Laplace Transforms¶

The classic first exercises in a controls course: work the Laplace transform table, run partial-fraction expansions, solve ODEs in the \(s\)-domain. This page takes three textbook problems — Homework 2, Problem 3.3(b) (forward transform), Homework 3, Problem 3.7(h) (inverse via partial-fraction expansion), and Homework 3, Problem 3.9(e) (a second-order ODE solved with Laplace) — and sends each to the model with a plain prompt.

Forward transforms — Homework 2, Problem 3.3(b)¶

Find the Laplace transform \(F(s) = \mathcal{L}\{f(t)\}\) for:

\[ f(t) = \cos 2t + 4\sin 5t + e^{-2t}\cos 7t \]

Textbook answer, from the standard Laplace table plus the frequency-shift property \(\mathcal{L}\{e^{-at}f(t)\} = F(s+a)\):

\[ F(s) = \frac{s}{s^2+4} + \frac{20}{s^2+25} + \frac{s+2}{(s+2)^2+49} \]

Pose the problem in the textbook's own wording — no custom system prompt, no formatting instructions, just the question as a student would ask it. The First Call pattern:

expr = "cos(2t) + 4 sin(5t) + e^(-2t) cos(7t)"

resp = client.chat.completions.create(
    model=model,
    messages=[
        {"role": "user", "content": f"Find the Laplace transform of the following time function: f(t) = {expr}"},
    ],
    temperature=0,
)
print(resp.choices[0].message.content)

Actual output — qwen-plus (2026-04-21)

We are asked to find the Laplace transform of

\[ f(t) = \cos(2t) + 4\sin(5t) + e^{-2t}\cos(7t) \]

Recall standard Laplace transforms (for \( \operatorname{Re}(s) > \text{appropriate value} \)):

\( \mathcal{L}\{\cos(at)\} = \dfrac{s}{s^2 + a^2}, \quad \text{Re}(s) > 0 \)
\( \mathcal{L}\{\sin(at)\} = \dfrac{a}{s^2 + a^2}, \quad \text{Re}(s) > 0 \)
\( \mathcal{L}\{e^{at}\cos(bt)\} = \dfrac{s - a}{(s - a)^2 + b^2}, \quad \text{Re}(s) > a \) (Note: For \( e^{-at}\cos(bt) \), replace \( a \to -a \), so it becomes \( \dfrac{s + a}{(s + a)^2 + b^2} \))

Let's compute term-by-term.

Term 1: \( \cos(2t) \)

\[ \mathcal{L}\{\cos(2t)\} = \frac{s}{s^2 + 2^2} = \frac{s}{s^2 + 4} \]

Term 2: \( 4\sin(5t) \)

\[ \mathcal{L}\{4\sin(5t)\} = 4 \cdot \frac{5}{s^2 + 5^2} = \frac{20}{s^2 + 25} \]

Term 3: \( e^{-2t}\cos(7t) \)

Use the s-shift property: If \( \mathcal{L}\{g(t)\} = G(s) \), then \( \mathcal{L}\{e^{at}g(t)\} = G(s - a) \). Here, \( g(t) = \cos(7t) \), so \( G(s) = \dfrac{s}{s^2 + 49} \), and \( a = -2 \), so:

\[ \mathcal{L}\{e^{-2t}\cos(7t)\} = G(s - (-2)) = G(s + 2) = \frac{s + 2}{(s + 2)^2 + 49} \]

Combine all terms:

\[ \boxed{ \mathcal{L}\{f(t)\} = \frac{s}{s^2 + 4} + \frac{20}{s^2 + 25} + \frac{s + 2}{(s + 2)^2 + 49} } \]

Region of convergence (ROC): - First term: Re(s) > 0 - Second term: Re(s) > 0 - Third term: Re(s) > −2 (since exponential decay rate is −2 ⇒ need Re(s) > −2 for convergence) So overall ROC is Re(s) > 0 (the intersection, since 0 > −2).

