Differential equation with delta dirac as the input

Question

I have this formula: $ C \frac{dv(t)}{d(t)} + Gv(t) = K \delta(t) $ and I have to calculate $v(0^{+})$ , given the $v(0^{-})$, without using the Laplace transformation. Could someone provide some help?

Answer 1

HINT:

For $t> 0$, it is easy to see that $v(t)=Ae^{-Gt/C}$.

where $A$ is a constant that will be found by enforcing the initial condition on $v(t)$.

Now, note that the presence of the Dirac Delta implies theta $v'(t)$ is discontinuous at $t=0$. Formally, we can write

$$\lim_{\epsilon \to 0^+}\int_{-\epsilon}^{\epsilon}\left(Cv'(t)+Gv(t)\right)\,dt=K\lim_{\epsilon\to 0^+}\int_{-\epsilon}^{\epsilon} \delta(t) \,dt \tag1 $$

The right-hand side of $(1)$ equals $K$.

Since $v$ is continuous, the contribution from integrating $v$ vanishes as $\epsilon \to 0$.

The left-hand side is then, $Cv(0^+)-Cv(0^-)=CA-Cv(0^-)$. Now, simply solve for $A$.