1+v2=e2 tan-1vPutting v=yx, we getCx1+yx2=e2 tan-1yxCx2+y2=e2 tan-1yxHence, x2+y2=Ke-2 tan-1yx is the required solution.

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Question 17:

dydx=yx-y2x2-1

Answer:

We have,dydx=yx-y2x2-1This is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we getv+xdvdx=v-v2-1xdvdx=-v2-11v2-1dv=-1xdxIntegrating both sides, we get1v2-1dv=-1xdx log v+v2-1=-log x +log Clog v+v2-1x=log C v+v2-1x= CPutting v=yx, we getyx+y2x2-1x=CHence, y+y2-x2 =C is the required solution.

Page No 21.83:

Question 18:

Solve the following differential equations:
dydx=yxlog y-log x+1

Answer:

We have,dydx=yxlog yx+1This is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we get v+xdvdx=vlog v+1xdvdx=vlog v+v-v1vlog vdv=1xdxIntegrating both sides, we get 1vlog vdv=1xdxPutting log v=t, we getdv= vdt 1v ×t×v dt=1xdxdtt=1xdxlog t=log x+log Ct=Cx          .....(1) Substituting the value of t in (1), we get log v=CxPutting v=yx, we getlog yx=CxHence, log yx=Cx is the required solution.

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Question 19:

dydx=yx+sinyx

Answer:

dydx=yx+sin yxThis is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we getv+xdvdx=v+ sin vxdvdx=v+ sin v-v1sin vdv=1xdxIntegrating both sides, we get1sin vdv=1xdxcosec v dv=1xdxlog tan v2=log x +log Clog tan v2=log x+log Clog tan v2=log Cxtan v2=CxPutting v=yx, we gettan y2x=CxHence, tan y2x=Cx is the required solution. 

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Question 20:

y2 dx + (x2xy + y2) dy = 0

Answer:

We have,y2 dx+x2-xy+y2 dy=0dydx=-y2x2-xy+y2This is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we get v+xdvdx=-v2x2x2-vx2+v2x2v+xdvdx=-v21-v+v2xdvdx=-v21-v+v2-vxdvdx=-v-v31-v+v21-v+v2v+v3dv=-1xdx1+v2-vv1+v2dv=-1xdxIntegrating both sides, we get 1+v2-vv1+v2dv=1xdx1+v2v1+v2dv-vv1+v2dv=-1xdx1vdv-11+v2dv=-1xdxlog v-tan -1v=-log x+log Clog vxC=tan-1vvxC= etan-1vPutting v=yx, we gety=Cetan-1vHence, y=Cetan-1v is the required solution.

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Question 21:

xx2+y2-y2 dx+xy dy=0

Answer:

We have,xx2+y2-y2dx+xy dy=0dydx=y2-xx2+y2xyThis is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we getv+xdvdx=v2x2-xx2+v2x2vx2v+xdvdx=v2-1+v2vv+xdvdx=v-1+v2vxdvdx=-1+v2v v1+v2dv=-1xdxPutting 1+v2=t, we getv dv=dt2 12tdt=-1xdxIntegrating both sides, we get  12tdt=-1xdxt=-log x+log C        .....(1)Substituting the value of t in (1), we get1+v2=log CxPutting v=yx, we gety2+x2=x log CxHence, y2+x2=x log Cx is the required solution.

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Question 22:

xdydx=y-x cos2yx

Answer:

xdydx=y-x cos2 yxdydx=y-x cos2 yxxdydx=yx- cos2 yxThis is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we get v+xdvdx=v-cos2vxdvdx=-cos2v 1cos2 vdv=-1xdxsec2 v=-1xdxIntegrating both sides, we get  sec2 v dv=-1xdxtan v=-log x+log C      tan v=log CxPutting v=yx, we gettan yx= log Cxtan yx= log Cx is the required solution.

