The Laplace Transform

MATH 341 — Notes 6

University of Scranton

2026-06-16

Goals

What We Will Cover

  1. The big picture — what the transform does and why it works

  2. Definition \(X(s)=\int_0^\infty x(t)e^{-st}dt\) and computing from it

  3. Linearity and the transform table

  4. Inverse transform and partial fractions

  5. Heaviside function — modeling switches

  6. Shift and switching properties

  7. A full IVP with discontinuous forcing

Note

Section 3.1 of (Logan 2015).

Tip

New vocabulary: time domain, \(s\)-domain, transform pair, exponential order, piecewise continuous.

The Big Picture

Why a New Method?

Undetermined coefficients and variation of parameters fail when \(f(t)\) is:

  • Discontinuous — a switch flipped at \(t=2\)
  • An impulse — a hammer blow at \(t=5\)

These are the most common real-world forcing functions in circuits, mechanics, and control.

The Laplace Strategy

\[\underbrace{ax''+bx'+cx = f(t)}_{\text{ODE (hard)}} \xrightarrow{\;\mathscr{L}\;} \underbrace{(as^2+bs+c)X = F(s)+\text{ICs}}_{\text{algebra (easy)}} \xrightarrow{\;\mathscr{L}^{-1}\;} \underbrace{x(t)}_{\text{solution}}\]

Differentiation \(\;\longrightarrow\;\) multiplication by \(s\).

ICs are automatically incorporated. No separate step.

Think of the logarithm analogy: \(\log(ab) = \log a + \log b\) converts multiplication to addition. The Laplace transform converts differentiation to multiplication.

The Definition

Definition and Notation

Definition

\[\mathscr{L}[x(t)](s) = X(s) = \int_0^\infty x(t)\,e^{-st}\,dt\] provided the improper integral converges.

Note

Notation: These slides use \(\mathscr{L}\) (script); Logan and the long-form notes use \(\mathcal{L}\) (calligraphic). Both denote the same operator. The script form is used here solely because it renders more cleanly in the slide environment — do not let the font difference confuse you when reading the textbook.

Time domain Transform domain
Variable \(t\) \(s\)
Function \(x(t)\) (lowercase) \(X(s)\) (uppercase)
Language “what happens” “how fast it oscillates/decays”

The damping kernel \(e^{-st}\) forces the integrand to zero as \(t\to\infty\) (when \(s>0\) is large enough), making the integral converge. Larger \(s\) damps the function more aggressively.

Computing from the Definition

\(\mathscr{L}[e^{at}]\): \[\int_0^\infty e^{at}e^{-st}dt = \int_0^\infty e^{(a-s)t}dt = \frac{1}{s-a}, \quad s>a.\]

\(\mathscr{L}[1]\) (set \(a=0\)): \(\quad 1/s\), \(\;s>0\).

\(\mathscr{L}[t]\) (integration by parts, \(u=t\), \(dv=e^{-st}dt\)): \[\int_0^\infty te^{-st}dt = \frac{1}{s^2}, \quad s>0.\]

\(\mathscr{L}[\sin kt]\) (using \(\sin kt = \frac{1}{2i}(e^{ikt}-e^{-ikt})\)): \[\frac{k}{s^2+k^2}, \qquad \mathscr{L}[\cos kt] = \frac{s}{s^2+k^2}.\]

Tip

Note: \(\mathscr{L}[\cos kt]\) has \(s\) in the numerator; \(\mathscr{L}[\sin kt]\) has the constant \(k\). This reflects \(\cos 0 = 1\) vs. \(\sin 0 = 0\).

Geometric Intuition

Figure 1

Linearity and Existence

Linearity

\[\mathscr{L}[\alpha x + \beta y] = \alpha\mathscr{L}[x] + \beta\mathscr{L}[y]\]

Example: \(\mathscr{L}[3e^{-2t}+5\sin 4t] = \dfrac{3}{s+2} + \dfrac{20}{s^2+16}\)

Existence conditions (sufficient):

  1. Piecewise continuous on every \([0,b]\) — finite number of finite-jump discontinuities.
  2. Exponential order — \(|x(t)| \leq Me^{rt}\) for large \(t\).

