JEE Main 2019 (Online) 10th April Evening Slot | Differential Equations Question 162 | Mathematics

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 :

$$y\left( {{\pi \over 4}} \right) - y\left( { - {\pi \over 4}} \right) = \sqrt 2 $$

$$y'\left( {{\pi \over 4}} \right) - y'\left( { - {\pi \over 4}} \right) = \pi - \sqrt 2 $$

$$y\left( {{\pi \over 4}} \right) + y\left( { - {\pi \over 4}} \right) = {{{\pi ^2}} \over 2} + 2$$

$$y'\left( {{\pi \over 4}} \right) + y'\left( { - {\pi \over 4}} \right) = - \sqrt 2 $$

JEE Main 2019 (Online) 10th April Morning Slot

MCQ (Single Correct Answer)

-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 :

$${1 \over 2} - e$$

$$e - 2$$

$$2 + {1 \over e}$$

$${1 \over e} - 2$$

JEE Main 2019 (Online) 9th April Evening Slot

MCQ (Single Correct Answer)

-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 :-

$$ - {{{\pi ^2}} \over {2 }}$$

$$ - {{{\pi ^2}} \over {4\sqrt 3 }}$$

$$ {{{\pi ^2}} \over {2\sqrt 3 }}$$

$$ - {{{\pi ^2}} \over {2\sqrt 3 }}$$

JEE Main 2019 (Online) 9th April Morning Slot

MCQ (Single Correct Answer)

-1

The solution of the differential equation

$$x{{dy} \over {dx}} + 2y$$ = x² (x $$ \ne $$ 0) with y(1) = 1, is :

$$y = {4 \over 5}{x^3} + {1 \over {5{x^2}}}$$

$$y = {3 \over 4}{x^2} + {1 \over {4{x^2}}}$$

$$y = {{{x^2}} \over 4} + {3 \over {4{x^2}}}$$

$$y = {{{x^3}} \over 5} + {1 \over {5{x^2}}}$$