Q246 / 270 · NDA 2019 (Phase 2) · Mathematics

The differential equation which represents the family of curves given by $\tan y=c\left(1-e^x\right)$ is

  • A
    $e^x \tan y d x+\left(1-e^x\right) d y=0$
  • Correct answer
    B
    $e^x \tan y d x+\left(1-e^x\right) \sec ^2 y d y=0$
  • C
    $e^x\left(1-e^x\right) d x+\tan y d y=0$
  • D
    $e^x \tan y d y+\left(1-e^x\right) d x=0$

Solution

$\begin{aligned} & \text { } \tan y=c\left(1-e^x\right) \\ & \Rightarrow \sec ^2 y \frac{d y}{d x}=c\left(-e^x\right) \\ & \Rightarrow c=\frac{1}{e^x} \sec ^2 y \frac{d y}{d x} \\ & \therefore \tan y=\frac{-1}{e^x} \sec ^2 y \cdot \frac{d y}{d x}\left(1-e^x\right) \\ & \Rightarrow e^x \cdot \tan y=\sec ^2 y\left(1-e^x\right) \frac{d y}{d x} \\ & \Rightarrow e^x \tan y d x+\left(1-e^x\right) \sec ^2 y d y=0\end{aligned}$

Hence, the correct answer is option: 'B'.

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