Failure examples: - \(e^{t^2}\): not exponential order (grows too fast) - \(1/t\): not piecewise continuous at \(t=0\) - If \(X(s)\not\to 0\) as \(s\to\infty\), it cannot be a Laplace transform

The Inverse Transform

\(\mathscr{L}^{-1}\) and Partial Fractions

\[x(t) = \mathscr{L}^{-1}[X(s)] \qquad \text{(also linear)}\]

In practice: partial fractions decompose \(X(s)\) into table-recognizable pieces.

Type 1 — Distinct real poles: \[\frac{P(s)}{(s-a)(s-b)} = \frac{A}{s-a}+\frac{B}{s-b} \;\Rightarrow\; Ae^{at}+Be^{bt}\]

Type 2 — Repeated pole: \[\frac{P(s)}{(s-a)^2} = \frac{A}{s-a}+\frac{B}{(s-a)^2} \;\Rightarrow\; Ae^{at}+Bte^{at}\]

Type 3 — Complex poles (complete the square): \[\frac{1}{s^2+2s+5} = \frac{1}{(s+1)^2+4} \;\Rightarrow\; \frac{1}{2}e^{-t}\sin 2t\]

Partial Fractions: Worked Example

Find \(\mathscr{L}^{-1}\!\left[\dfrac{2s+9}{(s+1)(s+3)}\right]\).

Write: \(\dfrac{2s+9}{(s+1)(s+3)} = \dfrac{A}{s+1}+\dfrac{B}{s+3}\)

Cover-up rule: \[A = \frac{2(-1)+9}{(-1+3)} = \frac{7}{2}, \qquad B = \frac{2(-3)+9}{(-3+1)} = -\frac{3}{2}.\]

Invert: \[x(t) = \frac{7}{2}e^{-t} - \frac{3}{2}e^{-3t}.\]

s_v, t_v = sym.symbols('s t', positive=True)
expr = (2*s_v+9)/((s_v+1)*(s_v+3))
display(Math(r"\text{PF: }" + sym.latex(sym.apart(expr,s_v))))
display(Math(r"x(t)=" + sym.latex(sym.inverse_laplace_transform(expr,s_v,t_v))))

\(\displaystyle \text{PF: }- \frac{3}{2 \left(s + 3\right)} + \frac{7}{2 \left(s + 1\right)}\)

\(\displaystyle x(t)=\frac{7 e^{- t}}{2} - \frac{3 e^{- 3 t}}{2}\)

Transform of Derivatives

The Key ODE Formula — Preview

The properties above let us compute transforms and inverses. To solve ODEs we need one more ingredient: what does \(\mathscr{L}\) do to \(x'\) and \(x''\)?

Transform of Derivatives (stated here; proved in Notes 7 as Theorem 3.11)

\[\mathscr{L}[x'(t)] = s\,X(s) - x(0)\] \[\mathscr{L}[x''(t)] = s^2 X(s) - s\,x(0) - x'(0)\]

Pattern: \(\mathscr{L}[x^{(n)}] = s^n X(s) - s^{n-1}x(0) - \cdots - x^{(n-1)}(0)\).

Why this is powerful. Applying \(\mathscr{L}\) to \(ax''+bx'+cx=f(t)\) with ICs \(x(0)=x_0\), \(x'(0)=x_1\): \[(as^2+bs+c)X(s) = F(s) + a(sx_0+x_1) + bx_0.\]

The ODE becomes an algebraic equation for \(X(s)\), and the initial conditions appear on the right automatically — no separate enforcement step needed.

First Look at the Three-Step Method

\[\boxed{\text{ODE in }x(t)} \xrightarrow{\;\mathscr{L}\;} \boxed{\text{Algebra in }X(s)} \xrightarrow{\;\text{solve}\;} \boxed{X(s)} \xrightarrow{\;\mathscr{L}^{-1}\;} \boxed{x(t)}\] Notes 7 develops this method fully, with proofs and multiple worked examples.

