Here's my suspicion but I'm not entirely sure: Don't let the partials intimidate you. It's still an ordinary DE in the variable z:

Alright let's down-grade it first: drop the a for now and consider just the ordinary-looking DE:

For constants a, b, and c. See the similarity? Ok, we can probably solve this and end up with a solution containing two arbitrary constants along with the a,b, and c (I'll call the solution p now):

Now, in terms of the PDE, the arbitrary constants are functions of the other two variables. Call them . Now, using the solution you get for y above (p), now make the substitutions . I bet a dollar this solution will then solve the PDE for a=1.

Yea, I know, it's not the one with variable a.. . . a journey begins with a first step.