\(\displaystyle \mathscr{L}^{-1}[\cdots] = \frac{\left(5 \sin{\left(3 t \right)} - 6 \cos{\left(3 t \right)}\right) e^{- t}}{3}\)
Solving ODEs via Laplace Transforms
MATH 341 — Notes 7
University of Scranton
2026-06-16
Goals
What We Will Cover
Derivative formulas (Theorem 3.11) — proof and bootstrapping applications
The Three-Step Method for IVPs — worked examples
Partial fractions — all three pole types (consolidation and drill)
Completing the square for complex poles
Piecewise forcing — Heaviside step approach for harder examples
Smoothing remark — why PWC input gives continuous output
Note
Section 3.2 of (Logan 2015).
Tip
Key principle: differentiation \(\to\) multiplication by \(s\). ICs appear automatically.
The Three-Step Method
Logan’s Figure 3.2
The Three Steps
\[\boxed{\text{ODE in }x(t)} \xrightarrow{\;\mathscr{L}\;} \boxed{\text{Algebraic equation in }X(s)} \xrightarrow{\;\text{solve}\;} \boxed{X(s)} \xrightarrow{\;\mathscr{L}^{-1}\;} \boxed{x(t)}\]
Step 1 — Transform: apply \(\mathscr{L}\) to both sides, use the derivative formulas, plug in ICs.
Step 2 — Solve: pure algebra — solve for \(X(s)\).
Step 3 — Invert: use the table + partial fractions + completing the square.
Note
The initial conditions enter automatically in Step 1 through the derivative formulas. You never need to apply them separately.
Relationship to Notes 6. Notes 6 introduced the same workflow as a detailed step-by-step checklist (five items covering how to handle each term in the ODE, how to group, and how to check the result). The three-step diagram here is the high-level framework — it names the three stages. The checklist in Notes 6 is more granular — it spells out what to do within each stage. Think of the diagram as the map and the checklist as the turn-by-turn directions. Both describe exactly the same algorithm.
Derivative Formulas
Theorem 3.11 (Logan) — Proof
Notes 6 stated these formulas to motivate the Laplace strategy. Here we prove them and see how they generate new transforms via a “bootstrapping” technique.
The Key Operational Formulas
\[\mathscr{L}[x'] = sX(s) - x(0) \tag{3.6}\] \[\mathscr{L}[x''] = s^2X(s) - sx(0) - x'(0) \tag{3.7}\]
Proof of (3.6): IBP with \(u=e^{-st}\), \(dv=x'dt\): \[\mathscr{L}[x'] = [x(t)e^{-st}]_0^\infty + s\int_0^\infty xe^{-st}dt = -x(0) + sX(s). \;\checkmark\]
Proof of (3.7): apply (3.6) to \(x'' = (x')'\): \[\mathscr{L}[x''] = s\mathscr{L}[x'] - x'(0) = s(sX-x(0)) - x'(0). \;\checkmark\]
Tip
Pattern: \(\mathscr{L}[x^{(n)}] = s^n X(s) - s^{n-1}x(0) - \cdots - x^{(n-1)}(0)\).
Each derivative adds one power of \(s\) and one initial condition.
Using Theorem 3.11 to Find New Transforms
Example 3.12 — \(\mathscr{L}[t^2]\).
\((t^2)' = 2t\), so \(\mathscr{L}[(t^2)'] = s\,\mathscr{L}[t^2] - 0 = \dfrac{2}{s^2}\), giving \(\mathscr{L}[t^2] = \dfrac{2}{s^3}\).
By induction: \(\mathscr{L}[t^n] = \dfrac{n!}{s^{n+1}}\).
Example 3.13 — \(\mathscr{L}[\cosh t]\).
