Find the continuous solution to the initial value problem (second order)

Question

$y''+4y = f(x)$

$f(x)=\cases{\sin x, &$0\le x\le 2\pi$\\ x, &$x>2\pi$}$

with $y(0)=0 , y'(0)=0$

My attempt is ... general solution $r^2 = -4 , r = \pm 2i$

$y(x) = c_1e^{2i} + c_2e^{-2i} = c_1(\cos(2t) +i\sin(2t))+c_2\cos(-2t) +i\sin(-2t))$

substitution with $0$ and it gets strange.. $y(x=0) = 0, y'(x=0) = 0$

$c1, c2 = 0...?$

I searched this ..Find the continuous solution to the initial value problem So, I went directly to integrate equation...and it seems doesn't right.

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Answer 1

I think your solution must be $$y(x)=C_1\cos(2x)+C_2\sin(2x)+\frac{\sin(x)}{3}$$ Check it!