Properties of Laplace transform (linearity, shift, derivative, convolution)

At a Glance

• Frequency: 3 sub-parts across 3 of 13 years (2020, 2023, 2025)
• Priority tier: T3
• Marks (count): 10 (2), 15 (1)
• Average solve time: ~13 min
• Difficulty mix: medium 3
• Section: B | Dominant type: derivation

Why This Chapter Matters

Laplace-transform properties have appeared in Paper 1 in three distinct flavours, each testing a different rule. The 2025 convolution theorem question (15 marks) requires a two-part answer: prove the theorem by Fubini and then solve a Volterra integral equation. The 2023 question (10 marks) uses the $\mathcal{L}\{f(t)/t\}$ integral property to evaluate a classical improper integral in four lines. The 2020 question (10 marks) applies $\mathcal{L}\{tf\}=-F'(s)$ to a variable-coefficient ODE, producing a first-order ODE in $Y(s)$. Each archetype recycles one property cold — the proof structure is nearly mechanical once you know which rule to invoke.

Minimum Theory

Standard transform pairs (for reference).

$\mathcal{L}\{1\} = \frac{1}{s},\qquad \mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}},\qquad \mathcal{L}\{e^{at}\} = \frac{1}{s-a},\qquad \mathcal{L}\{\sin at\} = \frac{a}{s^2+a^2},\qquad \mathcal{L}\{\cos at\} = \frac{s}{s^2+a^2}.$

Derivative rule. $\mathcal{L}\{y'\}=sY-y(0)$; $\mathcal{L}\{y''\}=s^2Y-sy(0)-y'(0)$.

Multiplication by $t$ rule. If $\mathcal{L}\{f(t)\}=F(s)$, then $\mathcal{L}\{t\,f(t)\}=-F'(s)=-dF/ds$. More generally, $\mathcal{L}\{t^n f(t)\}=(-1)^n F^{(n)}(s)$.

Division by $t$ rule. $\mathcal{L}\{f(t)/t\}=\displaystyle\int_s^\infty F(p)\,dp$, provided $\lim_{t\to0^+}f(t)/t$ exists. Equivalently, $\displaystyle\int_0^\infty f(t)/t\,dt=\int_0^\infty F(p)\,dp$ (Frullani form, obtained by setting $s=0$).

Convolution theorem. Define the convolution $(f*g)(t)=\displaystyle\int_0^t f(x)g(t-x)\,dx$. Then $\mathcal{L}\{f*g\}=F(s)G(s)$.

Question Archetypes

convolution-theorem (1 question(s); 2025)

Recognition Cues

• “Prove $\mathcal{L}\left(\int_0^t f(x)g(t-x)\,dx\right)=F(s)G(s)$.”
• “Solve the integral equation $y(t)=\ldots+\int_0^t y(x)h(t-x)\,dx$.”
• The integral on the right is a convolution $y * h$ — take Laplace transforms on both sides, apply the theorem, solve the resulting algebraic equation for $Y(s)$, invert.

Solution Template

Proof:

1. Write $\mathcal{L}\{(f*g)(t)\}=\int_0^\infty e^{-st}\int_0^t f(x)g(t-x)\,dx\,dt$.
2. Interchange the order of integration over the region $0\leq x\leq t<\infty$: outer integral over $x$ from $0$ to $\infty$, inner from $t=x$ to $\infty$.
3. Substitute $\tau=t-x$ in the inner integral: $\int_x^\infty e^{-st}g(t-x)\,dt=e^{-sx}\int_0^\infty e^{-s\tau}g(\tau)\,d\tau=e^{-sx}G(s)$.
4. Collect: $F(s)\cdot G(s)$.

Solving a Volterra integral equation:

1. Take $\mathcal{L}$ of both sides.
2. Apply the convolution theorem to the integral term.
3. Solve the algebraic equation for $Y(s)$.
4. Invert $Y(s)$ (partial fractions if needed).

Worked Example

2025 Paper 1, 2025-P1-Q6a (15 marks)

If $F(s)$ and $G(s)$ are Laplace transforms of $f(t)$ and $g(t)$, prove $\mathcal{L}\!\left(\int_0^t f(x)g(t-x)\,dx\right)=F(s)G(s)$. Hence solve $y(t)=t+\int_0^t y(x)\sin(t-x)\,dx$.

