Let y = y(x) be the solution of the differential equation, dy/dx + ytan x = 2x + x^2 tan x, x ∈ (-(π)/2, (π)/2 ), such that y(0) = 1. Then

y ((π)/4 ) - y (-(π)/4 ) = π - √(2)

y' ((π)/4 ) + y' (-(π)/4 ) = -√(2)

y ((π)/4 ) + y (-(π)/4 ) = (π^2)/2 + 2

y' ((π)/4 ) - y' (-(π)/4 ) = √(2)

Let y = y(x) be the solution of the differential equation,… — JEE Main 2019 Mathematics PYQ

JEE Main 2019 · 10 April · Shift II

Let

y = y(x)

be the solution of the differential equation,

\frac{dy}{dx} + y\tan x = 2x + x^2 \tan x

x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

, such that

y(0) = 1

. Then

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

Show answer

Correct option: A

[JEE Main 2019 · 10 April · Shift II]

Practise Differential Equations the way it is asked

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