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I have this equation: $$\frac{{\rm d^2}\psi(x)}{{\rm d}x^2}+\frac{2m}{\hbar^2}(E+V_0\delta(x))\psi(x)=0$$ here $E<0$ and $V_0>0$. This equation is for finding wave function of a particle under attractive delta potential.

I can easily solve for $x\ne0$. The boundary conditions are $\displaystyle\lim_{x\to\pm\infty}\psi(x)=0$

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Integrate both sides of the equation between $-\epsilon$ and $+\epsilon$ and then let $\epsilon\to 0$. Under assumption that $\psi$ remains continuous at $x=0$, this will lead (show it!) to $$\psi'(0^+)-\psi'(0^-)+\frac{2mV_0}{\hbar^2}\psi(0)=0.$$ More details can be found here.

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