Part 1 — Proof.

$\mathcal{L}\!\left(\int_0^t f(x)g(t-x)\,dx\right) = \int_0^\infty e^{-st}\!\left(\int_0^t f(x)g(t-x)\,dx\right)dt.$

Region: $0\leq x\leq t<\infty$. Interchange (Fubini, valid since both transforms exist absolutely):

$= \int_0^\infty f(x)\!\left(\int_x^\infty e^{-st}g(t-x)\,dt\right)dx.$

Inner integral: substitute $\tau=t-x$:

$\int_x^\infty e^{-st}g(t-x)\,dt = e^{-sx}\int_0^\infty e^{-s\tau}g(\tau)\,d\tau = e^{-sx}G(s).$

Therefore:

$\mathcal{L}\!\left(\int_0^t f(x)g(t-x)\,dx\right) = \int_0^\infty f(x)e^{-sx}\,dx\cdot G(s) = F(s)G(s).\qquad\blacksquare$

Part 2 — Solve the integral equation.

The equation $y(t)=t+(y*\sin)(t)$ transforms to:

$Y(s) = \frac{1}{s^2} + Y(s)\cdot\frac{1}{s^2+1}.$

Solve for $Y(s)$:

$Y(s)\!\left(1-\frac{1}{s^2+1}\right) = \frac{1}{s^2} \qquad\Longrightarrow\qquad Y(s)\cdot\frac{s^2}{s^2+1} = \frac{1}{s^2} \qquad\Longrightarrow\qquad Y(s) = \frac{s^2+1}{s^4} = \frac{1}{s^2}+\frac{1}{s^4}.$

Invert:

$y(t) = \mathcal{L}^{-1}\!\left\{\frac{1}{s^2}\right\} + \mathcal{L}^{-1}\!\left\{\frac{1}{s^4}\right\} = t + \frac{t^3}{3!} = t+\frac{t^3}{6}.$

$\boxed{\;y(t) = t+\frac{t^3}{6}.\;}$

Common Traps

• In the proof, the interchange of integration order is over the triangular region $\{0\leq x\leq t<\infty\}$: the outer integral runs over $x\in[0,\infty)$ and the inner integral runs over $t\in[x,\infty)$. Getting the limits wrong (e.g., keeping the outer integral from $0$ to $t$) invalidates the step.
• When solving $Y(s)$, the factor $1-\frac{1}{s^2+1}=\frac{s^2}{s^2+1}$ appears in the denominator: multiplying both sides by $(s^2+1)/s^2$ is the clean step.
• $\mathcal{L}^{-1}\{1/s^4\}=t^3/3!=t^3/6$, not $t^3/3$ (use $n!/s^{n+1}$ with $n=3$).

laplace-integral-property (1 question(s); 2023)

Recognition Cues

• “Show that $\int_0^\infty f(t)/t\,dt=\int_0^\infty F(p)\,dp$.”
• “Hence evaluate $\int_0^\infty (e^{-at}-e^{-bt})/t\,dt$.”
• The integrand has the form $f(t)/t$ where $f$ has a known transform and $f(0)=0$ (or the ratio converges as $t\to0^+$).

Solution Template

1. Prove the identity. Write $\int_0^\infty F(p)\,dp=\int_0^\infty\int_0^\infty e^{-pt}f(t)\,dt\,dp$; interchange (Fubini); inner integral $\int_0^\infty e^{-pt}\,dp=1/t$.
2. Compute $F(p)$ for the given $f(t)$ using standard transforms.
3. Integrate $F(p)$ from $0$ to $\infty$. For differences of exponentials: $F(p)=1/(p+a)-1/(p+b)$; antiderivative is $\ln(p+a)-\ln(p+b)$; evaluate at limits.

Worked Example

2023 Paper 1, 2023-P1-Q5b (10 marks)

Given $\mathcal{L}\{f(t);p\}=F(p)$, show $\int_0^\infty f(t)/t\,dt=\int_0^\infty F(p)\,dp$. Hence evaluate $\int_0^\infty(e^{-t}-e^{-3t})/t\,dt$.

