Trying to understand where the discontinuity comes from in this equation coming from an ODE course (link at the bottom): $$ \ddot{x}+3\dot{x}+2x = \delta(t), $$ Initial conditions: $\dot{x}(0)=x(0)=0$. Assume impulse occurs at $t = 0^+$.
The solution is: $$ x(t) = e^{-t} -e^{-2t} = e^{-t} (1 -e^{-t}) $$ And the derivative of the solution evaluated at IC: $$ \dot{x}(t=0) = -e^{-t} + 2e^{-2t} = -1 + 2 =1 $$
The course mentions that the $x(t)$ is continuous at $t = 0$, but the derivative $\dot{x}(t)$ is not because $\dot{x}(0^-)=0$ whilst $\dot{x}(0^+)=1$. How do I know that $\dot{x}(0^-)=0$? Does one need to read into when impulse occurs to arrive at the conclusion or can it be shown mathematically using the Dirac's delta definition?
The ODE is solved here https://www.youtube.com/watch?v=3AkX-cIg5wM and in https://www.math.hkust.edu.hk/~machas/differential-equations-for-engineers.pdf on p.117
The theory of distributions is needed to formalise the above, but this is overkill in this context. One can find some intuition by 'working under the integral sign'.
The net effect of the $\delta$ is essentially to introduce a step change in initial conditions.
That is, suppose the response of the above system given initial conditions $x(0), \dot{x}(0)$ is $t \mapsto s(x(0), \dot{x}(0) ; t)$. Then the response of the system above will be $s(0,0 ; 0)$ initially and then $s(0,1; t)$ for $t>0$ (note the $\dot{x}$ term changed to $1$).
To see informally how this comes about, integrate the system over a small interval to get $\int_0^\epsilon \ddot{x}(t) dt = \dot{x}(\epsilon)-\dot{x}(0) = \int_0^\epsilon -(3\dot{x}(t) + 2 x(t))dt+ \int_0^\epsilon \delta(t)dt$. Taking limits we get $\lim_{\epsilon \downarrow 0} (\dot{x}(\epsilon)- \dot{x}(0)) = 1$.
In particular, the response to the above system, for $t>0$ is the same as the response of the unforced system to the initial conditions $x(0)=0, \dot{x}(0)=1$.
(You could also use Laplace transforms to see the equivalence.)