1
JEE Main 2019 (Online) 10th April Evening Slot
+4
-1
Let y = y(x) be the solution of the differential equation,
$${{dy} \over {dx}} + y\tan x = 2x + {x^2}\tan x$$, $$x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$$, such that y(0) = 1. Then :
A
$$y\left( {{\pi \over 4}} \right) - y\left( { - {\pi \over 4}} \right) = \sqrt 2$$
B
$$y'\left( {{\pi \over 4}} \right) - y'\left( { - {\pi \over 4}} \right) = \pi - \sqrt 2$$
C
$$y\left( {{\pi \over 4}} \right) + y\left( { - {\pi \over 4}} \right) = {{{\pi ^2}} \over 2} + 2$$
D
$$y'\left( {{\pi \over 4}} \right) + y'\left( { - {\pi \over 4}} \right) = - \sqrt 2$$
2
JEE Main 2019 (Online) 10th April Morning Slot
+4
-1
If y = y(x) is the solution of the differential equation
$${{dy} \over {dx}} = \left( {\tan x - y} \right){\sec ^2}x$$, $$x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$$,
such that y (0) = 0, then $$y\left( { - {\pi \over 4}} \right)$$ is equal to :
A
$${1 \over 2} - e$$
B
$$e - 2$$
C
$$2 + {1 \over e}$$
D
$${1 \over e} - 2$$
3
JEE Main 2019 (Online) 9th April Evening Slot
+4
-1
If $$\cos x{{dy} \over {dx}} - y\sin x = 6x$$, (0 < x < $${\pi \over 2}$$)
and $$y\left( {{\pi \over 3}} \right)$$ = 0 then $$y\left( {{\pi \over 6}} \right)$$ is equal to :-
A
$$- {{{\pi ^2}} \over {2 }}$$
B
$$- {{{\pi ^2}} \over {4\sqrt 3 }}$$
C
$${{{\pi ^2}} \over {2\sqrt 3 }}$$
D
$$- {{{\pi ^2}} \over {2\sqrt 3 }}$$
4
JEE Main 2019 (Online) 9th April Morning Slot
+4
-1
The solution of the differential equation

$$x{{dy} \over {dx}} + 2y$$ = x2 (x $$\ne$$ 0) with y(1) = 1, is :
A
$$y = {4 \over 5}{x^3} + {1 \over {5{x^2}}}$$
B
$$y = {3 \over 4}{x^2} + {1 \over {4{x^2}}}$$
C
$$y = {{{x^2}} \over 4} + {3 \over {4{x^2}}}$$
D
$$y = {{{x^3}} \over 5} + {1 \over {5{x^2}}}$$
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