Particular solution of the differential equation $\log(\frac{dy}{dx})=x+y$, given that when $x = 0, y = 0$ is:

$e^x+e^{-y}=2$ $e^{-x}+e^{y}=2$ $e^x+e^{y}=2$ $e^{-x}+e^{-y}=2$

$e^x+e^{-y}=2$

$\log(\frac{dy}{dx})=x+y$ Taking out log on both sides $\frac{dy}{dx}=e^x+y$ $⇒\frac{dy}{dx}=e^xe^y⇒e^{-y}dy=e^xdx$ Integrating both side $\int e^{-y}dy=\int e^x\,dx$ $⇒-e^{-y}=e^x+c$   (for x = 0, y = 0) we get $-e^0=e^0+c$ $⇒-1=1+e⇒c=-2$ $⇒e^x+e^{-y}=2$