Solving ODEs with Laplace (including discontinuous forcing): Worked examples

The scenario matrix

Think of every linear-ODE-with-Laplace problem as a point in a grid. The two axes that decide how hard the algebra gets are:

What the denominator of $Y (s)$ looks like — distinct real roots, a repeated root, or a complex pair (which becomes an oscillation).
What the forcing does — nothing (homogeneous), a smooth push, a single switch $u (t - a)$ , a pulse (switch on then off), or a hammer-blow $δ (t - a)$ .

Below, each cell is a class of behaviour. Every cell is covered by at least one worked example.

Cell	Denominator type of $Y (s)$	Forcing type	Behaviour you must recognise	Example
A	distinct real roots	smooth (or none)	pure decaying exponentials	Ex 1
B	complex pair $s^{2} + ω^{2}$	none	pure oscillation	Ex 2
C	complex pair, root = forcing frequency	smooth sinusoid	resonance, amplitude grows like $t$	Ex 3
D	distinct real roots	single switch $u (t - a)$	gated relaxation, delay tag $e^{- a s}$	Ex 4
E	distinct real roots	pulse $u (t - a) - u (t - b)$	on-then-off, two delay tags	Ex 5
F	must rewrite forcing as $g (t - a)$	ramp $\times$ step	shifting trap $t = (t - a) + a$	Ex 6
G	distinct real roots	Dirac $δ (t - a)$	instant kick, no $1/ s$	Ex 7
H	repeated real root	smooth	terms with a factor $t e^{r t}$	Ex 8

Figure — Solving ODEs with Laplace (including discontinuous forcing)

Cell A — distinct real roots, smooth forcing

Worked example Example 1 ·

$y^{''} + 3 y^{'} + 2 y = e^{- 3 t}$ , $y (0) = 0, y^{'} (0) = 0$ Forecast: guess before working. The homogeneous roots come from $s^{2} + 3 s + 2 = (s + 1) (s + 2)$ , so $e^{- t}$ and $e^{- 2 t}$ . The forcing adds an $e^{- 3 t}$ . Expect a sum of three decaying exponentials, all $\to 0$ .

Step 1 — Stamp. With both ICs zero, $L {y^{''}} = s^{2} Y$ , $L {y^{'}} = s Y$ , and $L {e^{- 3 t}} = \frac{1}{s + 3}$ : $(s^{2} + 3 s + 2) Y = \frac{1}{s + 3} .$ Why this step? The recipe demands we transform term-by-term; zero ICs mean no boundary terms survive.

Step 2 — Solve. Divide (pure algebra): $Y = \frac{1}{( s + 1 ) ( s + 2 ) ( s + 3 )} .$ Why this step? $Y (s)$ is now a rational function — ready for Partial Fractions.

Step 3 — Stick-back. Split by cover-up (multiply by the factor, set $s$ to its root): $\frac{1}{( s + 1 ) ( s + 2 ) ( s + 3 )} = \frac{A}{s + 1} + \frac{B}{s + 2} + \frac{C}{s + 3},$ $A = \frac{1}{( 1 ) ( 2 )} = \frac{1}{2}, B = \frac{1}{( - 1 ) ( 1 )} = - 1, C = \frac{1}{( - 2 ) ( - 1 )} = \frac{1}{2} .$ Read each off the table $\frac{1}{s - r} \leftrightarrow e^{r t}$ : $y (t) = \frac{1}{2} e^{- t} - e^{- 2 t} + \frac{1}{2} e^{- 3 t}$ Why cover-up works: multiplying both sides by $(s + 1)$ then setting $s = - 1$ kills every other term because they still carry an $(s + 1)$ factor.

Verify: at $t = 0$ , $y (0) = \frac{1}{2} - 1 + \frac{1}{2} = 0$ ✔. Differentiate: $y^{'} = - \frac{1}{2} e^{- t} + 2 e^{- 2 t} - \frac{3}{2} e^{- 3 t}$ , so $y^{'} (0) = - \frac{1}{2} + 2 - \frac{3}{2} = 0$ ✔. Both ICs recovered.

Cell B — complex pair, no forcing (pure oscillation)

Worked example Example 2 ·

$y^{''} + 4 y = 0$ , $y (0) = 3, y^{'} (0) = - 2$ Forecast: $s^{2} + 4$ has roots $\pm 2 i$ → oscillation at $ω = 2$ . Non-zero ICs feed both a cosine (from position) and a sine (from velocity). Amplitude constant forever (no decay).

Step 1 — Stamp. Now ICs are not zero, so keep every boundary term: $(s^{2} Y - s \cdot 3 - (- 2)) + 4 Y = 0.$ Why this step? $L {y^{''}} = s^{2} Y - s y (0) - y^{'} (0)$ — dropping these is the classic mistake.

