Answer
$y=xe^{Cx}$
Work Step by Step
We are given:
$$y[\ln(\frac{y}{x}+1]dx-xdy=0$$
Rewrite as:
$$\frac{dy}{dx}=\frac{y}{x}(\ln\frac{y}{x}+1)$$
Assume that $V=\frac{y}{x} \rightarrow \frac{dy}{dx}=V+x\frac{dV}{dy}$
The equation becomes:
$$V+ x\frac{dV}{dx}=V(\ln V+1)$$
$$ x\frac{dV}{dx}=V\ln V$$
$$\frac{1}{V\ln V}dV=\frac{1}{x}dx$$
Integrating both sides:
$$\ln(\ln V)=\ln |xC|$$
$$V=e^{Cx}$$
$$\frac{y}{x}=e^{Cx}$$
Hence general solution is $y=xe^{Cx}$