✅ Final answer:

\[ \boxed{ F(s) = \frac{s}{s^2 + 4} + \frac{20}{s^2 + 25} + \frac{s + 2}{s^2 + 4s + 53}, \quad \text{Re}(s) > 0 } \]

The final boxed expression matches the textbook term-for-term. Note that the model even volunteers the region-of-convergence discussion — behavior you'd reward in a homework submission but didn't have to prompt for. At temperature=0 the model isn't deriving the transform from first principles; it's recalling it, and recalling deterministically.

Inverse transforms — Homework 3, Problem 3.7(h)¶

Same textbook, reversed direction. You're given \(F(s)\) and asked for \(f(t)\), using partial-fraction decomposition:

\[ F(s) = \frac{3s+2}{s^2-s-2} \]

Textbook answer — both the partial-fraction decomposition and the final \(f(t)\):

Partial-fraction form	\(f(t)\)
\(\dfrac{1/3}{s+1} + \dfrac{8/3}{s-2}\)	\(\dfrac{1}{3}e^{-t} + \dfrac{8}{3}e^{2t}\)

This is a different skill from Problem 3.3 — it requires actual algebra (factoring the denominator \(s^2 - s - 2 = (s+1)(s-2)\), solving for residues), not just a table lookup. Notice that the textbook's problem statement already says "using partial-fraction expansions" — so the method requirement is baked into the prompt itself, and no custom system prompt is needed to force it. Using the textbook wording verbatim:

Fs = "(3s + 2) / (s^2 - s - 2)"

resp = client.chat.completions.create(
    model=model,
    messages=[
        {"role": "user", "content": f"Find the time function corresponding to the following Laplace transform using partial-fraction expansion: F(s) = {Fs}"},
    ],
    temperature=0,
)
print(resp.choices[0].message.content)

Actual output — qwen-plus (2026-04-21)

We are given the Laplace transform:

\[ F(s) = \frac{3s + 2}{s^2 - s - 2} \]

We need to find the inverse Laplace transform, i.e., find \( f(t) = \mathcal{L}^{-1}\{F(s)\} \), using partial-fraction expansion.

Step 1: Factor the denominator

[ s^2 - s - 2 = (s - 2)(s + 1) ] (Since roots of \(s^2 - s - 2 = 0\) are \(s = 2\) and \(s = -1\).)

So,

\[ F(s) = \frac{3s + 2}{(s - 2)(s + 1)} \]

Step 2: Partial fraction decomposition

Assume:

\[ \frac{3s + 2}{(s - 2)(s + 1)} = \frac{A}{s - 2} + \frac{B}{s + 1} \]

Multiply both sides by \((s - 2)(s + 1)\):

\[ 3s + 2 = A(s + 1) + B(s - 2) \]

Expand RHS:

\[ A(s + 1) + B(s - 2) = As + A + Bs - 2B = (A + B)s + (A - 2B) \]

Match coefficients with LHS \(3s + 2\):

Coefficient of \(s\): \(A + B = 3\)
Constant term: \(A - 2B = 2\)

Solve the system:

From first equation: \(A = 3 - B\)

Substitute into second:

\[ (3 - B) - 2B = 2 \Rightarrow 3 - 3B = 2 \Rightarrow -3B = -1 \Rightarrow B = \frac{1}{3} \]

Then \(A = 3 - \frac{1}{3} = \frac{8}{3}\)

So,

\[ F(s) = \frac{8/3}{s - 2} + \frac{1/3}{s + 1} \]

Step 3: Inverse Laplace transform

Recall:

\[ \mathcal{L}^{-1}\left\{\frac{1}{s - a}\right\} = e^{at}, \quad \text{for } t \ge 0 \]

Thus:

\[ f(t) = \frac{8}{3} e^{2t} + \frac{1}{3} e^{-t}, \quad t \ge 0 \]

✅ Final Answer:

\[ \boxed{f(t) = \frac{8}{3}e^{2t} + \frac{1}{3}e^{-t}, \quad t \ge 0} \]

The response includes the PF decomposition alongside the final inverse transform — the model respects the textbook's wording and shows the work. Both the decomposition and \(f(t)\) match the textbook. Plain prompt, textbook wording verbatim, zero machinery.

