JEE Main 2024 — Differential Equations Question with Solution
JEE Main 2024 (04 Apr Shift 2)
Question
Let be the solution of the differential equation . If , then is equal to
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Show full solutionCorrect option: A
Correct answer
A
Step-by-step explanation
$\begin{aligned}
& \frac{d y}{d x}+y\left(\frac{2 x^3+8 x}{\left(x^2+4\right)^2}\right)=\frac{2}{\left(x^2+4\right)^2} \\
& \frac{d y}{d x}+y\left(\frac{2 x}{x^2+4}\right)=\frac{2}{\left(x^2+4\right)^2} \\
& \text { IF }=e^{\int \frac{2 x}{x^2+4} d x} \\
& \text { IF }=x^2+4 \\
& y \times\left(x^2+4\right)=\int \frac{2}{\left(x^2+4\right)^2} \times\left(x^2+4\right) \\
& y\left(x^2+4\right)=2 \int \frac{d x}{x^2+2^2} \\
& y\left(x^2+4\right)=\frac{2}{2} \tan ^{-1}\left(\frac{x}{2}\right)+c \\
& 0=0+c=c=0 \\
& y\left(x^2+4\right)=\tan ^{-1}\left(\frac{x}{2}\right) \\
& y \text { at } x=2 \\
& y(4+4)=\tan ^{-1}(1) \\
& y(2)=\frac{\pi}{32}
\end{aligned}$
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This is a previous-year question from JEE Main 2024, covering the Differential Equations chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.