#### Need Solution for R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 29 Maths Textbook Solution.

Answer: $y+\sqrt{y^{2}-x^{2}}=cx^{3}$

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

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

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

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

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

It is homogeneous equation.

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

So,

\begin{aligned} &\Rightarrow v+x \frac{d v}{d x}=\frac{2 \sqrt{v^{2} x^{2}-x^{2}}+v x}{x} \\ &\Rightarrow x \frac{d v}{d x}=2 \sqrt{v^{2}-1}+v-v \\ &\Rightarrow x \frac{d v}{d x}=2 \sqrt{v^{2}-1} \end{aligned}

Separating the variables and integrating both sides we get

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

$\Rightarrow \log \left|v+\sqrt{v^{2}-1}\right| \\$

$\qquad=2 \log x+\log c \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\therefore \int \frac{d v}{\sqrt{v^{2}-1}}=\log \left|v+\sqrt{v^{2}-1}\right|\right] \\$

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

$\Rightarrow v+\sqrt{v^{2}-1}=c x^{2} \\$

$\Rightarrow \frac{y}{x}+\sqrt{\left(\frac{y}{x}\right)^{2}-1}=c x^{2}\left[\therefore v=\frac{y}{x}\right]$

$\Rightarrow y+\sqrt{y^{2}-x^{2}}=c x^{3}$

This is required solution.