1
JEE Main 2021 (Online) 26th August Morning Shift
+4
-1
Let y = y(x) be a solution curve of the differential equation $$(y + 1){\tan ^2}x\,dx + \tan x\,dy + y\,dx = 0$$, $$x \in \left( {0,{\pi \over 2}} \right)$$. If $$\mathop {\lim }\limits_{x \to 0 + } xy(x) = 1$$, then the value of $$y\left( {{\pi \over 4}} \right)$$ is :
A
$$- {\pi \over 4}$$
B
$${\pi \over 4} - 1$$
C
$${\pi \over 4} + 1$$
D
$${\pi \over 4}$$
2
JEE Main 2021 (Online) 27th July Evening Shift
+4
-1
Let y = y(x) be the solution of the differential

equation (x $$-$$ x3)dy = (y + yx2 $$-$$ 3x4)dx, x > 2. If y(3) = 3, then y(4) is equal to :
A
4
B
12
C
8
D
16
3
JEE Main 2021 (Online) 27th July Morning Shift
+4
-1
Let y = y(x) be solution of the differential equation

$${\log _{}}\left( {{{dy} \over {dx}}} \right) = 3x + 4y$$, with y(0) = 0.

If $$y\left( { - {2 \over 3}{{\log }_e}2} \right) = \alpha {\log _e}2$$, then the value of $$\alpha$$ is equal to :
A
$$- {1 \over 4}$$
B
$${1 \over 4}$$
C
$$2$$
D
$$- {1 \over 2}$$
4
JEE Main 2021 (Online) 25th July Evening Shift
+4
-1
Let y = y(x) be the solution of the differential

equation xdy = (y + x3 cosx)dx with y($$\pi$$) = 0, then $$y\left( {{\pi \over 2}} \right)$$ is equal to :
A
$${{{\pi ^2}} \over 4} + {\pi \over 2}$$
B
$${{{\pi ^2}} \over 2} + {\pi \over 4}$$
C
$${{{\pi ^2}} \over 2} - {\pi \over 4}$$
D
$${{{\pi ^4}} \over 4} - {\pi \over 2}$$
EXAM MAP
Medical
NEET
GATE CSEGATE ECEGATE EEGATE MEGATE CEGATE PIGATE IN
CBSE
Class 12