#### Provide Solution For  R.D. Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 18 Maths Textbook Solution.

Answer: $log \left ( \frac{y}{x} \right )=cx$

Given:$\frac{dy}{dx}=\frac{y}{x}\left \{ logy -log x+1 \right \}$

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

Solution:$\frac{dy}{dx}=\frac{y}{x}\left \{ logy -log x+1 \right \}$

$\Rightarrow \frac{d y}{d x}=\frac{y}{x}\left\{\log \frac{y}{x}-1\right\}\left[\therefore \log y-\log x=\log \frac{y}{x}\right]$

It is homogeneous equation.

\begin{aligned} &\text { Put } y=v x \text { and } \frac{d y}{d x}=v+x \frac{d v}{d x} \\ &\text { So, } v+x \frac{d v}{d x}=\frac{v x}{x}\left\{\log \frac{v x}{x}+1\right\} \\ &\Rightarrow x \frac{d v}{d x}=v \log v \end{aligned}

Separating the variables we have,

$\Rightarrow \int \frac{1}{v l o g v} d v=\int \frac{d x}{d x} \\$

$\therefore \int \frac{1}{v \log v} d v \\$

$\text { Put } \log v=t \\$

$\Rightarrow \frac{1}{v} d v=d t \\$

$\therefore \int \frac{1}{t} d t=\log t+c \\$

$\Rightarrow \log |\log v|+c \\$

$\therefore \log |\log |=\log |x|+\log c \\$

$\Rightarrow \operatorname{logv}=x c \\$

$\Rightarrow \log \frac{y}{x}=C x$                                                                                                                                                                            $\left [ \therefore v=\frac{y}{x} \right ]$

This is required solution.