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Question 23:

yxcosyx dx-xysinyx+cosyx dy=0

Answer:

yxcos yxdx-xysin yx+cos yxdy=0xysin yx+cos yxdy=yxcos yxdxdydx=yxcos yxxysin yx+cos yxThis is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we get v+xdvdx=v cos v1vsin v+cos vxdvdx=v cos v1vsin v+cos v-vxdvdx=-sin v1vsin v+cos v1vsin v+cos vsin vdv=-1xdx1v+cot vdv=-1xdxIntegrating both sides, we get1v+cot vdv=-1xdx1vdv+ cot v dv=-1xdxlog v+log sin v=-log x+log Clog vxsin v=log Cv x sin v=CPutting v=yx, we getysin yx=CHence, ysin yx=C is the required solution.

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Question 24:

xy logxy dx+y2-x2 logxy dy=0

Answer:

Given:
xy logxy dx+y2-x2 logxydy=0xy logxy dx =-y2-x2 logxydydxdy=-y2-x2 logxyxy logxy = x2 logxy-y2xy logxyIt is a homogeneous equation.We put x=vydxdy=v+ydvdySo, v+ydvdy=v2y2 log(v)-y2vy2 log(v)
v+ydvdy=v2 log(v)-1v log(v)ydvdy=v2 log(v)-1v log(v)-vydvdy=v2 log(v)-1-v2 log(v)v log(v)ydvdy=-1v log(v)v log(v) dv=-1ydyOn integrating both sides we get,
v log(v) dv=-1ydyv22 log(v)-v2 dv=-log y + Cv22 log(v) - v24=-log y + Cv22log(v) -12=-log y+Cv2 log(v) -12=-2 log y + Cnow putting back the values of v as xy we get,x2y2log(v) -12 + log y2 = C
 

Page No 21.83:

Question 25:

1+ex/y dx+ex/y1-xy dy=0

Answer:

We have,1+exy dx+exy 1-xy dy=0dxdy=-exy 1-xy1+exyThis is a homogeneous differential equation.Putting x=vy and dxdy=v+ydvdy, we get v+ydvdy=-ev1-v1+evydvdy=-ev1-v1+ev-vydvdy=-ev+evv-v-vev1+evydvdy=-v+ev1+ev1+evv+evdv=-1ydyIntegrating both sides, we get 1+evv+evdv=-1ydylog v+ev=-log y+log Cv+ev=Cyv+ev=CyPutting v=xy, we getxy+exy=Cyx+yexy=CHence, x+yexy=C is the required solution.

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Question 26:

x2+y2dydx=8x2-3xy+2y2

Answer:

We have,x2+y2dydx=8 x2-3xy+2y2dydx=8 x2-3xy+2y2x2+y2This is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we getv+xdvdx=8 x2-3vx2+2v2x2x2+v2x2xdvdx=8-3v+2v21+v2-vxdvdx=8-4v+2v2-v31+v2xdvdx=42-v+v22-v1+v2xdvdx=4+v22-v1+v21+v24+v22-vdv=1xdxIntegrating both sides, we get 1+v24+v22-vdv=1xdx         .....(1)Let us consider the left hand side of (1).Using partial fraction,Let 1+v24+v22-v=Av+B4+v2+C2-v1+v2=Av2-v+B2-v+C 4+v21+v2=2Av-Av2+2B-Bv+4C+Cv2Comparing the coefficients of both sides, we get 2A-B=0 -A+C=1 & 2B+4C=1Solving these three equations, we getA=-38, B=-34and C=58   1+v24+v22-v=-38v-344+v2+582-v           .....(2)From (1) and (2), we get-38v-344+v2+582-v =1xdx   -38vv2+4dv-341v2+4dv+5812-vdv=1xdx-316log v2+4-34×2tan -1v2-58log 2-v=log x+log C-34×2tan -1v2=log Cx2-v58v2+4316e-38tan -1v2=Cx2-v58v2+4316Putting v=yx, we gete-38tan -1y2x=Cx2-yx58yx22+4316e-38tan -1y2x=Cx×1x2x-y58y2+4x2316e-38tan -1y2x=C2x-y58y2+4x2316Hence, e-38tan -1y2x=C2x-y58y2+4x2316 is the required solution.