Step 2 — Solve. $(s^{2} + 4) Y = 3 s - 2 \Rightarrow Y = \frac{3 s - 2}{s ^{2} + 4} .$

Step 3 — Stick-back. Split to match the two table entries $\frac{s}{s ^{2} + ω ^{2}} \leftrightarrow cos ω t$ and $\frac{ω}{s ^{2} + ω ^{2}} \leftrightarrow sin ω t$ with $ω = 2$ : $Y = 3 \cdot \frac{s}{s ^{2} + 4} - 1 \cdot \frac{2}{s ^{2} + 4} .$ Why write $- 2$ as $- 1 \cdot 2$ ? The sine table entry needs the numerator to be exactly $ω = 2$ ; we factor to make that visible. $y (t) = 3 cos 2 t - sin 2 t$ Verify: $y (0) = 3 \cdot 1 - 0 = 3$ ✔. $y^{'} = - 6 sin 2 t - 2 cos 2 t$ , so $y^{'} (0) = - 2$ ✔.

Cell C — resonance (forcing frequency = natural frequency)

Worked example Example 3 ·

$y^{''} + y = cos t$ , $y (0) = 0, y^{'} (0) = 0$ Forecast: natural frequency $ω = 1$ (from $s^{2} + 1$ ), forcing frequency also $1$ . Resonance! Expect the amplitude to grow like $t$ .

Step 1 — Stamp. $L {cos t} = \frac{s}{s ^{2} + 1}$ : $(s^{2} + 1) Y = \frac{s}{s ^{2} + 1} .$

Step 2 — Solve. $Y = \frac{s}{( s ^{2} + 1 ) ^{2}} .$ Why this is different: the factor $s^{2} + 1$ appears twice. No sum of plain $cos t, sin t$ can produce this; we need the growing pair.

Step 3 — Stick-back. Recall the pair (derivable, but memorise it): $L {t sin ω t} = \frac{2 ω s}{( s ^{2} + ω ^{2} ) ^{2}} .$ With $ω = 1$ : $L {t sin t} = \frac{2 s}{( s ^{2} + 1 ) ^{2}}$ , so $\frac{s}{( s ^{2} + 1 ) ^{2}} = \frac{1}{2} L {t sin t}$ . $y (t) = \frac{1}{2} t sin t$ The $t$ out front is the growing envelope — the physical signature of resonance.

Verify: $y (0) = 0$ ✔. $y^{'} = \frac{1}{2} sin t + \frac{1}{2} t cos t$ , $y^{'} (0) = 0$ ✔. Plug into the ODE: $y^{''} = cos t - \frac{1}{2} t sin t$ , so $y^{''} + y = cos t - \frac{1}{2} t sin t + \frac{1}{2} t sin t = cos t$ ✔.

Cell D — distinct real roots, single switch

Worked example Example 4 ·

$y^{'} + 2 y = 3 u (t - 1)$ , $y (0) = 0$ Forecast: nothing until $t = 1$ ; then the system relaxes toward its steady value $3/2$ (set $y^{'} = 0$ : $2 y = 3$ ). Root $- 2$ → time constant $\frac{1}{2}$ .

Step 1 — Stamp. $L {u (t - 1)} = \frac{e ^{- s}}{s}$ : $(s + 2) Y = \frac{3 e ^{- s}}{s} .$

Step 2 — Solve. $Y = e^{- s} \cdot H (s) \frac{3}{s ( s + 2 )} .$ Why factor out $e^{- s}$ ? The delay tag rides untouched through the algebra; we invert the plain $H (s)$ first, then delay-and-gate.

Step 3 — Stick-back. Cover-up on $H$ : at $s = 0$ residue $\frac{3}{2}$ ; at $s = - 2$ residue $\frac{3}{- 2} = - \frac{3}{2}$ : $H (s) = \frac{3/2}{s} - \frac{3/2}{s + 2} \Rightarrow h (t) = \frac{3}{2} - \frac{3}{2} e^{- 2 t} .$ Then $Y = e^{- s} H (s)$ means delay $h$ by $1$ AND gate with $u (t - 1)$ : $y (t) = u (t - 1) [\frac{3}{2} - \frac{3}{2} e^{- 2 (t - 1)}]$ Verify: for $t < 1$ , $y = 0$ ✔. As $t \to \infty$ , $y \to \frac{3}{2}$ = steady state ✔. Just after $t = 1$ : $y (1^{+}) = \frac{3}{2} - \frac{3}{2} = 0$ , continuous ✔.

Cell E — a pulse (on then off)

Cell F — the shifting trap (rewrite the forcing first)

Worked example Example 6 ·

$y^{'} + y = t u (t - 2)$ , $y (0) = 0$ Forecast: silent until $t = 2$ ; then driven by a ramp. Because the ramp value at the switch is already $2$ , expect a jump-like build-up, not a start from zero slope.

