#### Provide Solution For  R.D. Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21 .9 Question 20 Maths Textbook Solution.

Answer: $y=c e^{\tan ^{-1}\left(\frac{y}{x}\right)}$

Given: $y^{2}dx+\left ( x^{2}-xy+y^{2} \right )dy=0$

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

Solution: we have

$y^{2}dx+\left ( x^{2}-xy+y^{2} \right )dy=0$

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

It is homogeneous equation.

Putting $y=vx$ and $\frac{dy}{dx}=v+x\frac{dv}{dx}$

So,$v+x\frac{dv}{dx}=\frac{-v^{2}}{x^{2}-vx^{2}+v^{2}x^{2}}$

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

\begin{aligned} &\Rightarrow x \frac{d v}{d x}=\frac{-v^{2}-v+v^{2}-v^{3}}{1-v+v^{2}} \\ &\Rightarrow x \frac{d v}{d x}=\frac{-v-v^{3}}{1-v+v^{2}} \\ &\Rightarrow x \frac{d v}{d x}=\frac{-v\left(v^{2}+1\right)}{v^{2}-v+1} \end{aligned}

Separating the variables we have,

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

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

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

$\Rightarrow-\operatorname{logv}+\tan ^{-1} v=\log x=\log c \\$

$\Rightarrow-\log \left(\frac{y}{x}\right)+\tan ^{-1}\left(\frac{y}{x}\right)=\log x k \\$                                                                    $\left [ \therefore v=\frac{y}{x} \right ]$

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

$\Rightarrow \tan ^{-1}\left(\frac{y}{x}\right)=\log x k-\log \left(\frac{x}{y}\right) \\$

$\Rightarrow \tan ^{-1}\left(\frac{y}{x}\right)=\log \left(\frac{x k y}{x}\right) \\$

$\Rightarrow e^{\tan ^{-1}\left(\frac{y}{x}\right)}=k y \\$

$\Rightarrow \frac{1}{k} e^{\tan ^{-1}\left(\frac{y}{x}\right)}=k \\$

$\Rightarrow \mathrm{y}=c e^{\tan ^{-1}\left(\frac{y}{x}\right)} \quad \text { Where } c=\frac{1}{k}$

This is required solution.