1
JEE Main 2020 (Online) 5th September Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
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
2
JEE Main 2020 (Online) 4th September Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
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
3
JEE Main 2020 (Online) 4th September Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
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
$$1 + {\pi \over 2} + {{{\pi ^2}} \over 4}$$
4
JEE Main 2020 (Online) 3rd September Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
If x3dy + xy dx = x2dy + 2y dx; y(2) = e and
x > 1, then y(4) is equal to :
A
$${{\sqrt e } \over 2}$$
B
$${1 \over 2} + \sqrt e $$
C
$${3 \over 2} + \sqrt e $$
D
$${3 \over 2}\sqrt e $$
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