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:
$$ f(x) = \frac{\phi}{2} e^{-\phi |x|}. $$
