Please Solve R.D.Sharma class 12 Chapter 21  Differential Equations Exercise 21.9 Question 10 Maths textbook Solution.

Answer: $e^{x/y}=log\: y+c.$

Given:$ye^{x/y}dx=\left ( xe^{x/y}+y \right )dy\:$$ye^{x/y}dx=\left ( xe^{x/y}+y \right )dy\:$

To solve: we have to solve the given differential equation

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

Solution: we have,

$ye^{x/y}dx=\left ( xe^{x/y}+y \right )dy$

$\Rightarrow \frac{dy}{dx}=\frac{xe^{x/y}+y}{ye^{x/y}}$

It is a homogeneous equation.

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

So,

\begin{aligned} &v+y \frac{d v}{d y}=\frac{v y e^{v y} / y+y}{y e^{v y} / y} \\ &\Rightarrow v+y \frac{d v}{d y}=\frac{v e^{v}+1}{e^{v}} \\ &\Rightarrow y \frac{d v}{d y}=\frac{v e^{v}+1}{e^{v}}-v \\ &\Rightarrow y \frac{d v}{d y}=\frac{v e^{v}+1-v e^{v}}{e^{v}} \end{aligned}

$\Rightarrow y \frac{d v}{d y}=\frac{1}{e^{v}} \\$

$\Rightarrow \int e^{v} d v=\int \frac{d y}{e y} \\$                                                                                                $\left ( \therefore Integrating\: on\: both\: side \right )$

$\Rightarrow e^{v}=\log y+c \\$

$\Rightarrow e^{x} / y=\log y+c$                                                                                                                                    $\left [ \therefore v=y/x \right ]$

Hence this is required solution