Q42 / 100 · BITSAT 2022 · Mathematics

$\left(\frac{d y}{d x}\right) \tan x=y \sec ^2 x+\sin x$, find general solution

  • Correct answer
    A
    $y=\tan x(\log |\operatorname{cosec} x-\cot x|+\cos x+c)$
  • B
    $y=\sec ^2 x+\tan x+c$
  • C
    $y=\log |\sec x+\tan x|+\operatorname{cosec} x+c$
  • D
    $y=\tan ^2 x+\sin x+c$

Solution

Given that $$ \begin{aligned} & \left(\frac{d y}{d x}\right) \tan x=y \sec ^2 x+\sin x \\ & \tan x\left(\frac{d y}{d x}\right)-y \sec ^2 x=\sin x \\ & \frac{d y}{d x}+\left(\frac{-\sec ^2 x}{\tan x}\right) y=\frac{\sin x}{\tan x} \\ & \text { Here, } P=-\frac{\sec ^2 x}{\tan x} \text { and } Q=\cos x \\ & \text { IF }=e^{\int P d x}=e^{\int \frac{-\sec ^ 2 x}{\tan x} d x} \\ & \text { IF }=e^{-\log (\tan x)}=e^{\log \cot x}=\cot x \\ & \text { Now, } y \times \mathrm{IF}=\int(Q \times \mathrm{IF}) d x+C \\ & y \cdot \cot x=\int \cos x \cdot \cot x+C \\ & y \cdot \cot x=\int \frac{1-\sin ^2 x}{\sin x}+C \\ & y \cdot \cot x=\log |\operatorname{cosec} x-\cot x|+\cos x+C \end{aligned} $$

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

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