A hammer blow, a sudden voltage spike, a dose injected at once: forces that act for an instant but leave a lasting effect. The Dirac delta models them, and with step functions behind us, the Laplace transform handles them easily.
Key idea
$\delta(t - c)$ is an idealized spike at $t = c$ with total area 1: $$\int_{-\infty}^{\infty} f(t)\delta(t - c)\,dt = f(c),$$ $$\mathcal{L}\{\delta(t - c)\} = e^{-cs}.$$ In $my'' + \dots = K\delta(t - c)$ the velocity jumps by $\frac{K}{m}$ at $t = c$; in a first-order equation $y' + \dots = K\delta(t - c)$, $y$ itself jumps by $K$.
Why it works
Think of $\delta$ as the limit of tall narrow pulses $d_\varepsilon$, of height $\frac{1}{\varepsilon}$ on $[c, c + \varepsilon)$, each with area 1. Then $\int f d_\varepsilon\,dt$ is the average of $f$ over the pulse, which tends to $f(c)$, and $\mathcal{L}\{d_\varepsilon\} = \frac{e^{-cs} - e^{-(c + \varepsilon)s}}{\varepsilon s} \to e^{-cs}$. Physically, the integral of a force over time is the change of momentum, so an impulse $K$ changes $mv$ by $K$ instantly while the position has no time to move.
Worked example: Sifting
Evaluate $\int_0^{\infty} (t^{2} + 1)\delta(t - 3)\,dt$.
- The delta picks out the value at $t = 3$: $3^{2} + 1$.
Answer: 10
Worked example: A dose at $t = 1$
Solve $y' + 2y = 3\delta(t - 1)$, $y(0) = 1$.
- $(s + 2)Y - 1 = 3e^{-s}$, so $Y = \frac{1}{s + 2} + \frac{3e^{-s}}{s + 2}$.
- Invert: $e^{-2t}$, and the shifted $3u(t - 1)e^{-2(t - 1)}$.
Answer: $y = e^{-2t}$ for $t < 1$ and $y = e^{-2t} + 3e^{-2(t - 1)}$ for $t > 1$: a jump of 3 at $t = 1$
Worked example: A blow to a spring at rest
Solve $y'' + 4y = 2\delta(t - \pi)$, $y(0) = 0$, $y'(0) = 0$.
- $(s^{2} + 4)Y = 2e^{-\pi s}$, so $Y = e^{-\pi s}\cdot\frac{2}{s^{2} + 4}$.
- $\frac{2}{s^{2} + 4}$ inverts to $\sin 2t$; shift by $\pi$: $u(t - \pi)\sin(2(t - \pi))$ $= u(t - \pi)\sin 2t$.
Answer: $y = 0$ before $t = \pi$, then $y = \sin 2t$: the blow gives velocity 2 and the spring starts to swing
Common mistake
Treating $\delta(t - c)$ as an ordinary function equal to 1 at $t = c$. A single point carries no area, so that would do nothing. Its meaning is in integrals: area 1, concentrated at $c$. A second slip: in a second-order equation, the impulse makes the velocity jump; the position stays continuous.