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 :

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 :

$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 $

Solution

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