Solving ODEs with Laplace (including discontinuous forcing): Exercises

Before we start, one shared toolbox — every symbol you will see, in plain words:

Definition The symbols we use (nothing enters unexplained)

$t$ is time, running from $0$ upward. Our unknown answer is a function $y (t)$ : "where is the system at time $t$ ."
$s$ is the algebra-world variable the Laplace transform sends us into. Think of it as a knob; large $s$ means "focus on early, fast behaviour."
$L {y (t)} = Y (s)$ means "the transform of $y$ ." Capital letter = transformed (in $s$ -land); small letter = original (in $t$ -land).
$y^{'}$ is the slope of $y$ (rate of change); $y^{''}$ is the slope of the slope (acceleration).
$u (t - a)$ is the switch: $0$ before time $a$ , then jumps to $1$ and stays there. Picture a light turned on at $t = a$ .
$δ (t - a)$ is a hammer blow: an infinitely sharp kick delivered exactly at $t = a$ .
$e^{- a s}$ is the delay tag: whenever it multiplies something in $s$ -land, it means "in $t$ -land, start this piece $a$ seconds late."

The three rules we reuse constantly:

Level 1 — Recognition

L1.1 Write $L {y^{''} - 4 y^{'} + 3 y}$ in terms of $Y = Y (s)$ with general initial conditions $y (0) = y_{0}$ , $y^{'} (0) = y_{1}$ .

Recall Solution L1.1

Transform term by term. What we do: replace each derivative by its rule. Why: the transform is linear, so it passes through $+$ and constants. $L {y^{''}} = s^{2} Y - s y_{0} - y_{1}, L {- 4 y^{'}} = - 4 (s Y - y_{0}), L {3 y} = 3 Y .$ Add: $L {y^{''} - 4 y^{'} + 3 y} = (s^{2} - 4 s + 3) Y - s y_{0} - y_{1} + 4 y_{0} .$

L1.2 State $L {u (t - 3)}$ and $L {δ (t - 3)}$ .

Recall Solution L1.2

Special case $g = 1$ of the shift rule: $L {u (t - 3)} = \frac{e ^{- 3 s}}{s}$ . The impulse: $L {δ (t - 3)} = e^{- 3 s}$ . Why the difference? The step is "on forever after $t = 3$ " (a lasting $\frac{1}{s}$ ), the impulse is "one instant only" (no $\frac{1}{s}$ ).

L1.3 Invert $\frac{2 s}{s ^{2} + 9}$ and $\frac{4}{s ^{2} + 9}$ .

Recall Solution L1.3

Match $\frac{s}{s ^{2} + ω ^{2}} \leftrightarrow cos ω t$ with $ω = 3$ : $\frac{2 s}{s ^{2} + 9} \to 2 cos 3 t$ . Match $\frac{ω}{s ^{2} + ω ^{2}} \leftrightarrow sin ω t$ : we need $ω = 3$ on top, but we have $4$ . Write $\frac{4}{s ^{2} + 9} = \frac{4}{3} \cdot \frac{3}{s ^{2} + 9} \to \frac{4}{3} sin 3 t$ .

Level 2 — Application

L2.1 Solve $y^{'} + 3 y = 0$ , $y (0) = 5$ .

Recall Solution L2.1

Transform: $(s Y - 5) + 3 Y = 0$ . Solve: $(s + 3) Y = 5 \Rightarrow Y = \frac{5}{s + 3}$ . Invert: $\frac{1}{s - a} \leftrightarrow e^{a t}$ with $a = - 3$ , so $y (t) = 5 e^{- 3 t}$ . Check: $y (0) = 5$ ✔, and $y^{'} = - 15 e^{- 3 t} = - 3 y$ ✔.

L2.2 Solve $y^{''} + 4 y = 0$ , $y (0) = 1$ , $y^{'} (0) = 0$ .

Recall Solution L2.2

Transform: $(s^{2} Y - s \cdot 1 - 0) + 4 Y = 0$ . Solve: $(s^{2} + 4) Y = s \Rightarrow Y = \frac{s}{s ^{2} + 4}$ . Invert: $\frac{s}{s ^{2} + ω ^{2}} \leftrightarrow cos ω t$ with $ω = 2$ , so $y (t) = cos 2 t$ . Check: $y (0) = 1$ ✔, $y^{'} (0) = - 2 sin 0 = 0$ ✔.

