1
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
2
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
$$1 + {\pi \over 2} + {{{\pi ^2}} \over 4}$$
3
JEE Main 2020 (Online) 3rd September Evening Slot
+4
-1
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
$${3 \over 2}\sqrt e$$
4
JEE Main 2020 (Online) 3rd September Morning Slot
+4
-1
The solution curve of the differential equation,

(1 + e-x)(1 + y2)$${{dy} \over {dx}}$$ = y2,

which passes through the point (0, 1), is :
A
y2 + 1 = y$$\left( {{{\log }_e}\left( {{{1 + {e^{ - x}}} \over 2}} \right) + 2} \right)$$
B
y2 + 1 = y$$\left( {{{\log }_e}\left( {{{1 + {e^{ x}}} \over 2}} \right) + 2} \right)$$
C
y2 = 1 + $${y{{\log }_e}\left( {{{1 + {e^{ - x}}} \over 2}} \right)}$$
D
y2 = 1 + $${y{{\log }_e}\left( {{{1 + {e^{ x}}} \over 2}} \right)}$$
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