Class 12-NCERT Solutions-Chapter-09-Differential Equations-Ex 9.4

Solution:

\( \frac{dy}{dx} = \frac{x^2 + y^2}{x^2 + xy} \). This is a homogeneous equation.

Put \( y = vx \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx} \).

\( v + x\frac{dv}{dx} = \frac{x^2 + v^2x^2}{x^2 + x(vx)} = \frac{1+v^2}{1+v} \).

\( x\frac{dv}{dx} = \frac{1+v^2}{1+v} - v = \frac{1+v^2 - v - v^2}{1+v} = \frac{1-v}{1+v} \).

Separating variables:

\( \frac{1+v}{1-v} dv = \frac{dx}{x} \).

Integrating: \( \int \frac{2 - (1-v)}{1-v} dv = \int \frac{dx}{x} \Rightarrow -2\log|1-v| - v = \log|x| + C \).

Substituting \( v = y/x \):

\( (x+y)^2 = Cxe^{y/x} \) (after simplifying).

Solution:

\( \frac{dy}{dx} = 1 + \frac{y}{x} \).

Put \( y = vx \Rightarrow v + x\frac{dv}{dx} = 1 + v \).

\( x\frac{dv}{dx} = 1 \Rightarrow dv = \frac{dx}{x} \).

Integrating: \( v = \log|x| + C \).

\( \frac{y}{x} = \log|x| + C \Rightarrow y = x \log|x| + Cx \).

Solution:

\( \frac{dy}{dx} = \frac{x+y}{x-y} \).

Put \( y = vx \):

\( v + x\frac{dv}{dx} = \frac{1+v}{1-v} \Rightarrow x\frac{dv}{dx} = \frac{1+v - v(1-v)}{1-v} = \frac{1+v^2}{1-v} \).

\( \frac{1-v}{1+v^2} dv = \frac{dx}{x} \).

Integrating: \( \tan^{-1}v - \frac{1}{2}\log(1+v^2) = \log|x| + C \).

Substitute \( v = y/x \) and simplify: \( \tan^{-1}\left(\frac{y}{x}\right) = \frac{1}{2}\log(x^2+y^2) + C \).

Solution:

\( \frac{dy}{dx} = \frac{y^2 - x^2}{2xy} \).

Put \( y = vx \): \( v + x\frac{dv}{dx} = \frac{v^2-1}{2v} \).

\( x\frac{dv}{dx} = \frac{v^2-1-2v^2}{2v} = -\frac{1+v^2}{2v} \).

\( \frac{2v}{1+v^2} dv = -\frac{dx}{x} \).

Integrating: \( \log(1+v^2) = -\log|x| + \log C \).

\( x(1+v^2) = C \Rightarrow x^2 + y^2 = Cx \).

Solution:

\( \frac{dy}{dx} = 1 - 2\left(\frac{y}{x}\right)^2 + \frac{y}{x} \).

Put \( y = vx \): \( v + x\frac{dv}{dx} = 1 - 2v^2 + v \).

\( x\frac{dv}{dx} = 1 - 2v^2 \).

\( \frac{dv}{1 - (\sqrt{2}v)^2} = \frac{dx}{x} \).

Using \( \int \frac{dx}{a^2-x^2} = \frac{1}{2a} \log \frac{a+x}{a-x} \):

\( \frac{1}{2\sqrt{2}} \log \left| \frac{1+\sqrt{2}v}{1-\sqrt{2}v} \right| = \log|x| + C \).

Solution:

\( \frac{dy}{dx} = \frac{y + \sqrt{x^2+y^2}}{x} = \frac{y}{x} + \sqrt{1 + (y/x)^2} \).

Put \( y = vx \): \( v + x\frac{dv}{dx} = v + \sqrt{1+v^2} \).

\( x\frac{dv}{dx} = \sqrt{1+v^2} \Rightarrow \frac{dv}{\sqrt{1+v^2}} = \frac{dx}{x} \).

Integrating: \( \log|v + \sqrt{1+v^2}| = \log|x| + \log C \).

\( \frac{y + \sqrt{x^2+y^2}}{x} = Cx \Rightarrow y + \sqrt{x^2+y^2} = Cx^2 \).

Solution:

Rearranging to \( \frac{dy}{dx} \):

\( \frac{dy}{dx} = \frac{y}{x} \cdot \frac{x \cos(y/x) + y \sin(y/x)}{y \sin(y/x) - x \cos(y/x)} \).

Put \( y = vx \):

\( v + x\frac{dv}{dx} = v \left[ \frac{\cos v + v \sin v}{v \sin v - \cos v} \right] \).

Simplifying leads to: \( x\frac{dv}{dx} = \frac{2v \cos v}{v \sin v - \cos v} \).

Separating variables and integrating: \( \int \frac{v \sin v - \cos v}{v \cos v} dv = \int \frac{2dx}{x} \).

\( \int (\tan v - \frac{1}{v}) dv = 2\log|x| \).

\( \log|\sec v| - \log|v| = 2\log|x| + \log C \).

\( \sec(y/x) = Cxy \).

Solution:

\( \frac{dy}{dx} = \frac{y}{x} - \sin\left(\frac{y}{x}\right) \).

Put \( y = vx \): \( v + x\frac{dv}{dx} = v - \sin v \).

\( x\frac{dv}{dx} = -\sin v \Rightarrow -\text{cosec } v \, dv = \frac{dx}{x} \).

Integrating: \( \log|\text{cosec } v + \cot v| = \log|x| + \log C \).