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Question 27:

(x2 − 2xy) dy + (x2 − 3xy + 2y2) dx = 0

Answer:

We have,x2-2xy dy+x2-3xy+2y2 dx=0dydx=x2-3xy+2y22xy-x2This is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we get v+xdvdx=x2-3vx2+2v2x22vx2-x2v+xdvdx=1-3v+2v22v-1xdvdx=1-3v+2v22v-1-vxdvdx=1-2v2v-1xdvdx=-1dv=-1xdxIntegrating both sides, we getdv=-1xdxv=-log x+CPutting v=yx, we getyx=-log x+Cyx+log x=CHence, yx+log x=C is the required solution.

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Question 28:

xdydx=y-x cos2yx

Answer:

xdydx=y-x cos2 yxdydx=y-x cos2 yxxThis is a homogeneous differential equation. Putting y=vx and dydx=v+xdvdx, we getv+xdvdx=vx-x cos2 vxv+xdvdx=v-cos2vxdvdx=-cos2v  sec2v dv=-1xdxIntegrating both sides, we get sec2v dv=-1xdx       tan v =- log  x+log Ctan v =log CxPutting v=yx, we get tan yx =log CxHence, tan yx =log Cx is the required solution.   

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Question 29:

xdydx-y=2y2-x2

Answer:

We have,xdydx-y=2y2-x2dydx=2y2-x2+yxThis is a homogeneous differential equation. Putting y=vx and dydx=v+xdvdx, we getv+xdvdx=2v2x2-x2+vxxv+xdvdx=2v2-1+vxdvdx=2v2-1+v-vxdvdx=2v2-112v2-1dv=1xdxIntegrating both sides, we get12v2-1dv=1xdx1v2-1dv=21xdxlog v+v2-1=2 log x +log Cv+v2-1=Cx2Putting v=yx, we get yx+y2x2-1 =Cx2 y+y2-x2=Cx3Hence, y+y2-x2=Cx3 is the required solution.   

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Question 30:

x cosyx·y dx+x dy=y sinyx·x dy-y dx

Answer:

We have,x cosyxy dx+x dy=y sin yxx dy-y dxxy cos yx dx+x2cos yx dy=xy sin yx dy-y2sin yx dxxy cos yx+y2sin yx dx=xy sin yx-x2cos yx dydydx=xy cos yx+y2sin yxxy sin yx-x2cos yxThis is a homogeneous differential equation. Putting y=vx and dydx=v+xdvdx, we getv+xdvdx=vx2 cos v+v2x2sin vvx2 sin v-x2cos vv+xdvdx=v cos v+v2sin vv sin v-cos vxdvdx=v cos v+v2sin vv sin v-cos v-vxdvdx=v cos v+v2sin v-v2 sin v+ v cos vv sin v-cos vxdvdx=2v cos vv sin v-cos vvsin v-cos v2 v cos vdv=1xdxIntegrating both sides, we getvsin v-cos v2 v cos vdv=1xdxvsin v-cos vv cos vdv=21xdxv sin vv cos vdv-cos vv cos vdv=21xdxtan v dv-1vdv=21xdxlog sec v-log v=2 log x+log Clog sec vv=log Cx2sec vv=Cx2Putting v=yx, we getsec yx=yx×C×x2 sec yx=CxyHence, sec yx=Cxy is the required solution.   

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Question 31:

(x2 + 3xy + y2) dxx2 dy = 0

Answer:

We have, x2+3xy+y2 dx-x2 dy=0dydx=x2+3xy+y2x2This is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we getv+xdvdx=x2+3vx2+v2x2x2xdvdx=1+3v+v2-vxdvdx=1+v2+2v11+v2+2vdv=1xdxIntegrating both sides, we get11+v2+2vdv=1xdx11+v2dv=1xdx-11+v=log x+Clog x+11+v=-CPutting v=yx, we getlog x+xx+y=C1 whereC1=-CHence, log x+xx+y=C1 is the required solution.   

