NCERT Solutions for Class 12 Science Math Chapter 3 - Differential Equations

Answer:

The highest order derivative present in the differential equation is. Therefore, its order is four.

The given differential equation is not a polynomial equation in its derivatives. Hence, its degree is not defined.

Answer:

The given differential equation is:

The highest order derivative present in the differential equation is. Therefore, its order is one.

It is a polynomial equation in. The highest power raised tois 1. Hence, its degree is one.

Answer:

The highest order derivative present in the given differential equation is. Therefore, its order is two.

It is a polynomial equation inand. The power raised tois 1.

Hence, its degree is one.

Answer:

The highest order derivative present in the given differential equation is. Therefore, its order is 2.

The given differential equation is not a polynomial equation in its derivatives. Hence, its degree is not defined.

Answer:

The highest order derivative present in the differential equation is. Therefore, its order is two.

It is a polynomial equation inand the power raised tois 1.

Hence, its degree is one.

Answer:

The highest order derivative present in the differential equation is. Therefore, its order is three.

The given differential equation is a polynomial equation in.

The highest power raised tois 2. Hence, its degree is 2.

Answer:

The highest order derivative present in the differential equation is. Therefore, its order is three.

It is a polynomial equation in. The highest power raised tois 1. Hence, its degree is 1.

Answer:

The highest order derivative present in the differential equation is. Therefore, its order is one.

The given differential equation is a polynomial equation inand the highest power raised tois one. Hence, its degree is one.

Answer:

The highest order derivative present in the differential equation is. Therefore, its order is two.

The given differential equation is a polynomial equation inandand the highest power raised tois one.

Hence, its degree is one.

Answer:

The highest order derivative present in the differential equation is. Therefore, its order is two.

This is a polynomial equation inandand the highest power raised tois one. Hence, its degree is one.

Answer:

The given differential equation is not a polynomial equation in its derivatives. Therefore, its degree is not defined.

Hence, the correct answer is D.

Answer:

The highest order derivative present in the given differential equation is. Therefore, its order is two.

Hence, the correct answer is A.

Answer:

Differentiating both sides of this equation with respect to x, we get:

Now, differentiating equation (1) with respect to x, we get:

Substituting the values ofin the given differential equation, we get the L.H.S. as:

Thus, the given function is the solution of the corresponding differential equation.

Answer:

Differentiating both sides of this equation with respect to x, we get:

Substituting the value ofin the given differential equation, we get:

L.H.S. == R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

Answer:

Differentiating both sides of this equation with respect to x, we get:

Substituting the value ofin the given differential equation, we get:

L.H.S. == R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

Answer:

Differentiating both sides of the equation with respect to x, we get:

L.H.S. = R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

Answer:

Differentiating both sides with respect to x, we get:

Substituting the value ofin the given differential equation, we get:

Hence, the given function is the solution of the corresponding differential equation.

Answer:

Differentiating both sides of this equation with respect to x, we get:

Substituting the value ofin the given differential equation, we get:

Hence, the given function is the solution of the corresponding differential equation.

Answer:

Differentiating both sides of this equation with respect to x, we get:

L.H.S. = R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

Answer:

Differentiating both sides of the equation with respect to x, we get:

Substituting the value ofin equation (1), we get:

Hence, the given function is the solution of the corresponding differential equation.

Answer:

Differentiating both sides of this equation with respect to x, we get:

Substituting the value ofin the given differential equation, we get:

Hence, the given function is the solution of the corresponding differential equation.

Answer:

Differentiating both sides of this equation with respect to x, we get:

Substituting the value ofin the given differential equation, we get:

Hence, the given function is the solution of the corresponding differential equation.

Answer:

We know that the number of constants in the general solution of a differential equation of order n is equal to its order.

Therefore, the number of constants in the general equation of fourth order differential equation is four.

Hence, the correct answer is D.

Answer:

In a particular solution of a differential equation, there are no arbitrary constants.

Hence, the correct answer is D.

Answer:

Differentiating both sides of the given equation with respect to x, we get:

Again, differentiating both sides with respect to x, we get:

Hence, the required differential equation of the given curve is

Answer:

Differentiating both sides with respect to x, we get:

Again, differentiating both sides with respect to x, we get:

Dividing equation (2) by equation (1), we get:

This is the required differential equation of the given curve.

