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 :

Mathematics · Differential Equations

JEE Main 2024 — 4 April, Shift 1 — Question 4

If the solution $y=y(x)$ of the differential equation $\left(x^{4}+2 x^{3}+3 x^{2}+2 x+2\right) d y-\left(2 x^{2}+2 x+3\right) d x=0$ satisfies $y(-1)=-\frac{\pi}{4}$ , then $y(0)$ is equal to :

Option A:
$-\frac{\pi}{12}$
Option B:
$0$
Option C:
$\frac{\pi}{4}$
Correct
Option D:
$\frac{\pi}{2}$

Answer: C

Step-by-step solution

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

$y=\int \frac{\left(2 x^{2}+2 x+3\right)}{\left(x^{2}+1\right)\left(x^{2}+2 x+2\right)} d x$

$y=\int \frac{d x}{x^{2}+2 x+2}+\int \frac{d x}{x^{2}+1}$

$y=\tan ^{-1}(x+1)+\tan ^{-1} x+C$

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

$\frac{-\pi}{4}=0-\frac{\pi}{4}+\mathrm{C} \Rightarrow \mathrm{C}=0$

$\Rightarrow \mathrm{y}=\tan ^{-1}(\mathrm{x}+1)+\tan ^{-1} \mathrm{x}$

$y(0)=\tan ^{-1} 1=\frac{\pi}{4}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 4 April, Shift 1
Subject: Mathematics
Chapter: Differential Equations
Topic: Methods of solving a First Order,First Degree Differential

Let the solution y= y( x) of the differential equation d y/d x-y=1+4 sin x satisfy y(pi)=1 . Then y (pi/2 )+10 is equal to

Answer: C

Step-by-step solution

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