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Question 32:

x-ydydx=x+2y

Answer:

 x-y dydx=x+2ydydx=x+2yx-yThis is a homogeneous differential equatiuon. Putting y=vx and dydx=v+xdvdx, we getv+xdvdx=x+2vxx-vxv+xdvdx=1+2v1-vxdvdx=1+2v1-v-vxdvdx=1+2v-v+v21-vxdvdx=1+v+v21-v1-v1+v+v2dv=1xdxIntegrating both sides, we get 1-v1+v+v2dv=1xdx   
-(v-1)v2+v+1dv=dxx12×2v-2v2+v+1dv=-dxx(2v+1)-3v2+v+1dv=-2dxx(2v+1)v2+v+1dv-3v2+v+1dv=-2dxxLet I1 =(2v+1)v2+v+1dvand   I2 =3v2+v+1dvI=I1 + I2
I1= 2v+1v2+v+1dvlet v2+v+1=t   (2v2+1)dv=dttherefore, I1= 2v+1v2+v+1dv = dtt=logt=logv2+v+1hence,  I1 = logv2+v+1Also, I2= 3v2+v+1dv = 3v2+2v12+122-122+1=3v+122+322dv=323tan-1v+1232=23 tan-1v+1232I2=23 tan-1v+1232Hence, I=I1+I2=logv2+v+1 +23 tan-1v+1232therefore, logv2+v+1 +23 tan-1v+1232 =-2logx+Cputting the value of v in the above equation we get,logx2+y2+xy = 23 tan-1x+2yx3 +C

 



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Question 33:

(2x2 y + y3) dx + (xy2 − 3x3) dy = 0

Answer:

We have, 2x2y+y3 dx+xy2-3x3 dy=0dydx=2x2y+y33x3-xy2This is a homogeneous differential equation. Putting y=vx and dydx=v+xdvdx, we get v+xdvdx=2vx3+v3x33x3-v2x3v+xdvdx=2v+v33-v2xdvdx=2v+v33-v2-vxdvdx=2v+v3-3v+v33-v2xdvdx=2v3-v3-v23-v22v3-vdv=1xdxIntegrating both sides, we get 3-v22v3-vdv=1xdx312v3-vdv-v22v3-vdv=1xdx       .....(1)Considering 12v3-v=1v2v2-1,let 1v2v2-1=Av+Bv+C2v2-1      .....(2)1=A2v2-1+Bv+C v1=2Av2-A+Bv2+CvComparing the coeficients of both sides, we get 2A+B=0 , C=0 and A=-1-2+B=0B=2Substituting A=-1, B=2 and C=0 in (2), we get1v2v2-1=-1v+2v2v2-1         .....(3)From (2) and (3), we get3-1v+2v2v2-1dv-v2v2-1dv=1xdx-31vdv+5v2v2-1dv=1xdx-3 log v+54log 2v2-1=log x+log C12 log 1v+5 log 2v2-1 4=log Cxlog 1v12×2v2-15=log C4x41v12×2v2-15=C4x4Putting v=yx, we getx12y12×2y2x2-15=C4x42y2-x2x25=C4x4×y12x12Hence, C4x2y12=2y2-x25 is the required solution. 

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Question 34:

xdydx-y+x sinyx=0

Answer:

We have,xdydx-y+x sin yx=0dydx=y-x sin yxxThis is a homogenoeus differential equation.Putting y=vx and dydx=v+xdvdx, we getv+xdvdx=vx-x sin vxxdvdx=v-sin v-vxdvdx=-sin v cosec v dv=-1xdxIntegrating both sides, we getcosec v dv=-1xdx-cosec v dv=1xdx-log cosec v-cot v=log x+log Clog 1cosec v-cot v=log Cxlog cosec v+cot v=log Cxlog 1+cos vsin v=log Cx1+cos vsin v=Cxx sin v=1C1+cos vx sin v=K1+cos v         where, K=1CPutting v=yx, we get x sinyx=K1+cosyxHence, x sinyx=K1+cosyx is the required solution.

