Solving inhomogeneous PDEs when you can't separate variables 
$$4+U_y - U_{xy} =0, \quad U(x,0)=0, \quad U_y(0,y)=3y^2 .$$ 

Usually I can solve these kind of problems with separation of variables, so I tried
$$ U=XY, \quad U_y=XY',  \quad U_{xy}=X'Y' $$
$$ \Rightarrow 4+XY'-X'Y'=0$$
But, I can never separate variables X and Y at different sides of equation. And I dont know how else I can solve this. 
 A: Hint If we set $V := U_y$, we may rewrite the equation as an "upper triangular" system of what are effectively o.d.e.s:
$$\left\{
\begin{array}{rcl}
U_y &=& V \\
V_x &=& V + 4
\end{array}
\right.$$
The second equation has general solution $$V(x, y) = g(y) e^x - 4 .$$
A: If the method of separation of variables is required, first you have to transform the inhomogeneous PDE to an homogeneous PDE. Any particular solution can be used.
For example, obviously $U=-4y$ is a particular solution. So, let $U(x,y)=-4y+V(x,y)$
$U_y=-4+V_y \quad\to\quad U_{xy}=V_{xy}$
$4+U_y-U_{xy}=0 \quad\to\quad 4+(-4+V_y)+V_{xy}=0$
$$V_y+V_{xy}=0$$
The new PDE is homogeneous and the separation of variables can be used.
Of course, the new conditions are : $\begin{cases} U(x,0)=0 \quad\to\quad V(x,0)=0 \\ U_y(0,y)=3y^2 \quad\to\quad V_y(0,y)=4+3y^2 \end{cases}$
You can take it from here.
NOTE :
The PDE $4+U_y-U_{xy}=0$ can be integrated directly :
$$4y+U-U_x=f(x)$$
any differentiable function $f(x)$.
The integration relatively to $x$ leads to the general solution :
$$U=-4y+g(x)+h(y)e^x$$
any differentiable functions $g(x)$ and $h(y)$.
Condition $U(x,0)=0\quad\to\quad g(x)+h(0)e^x=0 \quad\to\quad g(x)=-h(0)e^x$
$$U=-4y+\left(h(y)-h(0)\right)e^x$$
Condition $U_y(0,y)=3y^2 \quad\to\quad -4+h'(y)=3y^2\quad\to\quad h(y)=4y+y^3+c$
$c=h(0)$ Bringing it into $U=-4y+\left(h(y)-h(0)\right)e^x$ and after simplification :
$$U(x,y)=-4y+\left(4y+y^3\right)e^x$$
which is the solution of the PDE according to the two conditions.