Answer:

Differentiating both sides with respect to x, we get:

Again, differentiating both sides with respect to x, we get:

Multiplying equation (1) with (2) and then adding it to equation (2), we get:

Now, multiplying equation (1) with 3 and subtracting equation (2) from it, we get:

Substituting the values of in equation (3), we get:

This is the required differential equation of the given curve.

Answer:

Differentiating both sides with respect to x, we get:

Multiplying equation (1) with 2 and then subtracting it from equation (2), we get:

y' − 2y = e^2x (2a + 2bx + b) − e^2x (2a + 2bx)

⇒ y' − 2y = be^2x                                ...(3)

Differentiating both sides with respect to x, we get:

y'' − 2y' = 2be^2x                               ...(4)

Dividing equation (4) by equation (3), we get:

This is the required differential equation of the given curve.

Answer:

Differentiating both sides with respect to x, we get:

Again, differentiating with respect to x, we get:

Adding equations (1) and (3), we get:

This is the required differential equation of the given curve.

Answer:

The centre of the circle touching the y-axis at origin lies on the x-axis.

Let (a, 0) be the centre of the circle.

Since it touches the y-axis at origin, its radius is a.

Now, the equation of the circle with centre (a, 0) and radius (a) is

Differentiating equation (1) with respect to x, we get:

Now, on substituting the value of a in equation (1), we get:

This is the required differential equation.

Answer:

The equation of the parabola having the vertex at origin and the axis along the positive y-axis is:

Differentiating equation (1) with respect to x, we get:

Dividing equation (2) by equation (1), we get:

This is the required differential equation.

Answer:

The equation of the family of ellipses having foci on the y-axis and the centre at origin is as follows:

Differentiating equation (1) with respect to x, we get:

Again, differentiating with respect to x, we get:

Substituting this value in equation (2), we get:

This is the required differential equation.

Answer:

The equation of the family of hyperbolas with the centre at origin and foci along the x-axis is:

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

Again, differentiating both sides with respect to x, we get:

Substituting the value ofin equation (2), we get:

This is the required differential equation.

Answer:

Let the centre of the circle on y-axis be (0, b).

The differential equation of the family of circles with centre at (0, b) and radius 3 is as follows:

Differentiating equation (1) with respect to x, we get:

Substituting the value of (y – b) in equation (1), we get:

This is the required differential equation.

Answer:

The given equation is:

Differentiating with respect to x, we get:

Again, differentiating with respect to x, we get:

This is the required differential equation of the given equation of curve.

Hence, the correct answer is B.

Answer:

The given equation of curve is y = x.

Differentiating with respect to x, we get:

Again, differentiating with respect to x, we get:

Now, on substituting the values of y, from equation (1) and (2) in each of the given alternatives, we find that only the differential equation given in alternative C is correct.

Hence, the correct answer is C.

Answer:

The given differential equation is:

Now, integrating both sides of this equation, we get:

This is the required general solution of the given differential equation.

Answer:

The given differential equation is:

Now, integrating both sides of this equation, we get:

This is the required general solution of the given differential equation.

Answer:

The given differential equation is:

Now, integrating both sides, we get:

$$\begin{gathered} \int \frac{dy}{1-y}=\int dx \\ \Rightarrow -\log \left(1-y\right)=x+\log C \\ \Rightarrow -\log C-\log \left(1-y\right)=x \\ \Rightarrow \log C\left(1-y\right)=-x \\ \Rightarrow C\left(1-y\right)=e^{-x} \\ \Rightarrow \left(1-y\right)=\frac{1}{C}{e}^{-x} \\ \Rightarrow y=1-\frac{1}{C}{e}^{-x} \\ \Rightarrow y=1+A{e}^{-x } \left(Where A=-\frac{1}{C}\right) \end{gathered}$$

This is the required general solution of the given differential equation.

Answer:

The given differential equation is:

Integrating both sides of this equation, we get:

Substituting these values in equation (1), we get:

This is the required general solution of the given differential equation.

Answer:

The given differential equation is:

Integrating both sides of this equation, we get:

Let (e^x + e^–x) = t.

Differentiating both sides with respect to x, we get:

Substituting this value in equation (1), we get:

This is the required general solution of the given differential equation.

Answer:

The given differential equation is:

Integrating both sides of this equation, we get:

This is the required general solution of the given differential equation.

Answer:

The given differential equation is:

Integrating both sides, we get:

Substituting this value in equation (1), we get:

This is the required general solution of the given differential equation.

Answer:

The given differential equation is:

Integrating both sides, we get:

This is the required general solution of the given differential equation.