L2.3 Solve $y^{''} - y = 0$ , $y (0) = 0$ , $y^{'} (0) = 1$ .

Recall Solution L2.3

Transform: $(s^{2} Y - 0 - 1) - Y = 0 \Rightarrow (s^{2} - 1) Y = 1 \Rightarrow Y = \frac{1}{s ^{2} - 1}$ . Partial fractions: $\frac{1}{s ^{2} - 1} = \frac{1}{( s - 1 ) ( s + 1 )} = \frac{1/2}{s - 1} - \frac{1/2}{s + 1}$ . Invert: $y (t) = \frac{1}{2} e^{t} - \frac{1}{2} e^{- t} = sinh t$ . Check: $y (0) = 0$ ✔, $y^{'} (0) = cosh 0 = 1$ ✔.

L2.4 Solve $y^{'} + 2 y = e^{- t}$ , $y (0) = 0$ .

Recall Solution L2.4

Transform: $s Y + 2 Y = \frac{1}{s + 1}$ . Solve: $Y = \frac{1}{( s + 1 ) ( s + 2 )}$ . Partial fractions: cover-up at $s = - 1$ gives $\frac{1}{( - 1 + 2 )} = 1$ ; at $s = - 2$ gives $\frac{1}{( - 2 + 1 )} = - 1$ . So $Y = \frac{1}{s + 1} - \frac{1}{s + 2}$ . Invert: $y (t) = e^{- t} - e^{- 2 t}$ . Check: $y (0) = 0$ ✔.

Level 3 — Analysis (switches and shifts)

L3.1 Solve $y^{'} + 2 y = u (t - 1)$ , $y (0) = 0$ .

Recall Solution L3.1

Transform: $s Y + 2 Y = \frac{e ^{- s}}{s}$ , so $Y = \frac{e ^{- s}}{s ( s + 2 )}$ . Invert the un-delayed core first. Let $H (s) = \frac{1}{s ( s + 2 )}$ . Cover-up: at $s = 0$ residue $\frac{1}{2}$ , at $s = - 2$ residue $- \frac{1}{2}$ . So $H = \frac{1/2}{s} - \frac{1/2}{s + 2}$ and $h (t) = \frac{1}{2} - \frac{1}{2} e^{- 2 t}$ . Apply the delay tag $e^{- s}$ : delay $h$ by $1$ AND gate it: $y (t) = u (t - 1) [\frac{1}{2} - \frac{1}{2} e^{- 2 (t - 1)}] .$ Meaning: nothing until $t = 1$ , then $y$ rises toward $\frac{1}{2}$ . See the figure.

Figure — Solving ODEs with Laplace (including discontinuous forcing)

L3.2 Find $L {t u (t - 2)}$ .

Recall Solution L3.2

The trick: the shift rule needs the forcing written as $g (t - 2)$ , a function of $(t - 2)$ . But we have plain $t$ . Rewrite: $t = (t - 2) + 2$ . So $g (t - 2) = (t - 2) + 2$ , i.e. $g (τ) = τ + 2$ . $G (s) = L {τ + 2} = \frac{1}{s ^{2}} + \frac{2}{s}$ . Multiply by the delay tag: $L {t u (t - 2)} = e^{- 2 s} (\frac{1}{s ^{2}} + \frac{2}{s}) .$

L3.3 A rectangular pulse: solve $y^{'} + y = u (t - 1) - u (t - 3)$ , $y (0) = 0$ . (Input is ON only between $t = 1$ and $t = 3$ .)

Recall Solution L3.3

Transform: $s Y + Y = \frac{e ^{- s} - e ^{- 3 s}}{s}$ , so $Y = \frac{e ^{- s} - e ^{- 3 s}}{s ( s + 1 )}$ . Core inverse: $H (s) = \frac{1}{s ( s + 1 )} = \frac{1}{s} - \frac{1}{s + 1} \Rightarrow h (t) = 1 - e^{- t}$ . Two delay tags → two gated copies (subtract the second): $y (t) = u (t - 1) [1 - e^{- (t - 1)}] - u (t - 3) [1 - e^{- (t - 3)}] .$ Meaning: flat $0$ until $t = 1$ ; charges up toward $1$ ; at $t = 3$ the input switches off and the second term makes it relax back toward $0$ . See the figure.