Using \((\cosh t)'' = \cosh t\), \(\cosh 0 = 1\), \(\sinh 0 = 0\): \[\mathscr{L}[\cosh t] = s^2\mathscr{L}[\cosh t] - s \;\Longrightarrow\; \mathscr{L}[\cosh t] = \frac{s}{s^2-1}.\]
Tip
These bootstrapping arguments — using the derivative formulas to find transforms rather than evaluate integrals — are elegant and reusable.
Solving IVPs
Example 3.14 — Pure Oscillator
Solve: \(x'' + k^2x = 0\), \(x(0)=0\), \(x'(0)=1\).
Step 1 — Transform: \[s^2X - s\cdot\underbrace{0}_{x(0)} - \underbrace{1}_{x'(0)} + k^2X = 0 \;\Longrightarrow\; (s^2+k^2)X = 1.\]
Step 2 — Solve: \[X(s) = \frac{1}{s^2+k^2} = \frac{1}{k}\cdot\frac{k}{s^2+k^2}.\]
Step 3 — Invert (table: \(\mathscr{L}[\sin kt] = k/(s^2+k^2)\)): \[\boxed{x(t) = \frac{1}{k}\sin kt.}\]
Note
The IC \(x'(0)=1\) appeared as the \(-1\) term in Step 1. No separate substitution needed.
Example 3.15 — First-Order Nonhomogeneous
Solve: \(x' + 2x = e^{-t}\), \(x(0)=0\).
Step 1 — Transform: \[sX - 0 + 2X = \frac{1}{s+1} \;\Longrightarrow\; (s+2)X = \frac{1}{s+1}.\]
Step 2 — Solve: \[X(s) = \frac{1}{(s+1)(s+2)}.\]
Step 3 — Invert (table entry \(1/[(s-a)(s-b)]\), or partial fractions): \[\boxed{x(t) = e^{-t} - e^{-2t}.}\]
Example 3.16 — PF Verification
\(\dfrac{1}{(s+1)(s+2)} = \dfrac{1}{s+1} - \dfrac{1}{s+2}\), giving \(e^{-t}-e^{-2t}\). \(\checkmark\)
Partial Fractions
Three Cases — Consolidation and Drill
Note
Notes 6 introduced all three pole types and the cover-up rule. This section consolidates that material with additional examples — the goal is fluency, not novelty. Work through these actively: cover the answer and try each step before it appears.
Case 1 — Distinct real poles \((s-a)(s-b)\): \[\frac{A}{s-a}+\frac{B}{s-b} \;\longrightarrow\; Ae^{at}+Be^{bt}\]
Cover-up: \(A = X(s)\cdot(s-a)\big|_{s=a}\), \(B = X(s)\cdot(s-b)\big|_{s=b}\).
Case 2 — Repeated pole \((s-a)^2\): \[\frac{A}{s-a}+\frac{B}{(s-a)^2} \;\longrightarrow\; Ae^{at}+Bte^{at}\]
Case 3 — Complex poles \((s-\alpha)^2+\beta^2\):
Complete the square, then: \[\frac{A(s-\alpha)+B\beta}{(s-\alpha)^2+\beta^2} \;\longrightarrow\; e^{\alpha t}(A\cos\beta t+B\sin\beta t)\]
Tip
Decision rule: factor the denominator first. Real root \(\to\) Case 1 or 2. Discriminant negative \(\to\) complete the square (Case 3).
Case 3: Completing the Square
Example 3.17 (Logan) — \(X(s) = \dfrac{1}{s^2+3s+6}\).
\[s^2+3s+6 = \left(s+\tfrac{3}{2}\right)^2 - \tfrac{9}{4}+6 = \left(s+\tfrac{3}{2}\right)^2 + \left(\tfrac{\sqrt{15}}{2}\right)^2\]
Match \(k/[(s-\alpha)^2+k^2]\) with \(\alpha=-3/2\), \(k=\sqrt{15}/2\):
\[X(s) = \frac{2}{\sqrt{15}}\cdot\frac{\sqrt{15}/2}{(s+3/2)^2+(\sqrt{15}/2)^2}\]
\[\boxed{x(t) = \frac{2}{\sqrt{15}}\,e^{-3t/2}\sin\frac{\sqrt{15}}{2}t}\]
General Recipe for \(s^2+bs+c\)
\[s^2+bs+c = \left(s+\frac{b}{2}\right)^2 + \underbrace{\left(c-\frac{b^2}{4}\right)}_{\beta^2}\] \(\alpha = -b/2\), \(\beta = \sqrt{c-b^2/4}\) (when \(4c>b^2\)).
