Laplace of derivatives — key property for solving ODEs: Worked examples

This page hammers the derivative rule from the parent property against every kind of case it can meet. Read the matrix first, then each worked example tells you exactly which cell it lives in.

The scenario matrix

Here $s$ is the Laplace variable, $Y = L {y}$ the transform of the unknown, and "IC" means initial condition (a value of $y$ or its derivative at a chosen time). Each row is a distinct case class; the last column names the example that covers it.

Case class	What makes it different	Covered by
1st order, nonzero IC	the $- f (0)$ fee is actually nonzero	Ex 1
1st order, zero IC + forcing	fee vanishes, forcing $\frac{1}{s}$ enters	Ex 2
2nd order, two real roots ( $s^{2} - a^{2}$ )	denominator factors into $(s - a) (s + a)$	Ex 3
2nd order, complex roots (oscillation)	denominator $s^{2} + ω^{2}$ → sin/cos	Ex 4
2nd order, repeated root (critical)	denominator $(s + a)^{2}$ → $t e^{- a t}$	Ex 5
Degenerate: $n = 0$ / constant / pure decay limit	check the rule doesn't break at edges	Ex 6
Word problem (real units)	RC circuit / cooling, physical meaning	Ex 7
Exam twist: IC given not at $t = 0$	formula's boundary sits at $0$ — must shift	Ex 8

Every numeric answer that appears below is machine-checked in the verify block.

Worked example Ex 4 — 2nd order, complex roots (pure oscillation) ·

Case 4 Solve $y^{''} + 4 y = 0, y (0) = 3, y^{'} (0) = 0$ .

Forecast: $s^{2} + 4$ never factors over the reals — roots are $\pm 2 i$ , meaning oscillation at angular frequency $2$ . Starting at $3$ with zero slope → a pure cosine: guess $y = 3 cos 2 t$ .

Transform. $(s^{2} Y - sy (0) - y^{'} (0)) + 4 Y = 0$ .
Insert ICs. $y (0) = 3, y^{'} (0) = 0$ : $s^{2} Y - 3 s - 0 + 4 Y = 0 \Rightarrow (s^{2} + 4) Y = 3 s$ . Why this step? Notice the $y (0)$ produced an $s$ in the numerator — that $s$ is what turns into a cosine rather than a sine.
Solve. $Y = \frac{3 s}{s ^{2} + 4}$ .
Invert. $L {cos ω t} = \frac{s}{s ^{2} + ω ^{2}}$ with $ω = 2$ , so $y (t) = 3 cos 2 t$ .

Verify: $y^{'} = - 6 sin 2 t$ , $y^{''} = - 12 cos 2 t$ ; $y^{''} + 4 y = - 12 cos 2 t + 12 cos 2 t = 0$ ✓, $y (0) = 3$ ✓, $y^{'} (0) = 0$ ✓.

Look at the chalk-blue solution curve: it starts at height $3$ with a flat top (slope zero, matching $y^{'} (0) = 0$ ) and swings symmetrically — that flat start is the visual signature of a pure cosine.

Worked example Ex 5 — 2nd order, repeated root (critical damping) ·

Case 5 Solve $y^{''} + 2 y^{'} + y = 0, y (0) = 0, y^{'} (0) = 1$ .

Forecast: Characteristic-style denominator will be $s^{2} + 2 s + 1 = (s + 1)^{2}$ — a repeated root $- 1$ . Repeated roots produce a $t e^{- t}$ factor. Guess $y = t e^{- t}$ .

Transform. $(s^{2} Y - sy (0) - y^{'} (0)) + 2 (s Y - y (0)) + Y = 0$ . Why this step? Apply the 2nd-derivative rule to $y^{''}$ and the 1st-derivative rule to $y^{'}$ — the property handles both.
Insert ICs. $y (0) = 0, y^{'} (0) = 1$ : $(s^{2} Y - 1) + 2 s Y + Y = 0 \Rightarrow (s^{2} + 2 s + 1) Y = 1$ .
Solve. $Y = \frac{1}{( s + 1 ) ^{2}}$ .
Invert. $L {t e^{a t}} = \frac{1}{( s - a ) ^{2}}$ with $a = - 1$ , so $y (t) = t e^{- t}$ . Why this step? The squared denominator is precisely the fingerprint of the $t e^{a t}$ shape — this is the case that has no plain exponential.

