How unit 4 is examined
Laplace transforms (definition, properties, inverse, convolution, ODEs) carry almost all the marks; periodic functions, integrals and Fourier transforms are one-off questions.
Laplace Transform
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Definition. ==For $t>0$, $L\{f(t)\}=F(s)=\int_0^\infty e^{-st}f(t)\,dt$, defined for $s$ large enough that the integral converges.==
Key points.
- The transform is linear: $L\{af+bg\}=aF(s)+bG(s)$, so a sum is transformed term by term.
- Standard results: $L\{1\}=1/s$, $L\{t^n\}=n!/s^{n+1}$, $L\{t^{n}\}=\Gamma(n+1)/s^{n+1}$ for non-integer $n$, $L\{e^{at}\}=1/(s-a)$.
- Also $L\{\sin at\}=a/(s^2+a^2)$, $L\{\cos at\}=s/(s^2+a^2)$, $L\{\sinh at\}=a/(s^2-a^2)$, $L\{\cosh at\}=s/(s^2-a^2)$.
- For a piecewise function, split the integral at the break points and integrate each piece; the unit step gives $L\{u(t-a)\}=e^{-as}/s$.
- Use $\Gamma(1/2)=\sqrt\pi$ and $\Gamma(n+1)=n\Gamma(n)$ for fractional powers: $\Gamma(5/2)=\tfrac34\sqrt\pi$.
Example (piecewise). $f=\sin t$ on $(0,2\pi)$, $0$ after: $\int_0^{2\pi}e^{-st}\sin t\,dt=\left[\dfrac{e^{-st}(-s\sin t-\cos t)}{s^2+1}\right]_0^{2\pi}$.
Answer. $\mathbf{L\{f\}=\dfrac{1-e^{-2\pi s}}{s^2+1}}$
Example (window). $t^2$ on $(1,2)$ only: $\int_1^2t^2e^{-st}dt=\left[-e^{-st}\left(\tfrac{t^2}{s}+\tfrac{2t}{s^2}+\tfrac{2}{s^3}\right)\right]_1^2$.
Answer. $\mathbf{\dfrac{e^{-s}(s^2+2s+2)-e^{-2s}(4s^2+4s+2)}{s^3}}$
Example (sum). $L\{\sin t+3\cos2t+t^3+t^{3/2}+3e^{-2t}-e^{3t}\}$:
$$\frac1{s^2+1}+\frac{3s}{s^2+4}+\frac6{s^4}+\frac{3\sqrt\pi}{4s^{5/2}}+\frac3{s+2}-\frac1{s-3}$$
Example (with $\sin t/t$). $L\{1+t^3+e^{-3t}\sin t+\tfrac{\sin t}{t}\}=\dfrac1s+\dfrac6{s^4}+\dfrac1{(s+3)^2+1}+\cot^{-1}s$.
Answer frame. Open with the definition integral; for sums, apply linearity term by term and list each standard result; for piecewise data, split the integral at the break points and integrate by parts; close with the simplified boxed $F(s)$.
Asked: [7 marks] (May 2019, Jun 2020) Find $L\{\sin t+3\cos2t+t^3+t^{3/2}+3e^{-2t}-e^{3t}\}$ Asked: [7 marks] (Dec 2020, Jun 2022) Find the Laplace transform of $f(t)=\sin t$ for $0<t<2\pi$, $0$ for $t>2\pi$ Asked: [7 marks] (Dec 2020) Find the Laplace transform of $f(t)=0\ (0<t<1),\ t^2\ (1<t<2),\ 0\ (t>2)$ Asked: [7 marks] (Jun 2020) Find $L\{1+t^3+e^{-3t}\sin t+\sin t/t\}$
Properties of Laplace Transform
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Definition. <mark>Multiplying $f(t)$ by $e^{at}$ shifts $F(s)$ to $F(s-a)$, multiplying by $t^n$ differentiates $F(s)$ $n$ times, and dividing by $t$ integrates $F(s)$ from $s$ to $\infty$.</mark>
Key points.
- Linearity: $L\{af+bg\}=aF+bG$.
- First shifting: $L\{e^{at}f(t)\}=F(s-a)$, so replace $s$ by $s-a$ after transforming $f$.
