JEE Mains · Maths · STD 12 - 9. differential equations
Let \(\mathrm{y}=\mathrm{y}(\mathrm{x})\) be a solution curve of the differential equation \((y+1) \tan ^{2} x d x+\tan x d y+y d x=0\) \(x \in\left(0, \frac{\pi}{2}\right)\). If \(\lim _{x \rightarrow 0+} x y(x)=1\), then the value of \(\mathrm{y}\left(\frac{\pi}{4}\right)\) is :
- A \(-\frac{\pi}{4}\)
- B \(\frac{\pi}{4}-1\)
- C \(\frac{\pi}{4}+1\)
- D \(\frac{\pi}{4}\)
Answer & Solution
Correct Answer
(D) \(\frac{\pi}{4}\)
Step-by-step Solution
Detailed explanation
\((y+1) \tan ^{2} x \,d x+\tan x\, d y+y \,d x=0\) \(\frac{d y}{d x}+\frac{\sec ^{2} x}{\tan x} \cdot y=-\tan x\) \(\mathrm{IF}=\mathrm{e}^{\int \frac{\sec ^{2} x}{\tan \mathrm{x}} \mathrm{dx}}=\mathrm{e}^{\operatorname{lntan} \mathrm{x}}=\tan \mathrm{x}\)…
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