Step 1 — Rewrite the forcing. Let $τ = t - 2$ , so $t = τ + 2$ : $t u (t - 2) = [(t - 2) + 2] u (t - 2) = g_{1} (t - 2) u (t - 2) + g_{2} 2 u (t - 2) .$ Why this step? Both pieces are now genuine functions of $(t - 2)$ , so the shift theorem applies cleanly. $L {(t - 2) u (t - 2)} = e^{- 2 s} \cdot \frac{1}{s ^{2}}, L {2 u (t - 2)} = 2 \cdot \frac{e ^{- 2 s}}{s} .$ So $L {t u (t - 2)} = e^{- 2 s} (\frac{1}{s ^{2}} + \frac{2}{s})$ .

Step 2 — Stamp & Solve. $(s + 1) Y = e^{- 2 s} (\frac{1}{s ^{2}} + \frac{2}{s}) \Rightarrow Y = e^{- 2 s} H (s) \frac{2 s + 1}{s ^{2} ( s + 1 )} .$ (Combined: $\frac{1}{s ^{2}} + \frac{2}{s} = \frac{1 + 2 s}{s ^{2}}$ , then $\div (s + 1)$ .)

Step 3 — Stick-back. Partial-fraction $H (s) = \frac{2 s + 1}{s ^{2} ( s + 1 )} = \frac{A}{s} + \frac{B}{s ^{2}} + \frac{C}{s + 1}$ . Clear denominators: $2 s + 1 = A s (s + 1) + B (s + 1) + C s^{2}$ .

$s = 0$ : $1 = B$ .
$s = - 1$ : $- 1 = C$ .
compare $s^{2}$ coefficient: $0 = A + C \Rightarrow A = 1$ .

So $H (s) = \frac{1}{s} + \frac{1}{s ^{2}} - \frac{1}{s + 1} \Rightarrow h (t) = 1 + t - e^{- t}$ . Delay-and-gate by $2$ : $y (t) = u (t - 2) [1 + (t - 2) - e^{- (t - 2)}] = u (t - 2) [t - 1 - e^{- (t - 2)}]$ Verify: for $t < 2$ , $y = 0$ ✔. At $t = 2^{+}$ : $y = 1 + 0 - 1 = 0$ , continuous ✔. Check the ODE for $t > 2$ : $y = t - 1 - e^{- (t - 2)}$ , $y^{'} = 1 + e^{- (t - 2)}$ , so $y^{'} + y = (1 + e^{- (t - 2)}) + (t - 1 - e^{- (t - 2)}) = t$ ✔ (matches forcing $t$ ).

Cell G — Dirac impulse (a hammer blow)

Worked example Example 7 ·

$y^{''} + 2 y^{'} + 2 y = δ (t - 1)$ , $y (0) = 0, y^{'} (0) = 0$ Forecast: system at rest is kicked at $t = 1$ . Roots of $s^{2} + 2 s + 2$ : complete the square $(s + 1)^{2} + 1$ → $- 1 \pm i$ → decaying oscillation $e^{- t}$ envelope, $ω = 1$ . So after $t = 1$ expect a damped wiggle.

Step 1 — Stamp. $L {δ (t - 1)} = e^{- s}$ : $(s^{2} + 2 s + 2) Y = e^{- s} .$

Step 2 — Solve. $Y = e^{- s} H (s) \frac{1}{( s + 1 ) ^{2} + 1} .$

Step 3 — Stick-back. Match the shifted-sine table entry $L {e^{- t} sin t} = \frac{1}{( s + 1 ) ^{2} + 1}$ (this is the standard $\frac{ω}{( s - α ) ^{2} + ω ^{2}}$ with $α = - 1, ω = 1$ ). So $h (t) = e^{- t} sin t$ . Delay-and-gate by $1$ : $y (t) = u (t - 1) e^{- (t - 1)} sin (t - 1)$ Verify: for $t < 1$ , $y = 0$ ✔. Just after the kick, $y (1^{+}) = 0$ but the velocity jumps: $y^{'} (1^{+}) = e^{0} (cos 0) = 1$ — a unit impulse injects one unit of velocity into a system starting from rest ✔ (consistent with the $δ$ delivering one unit of "kick").

Cell H — repeated real root

The whole matrix in one picture

Recall Quick self-test

Which cell has a linearly-growing amplitude and why? ::: Cell C (resonance): forcing frequency equals a natural frequency, giving a repeated complex factor $(s^{2} + ω^{2})^{2}$ → a $t sin ω t$ term. Why does $t u (t - 2)$ need rewriting before the shift theorem? ::: The theorem needs a function of $(t - 2)$ ; write $t = (t - 2) + 2$ so both pieces are shifted, giving $e^{- 2 s} (1/ s^{2} + 2/ s)$ . What is $L {δ (t - a)}$ and why is impulse forcing easy? ::: $e^{- a s}$ — just the bare delay tag, no $1/ s$ , so it multiplies the transfer factor directly. Signature of a repeated real root $r$ in the answer? ::: A factor $t e^{r t}$ (double root) or $t^{2} e^{r t}$ (triple), from $L {t^{n} e^{r t}} = n! / (s - r)^{n + 1}$ .