- Multiplication by $t^n$: $L\{t^nf(t)\}=(-1)^n\dfrac{d^n}{ds^n}F(s)$.
- Division by $t$: $L\{f(t)/t\}=\int_s^\infty F(u)\,du$, valid when $\lim_{t\to0}f(t)/t$ exists.
- Derivatives: $L\{y'\}=sY-y(0)$ and $L\{y''\}=s^2Y-sy(0)-y'(0)$.
- Second shifting: $L\{f(t-a)u(t-a)\}=e^{-as}F(s)$.
Example. $L\{t\cos at\}=-\dfrac{d}{ds}\dfrac{s}{s^2+a^2}=-\dfrac{(s^2+a^2)-2s^2}{(s^2+a^2)^2}$.
| Function | Rule used | Result |
|---|---|---|
| $t\cos at$ | $n=1$ on $s/(s^2+a^2)$ | $\dfrac{s^2-a^2}{(s^2+a^2)^2}$ |
| $t^2\sin t$ | $F''$ of $1/(s^2+1)$ | $\dfrac{2(3s^2-1)}{(s^2+1)^3}$ |
| $\sin t/t$ | $\int_s^\infty\frac{du}{u^2+1}=\frac\pi2-\tan^{-1}s$ | $\cot^{-1}s=\tan^{-1}(1/s)$ |
| $e^{3t}\cos2t$ | shift $s\to s-3$ | $\dfrac{s-3}{(s-3)^2+4}$ |
| $t^2e^{-3t}$ | $2/s^3$ shifted by $s\to s+3$ | $\dfrac{2}{(s+3)^3}$ |
For $t^2\sin t$: $F'=-2s/(s^2+1)^2$, then $F''=\dfrac{-2(s^2+1)+8s^2}{(s^2+1)^3}=\dfrac{6s^2-2}{(s^2+1)^3}$, and $(-1)^2=1$.
Answer frame. Open by naming the property and writing its formula; write the transform of the base function; substitute or differentiate in clear steps; close with the simplified result.
Pitfall: Forgetting $(-1)^n$ in the $t^n$ rule, or shifting the wrong way ($e^{-3t}$ means $s\to s+3$).
Asked: [7 marks] (Jun 2020, Jun 2023) Evaluate $L(t\cos at)$ Asked: [7 marks] (May 2019) Find $L\{\sin t/t\}$ Asked: [7 marks] (Jun 2020) Find $L\{t^2\sin t\}$ Asked: [7 marks] (Nov 2022) Evaluate $L(e^{3t}\cos2t)$ Asked: [7 marks] (Jun 2020, Jun 2023) Evaluate $L\{t^2e^{-3t}\}$
Laplace transform of periodic functions
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Definition. ==If $f(t+T)=f(t)$ for all $t$, then $L\{f(t)\}=\dfrac{1}{1-e^{-sT}}\int_0^Te^{-st}f(t)\,dt$.==
Key points.
- $T$ is the period, and only the integral over one period is needed.
- The factor $1/(1-e^{-sT})$ comes from summing the geometric series of the repeated periods.
- Use it for square, sawtooth and half-wave rectified waves, integrating over one period by parts.
- It needs $s>0$ so that $e^{-sT}<1$.
Finding inverse Laplace transform by different methods
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Definition. ==If $L\{f(t)\}=F(s)$, then $f(t)=L^{-1}\{F(s)\}$ is its inverse, and $L^{-1}$ is linear.==
Key points.
- Standard inverses: $L^{-1}\{1/s^n\}=t^{n-1}/(n-1)!$, $L^{-1}\{1/(s-a)\}=e^{at}$, $L^{-1}\{\frac{b}{s^2+b^2}\}=\sin bt$, $L^{-1}\{\frac{s}{s^2+b^2}\}=\cos bt$.
- Completing the square turns $s^2+ps+q$ into $(s-a)^2+b^2$, and first shifting gives $e^{at}$ times the sine or cosine.
- Partial fractions split a rational $F(s)$ into simple terms, each inverted by the standard table; factor quartics such as $s^4+s^2+1=(s^2+s+1)(s^2-s+1)$.
- Derivative method for logarithms: $L^{-1}\{F'(s)\}=-t\,f(t)$, so $f=-\frac1t L^{-1}\{F'\}$.
