Solving a first order differential equation involving y and its exponential

Question

I am trying to solve the differential equation $$y'(x)={2e^{y(x)/x}+\frac{y(x)}{x}}$$ I think it is homogeneous but I have no idea as to how to manipulate this to get it into the required form.

Answer 1

Hint. Let $z(x)=\exp(y(x)/x)$ then $y(x)=x\ln(z(x))$ and $y'(x)=\ln(z(x))+\frac{xz'(x)}{z(x)}$. Hence the ODE becomes $$\ln(z(x))+\frac{xz'(x)}{z(x)}=2z(x)+\ln(z(x)).$$ That is, after separating variables (note that $z(x)>0$), $$\frac{z'(x)}{z^2(x)}=\frac{2}{x}.$$ Can you take it from here?

Answer 2

Hint

Start with the clear $y=x z$ to make the equation $$x z'=2e^z$$ which is separable.