Q197 / 270 · NDA 2010 (Phase 1) · Mathematics
What is the solution of the differential equation $3 e^x$ tan $y~ d x+\left(1+e^x\right) \sec 2 y~ d y=0$ ?
where $c$ is a constant of integration.
- A$\left(1+e^x\right) \tan y=c$
- Correct answerB$\left(1+e^x\right)^3 \tan y=c$
- C$\left(1+e^x\right)^2 \tan y=c$
- D$\left(1+e^x\right) \sec ^2 y=c$
Solution
$\begin{aligned}
& \left(1+e^x\right) \sec ^2 y d y=-3 e^x \tan y d x \\
& \frac{\sec ^2 y}{\tan y} d y=\frac{-3 e^x}{1+e^x} d x
\end{aligned}$
Using log on both side
or $\log \tan y=-3 \log \left(1+e^x\right)+\log c$
or $\log \tan y=\log \frac{c}{\left(1+e^x\right)^3}$
or $\tan y=\frac{c}{\left(1+e^x\right)^3} \Rightarrow\left(1+e^x\right)^3 \tan y=c$
Hence, the correct answer is option: 'B'.