1
JEE Main 2020 (Online) 5th September Evening Slot
+4
-1
Let y = y(x) be the solution of the differential equation
cosx$${{dy} \over {dx}}$$ + 2ysinx = sin2x, x $$\in$$ $$\left( {0,{\pi \over 2}} \right)$$.
If y$$\left( {{\pi \over 3}} \right)$$ = 0, then y$$\left( {{\pi \over 4}} \right)$$ is equal to :
A
$${1 \over {\sqrt 2 }} - 1$$
B
$${\sqrt 2 - 2}$$
C
$${2 - \sqrt 2 }$$
D
$${2 + \sqrt 2 }$$
2
JEE Main 2020 (Online) 5th September Morning Slot
+4
-1
If y = y(x) is the solution of the differential
equation $${{5 + {e^x}} \over {2 + y}}.{{dy} \over {dx}} + {e^x} = 0$$ satisfying y(0) = 1, then a value of y(loge13) is:
A
-1
B
1
C
0
D
2
3
JEE Main 2020 (Online) 4th September Evening Slot
+4
-1
The solution of the differential equation

$${{dy} \over {dx}} - {{y + 3x} \over {{{\log }_e}\left( {y + 3x} \right)}} + 3 = 0$$ is:

(where c is a constant of integration)
A
$$x - {1 \over 2}{\left( {{{\log }_e}\left( {y + 3x} \right)} \right)^2} = C$$
B
$$y + 3x - {1 \over 2}{\left( {{{\log }_e}x} \right)^2} = C$$
C
x – loge(y+3x) = C
D
x – 2loge(y+3x) = C
4
JEE Main 2020 (Online) 4th September Morning Slot
+4
-1
Let y = y(x) be the solution of the differential equation,
xy'- y = x2(xcosx + sinx), x > 0. if y ($$\pi$$) = $$\pi$$ then
$$y''\left( {{\pi \over 2}} \right) + y\left( {{\pi \over 2}} \right)$$ is equal to
A
$$1 + {\pi \over 2}$$
B
$$2 + {\pi \over 2} + {{{\pi ^2}} \over 4}$$
C
$$2 + {\pi \over 2}$$
D
$$1 + {\pi \over 2} + {{{\pi ^2}} \over 4}$$
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