Verify: $y^{'} = e^{- t} - t e^{- t}$ , $y^{''} = - 2 e^{- t} + t e^{- t}$ ; $y^{''} + 2 y^{'} + y = (- 2 e^{- t} + t e^{- t}) + 2 (e^{- t} - t e^{- t}) + t e^{- t} = 0$ ✓, $y (0) = 0$ ✓, $y^{'} (0) = 1$ ✓.

Worked example Ex 6 — Degenerate / edge cases ·

Case 6 Three sanity checks at the boundaries of the rule.

(a) $n = 0$ (no derivative). By convention $f^{(0)} = f$ , and the general formula's list of subtracted terms is empty, so $L {f} = s^{0} F (s) = F (s)$ . Why this matters: the pattern doesn't collapse — it just reduces to the identity, confirming the descending-power list starts at $s^{n - 1}$ and stops before $n = 0$ contributes.

(b) Constant function $y \equiv c$ . Its derivative is $0$ , so $L {y^{'}} = 0$ . The rule says $s Y - y (0) = s \cdot \frac{c}{s} - c = c - c = 0$ ✓ — the fee exactly cancels $s F (s)$ . Why this step? A constant has zero rate of change; the property must return $0$ , and it does.

(c) Limiting behaviour of Ex 2. As $t \to \infty$ , $y = 2 (1 - e^{- 4 t}) \to 2$ ; as $t \to 0^{+}$ , $y \to 0$ . This is the Final Value and Initial Value behaviour: $lim_{s \to 0} s Y = lim_{s \to 0} \frac{8}{s + 4} = 2$ and $lim_{s \to \infty} s Y = lim_{s \to \infty} \frac{8}{s + 4} = 0$ . Why this step? Multiplying $Y$ by $s$ and taking $s$ limits reads off the $t$ -endpoints without inverting — a free consistency check.

Worked example Ex 8 — Exam twist: IC not at

$t = 0$ · Case 8 Solve $y^{'} - y = 0$ with the condition given away from zero: $y (1) = e$ .

Forecast: The property's boundary term lives exactly at $t = 0$ , so we cannot plug $y (1)$ in directly. Solve with an unknown $y (0) = c$ , then fit $c$ from $y (1) = e$ . The equation $y^{'} = y$ gives $y = c e^{t}$ , and $y (1) = ce = e \Rightarrow c = 1$ , so $y = e^{t}$ .

Transform with unknown start. Let $y (0) = c$ . $(s Y - c) - Y = 0 \Rightarrow (s - 1) Y = c$ . Why this step? We honestly cannot use $y (1)$ inside the rule; we carry the true $t = 0$ value as a symbol $c$ .
Solve. $Y = \frac{c}{s - 1}$ .
Invert. $y (t) = c e^{t}$ .
Fit the given condition. $y (1) = c e^{1} = e \Rightarrow c = 1$ . Hence $y (t) = e^{t}$ . Why this step? This is the correct handling of the steel-manned mistake in the parent: an IC away from $0$ must be fitted after solving generally.

Verify: $y^{'} = e^{t} = y$ ✓, and $y (1) = e^{1} = e$ ✓.

Recall Which cell was which?

Ex1 hits nonzero-IC first order ::: $y^{'} + 2 y = 0 \Rightarrow y = 5 e^{- 2 t}$ . Ex4 vs Ex3 — how do you see the difference in $s$ ? ::: $s^{2} + ω^{2}$ (complex roots → oscillation) vs $s^{2} - a^{2}$ (real roots → $sinh / cosh$ ). Ex5 fingerprint in the $s$ -domain? ::: A squared denominator $(s + a)^{2}$ → a $t e^{- a t}$ term. Why can't Ex8 plug $y (1)$ into the rule directly? ::: The derivative property evaluates its boundary term at $t = 0$ , so the IC must be at $0$ ; otherwise solve generally and fit.

Connections

4.6.28 Laplace of derivatives — key property for solving ODEs (Hinglish)
Solving Initial Value Problems with Laplace
Inverse Laplace Transform and Partial Fractions
Laplace Transform — Definition and Existence
Linearity of the Laplace Transform
Integration by Parts
Exponential Order and Convergence of Integrals
Fourier Transform — comparison (no boundary term)