Question

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

Answer 1

We have,

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

Now,

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

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

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

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

On integration and we get,

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

$\Rightarrow \log (e^x-1)=\log \tan y+\log C$

Hence, this is the answer.