#### Need Solution for R.D .Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 32 Maths Textbook Solution.

Answer: $\log \left|x^{2}+x y+y^{2}\right|=2 \sqrt{3} \tan ^{-1}\left(\frac{x+2 y}{\sqrt{3 x}}\right)+c$

Given:$\left ( x-y \right )\frac{dy}{dx}=x+2y$

To solve: have to solve the given differential equation.

Hint: Put $y=vx$ and $\frac{dy}{dx}=v+x\frac{dv}{dx}$

Solution: Here,

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

$\Rightarrow \frac{dy}{dx}=\frac{x+2y}{x-y}$

It is homogeneous equation

Put $y=vx$$\Rightarrow$$\frac{dy}{dx}=v+x\frac{dv}{dx}$

So, $v+x \frac{d y}{d x}=\frac{x+2 v x}{x-v x} \\$

$\Rightarrow x \frac{d v}{d x}=\frac{1+2 v}{1-v}-v \\$

$\Rightarrow x \frac{d v}{d x}=\frac{1+2 v-v+v^{2}}{1-v} \\$

$\Rightarrow x \frac{d v}{d x}=\frac{1+v+v^{2}}{1-v}$

Separating the variable and Integrating both side we get

$\int \frac{1-v}{1+v+v^{2}} d v=\int \frac{d x}{x} \\$

$\Rightarrow \frac{1}{2} \int \frac{2 v-2}{1+v+v^{2}}=-\int \frac{1}{x} d x \\$

$\Rightarrow \int \frac{(2 v+1)-3}{1+v+v^{2}} d v=-2 \int \frac{1}{x} d x \\$

$\Rightarrow \int \frac{(2 v+1)-3}{1+v+v^{2}} d v=-\int \frac{3}{v^{2}+2 v\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}+1} d x=-2 \int \frac{1}{x} d x \\$

$\Rightarrow \log \left|1+v+v^{2}\right|-3 x \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{v+\frac{1}{2}}{\sqrt{3 / 2}}\right)=-2 \log x+c \\$

$\Rightarrow \log \left|1+v+v^{2}\right|+\log x^{2}-2 \sqrt{3} \tan ^{-1}\left(\frac{v+\frac{1}{2}}{\sqrt{3 / 2}}\right)=c . \\$

$\Rightarrow \log \left[x^{2}\left(1+\frac{y}{x}+{ }^{y^{2}} /_{x^{2}}\right]-2 \sqrt{3} \tan ^{-1}\left(\frac{\frac{2 y}{x}+1}{\sqrt{3}}\right)=c . \quad\left[\therefore v=\frac{y}{x}\right]\right.$

$\Rightarrow \log \left(y^{2}+x y+x^{2}\right)=2 \sqrt{3} \tan ^{-1}\left(\frac{2 y+x}{\sqrt{3 x}}\right)+c$

This is required solution.