\( \text{cosec}(y/x) + \cot(y/x) = Cx \).

Or \( \tan(y/2x) = C/x \).

Solution:

Rearranging: \( \frac{dy}{dx} = \frac{y}{2x - x \log(y/x)} \).

Put \( y = vx \): \( v + x\frac{dv}{dx} = \frac{v}{2 - \log v} \).

\( x\frac{dv}{dx} = \frac{v - v(2-\log v)}{2-\log v} = \frac{v(\log v - 1)}{2-\log v} \).

\( \frac{2-\log v}{v(\log v - 1)} dv = \frac{dx}{x} \).

Integrating: \( \log|\log(y/x) - 1| = \log|x| - \log(y/x) + C \).

Solution:

Let \( x = vy \) (since \( e^{x/y} \) appears). Then \( \frac{dx}{dy} = v + y\frac{dv}{dy} \).

Rearranging equation: \( \frac{dx}{dy} = -\frac{e^{x/y}(1-x/y)}{1+e^{x/y}} \).

\( v + y\frac{dv}{dy} = \frac{e^v(v-1)}{1+e^v} \).

\( y\frac{dv}{dy} = \frac{ve^v - e^v - v - ve^v}{1+e^v} = -\frac{v+e^v}{1+e^v} \).

\( \frac{1+e^v}{v+e^v} dv = -\frac{dy}{y} \).

Integrating: \( \log|v+e^v| = -\log|y| + \log C \).

\( y(v+e^v) = C \Rightarrow x + ye^{x/y} = C \).

Solution:

\( \frac{dy}{dx} = \frac{y-x}{y+x} \).

Put \( y = vx \): \( v + x\frac{dv}{dx} = \frac{v-1}{v+1} \).

\( x\frac{dv}{dx} = -\frac{1+v^2}{v+1} \).

Integrating: \( \frac{1}{2}\log(1+v^2) + \tan^{-1}v = -\log|x| + C \).

Substitute \( y=1, x=1 \Rightarrow v=1 \). \( \frac{1}{2}\log 2 + \frac{\pi}{4} = 0 + C \).

Solution: \( \log(x^2+y^2) + 2\tan^{-1}(y/x) = \frac{\pi}{2} + \log 2 \).

Solution:

\( \frac{dy}{dx} = -\frac{xy+y^2}{x^2} = -v - v^2 \).

\( v + x\frac{dv}{dx} = -v - v^2 \Rightarrow x\frac{dv}{dx} = -(2v+v^2) \).

Integrating \( \int \frac{dv}{v(v+2)} = -\int \frac{dx}{x} \):

\( \frac{1}{2}\log|\frac{v}{v+2}| = -\log|x| + C \).

At \( x=1, y=1 \Rightarrow v=1 \): \( \frac{1}{2}\log(1/3) = C \).

Solution: \( x^2y = \frac{1}{3}(y+2x) \).

Solution:

\( \frac{dy}{dx} = \frac{y - x \sin^2(y/x)}{x} = \frac{y}{x} - \sin^2(y/x) \).

Put \( y = vx \): \( x\frac{dv}{dx} = -\sin^2 v \).

\( -\text{cosec}^2 v \, dv = \frac{dx}{x} \).

Integrating: \( \cot v = \log|x| + C \).

At \( x=1, y=\pi/4 \Rightarrow v=\pi/4 \): \( \cot(\pi/4) = 0 + C \Rightarrow C=1 \).

Solution: \( \cot(y/x) = \log|x| + 1 \).

Solution:

\( \frac{dy}{dx} = \frac{y}{x} - \text{cosec}(y/x) \).

Put \( y=vx \): \( v + x\frac{dv}{dx} = v - \text{cosec } v \).

\( x\frac{dv}{dx} = -\text{cosec } v \Rightarrow -\sin v \, dv = \frac{dx}{x} \).

Integrating: \( \cos v = \log|x| + C \).

At \( x=1, y=0 \Rightarrow v=0 \): \( \cos 0 = 0 + C \Rightarrow C=1 \).

Solution: \( \cos(y/x) = \log|x| + 1 \).

Solution:

\( \frac{dy}{dx} = \frac{2xy+y^2}{2x^2} = \frac{y}{x} + \frac{1}{2}(\frac{y}{x})^2 \).

Put \( y=vx \): \( x\frac{dv}{dx} = \frac{1}{2}v^2 \).

\( 2v^{-2} dv = \frac{dx}{x} \).

Integrating: \( -\frac{2}{v} = \log|x| + C \).

\( -2\frac{x}{y} = \log|x| + C \).

At \( x=1, y=2 \): \( -1 = 0 + C \Rightarrow C=-1 \).

Solution: \( y = \frac{2x}{1-\log|x|} \).

Answer: (C)

Solution:

For a homogeneous equation where \( \frac{dx}{dy} \) is a function of \( \frac{x}{y} \), the standard substitution is \( x = vy \).

Answer: (D)

Solution:

A differential equation is homogeneous if all terms in the coefficients \( M(x, y) \) and \( N(x, y) \) are of the same degree.

• (A) Has constants, not homogeneous.
• (B) Degree of \( xy \) is 2, \( x^3 \) is 3. Not homogeneous.
• (C) Degree of \( x^3 \) is 3, \( 2y^2 \) is 2. Not homogeneous.
• (D) \( y^2 \) (deg 2), \( x^2 \) (deg 2), \( xy \) (deg 2). All terms are degree 2. Homogeneous.