$\begin{aligned}
& \frac{d y}{d x}+y\left(\frac{2 x^3+8 x}{\left(x^2+4\right)^2}\right)=\frac{2}{\left(x^2+4\right)^2} \
& \frac{d y}{d x}+y\left(\frac{2 x}{x^2+4}\right)=\frac{2}{\left(x^2+4\right)^2} \
& \text { IF }=e^{\int \frac{2 x}{x^2+4} d x} \
& \text { IF }=x^2+4 \
& y \times\left(x^2+4\right)=\int \frac{2}{\left(x^2+4\right)^2} \times\left(x^2+4\right) \
& y\left(x^2+4\right)=2 \int \frac{d x}{x^2+2^2} \
& y\left(x^2+4\right)=\frac{2}{2} \tan ^{-1}\left(\frac{x}{2}\right)+c \
& 0=0+c=c=0 \
& y\left(x^2+4\right)=\tan ^{-1}\left(\frac{x}{2}\right) \
& y \text { at } x=2 \
& y(4+4)=\tan ^{-1}(1) \
& y(2)=\frac{\pi}{32}
\end{aligned}$
Let y=y(x) be the solution of the differential equation (x^2+4 )^2 d y+ (2 x^3 y+8 x y-2 ) d x=0. If y(0)=0, then y(2) is equal to
JEE Main 2024 — Mathematics Calculus
Let y=y(x) be the solution of the differential equation (x2+4)2dy+(2x3y+8xy−2)dx=0. If y(0)=0, then y(2) is equal to
Held on 4 Apr 2024 · Verified 6 Jul 2026.
Options
- A
32π
- B
2π
- C
8π
- D
16π
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