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Section 13.6 Laplace Method with Piecewise Forcing Terms

This is the moment where everything you’ve learned about Laplace transforms, step functions, and piecewise rewriting comes together. We’ll now use the Laplace method to solve differential equations whose forcing terms switch ON and OFF across time.
The big picture is the same familiar three-step Laplace method you already know:
  • Step 1: Forward transform the differential equation into the Laplace domain.
  • Step 2: Solve for \(Y(s)\) algebraically and prepare it for inversion.
  • Step 3: Apply the inverse Laplace transform to recover \(y(t)\text{.}\)
The only twist is that our forcing term is now piecewise, so the process begins with a setup step of rewriting the forcing term using step functions.

Subsection A Model Problem

Let’s solve this initial value problem to see the full method in action:
\begin{equation*} y'' + 4y = g(t), \qquad y(0) = 0,\ y'(0) = 0 \end{equation*}
where the forcing term switches ON for just one second:
\begin{equation*} g(t) = \begin{cases} 1, \amp 1 \le t \lt 2 \\ 0, \amp \text{otherwise}. \end{cases} \end{equation*}
You’ll recognize the roadmap: forward transform → solve in the Laplace domain → inverse transform. But watch how the step function representation drives every stage of the solution.

Setup: Piecewise Form to Step Form.

The piecewise function, \(g(t)\text{,}\) must be expressed in step function form before we can take its transform. Since \(g(t)\) has only one non-zero piece on the activation interval, \(1 \le t \lt 2\text{,}\) the only switch we need is \(u_1(t) - u_2(t)\text{:}\)
\begin{equation*} g(t) = 1 \cdot (u_1(t) - u_2(t)) = u_1(t) - u_2(t) \end{equation*}

Step 1 — Into the Laplace Domain.

We are now ready to apply \(\laplacesym\) to both sides, term-by-term, using linearity:
\begin{gather*} \us{\ds\text{R}\vphantom{R}_2}{\ul{\lap{y''}}} + 4\us{\ds Y(s)}{\ul{\lap{y}}} = \us{\ds\text{L}\vphantom{L}_9}{\ul{\lap{u_1(t)}}} - \us{\ds\text{L}\vphantom{L}_9}{\ul{\lap{u_2(t)}}} \end{gather*}
Using rules R\(_{2}\) with \(y(0) = 0,\ y'(0) = 0\) and L\(_{9}\), gives the Laplace-domain equation:
\begin{equation*} s^2Y(s) + 4Y(s) = \frac{e^{-s}}{s} - \frac{e^{-2s}}{s}. \end{equation*}

Step 2 — Solving in the Laplace Domain.

Factor out \(Y(s)\) from the left side and divide to isolate it:
\begin{gather*} (s^2 + 4)Y(s) = \frac{e^{-s}}{s} - \frac{e^{-2s}}{s} \quad\implies\quad Y(s) = \frac{1}{s^2 + 4} \left( \frac{e^{-s}}{s} - \frac{e^{-2s}}{s} \right) \end{gather*}
This is the Laplace-domain solution. Next, we rewrite \(Y(s)\) into a form that matches the inverse Laplace table.
In general, any exponential terms in \(Y(s)\) (i.e., \(e^{-s}\) and \(e^{-2s}\)) will return to the original domain as unit step functions using L\(_{11}\). Therefore, you should prioritize collecting exponentials whenever you see them. For our equation, this looks like:

Checkpoint 289. 📖❓ Matching Exponential Factors.

What do exponential terms in the Laplace domain, like \(e^{-4s}\text{,}\) correspond to in the original-domain solution, \(y(t)\text{?}\)
  • A term with the step function switch, \(u_4(t)\text{.}\)
  • Correct — \(e^{-cs}\) always brings in \(u_c(t)\) and a shift by \(c\text{.}\)
  • A term with a coefficient of \(4\text{.}\)
  • No — the exponential doesn’t change coefficients like that.
  • An exponential term \(e^{4t}\text{.}\)
  • No — \(e^{-4s}\) doesn’t translate to \(e^{4t}\text{.}\)
  • A term containing \(\dfrac{1}{t+4}\text{.}\)
  • No — perhaps you thought this looked like the forward transform
    \begin{equation*} \lap{e^{-4t}} = \frac{1}{s+4}\text{?} \end{equation*}
\begin{equation*} Y(s) = \frac{1}{s(s^2 + 4)}e^{-s} - \frac{1}{s(s^2 + 4)}e^{-2s} \end{equation*}
To match the form in L\(_{11}\), it’s a good habit to define the shorthand:
\begin{equation*} F(s) = \frac{1}{s(s^2 + 4)} \end{equation*}
so that our prepared \(Y(s)\) becomes:
\begin{equation*} \text{(prepared)}\qquad Y(s) = F(s)e^{-s} - F(s)e^{-2s}\text{.} \end{equation*}
Since we’ll also need \(f(t) = \ilap{F(s)}\) in step 3, let’s compute it now:
\begin{equation*} f(t) = \ilap{F(s)} = \frac14\left(1 - \cos(2t)\right). \end{equation*}

Step 3 — Leaving the Laplace Domain.