Level 4 — Synthesis (combine tools)

L4.1 Solve $y^{''} + 3 y^{'} + 2 y = u (t - 1)$ , $y (0) = 0$ , $y^{'} (0) = 0$ , then evaluate $y (2)$ .

Recall Solution L4.1

Transform: $(s^{2} + 3 s + 2) Y = \frac{e ^{- s}}{s}$ , so $Y = \frac{e ^{- s}}{s ( s + 1 ) ( s + 2 )}$ . Core inverse: $H (s) = \frac{1}{s ( s + 1 ) ( s + 2 )}$ . Cover-up: at $s = 0$ : $\frac{1}{( 1 ) ( 2 )} = \frac{1}{2}$ ; at $s = - 1$ : $\frac{1}{( - 1 ) ( 1 )} = - 1$ ; at $s = - 2$ : $\frac{1}{( - 2 ) ( - 1 )} = \frac{1}{2}$ . $H = \frac{1/2}{s} - \frac{1}{s + 1} + \frac{1/2}{s + 2} \Rightarrow h (t) = \frac{1}{2} - e^{- t} + \frac{1}{2} e^{- 2 t} .$ Delay + gate: $y (t) = u (t - 1) [\frac{1}{2} - e^{- (t - 1)} + \frac{1}{2} e^{- 2 (t - 1)}]$ . Evaluate at $t = 2$ (so $t - 1 = 1$ ): $y (2) = \frac{1}{2} - e^{- 1} + \frac{1}{2} e^{- 2} \approx 0.5 - 0.367879 + 0.067668 = 0.199789$ .

L4.2 Solve $y^{''} + y = sin t$ , $y (0) = 0$ , $y^{'} (0) = 0$ (resonance — the forcing matches the natural frequency).

Recall Solution L4.2

Transform: $(s^{2} + 1) Y = \frac{1}{s ^{2} + 1}$ , so $Y = \frac{1}{( s ^{2} + 1 ) ^{2}}$ . Why is this special? The forcing $sin t$ has the same frequency as the natural oscillation, so its transform shares the same denominator — squaring it. A repeated quadratic factor signals resonance, growth in $t$ . Invert using the known pair $\frac{1}{( s ^{2} + ω ^{2} ) ^{2}} \leftrightarrow \frac{1}{2 ω ^{3}} (sin ω t - ω t cos ω t)$ , with $ω = 1$ : $y (t) = \frac{1}{2} (sin t - t cos t) .$ The $- \frac{1}{2} t cos t$ term grows without bound — the amplitude climbs linearly, the fingerprint of resonance. Check: $y (0) = 0$ ; $y^{'} (t) = \frac{1}{2} (t sin t)$ (after simplification) so $y^{'} (0) = 0$ ✔.

L4.3 Solve $y^{''} + 4 y = u (t - π)$ , $y (0) = 0$ , $y^{'} (0) = 0$ ; find $y (\frac{3 π}{2})$ .

Recall Solution L4.3

Transform: $(s^{2} + 4) Y = \frac{e ^{- π s}}{s}$ , so $Y = \frac{e ^{- π s}}{s ( s ^{2} + 4 )}$ . Core inverse: $H (s) = \frac{1}{s ( s ^{2} + 4 )} = \frac{A}{s} + \frac{B s + C}{s ^{2} + 4}$ . Multiply out: $1 = A (s^{2} + 4) + (B s + C) s$ . At $s = 0$ : $A = \frac{1}{4}$ . Coeff of $s^{2}$ : $A + B = 0 \Rightarrow B = - \frac{1}{4}$ . Coeff of $s$ : $C = 0$ . $H = \frac{1/4}{s} - \frac{1}{4} \cdot \frac{s}{s ^{2} + 4} \Rightarrow h (t) = \frac{1}{4} - \frac{1}{4} cos 2 t .$ Delay + gate: $y (t) = u (t - π) [\frac{1}{4} - \frac{1}{4} cos 2 (t - π)]$ . At $t = \frac{3 π}{2}$ : $t - π = \frac{π}{2}$ , $cos 2 \cdot \frac{π}{2} = cos π = - 1$ . So $y = \frac{1}{4} - \frac{1}{4} (- 1) = \frac{1}{2}$ .

