Question

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

Answer 1

$\begin{array}{c}x\frac{dy}{dx}=y\left(\log y-\log x+1\right)\\ \frac{dy}{dx}=\frac{y}{x}\left(\log \frac{y}{x}+1\right)\\ y=vx\\ \frac{dy}{dx}=v+x\frac{dy}{dx}\\ v+x\frac{dv}{dx}=v\left(\log v+1\right)\\ x\frac{dv}{dx}=v\log v\\ \frac{dv}{v.\log v}=\frac{dx}{x}\\ \int \frac{dv}{v.\log x}=\int \frac{dx}{x}\\ log\left(\log v\right)=\log x+C\\ \log \left(\log \frac{y}{x}\right)=\log x+C\end{array}$