Completing the Square with Non-Trivial Numerator
Find \(\mathscr{L}^{-1}\!\left[\dfrac{3-2s}{s^2+2s+10}\right]\).
Complete square: \(s^2+2s+10 = (s+1)^2+9\).
Rewrite numerator: \(3-2s = -2(s+1)+5\).
\[\frac{3-2s}{(s+1)^2+9} = -2\cdot\frac{s+1}{(s+1)^2+9} + \frac{5}{3}\cdot\frac{3}{(s+1)^2+9}\]
\[\boxed{x(t) = e^{-t}\!\left(-2\cos 3t + \tfrac{5}{3}\sin 3t\right)}\]
ODE with Complex Poles
Solve: \(x''-2x'+2x=0\), \(x(0)=0\), \(x'(0)=1\).
Step 1: Substitute ICs into (3.7): \[\bigl[s^2X - s\cdot\underbrace{0}_{x(0)} - \underbrace{1}_{x'(0)}\bigr] - 2\bigl[sX - \underbrace{0}_{x(0)}\bigr] + 2X = 0 \;\Longrightarrow\; (s^2-2s+2)X = 1\]
Tip
The \(-1\) comes from \(-x'(0)=-1\). A common slip is to forget this term and write \((s^2-2s+2)X=0\) — always expand the derivative formula in full before collecting.
Step 2: \(X(s) = \dfrac{1}{s^2-2s+2}\)
Step 3: \((s-1)^2+1\) — shift property:
\[\boxed{x(t) = e^t\sin t}\]
Piecewise Continuous Forcing
The Heaviside Approach
Why do we care? Methods from Chapter 2 fail for PWC forcing because we’d need to solve on each sub-interval and match solutions at every junction.
The Laplace method handles everything in one calculation.
Example 3.18 (Logan) — non-constant piecewise pieces: \[f(t) = \begin{cases}t & 0<t<1 \\ 2 & 1\leq t\leq 3 \\ 0 & t>3\end{cases} = t + (2-t)H(t-1) - 2H(t-3)\]
For \(\mathscr{L}[tH(t-1)]\): use \(\mathscr{L}[f(t)H(t-a)] = e^{-as}\mathscr{L}[f(t+a)]\) with \(f(t)=t\): \[\mathscr{L}[tH(t-1)] = e^{-s}\mathscr{L}[t+1] = e^{-s}\!\left(\frac{1}{s^2}+\frac{1}{s}\right)\]
\[F(s) = \frac{1}{s^2}+\frac{e^{-s}}{s}-\frac{e^{-s}}{s^2}-\frac{2e^{-3s}}{s}\]
Example 3.19 — Oscillator, Half-Sine Forcing
Solve: \(x''+4x = \sin t - H(t-\pi)\sin t\), \(x(0)=x'(0)=0\).
