# Solving differential equation with trigonometric and exponential functions

I am trying to solve $e^x \sin(x) + e^y \cos(y) \cdot y' =0$.

Trying to solve it via separation of variables or treating it as an exact differential equation, I end up solving

$\frac{1}{2} (e^x (\cos x-\sin x) + e^y (\sin y+\cos y) + C = 0$ for $y(x)$ ($C$ being a constant).

I tried different things in order to come closer to a solution (or showing that there is none), but nothing worked (mathematica for example says it cannot be solved). Do you have any ideas?

• An equation defining the function implicitly can perfectly be the way to present the solution of a differential equation. – arts Nov 24 '17 at 18:16
• no Chance to find an explicit solution – Dr. Sonnhard Graubner Nov 24 '17 at 18:19
• @Dr.SonnhardGraubner Can I somehow proof this? – Vazrael Nov 24 '17 at 18:20

the solution is given by $$e^y(\cos(y)+\sin(y))=e^x(\cos(x)-\sin(x))+C$$