IVPs with Step Functions

An IVP with step functions is simply an equation whose push turns on or off partway through, like a pump that runs from hour 1 to hour 3. All we're really saying is: if you flip a switch later, the system reacts later, in exactly the same way. That's good for anything scheduled or switched, since you never have to solve the problem one interval at a time. The formula $\mathcal L^{-1}\{e^{-cs}F(s)\} = u_c(t)\,f(t - c)$ looks dense, but its pieces are simple: $F(s)$ (the usual response, in $s$), $e^{-cs}$ (a tag that says "wait $c$ seconds"), and $u_c(t)\,f(t - c)$ (the same response, started at time $c$). This puts the last two lessons together. You already know how to solve an IVP by transforming it, and that a switch at time $c$ turns into $e^{-cs}$, so a switched input just means $Y(s)$ comes out with some of those tags in it. Nothing else about the method changes.

After solving for $Y(s)$, collect the terms that share each exponential factor. Invert each group without the exponential, then shift it by $c$ and multiply by $u_c(t)$. That's the shift rule from the step functions lesson, read backwards.

Think about what that means physically. The system responds to each switch the same way it would have responded at time $0$, just later. Because the equation is linear, those delayed responses simply add.

The answer is piecewise: before $t = c$ only the unshifted terms contribute; after, both do.

A common mistake. It's tempting to invert $e^{-s}\left(\frac1s - \frac{1}{s + 1}\right)$ as $u_1(t)\left(1 - e^{-t}\right)$, because $1 - e^{-t}$ is the inverse of the part in parentheses. It isn't: the delay applies to the whole function, so every $t$ inside becomes $t - 1$, giving $u_1(t)\left(1 - e^{-(t - 1)}\right)$. Without the shift, the response would jump at $t = 1$ instead of starting from zero.

Turning a motor on at a scheduled time, a dose given at hour 6, a load applied to a beam partway through a test: step-function IVPs.