Question

Find the general solution of the differential equation
$e^x\tan ydx+(1-e^x){\sec }^{2}ydy=0$.

Answer 1

Given differential equation is

$e^x\tan ydx+(1-e^x){\sec }^{2}ydy=0$

$⟹(1-e^x){\sec }^{2}ydy=-e^x\tan ydx$

$⟹\frac{{\sec }^{2}y}{\tan y}dy=-\frac{e^x}{(1-e^x)}dx$

Taking integration on both the sides we get,

$\int \frac{{\sec }^{2}y}{\tan y}dy=\int \frac{-e^x}{(1-e^x)}dx$

$⟹\log \left(\tan y\right)=\log (1-e^x)$

$⟹\tan y=1-e^x$

$⟹y={\tan }^{-1}(1-e^x)$ is the general solution of given DE.