Question

Solve the following differential equation:
$3e^x\tan ydx+(1-e^x){\sec }^{2}ydy=0$

Answer 1

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

$\frac{3e^x}{(e^x-1)}dx=\frac{{\sec }^{2}ydy}{\tan y}$

Now on LHS, put $e^x-1=z$

and on RHS put $\tan y=p$

${\sec }^{2}ydy=dp$

$\frac{3dz}{z}=\frac{dp}{p}$

Now, by integrating we get

$3\log z=\log p+\log C$

$\frac{z^3}{p}=c$

As, $z=e^x-1$

and $p=\tan y$

$(e^x-1{)}^3=C\tan y$