= Solution
At a finite jump in <potential energy>, the <wavefunction> and its first <derivative> must be continuous. A jump in the derivative would create a <Dirac delta function> in $\psi''$, for which the finite potential has no balancing term. For the even solution, matching at $x=a$ gives
$$
B\cos\beta=De^{-\alpha},\qquad -B\kappa\sin\beta=-kDe^{-\alpha},\qquad \alpha=ka,\quad\beta=\kappa a.
$$
A nonzero solution cannot have $B=0$. It also cannot have $\cos\beta=0$: the first condition would give $D=0$, while the second would force $B\kappa\sin\beta=0$, a contradiction. Division is therefore valid and yields
$$
\boxed{\alpha=\beta\tan\beta.}
$$
<Parity> makes the conditions at $x=-a$ identical to those already imposed. Finally the definitions of the decay and oscillation constants give
$$
\boxed{\alpha^2+\beta^2=a^2\left(-\frac{2mE}{\hbar^2}+\frac{2m(E+V_0)}{\hbar^2}\right)=\frac{2mV_0a^2}{\hbar^2}.}
$$
The circle relation fixes the well strength; the tangent relation selects its even <bound states>.
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