#### Provide Solution For R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 17 Maths Textbook Solution.

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

Given:   $\frac{dy}{dx}=\frac{y}{x}-\sqrt{\frac{y^{2}}{x^{2}}-1}$

To solve: We have to solve the given differential equation.

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

Solution: $\frac{dy}{dx}=\frac{y}{x}-\sqrt{\frac{y^{2}}{x^{2}}-1}$

It is a homogeneous equation

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

So,$v+x\frac{dv}{dx}=\frac{vx}{x}-\sqrt{\frac{v^{2}x^{2}}{x^{2}}-1}$

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

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

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

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

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

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

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

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