How to solve $\frac{\mathrm dy}{\mathrm dx}=\frac{x+y}{x}$ I am looking for a help with the below equation
$$\frac{\mathrm dy}{\mathrm dx}=\frac{x+y}{x}$$
I don't get the suggested answer $$y = x\ln(x) + c x$$
My attempt
$$\mathrm dx(1+\frac{y}{x})=\mathrm dy$$
$$c+x+y\ln(x)=y$$
$$y=\frac{c+x}{1-\ln x}$$

We have:
$$\frac{dy}{dx}=\frac{x+y}{x}=1+\frac{y}{x}$$
Suppose $y=xv(x)$ with $v(x):\mathbb{R}\rightarrow\mathbb{R}$
Then $$\frac{dy}{dx}=v+x\frac{dv}{dx}=1+v$$
Now notice we have:
$$x\frac{dv}{dx}=1$$ which is separable.
$$v(x)=\ln(x)+C,$$ where $C\in \mathbb{R}$
Thus $$y=x\ln(x)+Cx$$
