#### Need Solution for R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 36 Sub Question 1 Maths Textbook Solution.

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

Given: $x^{2}+y^{2}=2xydy,y\left ( 1 \right )=0$

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

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

Solution: we have,

$x^{2}+y^{2}=2xydy,y\left ( 1 \right )=0$

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

It is homogeneous equation.

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

So,

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

Integrating both side we get

\begin{aligned} &\int \frac{2 v}{1-v^{2}} d v=\int \frac{d x}{x} \\ &\Rightarrow \log \left|1-v^{2}\right|=-\log |x|+\log |c| \\ &\Rightarrow \log \left|1-v^{2}\right|=\log \left|\frac{c}{x}\right| \\ &\Rightarrow\left|1-v^{2}\right|=\left|\frac{c}{x}\right| \quad[\text { Taking antilog on bothsides }] \end{aligned}

Putting $v=\frac{y}{x}$

\begin{aligned} &\Rightarrow\left|\frac{x^{2}-y^{2}}{x^{2}}\right|=\left|\frac{c}{x}\right| \\ &\Rightarrow\left|x^{2}-y^{2}\right|=|c x| \end{aligned}                                     .......(ii)

It is given that y(1)=0 i.e when x=1 , y=0

Putting  x=1 , y=0 in (ii) we get

$1-c=0$

$c=1$

Putting value of c in equation (ii) we get

\begin{aligned} &\left|x^{2}-y^{2}\right|=|x| \\ &\left(x^{2}-y^{2}\right)^{2}=x^{2} \end{aligned}

This is required solution.