Key step — transform \(H(t-\pi)\sin t\): \[\mathscr{L}[H(t-\pi)\sin t] = e^{-\pi s}\mathscr{L}[\sin(t+\pi)] = e^{-\pi s}\mathscr{L}[-\sin t] = -\frac{e^{-\pi s}}{s^2+1}\]
(s^2+4)X = \(\dfrac{1+e^{-\pi s}}{s^2+1}\); PF: \(A=1/3\), \(B=-1/3\):
\[\mathscr{L}^{-1}\!\left[\frac{1}{(s^2+1)(s^2+4)}\right] = \frac{1}{3}\sin t - \frac{1}{6}\sin 2t\]
Switching property for \(e^{-\pi s}\) term; \(\sin(t-\pi)=-\sin t\):
\[\boxed{x(t) = \tfrac{1}{3}\sin t - \tfrac{1}{6}\sin 2t + H(t-\pi)\!\left(-\tfrac{1}{3}\sin t - \tfrac{1}{6}\sin 2t\right)}\]
Example 3.19 — Solution Plot
Note
After \(t=\pi\) the oscillator runs freely at a modified amplitude — the forcing deposited energy that persists.
Example 3.21 — RC Circuit, Square Pulse
Solve: \(q'+3q = H(t-1)-H(t-2)\), \(q(0)=0\).
\((s+3)Q = \dfrac{e^{-s}-e^{-2s}}{s}\); \(\quad Q = \dfrac{e^{-s}}{s(s+3)} - \dfrac{e^{-2s}}{s(s+3)}\)
PF: \(\dfrac{1}{s(s+3)} = \dfrac{1/3}{s}-\dfrac{1/3}{s+3}\), so \(\mathscr{L}^{-1}\!\left[\dfrac{1}{s(s+3)}\right] = \dfrac{1}{3}(1-e^{-3t})\).
Switching property for each term:
\[\boxed{q(t) = \tfrac{1}{3}(1-e^{-3(t-1)})H(t-1) - \tfrac{1}{3}(1-e^{-3(t-2)})H(t-2)}\]
Smoothing
Remark 3.20 — PWC Input, Continuous Output
Question: if \(f(t)\) is discontinuous, how smooth is \(x(t)\)?
Answer (Remark 3.20): \(x(t)\) is always more regular than \(f(t)\).
| ODE order | Input regularity | Output regularity |
|---|---|---|
| 1st | PWC (jump) | Continuous |
| 2nd | PWC (jump) | Continuously differentiable |
| \(n\)th | PWC | \(C^{n-1}\) |
Example: \(x'' = H(t-3)\), \(x(0)=x'(0)=0\).
\(x'(t) = (t-3)H(t-3)\) — continuous, kink at \(t=3\).
\(x(t) = \frac{1}{2}(t-3)^2H(t-3)\) — smooth.
Tip
Integration smooths. Each integration adds one degree of regularity. The Laplace transform handles this automatically — you never need to patch solutions together at jump points.
Summary
Key Takeaways
Theorem 3.11 — \(\mathscr{L}[x''] = s^2X-sx(0)-x'(0)\) — converts the ODE to algebra with ICs built in.
Three steps: transform → solve for \(X(s)\) → invert using table + PF.
Partial fractions: distinct real poles → exponentials; repeated → \(te^{at}\); complex → complete the square → damped sinusoids.
PWC forcing: write \(f(t)\) as Heaviside steps; transform using \(\mathscr{L}[f(t)H(t-a)]=e^{-as}\mathscr{L}[f(t+a)]\); each \(e^{-as}\) factor in \(X(s)\) produces a time-delayed copy in \(x(t)\).
Smoothing (Remark 3.20): a discontinuous input produces a continuous output — integration gains regularity, no case-by-case patching needed.
| Pole type | \(X(s)\) form | \(x(t)\) |
|---|---|---|
| Distinct real | \(\frac{A}{s-a}+\frac{B}{s-b}\) | \(Ae^{at}+Be^{bt}\) |
| Repeated | \(\frac{A}{s-a}+\frac{B}{(s-a)^2}\) | \(Ae^{at}+Bte^{at}\) |
| Complex | \(\frac{A(s-\alpha)+B\beta}{(s-\alpha)^2+\beta^2}\) | \(e^{\alpha t}(A\cos\beta t+B\sin\beta t)\) |
References
Logan, J David. 2015. A First Course in Differential Equations, Third Edition.
MATH 341 Differential Equations — Notes 7