We now apply \(\laplacesym^{-1}\) to both sides of the “prepared” \(Y(s)\text{:}\)
\begin{equation*} \ilap{Y(s)} = \ilap{F(s)e^{-s}} - \ilap{F(s)e^{-2s}} \end{equation*}
Applying L\(_{11}\), our solution becomes:
\begin{equation*} y(t) = f(t - 1)\cdot u_1(t) - f(t - 2)\cdot u_2(t)\text{.} \end{equation*}
We use \(f(t)\) (from step 2) to plug in \(f(t - 1)\) and \(f(t - 2)\text{,}\) leading to the solution:
\begin{equation*} y(t) = \frac14\Big(1 - \cos\big(2(t - 1)\big)\Big)u_1(t) - \frac14\Big(1 - \cos\big(2(t - 2)\big)\Big)u_2(t) \end{equation*}
This problem demonstrates how the same Laplace method process can be applied to equations with piecewise forcing terms. With practice, you’ll see that problems with piecewise forcing terms are no different from any other Laplace method problem.

Subsection Additional Examples

To cement the idea, here are a few more examples. Notice in each, that the three-step strategy stays exactly the same.

🌌 Example 290.

Solve the initial value problem
\begin{equation*} y'' + 2y' = g(t), \qquad y(0) = 0,\quad y'(0) = 1 \end{equation*}
where the forcing term is given by:
\begin{equation*} g(t) = \begin{cases} 3, \amp t \ge 5\\ 0, \amp \text{otherwise}. \end{cases} \end{equation*}
Solution.
Setup: Piecewise Form to Step Form.
Since the only nonzero piece of \(g\) is \(3\) on \(t \ge 5\text{,}\) the equation becomes:
\begin{equation*} y'' + 2y' = 3u_5(t)\text{.} \end{equation*}
Step 1 — Into the Laplace Domain.
Take the Laplace transform of both sides:
\begin{equation*} \lap{y''} + 2\lap{y'} = 3\lap{u_5(t)}\text{.} \end{equation*}
Using the initial conditions with R\(_1\) and L\(_9\) we get:
\begin{equation*} s^2 Y(s) - 1 + 2s Y(s) = \frac{3e^{-5s}}{s}\text{.} \end{equation*}
Step 2 — Solve for \(Y(s)\).
Next, we isolate \(Y(s)\text{:}\)
\begin{equation*} Y(s) = \frac{1}{s(s + 2)}\left(\frac{3e^{-5s}}{s} + 1\right)\text{.} \end{equation*}
and prepare for inversion by collecting the exponential terms:
\begin{equation*} \text{(prepared)}\quad Y(s) = F(s)e^{-5s} + \frac{1}{s(s + 2)}, \quad\text{where}\quad F(s) = \frac{3}{s^2(s + 2)}\text{.} \end{equation*}
Lastly, we compute \(f(t) = \ilap{F(s)}\) so it will be ready for step 3:
\begin{equation*} f(t) = -\frac34 + \frac32t + \frac34e^{-2t}\text{,} \end{equation*}
Step 3 — Leave the Laplace Domain.
Apply \(\laplacesym^{-1}\) to each term of the prepared \(Y(s)\text{:}\)
\begin{gather*} \ilap{Y(s)} = \ilap{F(s)e^{-5s}} + \ilap{\frac{1}{s(s + 2)}}\\ y(t) = f(t-5)u_5(t) + \frac12\left(1 - e^{-2t}\right) \end{gather*}
Substituting in \(f(t-5)\text{,}\) gives the final solution:
\begin{equation*} y(t) = \left(-\frac34 + \frac32(t-5) + \frac34e^{-2(t-5)}\right)u_5(t) + \frac12\left(1 - e^{-2t}\right) \end{equation*}

🌌 Example 291.