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Question 35:

y dx+x logyx dy-2x dy=0

Answer:

We have,y dx+x log yx dy-2x dy=02x-x log yx dy=y dxdydx=y2x-x log yxThis is a homogenoeus differential equation.Putting y=vx and dydx=v+xdvdx, we get v+xdvdx=vx2x-x log vv+xdvdx=v2-log vxdvdx=v2-log v-vxdvdx=v-2v+v log v 2- log vxdvdx=v log v-v2-log v2-log vv log v-vdv=1xdxIntegrating both sides, we get2-log vv log v-vdv=1xdx1-log v-1v log v-1dv=1xdxPutting log v-1=t1vdv=dt1-ttdt=1xdx1t-1dt=1xdxlog t-t=log x+log Clog log v-1-log v-1=log x+log Clog log v-1-log v=log x+log C1      where,  log C1=log C-1log log v-1v=log C1xlog v-1v=C1xlog v-1=C1xvPutting v=yx, we getlog yx-1=C1x×yxlog yx-1=C1yHence, log yx-1=C1y is the required solution.

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Question 36:

Solve each of the following initial value problems:
(i) (x2 + y2) dx = 2xy dy, y (1) = 0

(ii) xey/x-y+xdydx=0, ye=0

(iii) dydx-yx+cosecyx=0, y1=0

(iv) (xyy2) dxx2 dy = 0, y(1) = 1

(v) dydx=yx+2yx2x+y, y1=2

(vi) (y4 − 2x3 y) dx + (x4 − 2xy3) dy = 0, y (1) = 1

(vii) x (x2 + 3y2) dx + y (y2 + 3x2) dy = 0, y (1) = 1

(viii) x sin2yx-ydx+x dy=0, y1=π4

(ix) xdydx-y+x sinyx=0, y2=x

Answer:

(i) (x2 + y2)dx = 2xy dy, y(1) = 0
We have,
(x2 + y2) dx = 2xy        .....(i)

This is a homogenous equation, so let us take y = vx

Then, dydx=v+xdvdxPutting y=vx in equation (i)x2+v2x2=2vx2v+xdvdxx21+v2=2vx2v+xdvdx1+v2=2v2+2vxdvdx1-v2=2vxdvdxdxx=2v dv1-v2
On integrating both sides, we get
1xdx=2v1-v2dvLet,  1-v2=t-2v dv=dtloge x=-dtt                       loge x=-loge t+cloge x=-loge 1-y2x2+cloge xx2-y2x2=cloge x2-y2x=cAs y1=0c=0loge x2-y2x=0x2-y2x=1x2-y2=x

(ii) xeyx-y+xdydx=0 ye=0
This is also a homogenous equation,

Put y = vx

dydx=v+xdvdxx ev-vx+xv+xdvdx=0x ev-vx+xv+x2dvdx=0x ev+x2dvdx=0ev=-xdvdxdxx=-1evdv
On integration both sides we get,
dxx=-1evdvloge x=-e-v dvlogex=e-yx+c                 y=vxAs given ye=0logee=e-0e+c1=1+cc=0logex=e-yx

(iii) dydx-yx+cosecyx=0, y1=0
This is an homogenous equation, put y = vx
dydx+v+xdvdxv+xdvdx-v+cosec v=0xdvdx=cosec vdvcosec v=dxxsin v dv=dxx
On integrating both sides, we get
 sin v dv=dxx-cos v=logex+c-cos v+logex=ccos v+logex=-ccos yx+logex=-cAs y1=0cos 01=0+loge1=-c1+0=-cc=-1cos yx+logex=1

(iv) (xyy2) dxx2 dy = 0, y(1) = 1
This is an homogenous equation, put y = vx

dydx=v+xdvdxxy-y2=x2dydxvx2-v2x2=x2v+xdvdxvx21-v=x2v+xdvdxv1-v=v+xdvdxv-v2=v+xdvdx-v2=xdvdx-1xdx=1v2dv
On integrating both sides we get,
-1xdx=1v2dv-logex=v-2+1-2+1+c-logex=v-1-1+c-logex=-1v+c-logex=-1v+cxy-logex=cAs y1=111-loge1=cc=1

