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

Correct answer: A. (π)/(32) aligned & (d y)/(d x)+y((2 x^3+8 x)/((x^2+4)^2))=(2)/((x^2+4)^2) & (d y)/(d x)+y((2 x)/(x^2+4))=(2)/((x^2+4)^2) & IF =e^ ∫ (2 x)/(x^2+4) d x & IF =x^2+4 & y ×(x^2+4)=∫ (2)/((x^2+4)^2) ×(x^2+4) & y(x^2+4)=2 ∫ (d x)/(x^2+2^2) & y(x^2+4)=(2)/(2) ^ -1 ((x)/(2))+c & 0=0+c=c=0 & y(x^2+4)= ^ -1 ((x)/(2)) & y at x=2 & y(4+4)= ^ -1 (1) & y(2)=(π)/(32) aligned

JEE Main 20244 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 4 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2024 · 4 April · Evening Shift · Mathematics · Differential Equations

Let y=y(x) be the solution of the differential equation (x^2+4)^2 d...

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

Options

A $\frac{\pi}{32}$
B ${2}{\pi}$
C $\frac{\pi}{8}$
D $\frac{\pi}{16}$

Correct answer

A. $\frac{\pi}{32}$

Step-by-step solution

\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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