Something feels wrong in my solution

Question

Solve the given differential equation. $$xdy-ydx=(x^2+y^2)e^ydy $$ . $$\text {Attempt} $$ We divide by $x^2$ and see that lhs is $d (\frac {y}{x})=(1+(\frac {y}{x})^2)e^ydy $ letting $y=vx $ I get $dv=(1+v^2)e^ydy $ .Thus is variable separable form. Thus the solution is $\arctan (v)=e^y+c $ resubstituting $v $ the solution is $\arctan (\frac {y}{x})=e^y+c $ . Is this correct?. The step after substitution seems fishy to me.If this approach is incorrect then how to go about it? Thanks!

Answer 1

Correct, also can do directly, for $x\neq 0,$

$$d(\tfrac{y}{x}) = \left(1+(\tfrac{y}{x})^2\right) e^y dy$$

to get same answer. Here we assumed $x \neq 0$. For $x = 0$ we have

$$-y = y^2 e^y y'_x$$

which again is variable separable and yields $-x = (y-1)e^y+c$