Use the Laplace transform to solve $ u_{tt}(x, t) − c^2u_{xx}(x, t) = 0$ Use the Laplace transform to solve the following initial boundary value problem for the wave equation $ u_{tt}(x, t) − c^2u_{xx}(x, t) = 0$
$u(x, 0) = 0$,
$u_{t}(x, 0) = 0  ∀x > 0$ , and
$u(0, t) =  \sin\left(2\pi t\right)$ $ 0 ≤ t < 1$, and $0$ otherwise.  (with $lim_{x→∞}u(x, t) = 0 $).
You may use standard results for the Laplace transform without proof. 
Then, sketch the solution profile at times $t = 0, t = 1$ and $ t = 2$. 
I'm a bit stuck on this question. Here's what I've done so far:
$ u_{tt}(x, t) − c^2u_{xx}(x, t) = 0$
Taking the laplace transform with respect to $t$ i got:
$s^2\tilde u - su(x,0)-u_{t}(x,0)- c^2u_{xx} = 0$
Substituting in the conditions = $s^2\tilde u (x,s) - c^2\tilde u_{xx}(x,s) = \tilde u_{xx} - s^2/c^2 \tilde u= 0$ which is a 2nd order constant coefficient ODE in x .
$\tilde u (x,s) = A(s)e^{sx/c} + B(s)e^{-sx/c}$.
Am i correct so far? Im not sure what to do with the initial conditions  $u(0,t) = sin(2πt)$. My guess is to the the Laplace transform of that. 
Would greatly appreciate the help.
Edit: I managed to complete the question, but im not certain it's correct:
Taking the Laplace transform of the 2 conditions:
(1) $u(0,t) = sin(2πt)$ which is $\tilde u (0,s) = 2π/(4π^2 +t^2)$
(2) $lim_{x→∞}u(x, t) = 0 $ gives $lim_{x→∞} \tilde u (x,s) =0$
So we have 2 constants A,B and 2 conditions. 2nd condition gives  $A (s) = 0 $ and 1st condition gives $B(s)= 2π/(4π^2 +t^2)$ which gives $\tilde u (x,s) = 2π/(4π^2 +t^2).e^{-sx/c}$
Therefore the solution is  $u(x,t) = H(t-x/c)sin2π(t-x/c)$. Where H is the heavy side function. For $x/c>t$ we have $0$ and for $t<x/c$ we have no solution. 
 A: For the boundary function, write
\begin{align} 
u(0,t) &= \sin(2\pi t)u(1-t) \\
& = \sin(2\pi t)\big[1 - u(t-1)\big] \\
&= \sin(2\pi t) - \sin(2\pi(t-1) + 2\pi)u(t-1) \\
&= \sin(2\pi t) - \sin(2\pi(t-1))u(t-1) 
\end{align}
where $u(t)$ is the unit step function
Using the time-shift property of the Laplace transform
$$ \tilde u(0,s) = \frac{2\pi}{s^2+4\pi^2} - e^{-s}\frac{2\pi}{s^2+4\pi^2} = \frac{2\pi(1-e^{-s})}{s^2+4\pi^2} $$
Hence
$$ \tilde u(x,s) = \frac{2\pi(1-e^{-s})}{s^2+4\pi^2}e^{-xs/c} $$
To transform back, write
$$ \tilde u(x,s) = \frac{2\pi}{s^2+4\pi^2}e^{-(x/c)s} - \frac{2\pi}{s^2+4\pi^2}e^{-(x/c+1)s} $$
Then use the shift theorem in reverse.
\begin{align}
u(x,t) &= \sin\big(2\pi(t-x/c)\big)u(t-x/c) - \sin\big(2\pi(t-x/c-1)\big)u(t-x/c-1) \\
&= \sin\big(2\pi(t-x/c)\big)u(t-x/c) - \sin\big(2\pi(t-x/c)\big)u(t-x/c-1) \\
&= \sin\big(2\pi(t-x/c)\big)\big[u(t-x/c) - u(t-x/c-1)\big] 
\end{align}
$$ \implies u(x,t) = \begin{cases} \sin\big(2\pi(t-x/c)\big), & x/c < t < x/c+1 \\
0, & t < x/c, \ t > x/c+1 \end{cases} $$
Edit: We can rewrite in terms of $x$
$$ u(x,t) = \begin{cases} -\sin\big(2\pi(x/c-t)\big), & c(t-1) < x < ct  \\
0, & x < c(t-1), \ x > ct \end{cases} $$
For $t=0$ $u$ is non-zero for $-c < x < 0$ but this isn't in the domain, so $u(x,0)=0, \forall x > 0$
For the other cases
\begin{align}
u(x,1) &= \begin{cases} -\sin\big(2\pi x/c\big), & 0 < x < c  \\
0, & x > c \end{cases} \\
u(x,2) &= \begin{cases} -\sin\big(2\pi x/c\big), & c < x < 2c  \\
0, & x < c, \ x > 2c \end{cases}
\end{align}
