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Differential Equations question

2019 · 10 Apr · Shift 2 · Q34

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JEE Main
Mathematics
Differential Equations
2019 · 10 Apr · Shift 2 · Q34

Differential Equations question

2019 · 10 Apr · Shift 2 · Q34

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

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