The particular solution of $\cos ydx+(1+2e^{-x})\sin ydy=0$ when $x=0,y=\frac{\pi}{4}$ is:
Answer & explanation
Correct answer: option 3
$\cos ydx+(1+2e^{-x})\sin ydy=0⇒-\int\frac{\sin y}{\cos y}dy=\int\frac{dx}{(1+2e^{-x})}$
$⇒log(\cos y)=log(e^x+2)+c⇒\frac{\cos y}{(e^x+2)}=k$ At $x=0,y=\frac{\pi}{4}$
$⇒k=\frac{1}{3\sqrt{2}}$
$⇒e^x+2=3\sqrt{2}\cos y$