Show z=ce^(ax+a^2y) and z=ce^(-a^2y)cos(ax+b) solve…
Show $z=ce^{ax+a^2y}$ and $z=ce^{-a^2y}\cos(ax+b)$ solve $z_{xx}=z_y$; show $H=e^{ax+a^2y}$ and all its $a$-derivatives also solve it.
Explanation of Method
Separate $z=f(x)g(y)$ to get $f''/f=g'/g=\lambda$; the two signs of $\lambda$ give the exponential and the oscillatory$\times$decaying families. Then use that the operator commutes with $\partial_a$ since $a$ doesn't appear in the operator's coefficients.
Step 1 · Separate variables
$f''-\lambda f=0$, $g'-\lambda g=0$. For $\lambda=a^2>0$: $f=e^{ax}$ (dropping the decaying branch), $g=e^{a^2y}$, giving $z=ce^{ax+a^2y}$. For $\lambda=-a^2<0$: $f=\cos(ax+b)$, $g=e^{-a^2y}$, giving $z=ce^{-a^2y}\cos(ax+b)$.
Step 2 · Verify both families directly
$z_{xx}=a^2ce^{ax+a^2y}=z_y$ for the first; $z_{xx}=-a^2ce^{-a^2y}\cos(ax+b)=z_y$ for the second. \checkmark
Step 3 · $a$-derivatives of $H$
Since $L=\partial_x^2-\partial_y$ involves no $a$-dependence, $L$ commutes with $\partial_a$: \[ L\!\left(\frac{\partial^nH}{\partial a^n}\right) = \frac{\partial^n}{\partial a^n}(LH)=0. \] E.g. $H_a=(x+2ay)e^{ax+a^2y}$, $H_{aa}=[(x+2ay)^2+2y]e^{ax+a^2y}$, both solutions.
Verification
$LH=a^2e^{ax+a^2y}-a^2e^{ax+a^2y}=0$. \checkmark Direct substitution confirms $H_a$ and $H_{aa}$ also satisfy $z_{xx}=z_y$.
Result
Conclusion: