1
JEE Main 2025 (Online) 7th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let y = y(x) be the solution of the differential equation $(x^2 + 1)y' - 2xy = (x^4 + 2x^2 + 1)\cos x$,

$y(0) = 1$. Then $ \int\limits_{-3}^{3} y(x) \, dx $ is :

A

36

B

24

C

18

D

30

2
JEE Main 2025 (Online) 7th April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

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

$x\left(x^2+e^x\right) d y+\left(\mathrm{e}^x(x-2) y-x^3\right) \mathrm{d} x=0, x>0$, passing through the point $(1,0)$. Then $y(2)$ is equal to :

A
$\frac{2}{2+e^2}$
B
$\frac{4}{4-e^2}$
C
$\frac{4}{4+e^2}$
D
$\frac{2}{2-e^2}$
3
JEE Main 2025 (Online) 4th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

If a curve $y=y(x)$ passes through the point $\left(1, \frac{\pi}{2}\right)$ and satisfies the differential equation $\left(7 x^4 \cot y-\mathrm{e}^x \operatorname{cosec} y\right) \frac{\mathrm{d} x}{\mathrm{~d} y}=x^5, x \geq 1$, then at $x=2$, the value of $\cos y$ is :

A
$\frac{2 \mathrm{e}^2+\mathrm{e}}{64}$
B
$\frac{2 \mathrm{e}^2-\mathrm{e}}{64}$
C
$\frac{2 \mathrm{e}^2-\mathrm{e}}{128}$
D
$\frac{2 \mathrm{e}^2+\mathrm{e}}{128}$
4
JEE Main 2025 (Online) 3rd April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let $y=y(x)$ be the solution of the differential equation

$\frac{d y}{d x}+3\left(\tan ^2 x\right) y+3 y=\sec ^2 x, y(0)=\frac{1}{3}+e^3$. Then $y\left(\frac{\pi}{4}\right)$ is equal to :

A
$\frac{4}{3}$
B
$\frac{2}{3}+e^3$
C
$\frac{4}{3}+e^3$
D
$\frac{2}{3}$
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