The Heaviside Function

Modeling On/Off Switches

\[H(t-a) = \begin{cases} 0, & t < a \\ 1, & t \geq a \end{cases}\]

A mathematical switch: off before \(t=a\), on at \(t=a\).

\[\mathscr{L}[H(t-a)] = \int_a^\infty e^{-st}dt = \frac{e^{-as}}{s}, \quad s>0.\]

Building piecewise functions. The pulse \(H(t-a)-H(t-b)\) equals 1 on \([a,b)\), zero elsewhere.

Example (Logan 3.8): \[f(t) = \begin{cases}3 & 0\leq t<2\\4 & 2\leq t<3\\2 & 3\leq t<6\\0 & t>6\end{cases} = 3H(t) + H(t-2) - 2H(t-3) - 2H(t-6)\]

\[F(s) = \frac{1}{s}\left(3 + e^{-2s} - 2e^{-3s} - 2e^{-6s}\right)\]

Heaviside Decomposition — Visual

Figure 2

Tip

Think of it as a running adjustment: start at \(3\), add \(1\) at \(t=2\), subtract \(2\) at \(t=3\), subtract \(2\) at \(t=6\).

Operational Properties

Shift Property

Shift Property

\[\mathscr{L}[f(t)e^{at}] = F(s-a)\]

Multiplying by \(e^{at}\) in the time domain shifts the transform \(a\) units right in the \(s\)-domain.

\(f(t)\) \(F(s)\) \(f(t)e^{at}\) \(\mathscr{L}[f(t)e^{at}]\)
\(t\) \(1/s^2\) \(te^{-2t}\) \(1/(s+2)^2\)
\(\sin 3t\) \(3/(s^2+9)\) \(e^{-t}\sin 3t\) \(3/((s+1)^2+9)\)
\(\cos 3t\) \(s/(s^2+9)\) \(e^{2t}\cos 3t\) \((s-2)/((s-2)^2+9)\)

Inverse direction. Whenever you see \((s-a)^2+k^2\) in the denominator, complete the square and apply the shift property in reverse.

Example: \(\mathscr{L}^{-1}\!\left[\dfrac{3}{(s+1)^2+9}\right] = e^{-t}\sin 3t.\)

Switching Property

Switching Property

\[\mathscr{L}[H(t-a)f(t-a)] = e^{-as}F(s)\] \[\mathscr{L}^{-1}[e^{-as}F(s)] = H(t-a)f(t-a)\]

A factor \(e^{-as}\) in the \(s\)-domain signals a time delay of \(a\).

Example. Find \(\mathscr{L}^{-1}\!\left[\dfrac{e^{-3s}}{s-2}\right]\).

Since \(\mathscr{L}^{-1}[1/(s-2)] = e^{2t}\), the switching property gives: \[\mathscr{L}^{-1}\!\left[\frac{e^{-3s}}{s-2}\right] = H(t-3)\,e^{2(t-3)}.\]

The exponential starts growing from \(t=3\), not from \(t=0\).

Tip

Signal: see an \(e^{-as}\) multiplying a transform? That’s a time delay. Strip off the exponential, invert the remaining \(F(s)\), replace \(t\) with \(t-a\), and multiply by \(H(t-a)\).

Discontinuous Forcing

Example: RC Circuit with Switch

Solve \(x' + 2x = f(t)\), \(x(0)=0\), where \(f(t)=1\) for \(0\leq t<3\), \(f(t)=0\) for \(t\geq 3\).

Step 1. \(f(t) = H(t) - H(t-3)\), so \(F(s) = \frac{1}{s} - \frac{e^{-3s}}{s}\).

Step 2. \((s+2)X = \frac{1-e^{-3s}}{s}\), so \(X(s) = \frac{1}{s(s+2)} - \frac{e^{-3s}}{s(s+2)}\).

