# PDE problem with initial and boundary conditions

I'm new to mathematica and I don't know what I do wrong, I have written the following PDE problem, with both initial and boundary (four in total)

eq = D[u[x, t], t, t] + D[u[x, t], t] - 2*D[u[x, t], x, x] == 0
bc = {u[0, t] == 0, u[2*Pi, t] == 0, u[x, 0] == 0, Derivative[0, 1][u][x, 0] == sin (2*x)}

DSolve[{eq, bc}, u[x,t], {t, x}]


I need it to solve the PDE, but it doesn't even try to make a solution... what to do. I hope you can help me

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Looking at initial conditions, hence $t$ can be removed for a second. Hence your IC is saying $u(x)=0$ and $u'(x)=\sin(2x)$. Right? How can this be possible for the function to be zero, yet its derivative be $\sin(2x)$ ? –  Nasser Oct 6 '13 at 20:43
There is also a syntax error: sin(2*x). –  Peltio Oct 6 '13 at 21:34
@Nasser, isn't the condition stating the the time derivative at t=0 is Sin[2x]? –  Peltio Oct 6 '13 at 21:46
@Peltio, yes. that is what I read it. But since $t$ is constant, therefore, it is just like saying $u'(x)=\sin(x)$. (when $t=0$). i.e. initially, everywhere, the derivative is $\sin(2x)$. But also, initially, everywhere, $u(x)=0$. This seems to contradict. (I could be wrong, have not done PDE for sometime). (it might also be just the syntax error you showed. I did not actually run the code :) –  Nasser Oct 6 '13 at 21:57
I was thinking of a function like this: -t Sin[t - 2x]. Time derivative at t=0 is Sin[2x], value at t=0 is 0. If I am not mistaken. –  Peltio Oct 6 '13 at 22:16