Level 5 — Mastery (impulses, reasoning, edge cases)

L5.1 Solve $y^{''} + 2 y^{'} + y = δ (t - 2)$ , $y (0) = 0$ , $y^{'} (0) = 0$ . (A single hammer blow on a critically-damped system.)

Recall Solution L5.1

Transform: $(s^{2} + 2 s + 1) Y = e^{- 2 s}$ . Note $s^{2} + 2 s + 1 = (s + 1)^{2}$ . So $Y = \frac{e ^{- 2 s}}{( s + 1 ) ^{2}}$ . Core inverse: $\frac{1}{( s + 1 ) ^{2}} \leftrightarrow t e^{- t}$ (from $\frac{n !}{( s - a ) ^{n + 1}} \leftrightarrow t^{n} e^{a t}$ with $n = 1, a = - 1$ ). So $h (t) = t e^{- t}$ . Delay + gate: $y (t) = u (t - 2) (t - 2) e^{- (t - 2)}$ . Meaning: nothing until the kick at $t = 2$ , then a bump that rises and decays back to $0$ — the system's impulse response, delayed. Sanity: at $t = 3$ , $y = 1 \cdot e^{- 1} \approx 0.367879$ .

L5.2 Degenerate check: solve $y^{'} = δ (t)$ , $y (0) = 0$ . What does the answer say physically, and why is the tiny $t = 0$ subtlety worth noticing?

Recall Solution L5.2

Transform: $s Y - 0 = e^{- 0 \cdot s} = 1$ , so $Y = \frac{1}{s}$ and $y (t) = 1$ for $t > 0$ . Physical meaning: a unit impulse at $t = 0$ instantly deposits one unit; afterward $y$ is stuck at $1$ (no restoring term). The impulse acts as an instantaneous "step-in" of the initial condition. The subtlety: the impulse sits exactly at the lower limit $t = 0$ . With the standard one-sided transform the kick is fully captured, giving $y = 1$ for all $t > 0$ ; right at $t = 0$ the value is undefined (it is jumping). This is the boundary/degenerate case where "just after $0$ " and "just before $0$ " differ — always ask which side you mean.

L5.3 Reasoning (no full inversion needed): for $y^{''} + 5 y^{'} + 6 y = f (t)$ with zero initial conditions, the transform factors as $Y (s) = \frac{F ( s )}{( s + 2 ) ( s + 3 )}$ . Predict the long-time behaviour when $f (t) = u (t)$ (a constant switch-on), using only the roots.

Recall Solution L5.3

Read the poles. The system's own decay modes come from the roots $s = - 2$ and $s = - 3$ , both negative → both give terms $e^{- 2 t}, e^{- 3 t}$ that die out. With $f = u (t)$ , $F = \frac{1}{s}$ adds a pole at $s = 0$ , whose residue gives a constant term. Compute the steady state directly by the value the constant term takes: as $t \to \infty$ the decaying modes vanish, leaving the $s = 0$ residue of $\frac{1}{s ( s + 2 ) ( s + 3 )}$ , which is $\frac{1}{( 2 ) ( 3 )} = \frac{1}{6}$ . Conclusion: $y (t) \to \frac{1}{6}$ . You never had to fully invert — the poles told the story. (Cross-check with the ODE: at steady state $y^{''} = y^{'} = 0$ , so $6 y = 1 \Rightarrow y = \frac{1}{6}$ ✔.)

Recap flashcards

Recall Quick self-test

How do you invert $e^{- a s} H (s)$ ? ::: Delay AND gate: $u (t - a) h (t - a)$ . Why rewrite $t = (t - 2) + 2$ before transforming $t u (t - 2)$ ? ::: The shift rule needs the forcing as a function of $(t - 2)$ ; then $G (s) = \frac{1}{s ^{2}} + \frac{2}{s}$ and you tag with $e^{- 2 s}$ . What denominator signals resonance? ::: A repeated quadratic like $(s^{2} + ω^{2})^{2}$ ; it yields a $t cos ω t$ growing term. $L {δ (t - a)}$ vs $L {u (t - a)}$ ? ::: $e^{- a s}$ vs $\frac{e ^{- a s}}{s}$ — impulse has no $\frac{1}{s}$ . Steady state of $\frac{1}{s ( s + 2 ) ( s + 3 )}$ as $t \to \infty$ ? ::: The $s = 0$ residue $\frac{1}{6}$ (decaying modes vanish).