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Alright, so this is what I've done so far:

  1. First, form the characteristic quadratic equation, and solve it.
  2. I solved it using the quadratic formula to get solutions

$-\frac{1}{2} \pm i\frac{\sqrt{3}}{2}$

  1. Then the answer should be $y(x) = e^{x/2}(A\cos(\sqrt{3}x/2) + B\sin(\sqrt{3}x/2))$ right?
  2. Webworks gives me two blanks to fill in: It has $y(x) = C_1$ ____ + $C_2$ _____ where $C_1$ and $C_2$ are constants.
  3. I just distributed the $e^{x/2}$ and put in $e^{x/2}\cos(\sqrt{3}x/2)$ and the same thing except replacing $\cos x$ with $\sin x$. I also tried swapping the answers. Also, I found out it doesn't accept answers with the imaginary unit in them.
  4. Any idea what format my professor could want the answers in? Am I even doing the problem correctly?

Any help is greatly appreciated! I'll respond quickly

The comment below just solved my problem I believe, thank you!

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    ^^^ I was really close to being correct, J. Mohr found out the problem. The e^(x/2) needs to be changed to e^(-x/2). Oh negative signs, how I love you2011-08-04

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Since the roots of characteristic equation are $-\frac{1}{2} \pm i\frac{\sqrt{3}}{2}$, the general solution in WeBWork-friendly form is $y(x) = C_1 e^{-x/2}\cos(\sqrt{3}x/2) + C_2 e^{-x/2}\sin(\sqrt{3}x/2))$ (Getting the question off the unanswered list.)