Differential Equation using Laplace Transform

Tell me where I screwed up for 10 points! :D

I know how to solve this but just made a mistake somewhere and I can't figure out where.

x''(t) - 4x'(t) + 13x(t) = e^t, x(0) = 0, x(0) = 1

Here is the answer from WolframAlpha:
http://www.wolframalpha.com/input/?i=x%2…

Here's a link to my workings:
http://img27.imageshack.us/img27/2329/an…

Ignore the smaller writing halfway down on the right, I forgot to remove it before uploading.

Laplace transform for x''(t) is s² X(s) − s x(0) − x'(0) and not s² X(s) + s x(0) + x'(0)
Laplace transform for −4 x'(t) is −4s X(s) + 4 x(0) and not −4s X(s) − 4 x(0)
http://www.intmath.com/laplace-transform…

This means that 3rd and 4th blue line should be
(s² − 4s + 13) X(s) = 1/(s−1) + 1 -----> (not 1/(s−1) − 1)
X(s) = s / ((s−1)(s² − 4s + 13))

Partial fraction decomposition then gives:
X(s) = 1 / (10(s−1)) − (s−13) / (10(s² − 4s + 13))
X(s) = 1/10 [ 1/(s−1) − (s−2−11)/((s−2)² + 3²) ]
X(s) = 1/10 [ 1/(s−1) − (s−2)/((s−2)² + 3²) − (−11)/((s−2)² + 3²) ]
X(s) = 1/10 [ 1/(s−1) − (s−2)/((s−2)² + 3²) + 11/3 * 3/((s−2)² + 3²) ]

and this gives us
x(t) = 1/10 [ e^t − e^(2t) cos(3t) + 11/3 e^(2t) sin(3t) ]

Note that the final correct solution you show in red should have positive coefficient for the sin(3t) term, just as it does in the WolframAlpha answer in link above