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

Answer:$\sqrt{x^{2}+y^{2}}=xlog|\frac{c}{x}|$

Given: $\left [ x\sqrt{x^{2}+y^{2}}-y^{2} \right ]dx+xydy=0$

To solve: we have to solve the given differential equation

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

Solution: $\left [ x\sqrt{x^{2}+y^{2}}-y^{2} \right ]dx+xydy=0$

$\Rightarrow \frac{dy}{dx}=\frac{\left [ y^{2}-x\sqrt{x^{2}+y^{2}} \right ]}{xy}$

It is a homogeneous equation.

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

So,$v+x \frac{d v}{d x}=\frac{v^{2} x^{2}-x \sqrt{x^{2}+v^{2}} x^{2}}{v x^{2}}$

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

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

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

$\Rightarrow \int \frac{v}{\sqrt{\left(1+v^{2}\right)}} d v=-\frac{d x}{x} \\$                                            $(seperating \: the \: variables\: and\: intersecting \: both \: sides)$

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

Let$1+v^{2}=t$

$2vdv=dt$

$\Rightarrow \frac{1}{2} \int \frac{1}{\sqrt{t}} d t=-\log x+\log c \\$

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

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

$\Rightarrow \sqrt{x^{2}+y^{2}}=x \log \left(\frac{c}{x}\right)$

This is required solution.