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 $\frac{d y}{d x} = \tan θ$ .
Here,

$θ = \tan^{- 1} (2 x + 3 y) ∴ \frac{d y}{d x} = \tan (\tan^{- 1} 2 x + 3 y) \Rightarrow \frac{d y}{d x} = 2 x + 3 y$

$\Rightarrow \frac{d y}{d x} - 3 y = 2 x . . . . . (1) Clearly, it is a linear differential equation of the form \frac{d y}{d x} + P y = Q where P = - 3 and Q = 2 x ∴ I . F . = e^{\int P d x} = e^{\int - 3 d x} = e^{- 3 x} Multiplying both sides of (1), by I . F . = e^{- 3 x}, we get e^{- 3 x} (\frac{d y}{d x} - 3 y) = e^{- 3 x} . 2 x \Rightarrow e^{- 3 x} (\frac{d y}{d x} - 3 y) = 2 x e^{- 3 x} Integrating both sides with respect to x, we get y e^{- 3 x} = 2 \int x e^{- 3 x} d x + C \Rightarrow y e^{- 3 x} = 2 \int \underset{I}{x} \underset{II}{e^{- 3 x}} d x + C \Rightarrow y e^{- 3 x} = 2 x \int e^{- 3 x} d x - 2 \int [\frac{d}{d x} (x) \int e^{- 3 x} d x] d x + C \Rightarrow y e^{- 3 x} = - 2 x \frac{e^{- 3 x}}{3} + 2 \times \frac{1}{3} \int e^{- 3 x} d x + C \Rightarrow y e^{- 3 x} = - \frac{2}{3} x e^{- 3 x} - 2 \times \frac{1}{9} e^{- 3 x} + C \Rightarrow y e^{- 3 x} = - \frac{2}{3} x e^{- 3 x} - \frac{2}{9} e^{- 3 x} + C Since the curve passes through (1, 2), it satisfies the above equation . ∴ 2 e^{- 3} = - \frac{2}{3} e^{- 3} - \frac{2}{9} e^{- 3} + C \Rightarrow C = 2 e^{- 3} + \frac{2}{3} e^{- 3} + \frac{2}{9} e^{- 3} \Rightarrow C = \frac{26}{9} e^{- 3} Putting the value of C, we get y e^{- 3 x} = (- \frac{2}{3} x - \frac{2}{9}) e^{- 3 x} + \frac{26}{9} e^{- 3}$

Page No 21.135:

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 $= x - y \frac{d x}{d y} = OT$

It is given, OT = 2x

$\begin{array}{l} ∴ x - y \frac{d x}{d y} = 2 x \\ - x = y \frac{d x}{d y} \\ - \int \frac{d x}{x} = \int \frac{d y}{y} \\ ∴ x y = k \end{array}$

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

$(i) {(\frac{d s}{d t})}^{4} + 3 s \frac{d^{2} s}{d t^{2}} = 0$
The highest order derivative in the given equation is $\frac{d^{2} s}{d t^{2}}$ 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 + e^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. 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 $\frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} - 6 y = 0$ .

Answer:

^{$We have, \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} - 6 y = 0 . . . . . (1) Now, y = e^{- 3 x} \Rightarrow \frac{d y}{d x} = - 3 e^{- 3 x} \Rightarrow \frac{d^{2} y}{d x^{2}} = 9 e^{- 3 x}$}
$Putting the values of \frac{d^{2} y}{d x^{2}}, \frac{d y}{d x} and y in (1), we get$

$LHS = 9 e^{- 3 x} - 3 e^{- 3 x} - 6 e^{- 3 x} = 0 = RHS$

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 = e^x + 1	y'' − y' = 0
(ii) y = x² + 2x + C	y' − 2x − 2 = 0
(iii) y = cos x + C	y' + sin x = 0
(iv) y = $\sqrt{1 + x^{2}}$	y' = $\frac{x y}{1 + x^{2}}$
(v) y = x sin x	xy' = y + x $\sqrt{x^{2} - y^{2}}$
(vi) $y = \sqrt{a^{2} - x^{2}}$	$x + y \frac{d y}{d x} = 0$

Differential equation	Function
(i) $x \frac{d y}{d x} = y$	y = ax
(ii) $x + y \frac{d y}{d x} = 0$	$y = \pm \sqrt{a^{2} - x^{2}}$
(iii) $x \frac{d y}{d x} + y = y^{2}$	$y = \frac{a}{x + a}$
(iv) $x^{3} \frac{d^{2} y}{d x^{2}} = 1$	$y = a x + b + \frac{1}{2 x}$
(v) $y = {(\frac{d y}{d x})}^{2}$	$y = \frac{1}{4} {(x \pm a)}^{2}$

Answer:

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