Let’s solve this initial value problem:
\begin{equation*} 2y' + y = g(t), \qquad y(0) = 0 \end{equation*}
with a piecewise forcing term:
\begin{equation*} g(t) = \begin{cases} 3, \amp 0 \le t \lt 2 \\ 1, \amp 2 \le t \lt 4 \\ t - 4, \amp t \ge 4 \end{cases} \end{equation*}
Solution.
We’ll follow the Laplace roadmap: forward transform → solve in the Laplace domain → inverse transform. The step function form of \(g(t)\) will guide us.
Setup — Write g(t) Using Step Functions.
First, express the piecewise forcing term in step function form:
\begin{gather*} g(t) = 3(u_0(t) - u_2(t)) + 1(u_2(t) - u_4(t)) + (t-4)u_4(t)\\ = 3u_0(t) - 2u_2(t) + (t-4)u_4(t). \end{gather*}
Step 1 — Into the Laplace Domain.
Apply \(\laplacesym\) to both sides:
\begin{equation*} 2\lap{y'} + \lap{y} = \lap{g(t)}. \end{equation*}
With \(y(0)=0\) and the derivative rule:
\begin{equation*} 2(sY(s)) + Y(s) = \lap{g(t)}. \end{equation*}
Transforming each piece of \(g(t)\text{:}\)
\begin{gather*} \lap{3u_0(t)} = \frac{3}{s}, \quad \lap{-2u_2(t)} = -\frac{2e^{-2s}}{s}, \quad \lap{(t-4)u_4(t)} = e^{-4s}\frac{1}{s^2}. \end{gather*}
The Laplace-domain equation becomes:
\begin{equation*} 2sY(s) + Y(s) = \frac{3}{s} - \frac{2e^{-2s}}{s} + \frac{e^{-4s}}{s^2}. \end{equation*}
Step 2 — Solve in the Laplace Domain.
Isolate \(Y(s)\text{:}\)
\begin{gather*} Y(s) = \frac{1}{2s + 1}\left(2 + \frac{3}{s} - \frac{2e^{-2s}}{s} + \frac{e^{-4s}}{s^2}\right). \end{gather*}
Distribute and write \(Y(s)\) as pieces of the form \(F(s)e^{-cs}\text{:}\)
\begin{equation*} Y(s) = \frac{2}{2s + 1} + \frac{3}{s(2s + 1)} - \frac{2}{s(2s + 1)}e^{-2s} + \frac{1}{s^2(2s + 1)}e^{-4s}\text{.} \end{equation*}
Identify the rational parts:
\begin{equation*} F_1(s) = \frac{2}{2s+1}, \quad F_2(s) = \frac{1}{s(2s+1)}, \quad F_3(s) = \frac{1}{s^2(2s+1)}\text{,} \end{equation*}
which completes the prepared \(Y(s)\text{:}\)
\begin{equation*} \text{(prepared)}\quad Y(s) = F_1(s) + 3F_2(s) - 2F_2(s)e^{-2s} + F_3(s)e^{-4s}\text{.} \end{equation*}
The partial fractions decomposition of \(F_2(s)\) and \(F_3(s)\) needed for step 3 are:
\begin{equation*} F_2(s) = \frac{1}{s} - \frac{2}{2s+1}, \qquad F_3(s) = \frac{-2}{s} + \frac{1}{s^2} + \frac{4}{2s+1}. \end{equation*}
Step 3 — Leave the Laplace Domain.
Apply \(\laplacesym^{-1}\) to both sides of the “prepared” \(Y(s)\text{:}\)
\begin{align*} \ilap{Y(s)} = \ilap{F_1(s)} \amp + 3\ilap{F_2(s)}\\ \amp - 2\ilap{F_2(s)e^{-2s}} + \ilap{F_3(s)e^{-4s}} \end{align*}
Using L\(_{11}\) updates our solution to:
\begin{equation*} y(t) = f_1(t) + 3f_2(t) - 2f_2(t-2)u_2(t) + f_3(t-4)u_4(t) \end{equation*}
Inverting each \(F_i(s)\) gives \(f_i(t)\text{:}\)
\begin{align*} f_1(t) = \ilap{F_1(s)} \amp = 2e^{-t/2}\\ f_2(t) = \ilap{F_2(s)} \amp = 1 - 2e^{-t/2}\\ f_3(t) = \ilap{F_3(s)} \amp = -2 + t + 2e^{-t/2} \end{align*}
Plugging in the \(f_i(t)\)’s, we get the final solution:
\begin{align*} y(t) = 2e^{-\frac12 t} \amp + 3 - 6e^{-\frac12 t} \\ \amp - 2\left(1 - 2e^{-\frac12(t-2)}\right)u_2(t) + \left((t-4) - 2 + 2e^{-\frac12(t-4)}\right)u_4(t). \end{align*}

🌌 Example 292.

