= Solution
Use a stationary <wavefunction> $\Psi(x,t)=\psi(x)e^{-iEt/\hbar}$ solving the <Time-independent Schrödinger equation>. For <energy> above the positive <step potential>, the positive <wavenumbers> are
$$
\boxed{k_1=\frac{\sqrt{2mE}}{\hbar},\qquad
k_2=\frac{\sqrt{2m(E-V_0)}}{\hbar}.}
$$
Incidence only from the left permits a right-moving incident <plane wave> and a left-moving reflected <plane wave> for $x\leq0$, but only a right-moving transmitted <plane wave> for $x>0$:
$$
\psi(x)=\begin{cases}
Ae^{ik_1x}+Be^{-ik_1x},&x\leq0,\\
Ce^{ik_2x},&x>0.
\end{cases}
$$
The omitted right-incident component would describe a different scattering experiment. <Wavefunction matching at a finite potential step> requires continuity of both $\psi$ and $d\psi/dx$ for a finite jump and the same <mass> on either side:
$$
A+B=C,\qquad ik_1(A-B)=ik_2C.
$$
Eliminate $B=C-A$ from the second condition:
$$
k_1(2A-C)=k_2C,
\qquad
\boxed{\frac CA=\frac{2k_1}{k_1+k_2},\qquad
\frac BA=\frac{k_1-k_2}{k_1+k_2}.}
$$
The first ratio is a <transmission amplitude>; a transmitted <probability current> also depends on its different propagation <velocity>.
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