= Solution
Put
$$
G_k(x-x')=e^{ik|x-x'|}.
$$
Away from $x=x'$, this satisfies $(d^2/dx^2+k^2)G_k=0$. Its first <derivative> has jump $2ik$, so, distributionally,
$$
\left(\frac{d^2}{dx^2}+k^2\right)G_k(x-x')=2ik\delta(x-x').
$$
Applying $d^2/dx^2+k^2$ to the proposed <integral> equation therefore gives
$$
\left(\frac{d^2}{dx^2}+k^2\right)\psi(x)
=\frac{2m}{\hbar^2}V(x)\psi(x).
$$
Since $k^2=2mE/\hbar^2$, this is precisely the stated <Schrödinger equation>. The term $e^{ikx}$ is the incident wave and the choice $e^{ik|x-x'|}$ gives outgoing waves on both sides of each source point.
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