Provide Solution For  R.D. Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 16 Maths Textbook Solution.

Answer: $\sqrt{x^{2}+y^{2}}=ce^{2\tan -1}\frac{y}{x},x\neq 0$

Given:$\left ( x+2y \right )dx-\left ( 2x-y \right )dy=0$

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

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

Solution: We have,

$\left ( x+2y \right )dx-\left ( 2x-y \right )dy=0$

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

It is a homogeneous equation.

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

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

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

Separating the variables

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

$\Rightarrow \int \frac{d v}{1+v^{2}}-\int \frac{v}{1+v^{2}}=\int \frac{d x}{x}[\therefore \text { Integrating }]\\$

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

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

$\Rightarrow 2 \tan ^{-1} v=(1+v)^{\frac{1}{2}} x c\\$

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

$\Rightarrow e^{2 \tan ^{-1}} \frac{y}{x}=\left\{\frac{x^{2}+y^{2}}{x}\right\}^{\frac{1}{2}} x c\\$

$\Rightarrow e^{2 \tan ^{-1}} \frac{y}{x}=\left(x^{2}+y^{2}\right)^{\frac{1}{2}} c\\$

$\Rightarrow \sqrt{x^{2}+y^{2}}=\frac{1}{c} e^{2 \tan ^{-1}} \frac{y}{x}\\$

$\Rightarrow \sqrt{x^{2}+y^{2}}=C e^{2 \tan ^{-1} \frac{y}{x}} ; \text { Where }=\frac{1}{c}\\$

This is required solution.