There are several approaches that can solve this inhomogeneous linear ODE.
Approach $1$: classical variables transformations
Let $\begin{cases}p=x+y\\q=x-y\end{cases}$ ,
Then $\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial p}\dfrac{\partial p}{\partial x}+\dfrac{\partial u}{\partial q}\dfrac{\partial q}{\partial x}=\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}$
$\dfrac{\partial^2u}{\partial x^2}=\dfrac{\partial}{\partial x}\left(\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}\right)=\dfrac{\partial}{\partial p}\left(\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}\right)\dfrac{\partial p}{\partial x}+\dfrac{\partial}{\partial q}\left(\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}\right)\dfrac{\partial q}{\partial x}=\dfrac{\partial^2u}{\partial p^2}+\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial q^2}=\dfrac{\partial^2u}{\partial p^2}+2\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial q^2}$
$\dfrac{\partial u}{\partial y}=\dfrac{\partial u}{\partial p}\dfrac{\partial p}{\partial y}+\dfrac{\partial u}{\partial q}\dfrac{\partial q}{\partial y}=\dfrac{\partial u}{\partial p}-\dfrac{\partial u}{\partial q}$
$\dfrac{\partial^2u}{\partial y^2}=\dfrac{\partial}{\partial y}\left(\dfrac{\partial u}{\partial p}-\dfrac{\partial u}{\partial q}\right)=\dfrac{\partial}{\partial p}\left(\dfrac{\partial u}{\partial p}-\dfrac{\partial u}{\partial q}\right)\dfrac{\partial p}{\partial y}+\dfrac{\partial}{\partial q}\left(\dfrac{\partial u}{\partial p}-\dfrac{\partial u}{\partial q}\right)\dfrac{\partial q}{\partial y}=\dfrac{\partial^2u}{\partial p^2}-\dfrac{\partial^2u}{\partial pq}-\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial q^2}=\dfrac{\partial^2u}{\partial p^2}-2\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial q^2}$
$\therefore\dfrac{\partial^2u}{\partial p^2}+2\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial q^2}-\left(\dfrac{\partial^2u}{\partial p^2}-2\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial q^2}\right)=\left(\dfrac{p+q}{2}\right)^2\left(\dfrac{p-q}{2}\right)^{\frac{p-q}{2}}$
$4\dfrac{\partial^2u}{\partial pq}=\dfrac{(p+q)^2(p-q)^{\frac{p-q}{2}}}{4\times2^{\frac{p-q}{2}}}$
$\dfrac{\partial^2u}{\partial pq}=\dfrac{(p+q)^2(p-q)^{\frac{p-q}{2}}}{16\times2^{\frac{p-q}{2}}}$
$u(p,q)=f(p)+g(q)+\dfrac{1}{16}\int_b^q\int_a^p\dfrac{(s+t)^2(s-t)^{\frac{s-t}{2}}}{2^{\frac{s-t}{2}}}ds~dt$
$u(x,y)=f(x+y)+g(x-y)+\dfrac{1}{16}\int_b^{x-y}\int_a^{x+y}\dfrac{(s+t)^2(s-t)^{\frac{s-t}{2}}}{2^{\frac{s-t}{2}}}ds~dt$
Approach $2$: Duhamel's principle
With reference to http://en.wikipedia.org/wiki/Duhamel%27s_principle#Wave_equation and http://en.wikipedia.org/wiki/Wave_equation#Inhomogeneous_wave_equation_in_one_dimension, we have $u(x,y)=f(x+y)+g(x-y)+\dfrac{1}{2}\int_0^x\int_{y-x-s}^{y+x-s}s^2t^t~dt~ds$ or $u(x,y)=f(x+y)+g(x-y)-\dfrac{1}{2}\int_0^y\int_{x-y-t}^{x+y-t}s^2t^t~ds~dt$
Approach $3$: See achille hui's answer