Why the PIs using different methods are different? What am I doing wrong? I have a PDE
$$
\text { Solve }\left(D^{2}+3 D D^{\prime}+2 D^{\prime 2}\right) z=x+y .
$$
where $D = \frac{\partial }{\partial x}$, $D^{\prime} = \frac{\partial }{ \partial y}, D^2 = \frac{\partial^2}{\partial x^2}, DD^{\prime}= \frac{\partial^2 }{\partial x \partial y} $.
I solved for Particular Integral as follows:
$$
\begin{array}{l}
P.I. &=\frac{1}{D^{2}\left(1+\frac{3 D^{\prime}}{D}+\frac{2 {D^{\prime}}^{2}}{D^{2}}\right)}[x+y] \\
&=\frac{1}{D^{2}}\left[1+\frac{3 D^{\prime}}{D}+\frac{2 D^{\prime 2}}{D^{2}}\right]^{-1}(x+y) \\
&=\frac{1}{D^{2}}\left[1-\frac{3 D^{\prime}}{D}-\frac{2 {D^{\prime}}^{2}}{D^{2}}+\left(\frac{3 D^{\prime}}{D}+\frac{2 {D^{\prime}}^{2}}{D^{2}}\right)^{2}-. .\right](x+y) \\
&=\frac{1}{D^{2}}\left[x+y-\frac{3}{D}(0+1)\right] \\
&=\frac{1}{D^{2}}[x+y-3 x] \\
&=\frac{1}{D^{2}}[y-2 x] \\
&=\frac{1}{D}\left(yx-\frac{2 x^{2}}{2}\right) \\
&=\frac{yx^2}{2}-\frac{x^{3}}{3}
\end{array}
$$
But the textbook uses a different method, I have seen the derivation that looks correct.
$$
\begin{array}{l}
\text { When } F(a, b) \neq 0 \text { and } F\left(D, D^{\prime}\right) \text { is a homogeneous function of degree } n \text { , then }\\
\text { P.I. }=\frac{1}{F\left(D, D^{\prime}\right)} \phi(a x+b y)=\frac{1}{F(a, b)} \iint \ldots \int f(v) d v d v \ldots d v\\
\text { where } \quad v=a x+b y
\end{array}
$$
Now, P.I. $=\frac{1}{D^{2}+3 D D^{\prime}+2 D^{\prime 2}}(x+y)$
$=\frac{1}{1^{2}+3 \times 1 \times 1+2 \times 1^{2}} \iint v d v d v$, where $v=x+y$,
$$
=(1/6) \times \int\left(v^{2} / 2\right) d v=(1 / 6) \times\left(v^{3} / 6\right)=(1 / 36) \times(x+y)^{3} .
$$
Hence the required general solution is $\quad z=$ C.F. + P.I., i.e.,
$$
z=\phi_{1}(y-x)+\phi_{2}(y-2 x)+(1 / 36) \times(x+y)^{3} .
$$
Can someone point out what am I doing wrong?

$$
\frac{(x+y)^3}{36} = \frac{(y-x)^3}{12} - \frac{(y-2x)^3}{18} + \frac{x^2 y}{2} -\frac{x^3}{3}
$$
So the two answers are equivalent.
