JEE Main 2019 — Math solution | Fundamenthol

JEE Main2019Math

Q.

Let y = y(x) be the solution of the differential equation, $\frac{d y}{d x} + y tan x = 2 x + x^{2} t an x, x \in (- \frac{π}{2}, \frac{π}{2})$ such that y(0) = 1.Then :

A
$y (\frac{π}{4}) + y (- \frac{π}{4}) = \frac{π ^{2}}{2} + 2$
B
$y (\frac{π}{4}) - y (- \frac{π}{4}) = 2$
C
$y^{'} (\frac{π}{4}) + y^{'} (- \frac{π}{4}) = - 2$
D
$y^{'} (\frac{π}{4}) - y^{'} (- \frac{π}{4}) = π - 2$

Solution

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