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BITSAT 2022
Exam
BITSAT
Year
2022
Subject
Mathematics
Chapter
Differential Equations

BITSAT 2022 · Mathematics · Differential Equations

((d y)/(d x)) x=y ^2 x+ x , find general solution

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

Options

  1. Ay=tan⁡x(log⁡∣cosec⁡x−cot⁡x∣+cos⁡x+c)y=\tan x(\log |\operatorname{cosec} x-\cot x|+\cos x+c)
  2. By=sec⁡2x+tan⁡x+cy=\sec ^2 x+\tan x+c
  3. Cy=log⁡∣sec⁡x+tan⁡x∣+cosec⁡x+cy=\log |\sec x+\tan x|+\operatorname{cosec} x+c
  4. Dy=tan⁡2x+sin⁡x+cy=\tan ^2 x+\sin x+c

Correct answer

A. y=tan⁡x(log⁡∣cosec⁡x−cot⁡x∣+cos⁡x+c)y=\tan x(\log |\operatorname{cosec} x-\cot x|+\cos x+c)

Step-by-step solution

Given that (dydx)tan⁡x=ysec⁡2x+sin⁡xtan⁡x(dydx)−ysec⁡2x=sin⁡xdydx+(−sec⁡2xtan⁡x)y=sin⁡xtan⁡x Here, P=−sec⁡2xtan⁡x and  Q=cos⁡x IF =e∫Pdx=e∫−sec⁡2xtan⁡xdx IF =e−log⁡(tan⁡x)=elog⁡cot⁡x=cot⁡x Now, y×IF=∫(Q×IF)dx+Cy⋅cot⁡x=∫cos⁡x⋅cot⁡x+Cy⋅cot⁡x=∫1−sin⁡2xsin⁡x+Cy⋅cot⁡x=log⁡∣cosec⁡x−cot⁡x∣+cos⁡x+C\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{\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}
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