Combining two results from partial integration I have a set of two PDEs:
$$\partial_{\tau}\theta+\partial_{\eta}\psi=0$$
$$\partial_{\tau}\psi=-\partial_{\eta}\theta+\alpha\partial_{\eta}^{2}\psi$$
These can be combined into a wave equation of the form:
$$\partial_{\tau}^{2}\theta=\partial_{\eta}^{2}\left[\theta+\alpha\partial_{\tau}\theta\right]
 $$
which with the ansatz:
$$\theta=\sin\left(\eta\right)\exp\left(-\omega\tau\right)$$
gives the dispersion relation:
$$\omega^2-\alpha\omega+1=0$$
The solution to the dispersion relation is $\omega=\omega_r+i\omega_i$ with $\omega_r=\frac{1}{2}\alpha$ and $\omega_i=\pm\sqrt{1-\frac{1}{4}\alpha^2}$. 
Now i would like to determine $\psi\left(\eta,\tau\right)$, I attempted to do this by using the original equations in a rewritten form:


*

*$$\partial_{\eta}\psi = -\partial_{\tau}\theta \rightarrow \psi\left(\eta,\tau\right) = -\omega\cos\left(\eta\right)\exp\left(-\omega\tau\right)+K_1$$

*$$\partial_{\tau}\psi = -\partial_{\eta}\theta-\alpha\partial_{\eta}\partial_{\tau}\theta  \rightarrow \psi\left(\eta,\tau\right) = \left(\frac{1}{\omega}-\alpha\right)\cos\left(\eta\right)\exp\left(-\omega\tau\right) + K_2$$
but as far as i can see these two partial solutions cannot be combined to satisfy both original equations simulateously. How do i go about getting a solution for $\psi$?
 A: As it turns out the two results for $\psi$ are actually equivalent as by the dispersion relation:
$$\frac{1}{\omega}-\alpha=\frac{1}{\omega}\left(1-\alpha\omega\right)=-\omega$$
Alternative method: First, redefine $\theta$ and $\psi$ using a scalar $\phi$:
$$\theta=\partial_{\eta}\phi\quad\psi=-\partial_{\tau}\phi$$
This ensures that:
$$\partial_{\tau}\theta+\partial_{\eta}\psi=0$$
is exactly satisfied and yields a wave equation for $\phi$ from the other equation:
$$\partial_{\tau}^{2}\phi=\partial_{\eta}^{2}\left[\phi+\alpha\partial_{\tau}\phi\right]$$
The solution for $\phi$ has a similar form as previously found for $\theta$:
$$\phi\left(\eta,\tau\right)=\left[A\sin\left(\eta\right)+B\cos\left(\eta\right)\right]\exp\left(-\omega\tau\right)
 $$ where the constants $A$ and $B$ are yet to be determined.
From the definitions for $\theta$ and $\psi$ I find:
$$\theta=\partial_{\eta}\phi=\left[A\cos\left(\eta\right)-B\sin\left(\eta\right)\right]\exp\left(-\omega\tau\right)$$
$$\psi=-\partial_{\tau}\phi=\omega\left[A\sin\left(\eta\right)+B\cos\left(\eta\right)\right]\exp\left(-\omega\tau\right)
 $$
Now if I am looking for a solution of $\psi$ which corresponds to a $\theta$ distribution:
$$\theta=\sin\left(\eta\right)\exp\left(-\omega\tau\right)
 $$
I require $A=0$ and $B=-1$ such that:
$$\psi=-\omega\cos\left(\eta\right)\exp\left(-\omega\tau\right)
 $$
