#### Please Solve R.D.Sharma class 12 Chapter 21  Differential Equations Exercise 21.9 Question 3 Maths textbook Solution.

Answer: $x^{2}+y^{2}=cx$

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

To find: we have to solve the given differential equation.

Hint: in homogeneous differential equation put  $y=vx$ and $\frac{dy}{dx}=v+2\frac{dv}{dx}$

Solution: Here, $\frac{dy}{dx}=\frac{y^{2}-x^{2}}{2xy}$

Clearly since each of the function $y^{2}-x^{2}$ and $2xy$  is a homogeneous function of degree 2, the given equation is homogeneous

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

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

Integrating on both side we get

\begin{aligned} &\Rightarrow \int \frac{2 v}{1+v^{2} }dv=-\int \frac{d x}{x} \\ &\Rightarrow \log \left|1+v^{2}\right|=-\log x+\log c \\ &\Rightarrow \log \left|1+v^{2}\right|+\log x=\log c \\ &\Rightarrow \log \left|x\left(1+v^{2}\right)\right|=\log c \\ &\Rightarrow x\left(1+v^{2}\right)=c \\ &\Rightarrow x\left(1+\frac{y^{2}}{x^{2}}\right)=c \end{aligned}                            $\left [ \therefore v=y/x \right ]$

$\Rightarrow x^{2}+y^{2}=cx$

This is required solution.