Suppose I am looking to solve the wavefunction for the following 1D potential: $$U(x) = \begin{cases}V_0\frac{a-|x|}{a}&\quad\text{for}\quad|x|<a \\\infty&\quad\text{for}\quad|x|>a\end{cases} \tag{1}$$ Since our potential is symmetric, we have even and odd solutions and so we can solve the system for $x\ge 0$ and afterward construct the full solution for either $\psi_{even}$ or $\psi_{odd}$. For the region inside the well, we can cast the time-independent Schrodinger eq. into a form that is the Airy differential equation (note that it is a linear potential), namely $$\frac{d^2\psi}{dz^2}-z\psi = 0, \tag{2}$$ where $$z\equiv Q(1 - \eta), \tag{3}$$
where
$$\eta = \frac{x}{a_0}, \ \ \ Q = \frac{2mV_0a_0}{\hbar^2a}, \ \ \ a_0 \equiv a-\frac{a}{V_0}E. \tag{4}$$
Therefore we can express $\psi$ as
$$\psi(z) = C_1Ai(z)+C_2Bi(z). \tag{5}$$
Our boundary conditions tell us: $$\psi(\zeta_a) = 0 \tag{6}$$ $$\psi(\zeta_0) = 0\quad\text{for}\quad\psi_{odd} \tag{7}$$ $$\psi'(\zeta_0) = 0\quad\text{for}\quad\psi_{even} \tag{8}.$$
where I am denoting $\zeta_0= z|_{x=0}$ and $\zeta_a= z|_{x=a}$.
Question: Is this system exactly solvable?
Typically we only have one zero boundary condition in these types of problems which allows us to quantize the energy levels through the zero's of the Airy function. But here we must simultaneously satisfy two zero boundary conditions. To clarify, here is our system of linear equations.
$\psi_{odd}$: $$C_1Ai(\zeta_0)+C_2Bi(\zeta_0) = 0 \tag{9}$$ $$C_1Ai(\zeta_a)+C_2Bi(\zeta_a) = 0 \tag{10}$$
$\psi_{even}$: $$C_1Ai'(\zeta_0)+C_2Bi'(\zeta_0) = 0, \tag{11}$$ $$C_1Ai(\zeta_a)+C_2Bi(\zeta_a) = 0. \tag{12}$$
My proposed idea would be to solve for the zero determinant of the homogeneous linear equations, but this would force me to a numerical solution. I would like to find the energies in terms of the zeros of the Airy functions. Ideas?