Answer
Eq. 38-27, $\psi=Ae^{ikx}$, is a solution of Schrödinger’s equation (proved). See in step by step work.
Work Step by Step
For a nonrelativistic particle of mass $m$ traveling along an $x$ axis, with energy $E$ and potential energy $U(x)=U_0$, the space-dependent part can be found by solving Schrödinger’s equation
$\frac{d^2\psi}{dtx^2}+\frac{8\pi^2m}{h^2}[E-U_0]\psi=0$
Let,
$k^2=\frac{8\pi^2m}{h^2}[E-U_0]$
or, $k=\frac{2\pi}{h}\sqrt {2m(E-U_0)}$
Therefore,
$\frac{d^2\psi}{dtx^2}+k^2\psi=0$ ......................$(1)$
Let $\psi=\psi_0e^{px}$ is the solution of eq, $1$, where $\psi_0$ and $p$ are two constants
Substituting $\psi=\psi_0e^{px}$ in eq. $1$, we get
$p^2\psi_0e^{px}=-k^2\psi_0e^{px}$
or, $p^2+k^2=0$ $(\because e^{px}\ne0 )$
or, $p=±ik$
Thus, the solutions of eq. $1$ are
$\psi_1=\psi_0^1e^{ikx}$ and $\psi_2=\psi_0^2e^{-ikx}$
Therefore, the general solution of eq. $1$ is given by
$\psi=\psi_1+\psi_2$
or, $\psi=\psi_0^1e^{ikx}+\psi_0^2e^{-ikx}$ ......................$(2)$
The first term on the right in Eq. 2 represents a wave traveling in the positive direction of $x$, and the second term represents a wave traveling in the negative direction of $x$. If we eliminate the negative motion by setting $\psi_0^2$ to zero, and then the solution at t 0 becomes
$\psi=\psi_0^1e^{ikx}\implies \psi=Ae^{ikx}$, where $A=\psi_0^1$
Thus, Eq. 38-27, $\psi=Ae^{ikx}$, is a solution of Schrödinger’s equation.
