Question

If $x\frac{dy}{dx}=y(\log y-\log x+1),$ then the solution of the equation is ?

Answer 1

Given, the equation $x\frac{dy}{dx}=y(\log y-\log x+1)$

$\Rightarrow \frac{dy}{dx}=\frac{y}{x}(\log \frac{y}{x}+1)⟶(1)$, which is homogeneous equation

Let, $y=vx$

$\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$

Then equation $\left(1\right)$ becomes, $v+x\frac{dv}{dx}=v(\log v+1)$

$\Rightarrow x\frac{dv}{dx}=v\log v$

$\Rightarrow \frac{dv}{v\log v}=\frac{dx}{x}$

Integrating both sides we get,

$\int \frac{dv}{v\log v}=\int \frac{dx}{x}$

$\Rightarrow \log \left|\log v\right|=\log \left|x\right|+\log \left|C\right|$

$\Rightarrow \log (\log v)=\log x+\log C$

$\Rightarrow \log (\log \frac{y}{x})=\log x+\log C$

$\Rightarrow \log (\log \frac{y}{x})=\log \left(xC\right)$

$\Rightarrow \log \frac{y}{x}=xC$

$\Rightarrow \log y-\log x=xC$

$\Rightarrow \log y=\log x+xC$

$\Rightarrow y=e^{(\log x+xC)}$.