Step 3. PF: \(\frac{1}{s(s+2)} = \frac{1/2}{s} - \frac{1/2}{s+2}\), so \(\mathscr{L}^{-1}\!\left[\frac{1}{s(s+2)}\right] = \frac{1}{2}(1-e^{-2t})\).

Step 4. Switching: \(\mathscr{L}^{-1}\!\left[\frac{e^{-3s}}{s(s+2)}\right] = H(t-3)\cdot\frac{1}{2}(1-e^{-2(t-3)})\).

\[\boxed{x(t) = \frac{1}{2}(1-e^{-2t}) - \frac{H(t-3)}{2}(1-e^{-2(t-3)})}\]

Discontinuous Example — Solution Plot

Figure 3

Note

The solution \(x(t)\) is continuous even though the forcing is discontinuous. Its derivative \(x'(t)\) has a jump at \(t=3\).

This is a first taste of discontinuous forcing — Notes 7 develops the full technique with more complex piecewise functions and second-order equations.

The Transform Table

Short Table of Laplace Transforms

\(x(t)\) \(X(s)\)
\(e^{at}\) \(\frac{1}{s-a}\)
\(t^n\) \(\frac{n!}{s^{n+1}}\)
\(\sin kt\) \(\frac{k}{s^2+k^2}\)
\(\cos kt\) \(\frac{s}{s^2+k^2}\)
\(e^{at}\sin kt\) \(\frac{k}{(s-a)^2+k^2}\)
\(e^{at}\cos kt\) \(\frac{s-a}{(s-a)^2+k^2}\)
\(H(t-a)\) \(\frac{e^{-as}}{s}\)
\(H(t-a)f(t-a)\) \(e^{-as}F(s)\)
\(f(t)e^{at}\) \(F(s-a)\)
\(x^{(n)}(t)\) \(s^n X - s^{n-1}x(0)-\cdots-x^{(n-1)}(0)\)
\(\delta_a(t)\) \(e^{-as}\)
\(\int_0^t f(\tau)g(t-\tau)d\tau\) \(F(s)G(s)\)

Summary: The Solution Checklist

Relationship to Notes 7

Notes 7 presents this as the Three-Step Method (transform → solve → invert), following Logan’s Figure 3.2. The checklist below is the same algorithm expanded with implementation details for each step.

Detailed Checklist (same algorithm, more granular)

  1. Write \(f(t)\) using Heaviside steps if piecewise.
  2. Apply \(\mathscr{L}\) to both sides; use the derivative formulas (Notes 7, Theorem 3.11).
  3. Solve for \(X(s)\) algebraically.
  4. Partial fractions to decompose \(X(s)\) into table forms.
  5. Invert term by term; use the switching property for each \(e^{-as}\) factor.
  • Each \(e^{-as}\) factor in \(X(s)\) signals a time delay: strip it, invert the rest as \(f(t)\), then replace \(t\to t-a\) and multiply by \(H(t-a)\).
  • Complex poles \((s-\alpha)^2+\beta^2\): complete the square, then use \(\mathscr{L}[e^{\alpha t}\sin\beta t]\) and \(\mathscr{L}[e^{\alpha t}\cos\beta t]\).
  • Repeated poles \((s-a)^2\): the inverse is \(Ae^{at}+Bte^{at}\).
  • Initial conditions are free: they appear automatically when you expand \(\mathscr{L}[x'']\) and \(\mathscr{L}[x']\) — no need to enforce them separately.

References

Logan, J David. 2015. A First Course in Differential Equations, Third Edition.

Looking Ahead

The convolution theorem (\(\mathscr{L}[f*g]=F(s)G(s)\)) and the Dirac delta (\(\mathscr{L}[\delta_a]=e^{-as}\)) — both already in the table — are the main tools of §3.2 and beyond. Convolution expresses solutions as integrals over arbitrary forcing; the delta models instantaneous impulses that no classical function can describe.

Next: Applying Laplace Transforms to Differential Equations — Logan §3.2.