Confused on why i'm missing this 2nd Order Diff EQ with complex roots

mr_coffee
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Hello everyone. I"m not getting this problem right. <insert sad face here>
Find y as a function of t if
6y'' + 33y = 0,
y(0) = 8, y'(0) = 5 .
y(t) =

hokay, here is my work, it is sloppy sorry. Can you see any obvious mistakes I made? Note: the sqraure root should be encompassing both the 11 and the 2 in 11/2.

http://img136.imageshack.us/img136/681/lastscan8wk.jpg I submitted this and it was wrong:>
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/90/5f593c3fcc529485c44d6db5c6c1ea1.png
 
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For your general solution, that is

y=Ae^{i\sqrt{\frac{11}{2}}t}+Be^{-i\sqrt{\frac{11}{2}}t}

put now instead

y=c_1\sin {\sqrt{\frac{11}{2}}t}+c_2\cos {\sqrt{\frac{11}{2}}t}
 
that comes from Euler's formula e^{i\alpha}=\cos\alpha+i\sin\alpha, substituted into the general solution with the e's, some constant renaming and some trickery to show that if the new solution (with the cos + isin stuff) is a complex solution to the DE, then the same thing without the i is also a solution. But that was mostly hand-waving.
 
Thanks for the help again benorin, but I submitted:
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/32/eb2319e057d3fc3b2523e3d817a55a1.png
Did i mess up finding my constants?
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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