- Convolution theorem handles products (see below).
| $F(s)$ | Working | $f(t)$ |
|---|---|---|
| $\dfrac{s+7}{s^2+4s+8}$ | $\dfrac{(s+2)+5}{(s+2)^2+4}$ | $e^{-2t}\left(\cos2t+\tfrac52\sin2t\right)$ |
| $\dfrac1{s^2-6s+18}$ | $\dfrac{1}{(s-3)^2+9}$ | $\tfrac13e^{3t}\sin3t$ |
| $\dfrac1{(s+1)^3}$ | $L^{-1}\{1/s^3\}=t^2/2$, shift by $-1$ | $\tfrac12t^2e^{-t}$ |
| $\dfrac{s}{s^4+s^2+1}$ | $\tfrac12\left[\dfrac1{s^2-s+1}-\dfrac1{s^2+s+1}\right]$ | $\tfrac{2}{\sqrt3}\sinh\tfrac t2\sin\tfrac{\sqrt3t}2$ |
| $\ln\dfrac{s^2}{s^2+4}$ | $F'=\frac2s-\frac{2s}{s^2+4}$, inverse $2-2\cos2t=-tf$ | $\dfrac{2\cos2t-2}{t}$ |
For $s/(s^4+s^2+1)$: $\dfrac1{s^2\mp s+1}=\dfrac1{(s\mp\frac12)^2+\frac34}$, giving $\tfrac{2}{\sqrt3}e^{\pm t/2}\sin\tfrac{\sqrt3t}{2}$; their difference over 2 is $\tfrac1{\sqrt3}(e^{t/2}-e^{-t/2})\sin\tfrac{\sqrt3t}{2}$.
Answer frame. Open by naming the method (complete the square, partial fractions or derivative rule); rewrite $F(s)$ in standard form; invert term by term with the shifting theorem; close with $f(t)$.
Pitfall: The log example is $(2\cos2t-2)/t$, which is negative, matching $F<0$; the sign is easy to lose when dividing by $-t$.
Asked: [7 marks] (Jun 2020, Nov 2022) Evaluate $L^{-1}[(s+7)/(s^2+4s+8)]$ Asked: [7 marks] (May 2019) Find $L^{-1}\{s/(s^4+s^2+1)\}$ Asked: [7 marks] (Dec 2020) Determine the inverse Laplace transform of $\ln[s^2/(s^2+4)]$ Asked: [7 marks] (Jun 2020) Evaluate $L^{-1}\{1/(s^2-6s+18)\}$ Asked: [7 marks] (Jun 2023) Evaluate $L^{-1}[1/(1+s)^3]$
Convolution theorem
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Definition. ==If $L^{-1}\{F\}=f$ and $L^{-1}\{G\}=g$, then $L^{-1}\{F(s)G(s)\}=f*g=\int_0^tf(u)\,g(t-u)\,du$.==
Key points.
- It inverts a product of two transforms as an integral over $0$ to $t$.
- Convolution is commutative: $f*g=g*f$.
- Proof of the asked result: $\frac1{p^3}\cdot\frac1{p^2+1}$ has $f=t^2/2$, $g=\sin t$, so $\int_0^t\frac{u^2}{2}\sin(t-u)\,du=\frac{t^2}2-\int_0^tu\cos(t-u)\,du=\frac{t^2}2-(1-\cos t)$.
- Hence the result is $\dfrac{t^2}2+\cos t-1$, as required.
Asked: [7 marks] (Jun 2026) Using the convolution theorem, prove $L^{-1}[1/(p^3(p^2+1))]=t^2/2+\cos t-1$
Evaluation of integrals by Laplace transform
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Definition. ==Since $\int_0^\infty e^{-st}f(t)\,dt=F(s)$, an improper integral is evaluated by setting $s$ to the given value in $F(s)$.==
Key points.
- For an integral of $\int_0^t f(u)du$ use $L\left\{\int_0^tf\right\}=F(s)/s$.
- $L\{\sin u/u\}=\cot^{-1}s$, so $\int_0^t\frac{\sin u}{u}du$ transforms to $\frac1s\cot^{-1}s$.
