Let y=y(x) be the solution curve of the differential equation y d y ~d x +2 x y=x^3 y, y(1)=0 . Then y( 3 ) is equal to :

Correct answer: D. (π)/(4) aligned & ^2 y (d y)/(d x)+2 x y y =x^3 y y & ^2 y (d y)/(d x)+2 x y=x^3 & y=t ^2 y (d y)/(d x)=(d t)/(d x) & (d t)/(d x)+2 x t=x^3, If =e^ ∫ 2 x d x =e^ x^2 & t x^ x^2 =∫ x^3 · e^ x^2 d x+c & x^2=Z t · e^Z=(1)/(2) ∫ e^Z · Z d Z=(1)/(2)[e^Z · Z-e^Z]+c & 2 y=(x^2-1)+2 e^ -x^2 & y(1)=0 c=0 y( 3 )=(π)/(4) aligned

JEE Main 20248 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 8 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2024 · 8 April · Evening Shift · Mathematics · Differential Equations

Let y=y(x) be the solution curve of the differential equation y d y...

Let

y=y(x)

be the solution curve of the differential equation

\sec y \frac{\mathrm{d} y}{\mathrm{~d} x}+2 x \sin y=x^3 \cos y, y(1)=0

. Then

y(\sqrt{3})

is equal to :

Options

A $\frac{\pi}{3}$
B $\frac{\pi}{6}$
C $\frac{\pi}{12}$
D $\frac{\pi}{4}$

Correct answer

D. $\frac{\pi}{4}$

Step-by-step solution

\begin{aligned} & \sec ^2 y \frac{d y}{d x}+2 x \sin y \text{ \sec y }=x^3 \cos y \text{ \sec y } \\ & \sec ^2 y \frac{d y}{d x}+2 x \tan y=x^3 \\ & \operatorname{\tan y}=t \Rightarrow \sec ^2 y \frac{d y}{d x}=\frac{d t}{d x} \\ & \frac{d t}{d x}+2 x t=x^3, \text{ If }=e^{\int 2 x d x}=e^{x^2} \\ & t x^{x^2}=\int x^3 \cdot e^{x^2} d x+c \\ & x^2=Z \Rightarrow t \cdot e^Z=\frac{1}{2} \int e^Z \cdot Z d Z=\frac{1}{2}\left[e^Z \cdot Z-e^Z\right]+c \\ & 2 \tan y=\left(x^2-1\right)+2 e^{-x^2} \\ & y(1)=0 \Rightarrow c=0 \Rightarrow y(\sqrt{3})=\frac{\pi}{4}\end{aligned}

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