Question

Solve the differential equation :
$\frac{dy}{dx}=e^{-y}sinx+e^{x-y}$

Answer 1

we have,

$\frac{dy}{dx}=e^{-y}sinx+e^{x-y}$

$\frac{dy}{dx}=e^{-y}(sinx+e^x)$

$e^{y}dy=(sinx+e^x)dx$

${\int e}^{y}dy=\int (sinx+e^x)dx$

$e^y=cosx+e^x+c$

$y=ln(cosx+e^x+c)$