Answer:

The given differential equation is:

Integrating both sides, we get:

Substituting this value in equation (1), we get:

This is the required general solution of the given differential equation.

Answer:

The given differential equation is:

Integrating both sides, we get:

Substituting the values of in equation (1), we get:

This is the required general solution of the given differential equation.

Answer:

The given differential equation is:

Integrating both sides, we get:

Comparing the coefficients of x² and x, we get:

A + B = 2

B + C = 1

A + C = 0

Solving these equations, we get:

Substituting the values of A, B, and C in equation (2), we get:

Therefore, equation (1) becomes:

$$\begin{gathered} \int dy=\frac{1}{2}\int \frac{1}{x+1}dx+\frac{1}{2}\int \frac{3x-1}{x^2+1}dx \\ \Rightarrow y=\frac{1}{2}\log \left(x+1\right)+\frac{3}{2}\int \frac{x}{x^2+1}dx-\frac{1}{2}\int \frac{1}{x^2+1}dx \\ \Rightarrow y=\frac{1}{2}\log \left(x+1\right)+\frac{3}{4}\int \frac{2x}{x^2+1}dx-\frac{1}{2}{\tan }^{-1}x+C \\ \Rightarrow y=\frac{1}{2}\log \left(x+1\right)+\frac{3}{4}\log \left(x^2+1\right)-\frac{1}{2}{\tan }^{-1}x+C \\ \Rightarrow y=\frac{1}{4}\left[2\log \left(x+1\right)+3\log \left(x^2+1\right)\right]-\frac{1}{2}{\tan }^{-1}x+C \\ \Rightarrow y=\frac{1}{4}\left[\log {\left(x+1\right)}^2+\log {\left(x^2+1\right)}^3\right]-\frac{1}{2}{\tan }^{-1}x+C \\ \Rightarrow y=\frac{1}{4}\log {\left(x+1\right)}^2{\left(x^2+1\right)}^3-\frac{1}{2}{\tan }^{-1}x+C ...\left(3\right) \end{gathered}$$
Now, y = 1 when x = 0
$$\begin{gathered} \Rightarrow 1=\frac{1}{4}\log \left(1\right)-\frac{1}{2}{\tan }^{-1}0+C \\ \Rightarrow 1=\frac{1}{4}\times 0-\frac{1}{2}\times 0+C \\ \Rightarrow C=1 \end{gathered}$$

Substituting C = 1 in equation (3), we get:

Answer:

Integrating both sides, we get:

Comparing the coefficients of x², x, and constant, we get:

Solving these equations, we get

Substituting the values of A, B, and C in equation (2), we get:

Therefore, equation (1) becomes:

Substituting the value of k² in equation (3), we get:

Answer:

Integrating both sides, we get:

Substituting C = 1 in equation (1), we get:

Answer:

Integrating both sides, we get:

Substituting C = 1 in equation (1), we get:

y = sec x

Answer:

The differential equation of the curve is:

Integrating both sides, we get:

Substituting this value in equation (1), we get:

Now, the curve passes through point (0, 0).

Substituting in equation (2), we get:

Hence, the required equation of the curve is

Answer:

The differential equation of the given curve is:

Integrating both sides, we get:

Now, the curve passes through point (1, –1).

Substituting C = –2 in equation (1), we get:

This is the required solution of the given curve.

Answer:

Let x and y be the x-coordinate and y-coordinate of the curve respectively.

We know that the slope of a tangent to the curve in the coordinate axis is given by the relation,

According to the given information, we get:

Integrating both sides, we get:

Now, the curve passes through point (0, –2).

∴ (–2)² – 0² = 2C

⇒ 2C = 4

Substituting 2C = 4 in equation (1), we get:

y² – x² = 4

This is the required equation of the curve.

Answer:

It is given that (x, y) is the point of contact of the curve and its tangent.

The slope (m₁) of the line segment joining (x, y) and (–4, –3) is

We know that the slope of the tangent to the curve is given by the relation,

According to the given information:

Integrating both sides, we get:

This is the general equation of the curve.

It is given that it passes through point (–2, 1).

Substituting C = 1 in equation (1), we get:

y + 3 = (x + 4)²

This is the required equation of the curve.

Answer:

Let the rate of change of the volume of the balloon be k (where k is a constant).