(v) dydx=yx+2yx2x+y, y1=1
This is an homogenous equation, put y = vx
dydx=v+xdvdx
v+xdvdx=vx+2vx2x+vxv+xdvdx=v1+2v2+vxdvdx=v1+2v-v2+v2+vxdvdx=v+2v2-2v-v22+v xdvdx=v2-v2+v2+vdvv2-v=dxx
On integrating both side of the equation we get,
2+vv2-vdv=dxx2vv-1dv+vvv-1dv=dxx211-vdv-1vdv+1v-1dv=logex+c2logev-1-logev+logev-1=logex+c2loge v-1v+logev-1=logex+c2logey-xy+logey-xx=logex+cAs y1=22 loge2-12+loge2-11=loge1+c2 loge12+loge1=loge1+c-2 loge2+0=0+c-2 loge2=c2 logey-xy+logey-xx=logex-2 loge2

(vi) (y4 − 2x3 y) dx + (x4 − 2xy3) dy = 0, y (1) = 1
This is an homogenous equation, put y= vx
v4x4-2vx4+x4-2v3x4 v+xdvdx=0v4x4-2vx4=2v3x4-x4 v+xdvdxvx4v3-2=x42v3-1 v+xdvdxvv3-2=2v3-1v+x2v3-1dvdxvv3-2-2v3+1=x2v3-1dvdxv-1-v3=x2v3-1dvdxv1+v3=x1-2v3dvdxdxx=1-2v3v1+v3dv
On integrating both side of the equation we get,
dxx=1-2v3v1+v3dvlogex=1+v3-3v3v1+v3dvlogex=1+v3v1+v3dv-3vv1+v3dvlogex=1vdv-3v21+v3dvlogex=logev-dttlogex=logev-loge1+v3+c                      let 1+v3=t, 3v2dv=dtlogex=logev1+v3+cAs v=yxlogex=logeyx1+y3x+clogex=logeyx2x3+y3+cAs y1=1loge1=loge11+1+c0=loge12+cc=-loge12c=loge2logex=logeyx2x3+y3+loge2

(vii)  x(x2+3y2)dx +y(y2+3x2)dy = 0, y(1) = 1 dydx=-x(x2+3y2)y(y2+3x2)it is a homogeneous equation. Put y=vxand dydx=v+xdvdx

So, v+xdvdx =-x(x2+3v2x2)vx(v2x2+3x2)xdvdx=-(1+3v2)v(v2+3)-3= -1-3v2-v4-3v2v(v2+3)xdvdx=-v4-6v2-1v(v2+3)v(v2+3)v4+6v2+1dv=-dxx4v3+12vv4+6v2+1dv=-4dxxlogv4+6v2+1=logcx4v4+6v2+1=cx4y4+6y2x2+x4=c                          ....(1)

put y=1, x=1(1+6+1)=c c=8put c=8 in equation (1),(y4+x4+6x2y2)=8

(viii) {x sin2yx-y}dx + x dy = 0, y(1) = π4it is a homogeneous equation. so, we put y= vxdydx=v+xdvdxso, v+xdvdx=-sin2vxx+vxxxdvdx=-sin2vdvsin2v=-dxxintegrating both sides, we getcotyx=logcx
putting the values of x=1 and y=π4cotπ4=log c1=log cc=eHence,  cotyx=log(ex)

(ix) xdydx-y+x sinyx=0, y(2)=πit is a homogeneous equation. put y=vxand dydx=v+xdvdxso, v+xdvdx=vxx-sinvxxxdvdx=-sinvdvsinv=-dxxcosec(v)dv=-dxxintegraing both sides we get,log(cosec(v)-cot(v))=-log x + log c
logcosecyx-cotyx=-log x + log cputting the values x=2 and y=π logcosecπ2-cotπ2=-log 2 + log cc=0logcosecyx-cotyx=-log x 
 

Page No 21.84:

Question 37:

Find the particular solution of the differential equation x cosyxdydx=y cosyx+x, given that when x = 1, y=π4.