Solving ODEs with Laplace — Homework 3, Problem 3.9(e)¶

Step up: solve an ODE using Laplace transforms. The pattern is longer — Laplace-transform both sides, apply the initial conditions, solve algebraically for \(Y(s)\), partial-fraction, then inverse-transform back to \(y(t)\):

\[ y''(t) + 2y'(t) + 2y(t) = 5\sin t, \quad y(0) = y'(0) = 0 \]

Textbook answer:

\[ y(t) = \sin t - 2\cos t + 2e^{-t}\cos t + e^{-t}\sin t \]

More steps than a transform lookup or a pure partial-fraction problem, but every step is still within the well-trodden textbook repertoire. Same prompt shape, textbook wording:

resp = client.chat.completions.create(
    model=model,
    messages=[
        {"role": "user", "content": "Solve the following ODE using Laplace transforms: y''(t) + 2 y'(t) + 2 y(t) = 5 sin(t), given y(0) = y'(0) = 0"},
    ],
    temperature=0,
)
print(resp.choices[0].message.content)

Actual output — qwen-plus (2026-04-21)

We are given the second-order linear nonhomogeneous ODE:

\[ y''(t) + 2y'(t) + 2y(t) = 5 \sin t, \quad y(0) = 0,\; y'(0) = 0 \]

We will solve it using Laplace transforms.

Step 1: Take Laplace transform of both sides

Recall Laplace transforms:

\(\mathcal{L}\{y'(t)\} = sY(s) - y(0)\)
\(\mathcal{L}\{y''(t)\} = s^2 Y(s) - s y(0) - y'(0)\)
\(\mathcal{L}\{\sin(at)\} = \frac{a}{s^2 + a^2}\)

Given \(y(0) = y'(0) = 0\), we have:

\(\mathcal{L}\{y'(t)\} = sY(s)\)
\(\mathcal{L}\{y''(t)\} = s^2 Y(s)\)

Also, \(\mathcal{L}\{5 \sin t\} = 5 \cdot \frac{1}{s^2 + 1} = \frac{5}{s^2 + 1}\)

Now apply Laplace transform to the ODE:

\[ s^2 Y(s) + 2s Y(s) + 2 Y(s) = \frac{5}{s^2 + 1} \]

Factor left-hand side:

\[ (s^2 + 2s + 2) Y(s) = \frac{5}{s^2 + 1} \]

So,

\[ Y(s) = \frac{5}{(s^2 + 2s + 2)(s^2 + 1)} \]

Step 2: Partial fraction decomposition

Complete the square: \(s^2 + 2s + 2 = (s+1)^2 + 1\).

Decompose:

\[ \frac{5}{(s^2 + 1)\big((s+1)^2 + 1\big)} = \frac{As + B}{s^2 + 1} + \frac{Cs + D}{(s+1)^2 + 1} \]

Equate coefficients; solving the 4×4 system gives \(A = -2,\; B = 1,\; C = 2,\; D = 3\).

\[ Y(s) = \frac{-2s + 1}{s^2 + 1} + \frac{2s + 3}{(s+1)^2 + 1} \]

Step 3: Inverse Laplace transform

\(\frac{-2s + 1}{s^2 + 1} \;\rightarrow\; -2\cos t + \sin t\)
For the second term, substitute \(u = s+1\): \(\frac{2s + 3}{(s+1)^2 + 1} = \frac{2u + 1}{u^2 + 1}\), whose inverse (with the \(e^{-t}\) shift) is \(e^{-t}(2\cos t + \sin t)\).

Combine:

\[ \boxed{y(t) = -2 \cos t + \sin t + e^{-t}(2 \cos t + \sin t)} \]

Verify initial conditions:

\(y(0) = -2(1) + 0 + 1 \cdot (2 + 0) = 0\) ✅
\(y'(0)\): after differentiating and evaluating at 0, get \(1 - 1 = 0\) ✅

Both initial conditions satisfied.

The model chains all three stages (Laplace → partial fractions → inverse Laplace) and arrives at the textbook answer, then volunteers a verification that both initial conditions are satisfied — a detail you'd want on a homework submission but didn't prompt for. Still a plain prompt, still deterministic at temperature=0.

Next¶

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