Step 1 — Prove the identity.

$\int_0^\infty F(p)\,dp = \int_0^\infty\int_0^\infty e^{-pt}f(t)\,dt\,dp.$

Interchange the order of integration (Fubini, valid since $f(t)/t$ is integrable):

$= \int_0^\infty f(t)\!\left[\int_0^\infty e^{-pt}\,dp\right]dt = \int_0^\infty f(t)\cdot\frac{1}{t}\,dt. \qquad\blacksquare$

(The inner integral: $\int_0^\infty e^{-pt}\,dp = \bigl[-e^{-pt}/t\bigr]_0^\infty = 1/t$ for $t>0$.)

Step 2 — Apply to $f(t)=e^{-t}-e^{-3t}$.

$F(p) = \mathcal{L}\{e^{-t}\}-\mathcal{L}\{e^{-3t}\} = \frac{1}{p+1}-\frac{1}{p+3}.$

By the identity:

$\int_0^\infty\frac{e^{-t}-e^{-3t}}{t}\,dt = \int_0^\infty\!\left(\frac{1}{p+1}-\frac{1}{p+3}\right)dp = \Big[\ln(p+1)-\ln(p+3)\Big]_0^\infty.$

At $p=\infty$: $\ln\frac{p+1}{p+3}\to\ln 1=0$. At $p=0$: $\ln\frac{1}{3}=-\ln 3$.

$\int_0^\infty\frac{e^{-t}-e^{-3t}}{t}\,dt = 0-(-\ln 3) = \ln 3.$

$\boxed{\;\int_0^\infty\frac{e^{-t}-e^{-3t}}{t}\,dt = \ln 3.\;}$

Common Traps

• The identity $\int_0^\infty f(t)/t\,dt=\int_0^\infty F(p)\,dp$ requires $f(0)=0$ (more precisely, $\lim_{t\to0^+}f(t)/t$ must be finite). Here $e^{-t}-e^{-3t}=0$ at $t=0$, so the ratio $(e^{-t}-e^{-3t})/t\to 3-1=2$ as $t\to0^+$ — the integral is well-defined.
• The antiderivative of $1/(p+a)$ is $\ln(p+a)$, evaluated from $p=0$ to $p\to\infty$. At the upper limit, $\ln(p+a)-\ln(p+b)\to\ln(a/b)$ is the contribution (but here both logs go to $\infty$ separately; only the difference vanishes — take the ratio $\ln\frac{p+1}{p+3}\to0$ as $p\to\infty$).

laplace-variable-coefficient (1 question(s); 2020)

Recognition Cues

• The ODE has a factor of $t$ multiplying $y''$, $y'$, or $y$: e.g., $ty''+2ty'+2y=\ldots$
• The instruction “using Laplace transform” is explicit.
• After applying $\mathcal{L}\{tf\}=-F'(s)$, the resulting equation in $Y(s)$ is a first-order ODE (not algebraic) — expect to solve it with an integrating factor.

Solution Template

1. Write $Y(s)=\mathcal{L}\{y\}$. State the initial conditions.
2. Transform each term using $\mathcal{L}\{y'\}=sY-y(0)$, $\mathcal{L}\{y''\}=s^2Y-sy(0)-y'(0)$, and $\mathcal{L}\{tf\}=-d/ds[\mathcal{L}\{f\}]$.
3. Assemble the subsidiary equation in $Y$ and $Y'=dY/ds$. It will be a first-order linear ODE in $Y(s)$.
4. Solve with an integrating factor.
5. Partial-fraction decompose $Y(s)$.
6. Invert using standard pairs.
7. Discuss uniqueness if asked: if $y'(0)$ dropped out, the solution carries a free constant regardless of the initial slope — the ODE has a singular point at the initial time.

Worked Example

2020 Paper 1, 2020-P1-Q7b (10 marks)

Using Laplace transform, solve $ty''+2ty'+2y=2$; $y(0)=1$, $y'(0)$ arbitrary. Does this problem have a unique solution?

