Suppose there exists a solution to *$u_t+\Delta^2u+\Delta u=0$ of the form $u(x,y,t)=c(t)e^{i\pi(x/4\pi+y/4\pi)}$. I need to find such a function $c(t)$.
Plugging $u(x,y,t)$ into *, I got $c'(t)e^{i\pi(x/4\pi+y/4\pi)}-7/64c(t)e^{i\pi(x/4\pi+y/4\pi)}=0\implies c'(t)=7/64c(t)$. This is a first order ODE in $c(t)$, so the solution is $c(t)=u_0e^{(7/64)t}$, for a constant $u_0$.
So does my work look alright? And if so, since I got $c(t)=u_0e^{(7/64)t}$, does this mean that my solution $u(x,y,t)=u_0e^{(7/64)t}e^{i\pi(x/4\pi+y/4\pi)}$ grows in time, as opposed to decaying?
Let's check: $$u_t = \frac{7}{64} u$$ $$\Delta u = -\frac{1}{8}u$$ hence $$\Delta\Delta u = \frac{1}{64}u$$ Yes, the sum is $0$.
Yes, the solution increases in amplitude with time. Without the biLaplacian you would have time-reversed diffusion equation, which typically leads to concentration $u$ blowing up.