How to solve this 2nd order ODE with Dirac delta?

I need to solve the following ODE

$$f''(x) - \zeta f(x) + \zeta\delta(x-b) = 0,$$

where $$x\in(-\infty,\infty)$$ and where $$f(x)\rightarrow 0$$ as $$x \rightarrow \mp \infty$$. The solution ignoring the Dirac impulse is given by

$$f(x) = c_1 e^{\sqrt{\zeta}x} + c_2 e^{-\sqrt{\zeta}x}.$$

Since I have a Dirac impulse at $$x=b$$, I should be solving for two ODEs, one below $$x=b$$ and another above $$x=b$$. Then I have to put together both solutions at $$x=b$$. This is where I am confused, how can I do this part?

A bit more on the intuition behind the problem. The ODE in question is a steady state Fokker-Planck (or Kolmogorov Forward) Equation. Mass is injected at $$x=b$$ and dissipates both to the left and right of $$x=b$$. Then, mass anywhere in $$x\in(-\infty,\infty)$$ is taken out at a rate $$\zeta$$ and reinjected back to $$x=b$$.

You have two solutions as you noted for: $$f''(x) - \zeta f(x) + \zeta\delta(x-b) = 0,$$ given by (denoting $$\phi^2 = \zeta$$). For simplicity, I will let $$b = 0$$ (you can also do this via a shift). $$f(x) = \begin{cases} A e^{\phi x} & \text{if } x \le 0 \\ B e^{-\phi x} & \text{if } x > 0. \end{cases}$$ Continuity is required and so we must have $$A = B$$. Finally, we will see what is required in the neighborhood of $$x = 0$$ by looking at a small integral containing $$0$$ of the ODE and take the limit $$0 = \lim_{\epsilon \to 0} \int_{-\epsilon}^\epsilon f''(x) - \phi^2 f(x) + \phi^2\delta(x) \ dx = f'(0^+) - f'(0^-) + \phi^2$$ Hence, $$0 = -\phi B - \phi A + \phi^2$$. The two conditions imply that $$A = B = \phi/2$$. Hence the solution: $$f(x) = \frac{\phi}{2} e^{-\phi |x|}.$$
• Thank you! Before asking the question I actually tried integrating the ODE around $x=b=0$, but since I forgot that $f(x)$ is continuous at $b$ I got confused with the integral of $f(x)$ (I didnt recognize that this term would vanish). Commented Sep 20, 2022 at 21:39