Laplace Transform of the Wave Equation I am given a damped wave equation
$u_{tt}(t,x)+2u_t(t,x)=u_{xx}(t,x); \forall t>0$
Now I know the laplace transform of this given the initial conditions,
$u(0,x)=\sin x, u_t(0,x)=0;$
is 
$\tilde u_{xx}(s,x)-(s^2+2s)\tilde u(s,x)+(2+s)\sin x=0$
Now to solve this, I will use method of second order linear homogenous with constant coefficients, however my question is how can solve the wave equation if I am not given any boundary conditions?
All help appreciated!
 A: Let $u=e^{-t}v$ ,
Then $u_t=e^{-t}v_t-e^{-t}v$
$u_{tt}=e^{-t}v_{tt}-e^{-t}v_t-e^{-t}v_t+e^{-t}v=e^{-t}v_{tt}-2e^{-t}v_t+e^{-t}v$
$u_x=e^{-t}v_x$
$u_{xx}=e^{-t}v_{xx}$
$\therefore e^{-t}v_{tt}-2e^{-t}v_t+e^{-t}v+2e^{-t}v_t-2e^{-t}v=e^{-t}v_{xx}$
$v_{tt}=v+v_{xx}$
Similar to Solve initial value problem for $u_{tt} - u_{xx} - u = 0$ using characteristics,
Consider $v(a,x)=f(x)$ and $v_t(a,x)=g(x)$ ,
Let $v(t,x)=\sum\limits_{n=0}^\infty\dfrac{(t-a)^n}{n!}\dfrac{\partial^nv(a,x)}{\partial t^n}$ ,
Then $v(t,x)=\sum\limits_{n=0}^\infty\dfrac{(t-a)^{2n}}{(2n)!}\dfrac{\partial^{2n}v(a,x)}{\partial t^{2n}}+\sum\limits_{n=0}^\infty\dfrac{(t-a)^{2n+1}}{(2n+1)!}\dfrac{\partial^{2n+1}v(a,x)}{\partial t^{2n+1}}$
$v_{tttt}=v_{tt}+v_{xxtt}=v+v_{xx}+v_{xx}+v_{xxxx}=v+2v_{xx}+v_{xxxx}$
Similarly, $\dfrac{\partial^{2n}v}{\partial t^{2n}}=\sum\limits_{k=0}^nC_k^n\dfrac{\partial^{2k}v}{\partial x^{2k}}$
$v_{ttt}=v_t+v_{xxt}$
$v_{ttttt}=v_{ttt}+v_{xxttt}=v_t+v_{xxt}+v_{xxt}+v_{xxxxt}=v_t+2v_{xxt}+v_{xxxxt}$
Similarly, $\dfrac{\partial^{2n+1}v}{\partial t^{2n+1}}=\sum\limits_{k=0}^nC_k^n\dfrac{\partial^{2k+1}v}{\partial x^{2k}\partial t}$
$\therefore v(t,x)=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{C_k^nf^{(2k)}(x)(t-a)^{2n}}{(2n)!}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{C_k^ng^{(2k)}(x)(t-a)^{2n+1}}{(2n+1)!}$
Hence $u(t,x)=e^{-t}\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{C_k^nf^{(2k)}(x)(t-a)^{2n}}{(2n)!}+e^{-t}\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{C_k^ng^{(2k)}(x)(t-a)^{2n+1}}{(2n+1)!}$
For the case of conditions specific at $u(0,x)=\sin x$ and $u_t(0,x)=0$ ,
the corresponding conditions of $v$ are $v(0,x)=\sin x$ and $v_t(0,x)=\sin x$
$\therefore u(t,x)=e^{-t}\sin x\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kC_k^nt^{2n}}{(2n)!}+e^{-t}\sin x\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kC_k^nt^{2n+1}}{(2n+1)!}$