Integrating both sides, we get:

⇒ 4π × 3³ = 3 (k × 0 + C)

⇒ 108π = 3C

⇒ C = 36π

At t = 3, r = 6:

⇒ 4π × 6³ = 3 (k × 3 + C)

⇒ 864π = 3 (3k + 36π)

⇒ 3k = –288π – 36π = 252π

⇒ k = 84π

Substituting the values of k and C in equation (1), we get:

Thus, the radius of the balloon after t seconds is.

Answer:

Let p, t, and r represent the principal, time, and rate of interest respectively.

It is given that the principal increases continuously at the rate of r% per year.

Integrating both sides, we get:

It is given that when t = 0, p = 100.

⇒ 100 = e^k … (2)

Now, if t = 10, then p = 2 × 100 = 200.

Therefore, equation (1) becomes:

Hence, the value of r is 6.93%.

Answer:

Let p and t be the principal and time respectively.

It is given that the principal increases continuously at the rate of 5% per year.

Integrating both sides, we get:

Now, when t = 0, p = 1000.

⇒ 1000 = e^C … (2)

At t = 10, equation (1) becomes:

Hence, after 10 years the amount will worth Rs 1648.

Answer:

Let y be the number of bacteria at any instant t.

It is given that the rate of growth of the bacteria is proportional to the number present.

Integrating both sides, we get:

Let y₀ be the number of bacteria at t = 0.

⇒ log y₀ = C

Substituting the value of C in equation (1), we get:

Also, it is given that the number of bacteria increases by 10% in 2 hours.

Substituting this value in equation (2), we get:

Therefore, equation (2) becomes:

Now, let the time when the number of bacteria increases from 100000 to 200000 be t₁.

⇒ y = 2y₀ at t = t₁

From equation (4), we get:

Hence, in hours the number of bacteria increases from 100000 to 200000.

Answer:

Integrating both sides, we get:

Hence, the correct answer is A.

Answer:

The given differential equation i.e., (x² + xy) dy = (x² + y²) dx can be written as:

This shows that equation (1) is a homogeneous equation.

To solve it, we make the substitution as:

y = vx

Differentiating both sides with respect to x, we get:

Substituting the values of v and in equation (1), we get:

Integrating both sides, we get:

This is the required solution of the given differential equation.

Answer:

The given differential equation is:

Thus, the given equation is a homogeneous equation.

To solve it, we make the substitution as:

y = vx

Differentiating both sides with respect to x, we get:

Substituting the values of y and in equation (1), we get:

Integrating both sides, we get:

This is the required solution of the given differential equation.

Answer:

The given differential equation is:

Thus, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

y = vx

Substituting the values of y and in equation (1), we get:

Integrating both sides, we get:

This is the required solution of the given differential equation.

Answer:

The given differential equation is:

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

y = vx

Substituting the values of y and in equation (1), we get:

Integrating both sides, we get:

This is the required solution of the given differential equation.

Answer:

The given differential equation is:

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

y = vx

Substituting the values of y and in equation (1), we get:

Integrating both sides, we get:

This is the required solution for the given differential equation.

Answer:

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

y = vx

Substituting the values of v and in equation (1), we get:

Integrating both sides, we get:

This is the required solution of the given differential equation.

Answer:

The given differential equation is:

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

y = vx

Substituting the values of y and in equation (1), we get:

Integrating both sides, we get:

This is the required solution of the given differential equation.

Answer:

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

y = vx

Substituting the values of y and in equation (1), we get:

Integrating both sides, we get:

This is the required solution of the given differential equation.

Answer:

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

y = vx

Substituting the values of y and in equation (1), we get:

Integrating both sides, we get:

Therefore, equation (1) becomes:

This is the required solution of the given differential equation.

Answer:

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

x = vy

Substituting the values of x and in equation (1), we get:

Integrating both sides, we get:

This is the required solution of the given differential equation.

Answer:

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

y = vx

Substituting the values of y and in equation (1), we get:

Integrating both sides, we get:

Now, y = 1 at x = 1.

Substituting the value of 2k in equation (2), we get:

This is the required solution of the given differential equation.

Answer:

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

y = vx

Substituting the values of y and in equation (1), we get:

Integrating both sides, we get:

Now, y = 1 at x = 1.

Substituting in equation (2), we get:

This is the required solution of the given differential equation.

Answer:

Therefore, the given differential equation is a homogeneous equation.

To solve this differential equation, we make the substitution as:

y = vx

Substituting the values of y and in equation (1), we get:

Integrating both sides, we get:

Now, .

Substituting C = e in equation (2), we get:

This is the required solution of the given differential equation.

