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

Mathematics · Differential Equations

JEE Main 2024 — 4 April, Shift 2 — Question 18

Let $y=y(x)$ be the solution of the differential equation $\left(x^{2}+4\right)^{2} d y+\left(2 x^{3} y+8 x y-2\right) d x=0$ . If $y(0)=0$ ,

then $y(2)$ is equal to

$\frac{d y}{d x}+y\left(\frac{2 x^{3}+8 x}{\left(x^{2}+4\right)^{2}}\right)=\frac{2}{\left(x^{2}+4\right)^{2}}$

$\frac{d y}{d x}+y\left(\frac{2 x}{x^{2}+4}\right)=\frac{2}{\left(x^{2}+4\right)^{2}}$

$I F=e^{\int \frac{2 x}{x^{2}+4} d x}$

$\mathrm{IF}=\mathrm{x}^{2}+4$

$y \times\left(x^{2}+4\right)=\int \frac{2}{\left(x^{2}+4\right)^{2}} \times\left(x^{2}+4\right)$

$y\left(x^{2}+4\right)=2 \int \frac{d x}{x^{2}+2^{2}}$

$y\left(x^{2}+4\right)=\frac{2}{2} \tan ^{-1}\left(\frac{x}{2}\right)+c$

$0=0+\mathrm{c}=\mathrm{c}=0$

$y\left(x^{2}+4\right)=\tan ^{-1}\left(\frac{x}{2}\right)$

$y$ at $x=2$ ; $y(4+4)=\tan ^{-1}(1)$ ; $y(2)=\frac{\pi}{32}$

Answer key and solution verified before publishing.

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