Define
$$y(x,s) = \int_0^{\infty} dt \, Y(x,t) \, e^{-s t}$$
Then, integrating by parts:
$$\int_0^{\infty} dt \, Y_t(x,t) \, e^{-s t} = -Y(x,0) + s y(x,s)$$
$$\int_0^{\infty} dt \, Y_{tt}(x,t) \, e^{-s t} = -Y_t(x,0) + s Y(x,0) + s^2 y(x,s)$$
Then using the initial conditions $Y(x,0)=Y_y(x,0)=0$, the PDE becomes the following ODE:
$$y''-2 s y' + s^2 y = 0$$
where $y(0,s)=0$ and $y(1,s)=f(s)$, where the prime represents derivative with respect to $x$, and where
$$f(s) = \int_0^{\infty} dt \, F(t) \, e^{-s t} $$
The general solution of the ODE is (I will not derive here)
$$y(x,s) = (A + B x) \, e^{s x}$$
Using the boundary conditions, we may find $A$ and $B$ and therefore the LT of the solution to the PDE:
$$y(x,s) = x \, f(s) \, e^{-s (1-x)} $$
We may find the inverse LT by convolution, as we know the individual LT's. The ILT of $f(s)$ is obviously $F(t)$ by definition, and the ILT of $e^{-s (1-x)}$ is $\delta(t-(1-x))$. Therefore, the ILT, and the solution to the equation, is
$$y(x,t) = x \int_0^t dt' \, F(t') \delta(t-t'-(1-x)) = x F(t-(1-x)) \theta(t-(1-x))$$
where $\theta$ is the Heaviside step function, which is necessary because the contribution to the integral from the $\delta$ function for $t < 1-x$ is zero (i.e., $t' \gt 0$ in the integral.)
We're considering the problem $$u_{xx}=u_{tt} ,\qquad t>0, \quad x\in \mathbb R$$ with $u(x,0) = 3x$ and $u_t(x,0) = \sin 2x$. The Laplace transform works almost the same way on PDEs as it does with ODEs, the main difference is that you have to pick which variable to perform the transform with. Since the conditions we're given are at $t=0$, take our transform with respect to the variable $t$ as the Laplace transform of a derivative can be expressed in terms of the functions initial value. I'll use the variable $\mathcal L$ to denote the Laplace transform operator with respect to the variable $t$, and I'll call our transformed variable $s$. Finally, I'll use capital letters to denote Laplace transform of it's lowercase counterpart. i.e. $$\mathcal L \{f(x,t) \} \equiv F(x,s)\equiv\int_0^\infty f(x,t)e^{-st} \,dt .$$
With our notation in order, let's apply the transform to the wave equation. We have that $$\mathcal L \{ u_{tt}(x,t) \} = s^2 \mathcal L \{u(x,t) \} -s u(x,0)- u_t(x,0)$$ which is the standard expansion of the Laplace transform of a derivative. Hopefully it is clear now why we are transforming with respect to $t$ as the functions $u(x,0)$ and $u_t(x,0)$ are already given to us. Plugging these in we have $$\mathcal L \{ u_{tt}(x,t) \} = s^2 U(x,s) - 3sx- \sin(2x).$$ On the other hand, we have that $$\mathcal L \{ u_{xx}(x,t) \} = \frac{\partial^2}{\partial x^2}L \{ u(x,t) \} = \frac{\partial^2 U}{\partial x^2}.$$ We can assume that the partial in $x$ commutes with the transform as this variable is independent to $t$. The wave equation then becomes $$\frac{\partial^2 U}{\partial x^2}=s^2 U - 3sx- \sin(2x)$$ which is effectively an ODE as the only derivatives are with respect to $x$. You can use your favourite method to solve this (variation of parameters, undetermined coefficients, etc..). I get that $$U(x,s) = c_1 e^{sx} + c_2 e^{-sx} + \frac{\sin(2x)}{s^2+4} + \frac{3x}{s}$$ for some constants $c_1,c_2$. Next we impose that our solution must be bounded as $x \to \pm \infty$, in particular that tells us that $c_1=c_2=0$ since the exponentials will be unbounded. So we have $$U(x,s) = \frac{\sin(2x)}{s^2+4} + \frac{3x}{s}.$$ All that we have to do now is invert the Laplace transform with respect to the variable $s$ $$u(x,t) = \mathcal L^{-1} \{ U(x,s) \} = \mathcal L^{-1} \left\{ \frac{\sin(2x)}{s^2+4} + \frac{3x}{s} \right\} = \sin(2x) \mathcal L^{-1} \left\{ \frac{1}{s^2+4} \right\} + 3x \mathcal L^{-1} \left\{ \frac 1s \right\} = \sin(2x) \left( \frac 12 \sin(2t) \right) + 3x (1).$$
Therefore $$u(x,t) = \frac 12 \sin(2x) \sin(2t)+3x.$$
Best Answer
I think initial conditions is $$u(x,0)=0,u_t(x,0)=\frac{1}{1+x^2}$$ Solve using d’Alembert’s formula http://www.math.usm.edu/lambers/mat606/lecture12.pdf
I get solution $$u(x,t) =\frac{\operatorname{atan}\left( x+t\right) -\operatorname{atan}\left( x-t\right) }{2}+e^{-t}+t-1$$