Cauchy's problem. Equation of mathematical physics 
$$U_{tt} = \Delta U + x^3 - 3xy^2$$
$$U|_{t=0} = e^x \cos y$$
$$U_t|_{t=0} = e^y \sin x$$

Help me, please, with solution of this equation. Can you prompt me algorithm to find the solution? I know, that I have to use Poisson's formula, but I can't integrate it.
 A: I figure out that the solution of this problem
\begin{align*}
&\text{PDE:}~~u_{tt}=u_{xx}+u_{yy}+x^3-3xy^2, \qquad (x,y)\in\mathbb{R}^2, t>0,\\
&\text{IC:}~~ u(x,y,0)=\operatorname{e}^x\cos(y),\quad u_t(x,y,0)=\operatorname{e}^y\sin(x),\qquad (x,y)\in\mathbb{R}^2
\end{align*}
is 
\begin{gather*}
u(x,y,t)=\operatorname{e}^x\cos(y)+t\cdot \operatorname{e}^y\sin(x)+\frac{t^2}{2}\cdot \big(x^3-3xy^2\big).
\end{gather*}
Although you can evaluate this solution by using directly the Poisson's formula, it is very tricky, and easy, to break it into three simple problems, for the given problem is linear, and then all the things are simple. 
Indeed, let 
\begin{gather*}
u(x,y,t)=v(x,y,t)+w(x,y,t)+q(x,y,t),
\end{gather*}
where $v$ solves
\begin{align*}
&v_{tt}(x,y,t)=v_{xx}(x,y,t)+v_{yy}(x,y,t),\\
&v(x,y,0)=\operatorname{e}^x\cos(y), \quad v_t(x,y,0)=0,
\end{align*}
and $w$ solves
\begin{align*}
&w_{tt}(x,y,t)=w_{xx}(x,y,t)+w_{yy}(x,y,t),\\
&w(x,y,0)=0, \quad v_t(x,y,0)=\operatorname{e}^y\sin(x),
\end{align*}
and as well $q$ solves
\begin{align*}
&q_{tt}(x,y,t)=q_{xx}(x,y,t)+x^3-3xy^2,\\
&q(x,y,0)=0, \quad q_t(x,y,0)=0.
\end{align*}
Observe that the three mappings 
\begin{gather*}
(x,y)\mapsto x^3-3xy^2,\\
(x,y)\mapsto \operatorname{e}^x\cos(y),\\
(x,y)\mapsto \operatorname{e}^y\sin(x)
\end{gather*}
are harmonic on $\mathbb{R}^2,$ so it is promising to assume that 
\begin{gather*}
\frac{\partial v(x,y,t)}{\partial t}\equiv0,\\
w(x,y,t)=w^*(x,y)\cdot t,\\
q(x,y,t)=q^*(t)\cdot(x^3-3xy^2). 
\end{gather*}
Then to determine $w^*, q^*$ is trivial. Thus we would obtain that 
$v(x,y,t)=\operatorname{e}^x\cos(y), $ and $w(x,y,t)=\operatorname{e}^y\sin(x)\cdot t,$ and as well $q(x,y,t)=\frac{t^2}{2}\cdot (x^3-3xy^2).$
