To solve a non-homogeneous linear PDE To solve a non-homogeneous linear PDE $\displaystyle \frac{\partial^2 z}{\partial x^2}+\frac{\partial^2 z}{\partial x \, \partial y}+\frac{\partial z}{\partial y}-z=e^{-x}$
My Attempt: 
Putting $\displaystyle D=\frac{\partial z}{\partial x}$ and$\displaystyle D’=\frac{\partial z}{\partial y}$, we have $\displaystyle (D^2+DD’+D’-1)z=e^{-x}$
$$(D+1)(D+D’-1)=e^{-x}$$
C.F. $\displaystyle \implies z=e^{-x}\phi_1(y) +e^x\phi_2(y-x))$
$$ \mathrm{P.I.}=\frac{1}{f(D,D')}e^{-x}=\frac{1}{(1+1)(D+D'-1)}$$
$$ P.I.=\frac{1}{2(D+D'-1)}e^{-x}$$  (Putting $D=1$ into first factor)
Putting $D=1$ in next factor will get us zero, hence, we move to 
$$ \frac{1}{(D-mD')}F(x,y)=\int F(x,c-mx)dx=\int e^{-x}dx=-e^{-x}$$
So we finally get, C.F. + P.I. $=\displaystyle e^{-x}\phi_1(y) +e^x\phi_2(y-x))- \frac{e^{-x}}{2}$
The given answer:  $$ e^{-x}\phi_1(y) +e^x\phi_2(y-x))- \frac{xe^{-x}}{2}$$
Where does the x come in here as per the given answer? Where am I going wrong?
 A: The pde 
\begin{align}
\psi_{xx} + \psi_{xy} + \psi_{y} - \psi = e^{-x}
\end{align}
can be solved by first making the change 
\begin{align}
\psi(x,y) = f(x,y) - \frac{x}{2} e^{-x}
\end{align}
which leads to the equation
\begin{align}
f_{xx} + f_{xy} + f_{y} - f = 0.
\end{align}
Since the derivative with respect to $y$ is of first order then it is expected to be of an exponential form. This suggests
\begin{align}
f(x,y) = g(x) e^{-\alpha y}.
\end{align}
The equation for $g$ is given by
\begin{align}
g'' - \alpha g' -(\alpha + 1) g = 0
\end{align}
and has the solution
\begin{align}
g(x) = A e^{(\alpha +1)x } + B e^{-x}.
\end{align}
Combining all the factors together the solution to the pde is
\begin{align}
\psi(x,y) = A e^{(\alpha + 1)x - \alpha y} + B e^{-x -\alpha y} - \frac{x}{2} e^{-x}.
\end{align}
This does have a similar functional form as proposed in the problem, namely,
\begin{align}
\psi(x,y) = e^{x} \phi(x-y) + e^{-x} \phi(y) - \frac{x}{2} e^{-x}
\end{align}
where 
\begin{align}
\phi(z) = e^{\alpha z}.
\end{align}
