Question

The soultion of $3e^x{\cos }^{2}ydx+(1-e^x)\cot ydy=0$ is

Answer 1

Given,

$3e^x{\cos }^{2}ydx+(1-e^x)\cot ydy=0$

$3e^x{\cos }^{2}ydx=(e^x-1)\cot ydy$

$\frac{3e^x}{e^x-1}dx=\frac{\cot y}{{\cos }^{2}y}dy$

integrating on both sides, we get,

$\int \frac{3e^x}{e^x-1}dx=\int \frac{\cot y}{{\cos }^{2}y}dy$

$\int \frac{3e^x}{e^x-1}dx=\int \frac{\sec y}{\sin y}dy$

$3\log (e^x-1)=\log \left(\tan y\right)+c$