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?
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}