Consider the initial value problem:
\begin{equation*} \frac{d^2y}{dt^2} - 4\frac{dy}{dt} - 5y = \begin{cases} e^{-t}, \amp 0 \le t \lt 1 \\ 1, \amp t \ge 1 \end{cases}, \qquad y(0) = 0,\ \frac{dy}{dt}(0) = 0. \end{equation*}
Solution.
The forcing function starts as \(e^{-t}\) and then abruptly switches to a constant at \(t=1\text{.}\) We’ll lean on the Laplace roadmap: forward transform → solve for \(Y(s)\) → return to \(y(t)\text{.}\)
Setup — Step Function Form.
First, we rewrite the forcing term with step functions so we can apply Laplace rules directly:
\begin{equation*} g(t) = e^{-t}(u_0(t) - u_1(t)) + 1 \cdot u_1(t). \end{equation*}
Combine the terms grouped by their switches:
\begin{equation*} g(t) = e^{-t}u_0(t) + \big(1 - e^{-t}\big)u_1(t). \end{equation*}
Step 1 — Into the Laplace Domain.
Apply \(\laplacesym\) to both sides:
\begin{equation*} \lap{\frac{d^2y}{dt^2}} - 4\lap{\frac{dy}{dt}} - 5\lap{y} = \lap{e^{-t}u_0(t)} + \lap{(1 - e^{-t})u_1(t)}. \end{equation*}
Use the derivative rules with \(y(0)=0, \sfrac{dy}{dt}(0)=0\) on the left, and the step/exponential rules on the right:
\begin{gather*} (s^2 - 4s - 5)Y(s) = \frac{1}{s + 1} + \left(\frac{1}{s} - \frac{1}{s + 1}\right)e^{-s}. \end{gather*}
Step 2 — Solve in the Laplace Domain.
Factor and isolate \(Y(s)\text{:}\)
\begin{equation*} Y(s) = \frac{1}{s^2 - 4s - 5} \left[\frac{1}{s + 1} + \left(\frac{1}{s} - \frac{1}{s + 1}\right)e^{-s}\right]. \end{equation*}
Since \(s^2 - 4s - 5 = (s + 1)(s - 5)\text{,}\) distribute and reorganize:
\begin{equation*} Y(s) = \frac{1}{(s + 1)(s - 5)} \left[\frac{1}{s + 1} + \frac{1}{s(s + 1)}e^{-s}\right]. \end{equation*}
Separate the terms into rational pieces to prepare for the inverse transform:
\begin{gather*} Y(s) = \ub{\frac{1}{(s + 1)^2(s - 5)}}_{\large F_1(s)} + \ub{\frac{1}{s(s + 1)^2(s - 5)}}_{\large F_2(s)} e^{-s}. \end{gather*}
Now
\begin{equation*} \text{(prepared)}\qquad Y(s) = F_1(s) + F_2(s)e^{-s}\text{.} \end{equation*}
Both \(F_1(s)\) and \(F_2(s)\) need partial fraction decomposition.
Step 3 — Leave the Laplace Domain.
Apply \(\laplacesym^{-1}\) to both sides of the “prepared” \(Y(s)\text{:}\)
\begin{equation*} \ilap{Y(s)} = \ilap{F_1(s)} + \ilap{F_2(s)e^{-s}} \end{equation*}
Using L\(_{11}\) updates our solution to:
\begin{equation*} y(t) = f_1(t) + f_2(t-1)u_1(t) \end{equation*}
Inverting \(F_1(s)\) and \(F_2(s)\) gives:
\begin{align*} f_1(t) \amp = \ilap{F_1(s)} = -\frac{1}{36}e^{-t} - \frac{1}{6}te^{-t} + \frac{1}{36}e^{5t}\\ f_2(t) \amp = \ilap{F_2(s)} = -\frac{1}{5} + \frac{1}{6}e^{-t} + \frac{2}{9}te^{-t} + \frac{1}{180}e^{5t} \end{align*}
Finally, substitute \(f_1(t)\) and \(f_2(t-2)\) to get the final solution:
\begin{align*} y(t) = - \amp\frac{1}{36}e^{-t} - \frac{1}{6}te^{-t} + \frac{1}{36}e^{5t} \\ \amp + \left[-\frac{1}{5} + \frac{1}{6}e^{-(t - 1)} + \frac{2}{9}(t - 1)e^{-(t - 1)} + \frac{1}{180}e^{5(t - 1)}\right] u_1(t). \end{align*}
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