JEE Math Practice Question

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 :

1. A.$$y\left( {{\pi \over 4}} \right) - y\left( { - {\pi \over 4}} \right) = \sqrt 2 $$
2. B.$$y'\left( {{\pi \over 4}} \right) - y'\left( { - {\pi \over 4}} \right) = \pi - \sqrt 2 $$
3. C.$$y\left( {{\pi \over 4}} \right) + y\left( { - {\pi \over 4}} \right) = {{{\pi ^2}} \over 2} + 2$$
4. D.$$y'\left( {{\pi \over 4}} \right) + y'\left( { - {\pi \over 4}} \right) = - \sqrt 2 $$