Question

For the differential equation $3e^x\tan ydx+1(1-e^x){\sec }^{2}ydy=0,$ it gives ${(1-e^x)}^k=C\tan y$. What is the value of k?

Answer 1

$3e^x\tan ydx+(1-e^x){\sec }^{2}ydy=0$
$\Rightarrow \frac{dy}{dx}=\frac{3e^x\cos y\sin y}{e^x-1}$

$\Rightarrow \csc y\sec y\frac{dy}{dx}=\frac{3e^x}{e^x-1}$
Integrating both sides, we get
$\int \csc y\sec y\frac{dy}{dx}dx=\int \frac{3e^x}{e^x-1}dx\Rightarrow -\log \left(\cos y\right)+\log \left(\sin y\right)=3\log (1-e^x)+c\Rightarrow {(1-e^x)}^3=C\tan y$

$\therefore k=3$