- Asked integral: $\int_0^\infty e^{-t}\int_0^t\frac{\sin u}{u}du\,dt$ is that transform at $s=1$, which is $\cot^{-1}1=\dfrac\pi4$.
- Also $L\{\cosh at\sin bt\}=\tfrac12\left[\dfrac{b}{(s-a)^2+b^2}+\dfrac{b}{(s+a)^2+b^2}\right]=\dfrac{b(s^2+a^2+b^2)}{(s^2+a^2+b^2)^2-4a^2s^2}$.
Asked: [7 marks] (Jun 2026) Prove $\int_0^\infty\int_0^te^{-t}\frac{\sin u}{u}du\,dt=\frac\pi4$ and find $L\{\cosh at\sin bt\}$
solving ODEs by Laplace Transform method
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Definition. ==Transform the ODE using $L\{y'\}=sY-y(0)$ and $L\{y''\}=s^2Y-sy(0)-y'(0)$, solve the algebraic equation for $Y(s)$, then invert to get $y(t)$.==
Steps.
Step 1: Take the Laplace transform of both sides.
Step 2: Substitute the initial conditions y(0), y'(0).
Step 3: Solve for Y(s).
Step 4: Split Y(s) into partial fractions or shifted forms.
Step 5: Invert term by term to get y(t).
Key points.
- The method turns a differential equation with initial conditions into algebra, and the initial conditions enter automatically.
- The right side is transformed with standard results: $4t\to4/s^2$, $e^{3t}\to1/(s-3)$, $\sin3t\to3/(s^2+9)$.
- Complex-root denominators are completed to squares, not factorised.
- Check by putting $t=0$ into $y(t)$ and comparing with the given initial value.
| Problem | $Y(s)$ | $y(t)$ |
|---|---|---|
| $y''-3y'+2y=4t+e^{3t}$, $y(0)=1$, $y'(0)=-1$ | $\dfrac3s+\dfrac2{s^2}-\dfrac{1/2}{s-1}-\dfrac2{s-2}+\dfrac{1/2}{s-3}$ | $3+2t-\tfrac12e^t-2e^{2t}+\tfrac12e^{3t}$ |
| $y''-2y'+2y=0$, $y(0)=y'(0)=1$ | $\dfrac{s-1}{(s-1)^2+1}$ | $e^t\cos t$ |
| $y'+2y=26\sin3t$, $y(0)=3$ | $\dfrac9{s+2}+\dfrac{12-6s}{s^2+9}$ | $9e^{-2t}-6\cos3t+4\sin3t$ |
| $(D^2-2D+1)y=e^t$, $y(0)=2$, $y'(0)=-1$ | $\dfrac1{(s-1)^3}+\dfrac2{s-1}-\dfrac3{(s-1)^2}$ | $e^t\left(\tfrac{t^2}2-3t+2\right)$ |
Working for the first: $(s^2-3s+2)Y=\dfrac4{s^2}+\dfrac1{s-3}+s-4$, and $s^2-3s+2=(s-1)(s-2)$. For the third: $(s+2)Y=3+\dfrac{78}{s^2+9}$, and $\dfrac{78}{(s+2)(s^2+9)}=\dfrac6{s+2}+\dfrac{12-6s}{s^2+9}$.
Answer frame. Open by transforming both sides with the derivative formulas; substitute the initial conditions; solve for $Y(s)$ and show the partial fractions; close with $y(t)$ boxed.
Asked: [7 marks] (May 2019, Jun 2020, Jun 2026) Solve $y''-3y'+2y=4t+e^{3t}$, $y(0)=1$, $y'(0)=-1$, and $(D^2-2D+1)y=e^t$, $y(0)=2$, $y'(0)=-1$ Asked: [7 marks] (Jun 2020) Solve $y''-2y'+2y=0$, $y(0)=y'(0)=1$ Asked: [7 marks] (Jun 2023) Solve $y'+2y=26\sin3t$, $y(0)=3$
Fourier transforms
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Definition. ==The Fourier cosine transform of $f(x)$ is $F_c(s)=\int_0^\infty f(x)\cos sx\,dx$, with inverse $f(x)=\frac2\pi\int_0^\infty F_c(s)\cos sx\,ds$.==
Key points.