Answer:

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

y = vx

Substituting the values of y and in equation (1), we get:

Integrating both sides, we get:

This is the required solution of the given differential equation.

Now, y = 0 at x = 1.

Substituting C = e in equation (2), we get:

This is the required solution of the given differential equation.

Answer:

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

y = vx

Substituting the value of y and in equation (1), we get:

Integrating both sides, we get:

Now, y = 2 at x = 1.

Substituting C = –1 in equation (2), we get:

This is the required solution of the given differential equation.

Answer:

For solving the homogeneous equation of the form, we need to make the substitution as x = vy.

Hence, the correct answer is C.

Answer:

Function F(x, y) is said to be the homogenous function of degree n, if

F(λx, λy) = λⁿ F(x, y) for any non-zero constant (λ).

Consider the equation given in alternativeD:

Hence, the differential equation given in alternative D is a homogenous equation.

Answer:

The given differential equation is

This is in the form of

The solution of the given differential equation is given by the relation,

Therefore, equation (1) becomes:

This is the required general solution of the given differential equation.

Answer:

The given differential equation is

The solution of the given differential equation is given by the relation,

This is the required general solution of the given differential equation.

Answer:

The given differential equation is:

The solution of the given differential equation is given by the relation,

This is the required general solution of the given differential equation.

Answer:

The given differential equation is:

The general solution of the given differential equation is given by the relation,

Answer:

The given differential equation is:

This equation is in the form of:

The general solution of the given differential equation is given by the relation,

Therefore, equation (1) becomes:

Answer:

The given differential equation is:

This equation is in the form of a linear differential equation as:

The general solution of the given differential equation is given by the relation,

Answer:

The given differential equation is:

This equation is the form of a linear differential equation as:

The general solution of the given differential equation is given by the relation,

Substituting the value of in equation (1), we get:

This is the required general solution of the given differential equation.

Answer:

This equation is a linear differential equation of the form:

The general solution of the given differential equation is given by the relation,

Answer:

This equation is a linear differential equation of the form:

The general solution of the given differential equation is given by the relation,

Answer:

This is a linear differential equation of the form:

The general solution of the given differential equation is given by the relation,

Answer:

This is a linear differential equation of the form:

The general solution of the given differential equation is given by the relation,

Answer:

This is a linear differential equation of the form:

The general solution of the given differential equation is given by the relation,

Answer:

The given differential equation is

This is a linear equation of the form:

The general solution of the given differential equation is given by the relation,

Now,

Therefore,

Substituting C = –2 in equation (1), we get:

Hence, the required solution of the given differential equation is

Answer:

This is a linear differential equation of the form:

The general solution of the given differential equation is given by the relation,

Now, y = 0 at x = 1.

Therefore,

Substituting in equation (1), we get:

This is the required general solution of the given differential equation.

Answer:

The given differential equation is

This is a linear differential equation of the form:

The general solution of the given differential equation is given by the relation,

Now,

Therefore, we get:

Substituting C = 4 in equation (1), we get:

This is the required particular solution of the given differential equation.

Answer:

Let F (x, y) be the curve passing through the origin.

At point (x, y), the slope of the curve will be

According to the given information:

This is a linear differential equation of the form:

The general solution of the given differential equation is given by the relation,

Substituting in equation (1), we get:

The curve passes through the origin.

Therefore, equation (2) becomes:

1 = C

⇒ C = 1

Substituting C = 1 in equation (2), we get:

Hence, the required equation of curve passing through the origin is

Answer:

Let F (x, y) be the curve and let (x, y) be a point on the curve. The slope of the tangent to the curve at (x, y) is

According to the given information:

This is a linear differential equation of the form:

The general equation of the curve is given by the relation,

Therefore, equation (1) becomes:

The curve passes through point (0, 2).

Therefore, equation (2) becomes:

0 + 2 – 4 = Ce⁰

⇒ – 2 = C

⇒ C = – 2

Substituting C = –2 in equation (2), we get:

This is the required equation of the curve.

Answer:

The given differential equation is:

This is a linear differential equation of the form:

The integrating factor (I.F) is given by the relation,

Hence, the correct answer is C.

Answer:

The given differential equation is:

This is a linear differential equation of the form:

The integrating factor (I.F) is given by the relation,

Hence, the correct answer is D.

Answer:

(i) The differential equation is given as:

The highest order derivative present in the differential equation is. Thus, its order is two. The highest power raised to is one. Hence, its degree is one.

