Real forcing often switches on and off: a motor starts at $t = 2$, a voltage is applied for 5 seconds. The unit step function writes such inputs in one formula, and the Laplace transform handles them without splitting the problem into pieces.
Idea clave
$u(t - c) = 0$ for $t < c$ and $1$ for $t \ge c$. $$\mathcal{L}\{u(t - c)\} = \frac{e^{-cs}}{s},$$ $$\mathcal{L}\{u(t - c)f(t - c)\} = e^{-cs}F(s).$$ Read backwards: a factor $e^{-cs}$ in $Y(s)$ means “invert without it, then shift right by $c$ and switch on at $t = c$.”
Por qué funciona
The step switches the integrand off before $t = c$: $$\int_0^{\infty} e^{-st}u(t - c)f(t - c)\,dt = \int_c^{\infty} e^{-st}f(t - c)\,dt.$$ Substitute $\tau = t - c$: $\int_0^{\infty} e^{-s(\tau + c)}f(\tau)\,d\tau$ $= e^{-cs}F(s)$. Delaying a signal by $c$ multiplies its transform by $e^{-cs}$. A piecewise function that switches from $f_1$ to $f_2$ at $t = c$ is $f_1 + (f_2 - f_1)u(t - c)$.
Ejemplo resuelto: Already shifted
Find $\mathcal{L}\{u(t - 2)(t - 2)^{2}\}$.
- It is $u(t - 2)f(t - 2)$ with $f(t) = t^{2}$ and $F = \frac{2}{s^{3}}$.
Respuesta: $\frac{2e^{-2s}}{s^{3}}$
Ejemplo resuelto: Not yet shifted
Find $\mathcal{L}\{u(t - 1)\,t\}$.
- Write $t$ in terms of $t - 1$: $t = (t - 1) + 1$.
- $u(t - 1)t$ $= u(t - 1)(t - 1) + u(t - 1)$.
- Transform each: $e^{-s}\cdot\frac{1}{s^{2}} + e^{-s}\cdot\frac{1}{s}$.
Respuesta: $e^{-s}\left(\frac{1}{s^{2}} + \frac{1}{s}\right)$
Ejemplo resuelto: A force that switches on
Solve $y' + y = u(t - 2)$, $y(0) = 0$.
- $(s + 1)Y = \frac{e^{-2s}}{s}$, so $Y = e^{-2s}\cdot\frac{1}{s(s + 1)}$.
- $\frac{1}{s(s + 1)} = \frac{1}{s} - \frac{1}{s + 1}$ inverts to $1 - e^{-t}$.
- The factor $e^{-2s}$ shifts it: $y = u(t - 2)\left(1 - e^{-(t - 2)}\right)$.
Respuesta: $y = 0$ for $t < 2$ and $y = 1 - e^{-(t - 2)}$ for $t \ge 2$
Error común
Writing $\mathcal{L}\{u(t - c)f(t)\} = e^{-cs}F(s)$. The shift rule needs $f(t - c)$, the function delayed by $c$. If you are given $u(t - c)f(t)$, first rewrite $f(t)$ in powers of $t - c$ (as in the second example). Check with $f = 1$ and $f = t$: $u(t - 1)t$ is not the same as $u(t - 1)(t - 1)$; at $t = 1$ one is 1 and the other is 0.
A box (or pulse) that is on only for $a \le t < b$ is the difference of two steps, $u(t - a) - u(t - b)$. For example, 5 volts applied for the first 5 seconds is $5\left(1 - u(t - 5)\right)$, with transform $\frac{5(1 - e^{-5s})}{s}$. Each step contributes its own $e^{-cs}$ factor, so the solution comes in pieces that switch on at $t = a$ and at $t = b$. A function that changes formula several times is written as a sum of such switches, one for each change.