Solution of $y''(t) + 4y'(t)+5y(t) = f(t) $ using Laplace transform and Dirac delta function. As the title says, I need to solve the initial value problem $$y''(t) + 4y'(t)+5y(t) = f(t);\quad y(0) = 1, y'(0) = 0 ,$$ where $ f(t) $ is an impulsive force which acts on the interval $ 1 \leq t \leq 1 + \tau ,$ and $$ \int_{1}^{1+\tau} f(t)\ dt = 2 .$$
I really don't know what to do with the RHS of the equation, but something tells me that $ f(t) $ is a piecewise function that may be defined with the Dirac delta in some way, something like $\delta (t - 1) $ on the interval $ 0 \leq t \leq 2 ,$ but that's just a guess by looking at the integral itself.
 A: Well, when we take the Laplace transform of both sides we get:


*

*LHS (and applying the initial conditions):
$$\mathscr{L}_t\left[\text{y}''\left(t\right)+4\cdot\text{y}'\left(t\right)+5\cdot\text{y}\left(t\right)\right]_{\left(\text{s}\right)}=\text{s}^2\cdot\text{Y}\left(\text{s}\right)+4\cdot\text{s}\cdot\text{Y}\left(\text{s}\right)+5\cdot\text{Y}\left(\text{s}\right)\tag1$$

*RHS:
$$\mathscr{L}_t\left[\text{f}\left(t\right)\right]_{\left(\text{s}\right)}=\text{F}\left(\text{s}\right)\tag2$$


So, we get:
$$\text{s}^2\cdot\text{Y}\left(\text{s}\right)+4\cdot\text{s}\cdot\text{Y}\left(\text{s}\right)+5\cdot\text{Y}\left(\text{s}\right)=\text{F}\left(\text{s}\right)\space\Longleftrightarrow\space\text{Y}\left(\text{s}\right)=\frac{\text{F}\left(\text{s}\right)}{\text{s}^2+4\cdot\text{s}+5}\tag3$$
Now, using the convolution theorem:
$$\text{y}\left(t\right)=\mathscr{L}_\text{s}^{-1}\left[\frac{\text{F}\left(\text{s}\right)}{\text{s}^2+4\cdot\text{s}+5}\right]_{\left(\text{t}\right)}=\mathscr{L}_\text{s}^{-1}\left[\text{F}\left(\text{s}\right)\right]_{\left(\text{t}\right)}\space*\space\mathscr{L}_\text{s}^{-1}\left[\frac{1}{\text{s}^2+4\cdot\text{s}+5}\right]_{\left(\text{t}\right)}=$$
$$\text{f}\left(t\right)\space*\space e^{-2t}\cdot\sin\left(t\right)\tag4$$
