Explain Solution R.D.Sharma Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 36 Sub Quetion 2 Maths Textbook Solution.

Answer: $y=xlog\left ( log|x| \right )$

Given:$xe^{\frac{y}{x}}-y+x\frac{dy}{dx}=0,y\left ( e \right )=0$

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

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

Solution: we have,

$xe^{\frac{y}{x}}-y+x\frac{dy}{dx}=0,y\left ( e \right )=0$

$\frac{dy}{dx}=\frac{y-xe^{\frac{x}{y}}}{x}$                            ......(i)

It is homogeneous equation.

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

So,

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

Separating the variables and integrating both side we get

\begin{aligned} &\int-e^{-v} d v=\int \frac{d x}{x} \\ &\Rightarrow e^{r}=\log |x|+\log |c| \\ &\Rightarrow v=\log (\log |x|)+\log (\log |c|) \end{aligned}

Putting $v=\frac{y}{x}$

$\Rightarrow \frac{y}{x}=log\left ( log|x| \right )+k$        ....(ii)

it is given that $y=0$ when $x=e$

Putting $x=e$,$y=0$ in $\left ( ii \right )$ we get

\begin{aligned} &\Rightarrow y=x \log (\log |x|)+k \\ &\Rightarrow 0=e \log (\log |x|)+k \\ &\Rightarrow k=0 \end{aligned}

Putting value of c in equation (ii) we get

\begin{aligned} &\Rightarrow \frac{y}{x}=\log (\log |x|)+0 \\ &\Rightarrow y=x \log (\log |x|) \end{aligned}

This is required solution.