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The answer to this is:

$ y(x) = c_2 \sin(2x) + c_1 \cos(2x)$ [1]

However, if I work this out.

$y'' + 4y = 0$ becomes $r^2 + 4r = 0$ which I believes gives you an entirely real root. If its not imaginary how can the solution be in in such form [1]

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7  
Your auxiliary equation should actually be $r^2+4=0$. –  Jp McCarthy Feb 28 '13 at 23:12
    
thank you . Why don't you post this as an answer and I'll mark it as correct. –  40Plot Feb 28 '13 at 23:15
    
Did you create a tag just for this question? –  Daniel Robert-Nicoud Feb 28 '13 at 23:15
    
I replaced your tag with the appropriate one we already had. (You accidentally added an extra "f" in there, which is why the tag search didn't find it for you.) –  Cameron Buie Feb 28 '13 at 23:18
2  
The general solution is "really" $Ce^{2ix}+De^{-2ix}$, where $C$ and $D$ range over the complex numbers. But for those who insist on living in the real world, we express it in terms of sines and cosines. –  André Nicolas Feb 28 '13 at 23:27

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