Answer
As given below
Work Step by Step
Take the derivatives of the function y. Remember to use chain rule.
$y'(x)=-2c_1\sin2x+2c_2\cos2x$
$y''(x)=-4c_1\cos2x-4c_2\cos2x$
Substitute these equations into the differential equation to get:
$y''+4y=0$
$-4c_1\cos2x-4c_2\cos2x +4(c_1\cos2x + c_2\sin2x)=0$
$-4c_1\cos2x-4c_2\cos2x+4c_1\cos2x + 4c_2\sin2x=0$
$0=0$
This satisfies the differential equation, so it must be a solution.