- For $f=e^{-x^2}$ let $I(s)=\int_0^\infty e^{-x^2}\cos sx\,dx$; integrating by parts gives $I'(s)=-\frac s2I(s)$.
- With $I(0)=\frac{\sqrt\pi}2$ (Gaussian integral), $I(s)=\dfrac{\sqrt\pi}{2}e^{-s^2/4}$.
- With the $\sqrt{2/\pi}$ normalisation the answer is $\dfrac1{\sqrt2}e^{-s^2/4}$.
- The sine transform is $F_s(s)=\int_0^\infty f(x)\sin sx\,dx$.
- Fourier series of the half-wave $\sin t$ ($0\le t\le\pi$), $0$ ($\pi\le t\le2\pi$), period $2\pi$: $a_0=\frac2\pi$, $b_1=\frac12$, $a_n=-\frac{2}{\pi(n^2-1)}$ for even $n$, and other coefficients $0$.
- So $f(t)=\dfrac1\pi+\dfrac12\sin t-\dfrac2\pi\sum_{n=2,4,6,\dots}\dfrac{\cos nt}{n^2-1}$.
Asked: [7 marks] (Jun 2026) Find the Fourier cosine transform of $e^{-x^2}$ Asked: [7 marks] (Jun 2022) Find the Fourier series for the periodic extension of $f(t)=\sin t$ ($0\le t\le\pi$), $0$ ($\pi\le t\le2\pi$)
Last-minute revision
- $L\{f\}=\int_0^\infty e^{-st}f\,dt$; $L\{t^n\}=n!/s^{n+1}$; $L\{e^{at}\}=1/(s-a)$.
- $L\{\sin at\}=a/(s^2+a^2)$; $L\{\cos at\}=s/(s^2+a^2)$.
- $\Gamma(5/2)=\tfrac34\sqrt\pi$, so $L\{t^{3/2}\}=3\sqrt\pi/(4s^{5/2})$.
- First shifting: $L\{e^{at}f\}=F(s-a)$; $t^n$ rule: $(-1)^nF^{(n)}(s)$; division by $t$: $\int_s^\infty F$.
- $L\{\sin t/t\}=\cot^{-1}s$; $L\{t\cos at\}=(s^2-a^2)/(s^2+a^2)^2$.
- $L\{y'\}=sY-y(0)$; $L\{y''\}=s^2Y-sy(0)-y'(0)$.
- Periodic: $\dfrac{1}{1-e^{-sT}}\int_0^Te^{-st}f\,dt$.
- Convolution: $L^{-1}\{FG\}=\int_0^tf(u)g(t-u)du$.
- $\int_0^\infty e^{-t}\int_0^t\frac{\sin u}{u}du\,dt=\frac\pi4$; $F_c\{e^{-x^2}\}=\frac{\sqrt\pi}2e^{-s^2/4}$.
- Answers: $e^{-2t}(\cos2t+\frac52\sin2t)$, $\frac13e^{3t}\sin3t$, $\frac12t^2e^{-t}$.
Memory hooks
- Shift the exponent, shift the $s$: $e^{at}$ means $s\to s-a$.
- Multiply by $t$, differentiate; divide by $t$, integrate.
- Complete the square first, then peel off the $(s-a)$ numerator.
- ODE recipe: transform, insert initial values, solve, partial fractions, invert.
- Periodic: one period over $1-e^{-sT}$.
Coverage checklist
- Laplace Transform: sum with $t^{3/2}$, $\sin t$ window, $t^2$ window, $1+t^3+e^{-3t}\sin t+\sin t/t$.
- Properties of Laplace Transform: $t\cos at$, $\sin t/t$, $t^2\sin t$, $e^{3t}\cos2t$, $t^2e^{-3t}$.
- Laplace transform of periodic functions: definition and formula.
- Finding inverse Laplace transform by different methods: five inverses.
- convolution theorem: proof of $t^2/2+\cos t-1$.
- Evaluation of integrals by Laplace transform: $\pi/4$ integral and $\cosh at\sin bt$.
- solving ODEs by Laplace Transform method: four IVPs.
- Fourier transforms: cosine transform of $e^{-x^2}$, Fourier series of half-wave $\sin t$.