Let $Y(s)=\mathcal{L}\{y\}$, $y(0)=1$, $y'(0)=a$ (left free).

Step 1 — Transform each term.

$\mathcal{L}\{ty''\} = -\frac{d}{ds}(s^2Y-s-a) = -(2sY+s^2Y'-1),$

$\mathcal{L}\{2ty'\} = -2\frac{d}{ds}(sY-1) = -2(Y+sY'),$

$\mathcal{L}\{2y\} = 2Y, \qquad \mathcal{L}\{2\} = \frac{2}{s}.$

Step 2 — Assemble.

$-(2sY+s^2Y'-1) - 2(Y+sY') + 2Y = \frac{2}{s}.$

$-s^2Y' - 2sY + 1 - 2sY' - 2Y + 2Y = \frac{2}{s}.$

Note the $a$ term from $y'(0)$ cancelled. Simplify and multiply by $-1$:

$(s^2+2s)Y' + 2sY = 1 - \frac{2}{s}.\quad(\ast)$

Step 3 — Solve as a first-order ODE in $Y$. Rewrite:

$Y' + \frac{2}{s+2}Y = \frac{s-2}{s^2(s+2)}.$

Integrating factor: $\mu=(s+2)^2$. Then:

$\frac{d}{ds}\!\big[(s+2)^2Y\big] = \frac{(s+2)(s-2)}{s^2} = \frac{s^2-4}{s^2} = 1 - \frac{4}{s^2}.$

Integrate:

$(s+2)^2Y = s + \frac{4}{s} + C_1.$

$Y(s) = \frac{s^2+C_1 s+4}{s(s+2)^2}.$

Step 4 — Partial fractions.

$Y(s) = \frac{1}{s} + \frac{C}{(s+2)^2}, \qquad C = C_1-4 \text{ (arbitrary)}.$

Step 5 — Invert.

$\boxed{\;y(t) = 1 + C\,t\,e^{-2t},\quad C\text{ arbitrary}.\;}$

Step 6 — Uniqueness. The arbitrary constant $C$ cannot be fixed because $y'(0)=C$ (which was left free, and dropped from the equation). The coefficient of $y''$ vanishes at $t=0$ — a regular singular point — so standard existence-uniqueness does not apply: specifying only $y(0)$ does not determine a unique solution.

$\boxed{\;\text{The solution is NOT unique.}\;}$

Verify: $y'=C(1-2t)e^{-2t}$, $y''=C(4t-4)e^{-2t}$. Then $ty''+2ty'+2y=Ce^{-2t}[t(4t-4)+2t(1-2t)+2t]+2=0+2=2$ ✓.

Common Traps

• $\mathcal{L}\{ty''\}$ and $\mathcal{L}\{ty'\}$ require differentiating the Laplace transforms in $s$; do not just multiply $s^2Y$ by $-t$ symbolically.
• The vanishing of $y'(0)=a$ from the equation is the key diagnostic for non-uniqueness; mention it explicitly in the answer.
• After computing $(s+2)^2Y$, the partial-fraction step requires matching numerator $s^2+C_1 s+4$ against $s\cdot\frac{(s+2)^2}{s}+(C_1-4)\cdot\frac{s(s+2)^2}{(s+2)^2}$ — check the algebra by cross-multiplying.
• The transformed equation $(\ast)$ is a first-order ODE in $s$, not an algebraic equation. Do not try to solve it algebraically.

Marks-Aware Writing

10-mark questions (2023, 2020): For the integral-property question (2023): the proof via Fubini (3 marks) and the evaluation of the definite integral (7 marks). For the variable-coefficient question (2020): setting up the transformed ODE $(\ast)$ (4 marks), solving it with integrating factor (3 marks), inverting (2 marks), uniqueness discussion (1 mark).

15-mark question (2025): Part 1 (proof, 6–7 marks): state Fubini, write the interchange, the $\tau=t-x$ substitution, and the conclusion. Part 2 (integral equation, 7–8 marks): take transform, apply convolution theorem, solve for $Y$, invert. Missing the proof of the theorem costs all Part 1 marks; a correct Part 2 without invoking Part 1 earns partial credit.

Practice Set