If the solution y=y(x) of the differential equation (x^4+2 x^3+3 x^2+2 x+2) d y-(2 x^2+2 x+3) d x=0 satisfies y(-1)=-(π)/(4) , then y(0) is equal to :

Correct answer: D. (π)/(4) aligned & ∫ d y=∫ ((2 x^2+2 x+3))/(x^4+2 x^3+3 x^2+2 x+2) d x & y=∫ ((2 x^2+2 x+3))/((x^2+1)(x^2+2 x+2)) d x & y=∫ (d x)/(x^2+2 x+2)+∫ (d x)/(x^2+1) & y= ^ -1 (x+1)+ ^ -1 x+C & y(-1)=(-π)/(4) & (-π)/(4)=0-(π)/(4)+C C=0 & y= ^ -1 (x+1)+ ^ -1 x & y(0)= ^ -1 1=(π)/(4) aligned

JEE Main 20244 AprilMorning Shift

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

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

If the solution y=y(x) of the differential equation (x^4+2 x^3+3...

If the solution

y=y(x)

of the differential equation \left(x^4+2 x^3+3 x^2+2 x+2\right) \mathrm{d} y-\left(2 x^2+2 x+3\right) \mathrm{d} x=0

satisfies

y(-1)=-\frac{\pi}{4}

, then

y(0) is equal to :

Options

A $\frac{\pi}{2}$
B $-\frac{\pi}{2}$
C0
D $\frac{\pi}{4}$

Correct answer

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

Step-by-step solution

\begin{aligned} & \int d y=\int \frac{\left(2 x^2+2 x+3\right)}{x^4+2 x^3+3 x^2+2 x+2} d x \\ & y=\int \frac{\left(2 x^2+2 x+3\right)}{\left(x^2+1\right)\left(x^2+2 x+2\right)} d x \\ & y=\int \frac{d x}{x^2+2 x+2}+\int \frac{d x}{x^2+1} \\ & y=\tan ^{-1}(x+1)+\tan ^{-1} x+C \\ & y(-1)=\frac{-\pi}{4} \\ & \frac{-\pi}{4}=0-\frac{\pi}{4}+C \Rightarrow C=0 \\ & \Rightarrow y=\tan ^{-1}(x+1)+\tan ^{-1} x \\ & y(0)=\tan ^{-1} 1=\frac{\pi}{4}\end{aligned}

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