(ii) The differential equation is given as:

The highest order derivative present in the differential equation is. Thus, its order is one. The highest power raised to is three. Hence, its degree is three.

(iii) The differential equation is given as:

The highest order derivative present in the differential equation is. Thus, its order is four.

However, the given differential equation is not a polynomial equation. Hence, its degree is not defined.

Answer:

(i)

Differentiating both sides with respect to x, we get:

Again, differentiating both sides with respect to x, we get:

Now, on substituting the values of and in the differential equation, we get:

⇒ L.H.S. ≠ R.H.S.

Hence, the given function is not a solution of the corresponding differential equation.

(ii)

Differentiating both sides with respect to x, we get:

Again, differentiating both sides with respect to x, we get:

Now, on substituting the values of and in the L.H.S. of the given differential equation, we get:

Hence, the given function is a solution of the corresponding differential equation.

(iii)

Differentiating both sides with respect to x, we get:

Again, differentiating both sides with respect to x, we get:

Substituting the value of in the L.H.S. of the given differential equation, we get:

Hence, the given function is a solution of the corresponding differential equation.

(iv)

Differentiating both sides with respect to x, we get:

Substituting the value of in the L.H.S. of the given differential equation, we get:

Hence, the given function is a solution of the corresponding differential equation.

Answer:

Differentiating with respect to x, we get:

From equation (1), we get:

On substituting this value in equation (3), we get:

Hence, the differential equation of the family of curves is given as

Answer:

This is a homogeneous equation. To simplify it, we need to make the substitution as:

Substituting the values of y and in equation (1), we get:

Integrating both sides, we get:

Substituting the values of I₁ and I₂ in equation (3), we get:

Therefore, equation (2) becomes:

Hence, the given result is proved.

Answer:

The equation of a circle in the first quadrant with centre (a, a) and radius (a) which touches the coordinate axes is:

Differentiating equation (1) with respect to x, we get:

Substituting the value of a in equation (1), we get:

Hence, the required differential equation of the family of circles is

Answer:

Integrating both sides, we get:

Answer:

Integrating both sides, we get:

Hence, the given result is proved.

Answer:

The differential equation of the given curve is:

Integrating both sides, we get:

The curve passes through point

On substituting in equation (1), we get:

Hence, the required equation of the curve is

Answer:

Integrating both sides, we get:

Substituting these values in equation (1), we get:

Now, y = 1 at x = 0.

Therefore, equation (2) becomes:

Substituting in equation (2), we get:

This is the required particular solution of the given differential equation.

Answer:

Differentiating it with respect to y, we get:

From equation (1) and equation (2), we get:

Integrating both sides, we get:

Answer:

Substituting the values of x – y and in equation (1), we get:

Integrating both sides, we get:

Now, y = –1 at x = 0.

Therefore, equation (3) becomes:

log 1 = 0 – 1 + C

⇒ C = 1

Substituting C = 1 in equation (3) we get:

This is the required particular solution of the given differential equation.

Answer:

This equation is a linear differential equation of the form

The general solution of the given differential equation is given by,

Answer:

The given differential equation is:

This equation is a linear differential equation of the form

The general solution of the given differential equation is given by,

Now,

Therefore, equation (1) becomes:

Substituting in equation (1), we get:

This is the required particular solution of the given differential equation.

Answer:

Integrating both sides, we get:

Substituting this value in equation (1), we get:

Now, at x = 0 and y = 0, equation (2) becomes:

Substituting C = 1 in equation (2), we get:

This is the required particular solution of the given differential equation.

Answer:

Let the population at any instant (t) be y.

It is given that the rate of increase of population is proportional to the number of inhabitants at any instant.

Integrating both sides, we get:

log y = kt + C … (1)

In the year 1999, t = 0 and y = 20000.

Therefore, we get:

log 20000 = C … (2)

In the year 2004, t = 5 and y = 25000.

Therefore, we get:

In the year 2009, t = 10 years.

Now, on substituting the values of t, k, and C in equation (1), we get:

Hence, the population of the village in 2009 will be 31250.

Answer:

The given differential equation is:

Integrating both sides, we get:

Hence, the correct answer is C.

Answer:

The integrating factor of the given differential equation

The general solution of the differential equation is given by,

Hence, the correct answer is C.

Answer:

The given differential equation is:

This is a linear differential equation of the form

The general solution of the given differential equation is given by,

Hence, the correct answer is C.