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NDA 2010
Exam
NDA
Year
2010
Subject
Mathematics
Chapter
Differential Equations

NDA 2010 · Mathematics · Differential Equations

What is the solution of the differential equation 3 e^x tan y d...

What is the solution of the differential equation 3ex3 e^x tan ydx+(1+ex)sec⁡2ydy=0y d x+\left(1+e^x\right) \sec 2 y d y=0 ? where cc is a constant of integration.

Options

  1. A(1+ex)tan⁡y=c\left(1+e^x\right) \tan y=c
  2. B(1+ex)3tan⁡y=c\left(1+e^x\right)^3 \tan y=c
  3. C(1+ex)2tan⁡y=c\left(1+e^x\right)^2 \tan y=c
  4. D(1+ex)sec⁡2y=c\left(1+e^x\right) \sec ^2 y=c

Correct answer

B. (1+ex)3tan⁡y=c\left(1+e^x\right)^3 \tan y=c

Step-by-step solution

(1+ex)sec⁡2ydy=−3extan⁡ydxsec⁡2ytan⁡ydy=−3ex1+exdx\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=−3log⁡(1+ex)+log⁡corlog⁡tan⁡y=log⁡c(1+ex)3ortan⁡y=c(1+ex)3⇒(1+ex)3tan⁡y=c\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
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