Question

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

• A. $\sin y=c(1-e^x)$ where $c$ is the constant of integration
• B. $\cos y=c(1-e^x)$ where $c$ is the constant of integration
• C. $\cot y=c(1-e^x)$ where $c$ is the constant of integration
• D. None of the above

Answer 1

The correct option is D None of the above

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

$\Rightarrow \frac{e^x}{1-e^x}dx+\frac{{\sec }^{2}y}{\tan y}dy=0$ ..... $\left(i\right)$

Now, to calculate $\int \frac{e^x}{1-e^x}dx$

Put $1-e^x=t\Rightarrow -e^{x}dx=dt$

$\therefore \int \frac{e^x}{1-e^x}dx=\int -\frac{dt}{t}=-\ln t=-\ln (1-e^x)$

Also, $\int \frac{{\sec }^{2}y}{\tan y}dy$

Put $p=\tan y\Rightarrow dp={\sec }^{2}ydy$

$\therefore \int \frac{{\sec }^{2}y}{\tan y}dy=\int \frac{dp}{p}=\ln p=\ln \left(\tan y\right)$

On integrating $\left(i\right)$, we get

$-\ln (1-e^x)+\ln \left(\tan y\right)=\ln k\Rightarrow \ln \left(\tan y\right)=\ln \left(k\right(1-e^x\left)\right)\Rightarrow \tan y=k(1-e^x)$

Hence, D is correct.