#### Provide Solution For  R.D. Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 24 Maths Textbook Solution.

Answer: $\frac{x^{2}}{y^{2}}\left \{ \left ( \frac{x}{y} \right )-\frac{1}{2} \right \}+logy^{2}=c$

Given:$xylog \left ( \frac{x}{y} \right )dx+\left \{ y^{2}-x^{2}log\left ( \frac{x}{y} \right ) \right \}dy=0$

To solve:We have to solve the given differential equation

Hint:$\left ( 1 \right )$Put $y=vx$ and $\frac{dy}{dx}=v+x\frac{dv}{dx}or$

$\left ( 2 \right )$ Put $y=vx$ and  $\frac{dy}{dx}=v+y\frac{dv}{dx}$

Solution: We have,

\begin{aligned} &x y \log \left(\frac{x}{y}\right) d x+\left\{y^{2}-x^{2} \log \left(\frac{x}{y}\right)\right\} d y=0 \\ &\Rightarrow \frac{d x}{d y}=\frac{x^{2} \log \left(\frac{x}{y}\right)-y^{2}}{x y \log \left(\frac{x}{y}\right)} \end{aligned}

It is a homogeneous equation.

We put  $x=vy$

$\Rightarrow \frac{d x}{d y}=v+y \frac{d v}{d y} \\$

So,$v+y \frac{d v}{d y}=\frac{v^{2} y^{2} \log (v)-y^{2}}{v y^{2} \log v} \\$

$\Rightarrow y \frac{d v}{d y}=\frac{v^{2} \log (v)-1}{v \log v}-v \\$

$\Rightarrow y \frac{d v}{d y}=\frac{v^{2} \log (v)-1-v^{2} \log v}{v \log v} \\$

$\Rightarrow y \frac{d v}{d y}=\frac{-1}{v \log v}$

Separating the variables, we get

$\Rightarrow vlogv=\frac{1}{y}dy$

On integrating both sides, we get

$\Rightarrow \int \Rightarrow vlogdv=-\int \frac{1}{y}dy$

$\Rightarrow \log v \int v \log v-\int\left(\frac{d}{d v}(\log v) \int v d v\right)=-\log y+\log c \\$  $[Integrating \: using \: by \: parts]$

$\Rightarrow \frac{v^{2}}{2} \log v-\int \frac{v}{2} d v=-\log y+c \\$

$\Rightarrow \frac{v^{2}}{2} \log v-\frac{v^{2}}{4}=-\log y+c \\$

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

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

Now, putting back the value of $v$  as $\frac{x}{y}$  , we get

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

Hence this is required solution.