Answer:

 x cos yxdydx=y cos yx+xdydx=y cos yx+xx cos yxThis is a homogeneous differential equation. Putting y=vx and dydx=v+xdvdx, we getv+xdvdx=vx cos v+xx cos vv+xdvdx=v cos v+1cos vxdvdx=v cos v+1-v cos vcos vxdvdx=1cos vcos v dv=1xdxIntegrating both sides, we getcos v dv=1xdxsin v=log x+CPutting v=yx, we getsinyx=log x+C        .....1At x=1, y=π4                     GivenPutting x=1 and y=π4 in (1), we getC=12Putting C=12 in (1), we getsin yx=log x+12Hence, sin yx=log x+12 is the required solution.   

Page No 21.84:

Question 38:

Find the particular solution of the differential equation x-ydydx=x+2y, given that when x = 1, y = 0.

Answer:

x-ydydx=x+2ydydx=x+2yx-yThis is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we getv+xdvdx=x+2vxx-vxxdvdx=1+2v1-v-vxdvdx=1+2v-v+v21-vxdvdx=1+v+v21-v1-v1+v+v2 dv=1xdxIntegrating both sides, we get 1-v1+v+v2 dv=1xdx11+v+v2dv-122v+1-11+v+v2=1xdx11+v+v2dv-122v+11+v+v2dv+1211+v+v2dv=1xdx      3211+v+v2dv-122v+11+v+v2dv=1xdx    3211+v+v2+14-14dv-122v+11+v+v2dv=1xdx  321v+122+322dv-122v+11+v+v2dv=1xdx3tan -1 v+1232-12log 1+v+v2=log x+CPutting v=yx, we get3tan-12y+x3x -12log x2+xy+y2x2=log x+C3tan-12y+x3x -12log x2+xy+y2+log x=log x+C 3tan-12y+x3x -12log x2+xy+y2=C      .....(1) At x=1, y=0             GivenPutting x=1 and y=0 in (1), we get3 tan-113 -12log 1=CC=3 tan-113C=3 ×π6C=π23Substituting the value of C in (1), we get 3tan-12y+x3x -12log x2+xy+y2=π2323tan-12y+x3x -log x2+xy+y2=π3log x2+xy+y2=23tan-12y+x3x-π3Hence, log x2+xy+y2=23tan-12y+x3x-π3 is the required solution.   

Page No 21.84:

Question 39:

Find the particular solution of the differential equation dydx=xyx2+y2given that y = 1 when x = 0. [CBSE2015]

Answer:

dydx=xyx2+y2          .....(1)
Let y = xv
dydx=v+xdvdx

Substituting the value of y = xv and dydx=v+xdvdxin (1), we get
v+xdvdx=x2vx2+x2v2xdvdx=v1+v2-vxdvdx=-v31+v21+v2-v3dv=1xdx1+v2-v3dv=1xdx12v2-logv=logx+C
12yx2-logyx=logx+Cx22y2-logyx=logx+C       .....202-log10=log0+CC=0
Substituting the value of C in (2), we get
x22y2-logyx=logxx22y2=logx+logyxx22y2=logy

Page No 21.84:

Question 40:

Show that the family of curves for which dydx = x2+y22xy, is given by x2-y2=Cx

Answer:


The given differential equation is

 dydx=x2+y22xy        .....(1)

This is a homogeneous differential equation.

Putting yvx and dydx=v+xdvdx in (1), we get

v+xdvdx=x2+v2x22vx2v+xdvdx=1+v22v
1+v22v-v=xdvdx1-v22v=xdvdx2v1-v2dv=dxx
Integrating on both sides, we get

2v1-v2dv=dxx-2v1-v2dv=-dxxlog1-v2=-logx+logClog1-v2+logx=logC
log1-v2x=logC1-v2x=C1-y2x2x=Cx2-y2=Cx
Thus, the family of curves for which dydx = x2+y22xy is given by x2-y2=Cx.

 



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