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Differential Equations question

2021 · 22 Jul · Shift 2 · Q26

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JEE Main
Mathematics
Differential Equations
2021 · 22 Jul · Shift 2 · Q26

Differential Equations question

2021 · 22 Jul · Shift 2 · Q26

JEE MainMathematicsDifferential EquationsMCQ+4 / −1

Let y = y(x) be the solution of the differential equation

\cos e{c^2}xdy + 2dx = (1 + y\cos 2x)\cos e{c^2}xdx

, with

y\left( {{\pi \over 4}} \right) = 0

. Then, the value of

{(y(0) + 1)^2}

is equal to :

A
e1/2
B
e $-$ 1/2
C
e $-$ 1
D
e

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