= Solution
The hard wall imposes $\psi(0)=0$, while the increasing linear potential requires decay as $x\to\infty$. Set
$$
\ell=\left(\frac{\hbar^2}{2m\beta}\right)^{1/3},\qquad
\xi=\frac x\ell-\frac E{\beta\ell}.
$$
The stationary <Schrödinger equation> becomes $d^2\psi/d\xi^2-\xi\psi=0$. The growing <Airy function> solution is excluded, leaving
$$
\boxed{\psi_n(x)\propto\operatorname{Ai}(x/\ell-z_n),\quad
E_n=\beta\ell z_n,\quad\operatorname{Ai}(-z_n)=0.}
$$
This is the <quantum bouncer>. For the first four states, $z_n\simeq2.3381,4.0879,5.5206,6.7867$. Their interior node counts are respectively \b[zero, one, two and three]. Each <wavefunction> starts at zero at the wall, oscillates in the classically allowed interval, and crosses smoothly into an exponentially decaying tail beyond the turning point $x_n=E_n/\beta=z_n\ell$. The turning point is not an extra hard wall: the <wavefunction> generally remains nonzero there and beyond it.
\Image[/past-exam-of-the-physics-course-of-the-year-4-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-1-quantum-bouncer.png]
{title=The first four quantum-bouncer wavefunctions and their classical turning points}
{description=Original normalized Airy-function calculation. The vertical dashed lines mark where the potential equals each energy; the decaying tail is retained.}
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