# Differential Equation with Delta

I'm studying differential equations and I was trying to solve the following first order differential equation:

$\dot y(t) + \frac{1}{a}y(t)=\sum_i w_i \sum_{t_i} \delta(t-t_i)$

where $w_i$ are constant coefficients. I proceeded solving first the homogeneous equation

$\dot y(t) = -\frac{1}{a}y(t) \implies y(t) = y_0 exp(- \frac{t-t_0}{a})$

To obtain a particular integral I need to solve

$y_p(t) = e^{-\frac{t}{a}} \int_{t_0}^t exp(\frac{\tau}{a}) \sum_i w_i \sum_{t_i} \delta(\tau-t_i) d\tau = e^{-\frac{t}{a}} \sum_i w_i \sum_{t_i}\int_{t_0}^t\delta(\tau-t_i)exp(\frac{\tau}{a})d\tau$ $= e^{-\frac{t}{a}}\sum_i w_i \sum_{t_i}exp(\frac{t_i}{a}) = \sum_i w_i \sum_{t_i}exp(-\frac{t-t_i}{a})$

assuming that $t_0 <t_i < t ~~~\forall i$.

The overall solution should then be

$y(t) = y_0 exp(- \frac{t-t_0}{a}) + \sum_i w_i \sum_{t_i}exp(-\frac{t-t_i}{a})$

Is this correct? What confuses me is that if I try to substitute the solution I obtained in the original equation I don't get the identity. Sorry if the problem is trivial.

To begin with, $\sum_i w_i \sum_{t_i} \delta(t-t_i)$ does not look like a valid formula (outer summation over $i$, and inside of it, summation over $t_i$?). Perhaps $\sum_i w_i \delta(t-t_i)$ is what was intended.
My preferred interpretation of $\delta(t-t_i)$ in a basic ODE course is that it is the derivative of Heaviside function $H(t-t_i)$. More generally, for any continuous function $f$ we have $$f(t) \delta_(t-t_i) = f(t_i) \delta_(t-t_i) = \frac{d}{dt}(f(t_i) H(t-t_i)) \tag{1}$$ where the first step is justified by $\delta$ being concentrated at $t_i$, and the second step is linearity of derivative.
$$e^{\tau/a} w_i \delta(\tau-t_i) = \frac{d}{dt}\left(e^{t_i/a} w_i H(t-t_i)\right)\tag{2}$$ Therefore,
$$\begin{split} e^{-t/a} \int e^{\tau/a} \sum_i w_i \delta(\tau-t_i) d\tau &= e^{-t/a}\left( C + \sum_i e^{t_i/a} w_i H(t-t_i)\right) \\ & = Ce^{-t/a} + \sum_i e^{(t_i-t)/a} w_i H(t-t_i) \end{split}\tag{3}$$ Formula (3) gives the general solution of your ODE. You can satisfy the initial condition by choosing $C$.
You can also use definite integral ($t_0$ to $t$) in (3), but I thought that indefinite integral is easier to write down.