#### Please Solve R.D.Sharma class 12 Chapter 21  Differential Equations Exercise 21.9 Question 7 Maths textbook Solution.

Answer: $x=c\left ( x^{2}+y^{2} \right )$

Given:$2xy \frac{dy}{dx}= x^{2}+y^{2}$

To find: we have to find the solution of given differential equation.

Hint: IN homogeneous differential equation

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

Solution: We have

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

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

It is homogeneous equation

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

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

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

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

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

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

$\Rightarrow \int \frac{-2 v}{1-v^{2}} d v=-\int \frac{d x}{x} \\$                                        $[Integrating\; on \; both\; side]$

$\Rightarrow \log \left|1-v^{2}\right|=-\log x+\log c \mid \\$

$\Rightarrow 1-v^{2}=\frac{c}{x}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \left[\therefore \log c-\log x=\log \frac{c}{x}\right] \\$

$\Rightarrow x\left(1-\frac{y^{2}}{x^{2}}\right)=C \; \; \; \; \; \; \; \; \quad\left[\therefore \text { put } v=\frac{y}{x}\right] \\$

$\Rightarrow x\left(\frac{x^{2}-y^{2}}{x^{2}}\right)=C \\$

$\Rightarrow x^{2}-y^{2}=c x$                        which is required solution.