Question

Solve $x\frac{dy}{dx}=y(\log y-\log x+1)$

Answer 1

The given equation is,

$x\frac{dy}{dx}=y\left(\log y-\log x+1\right)$

$\frac{dy}{dx}=\frac{y}{x}\left(\log \frac{y}{x}+1\right)$

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

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

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

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

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

Integrate both sides,

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

$\log \left(\log v\right)=\log x+\log c$

$\log \left(\log v\right)=\log \left(xc\right)$

$\log \frac{y}{x}=xc$