Particular solution of the differential equation $\log(\frac{dy}{dx})=x+y$, given that when $x = 0, y = 0$ is:
Answer & explanation
Correct answer: option 1
$\log(\frac{dy}{dx})=x+y$ 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 $-e^0=e^0+c$
$⇒-1=1+e⇒c=-2$
$⇒e^x+e^{-y}=2$