1
JEE Main 2024 (Online) 6th April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let $$y=y(x)$$ be the solution of the differential equation $$\left(1+x^2\right) \frac{d y}{d x}+y=e^{\tan ^{-1} x}$$, $$y(1)=0$$. Then $$y(0)$$ is

A
$$\frac{1}{4}\left(e^{\pi / 2}-1\right)$$
B
$$\frac{1}{2}\left(1-e^{\pi / 2}\right)$$
C
$$\frac{1}{4}\left(1-e^{\pi / 2}\right)$$
D
$$\frac{1}{2}\left(e^{\pi / 2}-1\right)$$
2
JEE Main 2024 (Online) 5th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

The differential equation of the family of circles passing through the origin and having centre at the line $$y=x$$ is :

A
$$\left(x^2-y^2+2 x y\right) \mathrm{d} x=\left(x^2-y^2+2 x y\right) \mathrm{d} y$$
B
$$\left(x^2+y^2-2 x y\right) \mathrm{d} x=\left(x^2+y^2+2 x y\right) \mathrm{d} y$$
C
$$\left(x^2+y^2+2 x y\right) \mathrm{d} x=\left(x^2+y^2-2 x y\right) \mathrm{d} y$$
D
$$\left(x^2-y^2+2 x y\right) \mathrm{d} x=\left(x^2-y^2-2 x y\right) \mathrm{d} y$$
3
JEE Main 2024 (Online) 5th April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

If $$y=y(x)$$ is the solution of the differential equation $$\frac{\mathrm{d} y}{\mathrm{~d} x}+2 y=\sin (2 x), y(0)=\frac{3}{4}$$, then $$y\left(\frac{\pi}{8}\right)$$ is equal to :

A
$$\mathrm{e}^{-\pi / 8}$$
B
$$\mathrm{e}^{\pi / 4}$$
C
$$\mathrm{e}^{-\pi / 4}$$
D
$$\mathrm{e}^{\pi / 8}$$
4
JEE Main 2024 (Online) 4th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

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

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