The question to solve the homogenous linear PDE: $(D^{2}+2DD'-D'^{2})z=\cos(3x+y)$ where $D=\partial /\partial x$ and $D'=\partial/\partial y$ appeared on my exam, and I'm trying to check whether the answer I've submitted is correct or not. Here is what I did:
For the complimentary function, first we need to solve $(D^{2}+2DD'-D'^{2})z=0$, which we'll get by putting $D=m$ and $D'=1$, so that the auxiliary equation becomes $m^{2}+2m-1=0$ whose solutions are $m_{1,2}=-1\pm\sqrt{2}$, so that the complimentary function is $\phi_{1}(y+(-1+\sqrt{2})x)+\phi_{2}(y+(-1-\sqrt{2})x)$, where $\phi_{1,2}$ are arbitrary functions.
For the particular integral: $$\begin{aligned}z&=\frac{1}{D^{2}+2DD'-D'^{2}}\cos(3x+y) \\ &= \frac{1}{-9-6+1}\cos(3x+y) \\ &=\frac{-1}{14}\cos(3x+y) \end{aligned}$$
So the complete solution, should be as under:
$$\boxed{z=\phi_{1}(y+(-1+\sqrt{2})x)+\phi_{2}(y+(-1-\sqrt{2})x)+\frac{-1}{14}\cos(3x+y)}$$
Since, there was no option as this, I went with the "None of These" option, there were some close options but they had the complimentary solution of the format $\phi_{1}(y-(-1+\sqrt{2})x)+\phi_{2}(y-(-1-\sqrt{2})x)$, so I just want to check if my reasoning is correct. Any insights are appreciated. Thanks.