How unit 4 is examined
Laplace transform, its properties, inverse transform, convolution and ODE solving, plus the Fourier transform; the marks sit in solving ODEs (14 marks), then convolution and inverse Laplace (7 marks each).
Laplace Transform
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Definition. ==For a function $f(t)$ defined for $t \ge 0$, the Laplace transform is $L\{f(t)\} = F(s) = \int_0^\infty e^{-st} f(t)\,dt$, provided the integral converges.==
Key points.
- The transform turns a differential equation in $t$ into an algebraic equation in $s$, which is why it is used for ODEs.
- It is linear: $L\{af + bg\} = aF(s) + bG(s)$.
- Standard results: $L\{1\} = \frac1s$, $L\{t^n\} = \frac{n!}{s^{n+1}}$ (and $\frac{\Gamma(n+1)}{s^{n+1}}$ for non-integer $n$), $L\{e^{at}\} = \frac{1}{s-a}$, $L\{\sin at\} = \frac{a}{s^2+a^2}$, $L\{\cos at\} = \frac{s}{s^2+a^2}$, $L\{\sinh at\} = \frac{a}{s^2-a^2}$, $L\{\cosh at\} = \frac{s}{s^2-a^2}$.
- Proof method for $L\{\cos\sqrt t/\sqrt t\}$: expand in a power series and transform term by term.
Example. Show $L\left\{\frac{\cos\sqrt t}{\sqrt t}\right\} = \sqrt{\frac{\pi}{s}}e^{-1/(4s)}$.
$$\frac{\cos\sqrt t}{\sqrt t} = \sum_{n=0}^\infty \frac{(-1)^n t^{\,n-1/2}}{(2n)!}, \qquad L\{t^{n-1/2}\} = \frac{\Gamma(n+\frac12)}{s^{n+1/2}}, \qquad \Gamma\!\left(n+\tfrac12\right) = \frac{(2n)!\sqrt\pi}{4^n\, n!}$$
$$L = \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}\cdot\frac{(2n)!\sqrt\pi}{4^n n!\, s^{n+1/2}} = \sqrt{\frac{\pi}{s}}\sum_{n=0}^\infty \frac{1}{n!}\left(\frac{-1}{4s}\right)^n$$
$= \sqrt{\pi/s}\; e^{-1/(4s)}$ (the sum is the series of $e^x$ with $x=-\frac1{4s}$).
Asked: [7 marks] (Dec 2025) Show that $L\left\{\frac{\cos \sqrt{t}}{\sqrt{t}}\right\} = \sqrt{\frac{\pi}{s}} e^{-\frac{1}{4s}}$
Properties of Laplace Transform
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Definition. ==If $L\{f(t)\}=F(s)$, then the first shifting theorem says $L\{e^{at}f(t)\} = F(s-a)$, and multiplying by $t^n$ differentiates the transform: $L\{t^n f(t)\} = (-1)^n \frac{d^n}{ds^n}F(s)$.==
Key points.
- Linearity: $L\{af+bg\} = aF+bG$.
- First shifting: multiplying by $e^{at}$ replaces $s$ by $s-a$ in $F(s)$.
- Multiplication by $t^n$: differentiate $F(s)$ $n$ times and multiply by $(-1)^n$.
- Division by $t$: $L\{f(t)/t\} = \int_s^\infty F(u)\,du$.
- Derivatives: $L\{y'\} = sY - y(0)$ and $L\{y''\} = s^2Y - sy(0) - y'(0)$; integral: $L\left\{\int_0^t f\right\} = F(s)/s$.
Example. Find (i) $L\{t^2\cos at\}$, (ii) $L\{t^2e^t\sin 4t\}$.
(i) $F = \frac{s}{s^2+a^2}$; $F' = \frac{a^2-s^2}{(s^2+a^2)^2}$; $F'' = \frac{2s(s^2-3a^2)}{(s^2+a^2)^3}$, so $L\{t^2\cos at\} = \frac{2s(s^2-3a^2)}{(s^2+a^2)^3}$.
(ii) $L\{\sin 4t\} = \frac{4}{s^2+16}$; twice differentiated gives $L\{t^2\sin4t\} = \frac{8(3s^2-16)}{(s^2+16)^3}$; now put $s \to s-1$:
$L\{t^2e^t\sin4t\} = \dfrac{8\,[3(s-1)^2-16]}{[(s-1)^2+16]^3}$
Asked: [7 marks] (Jun 2023) Find Laplace transform of the following data: i) $t^2 \cos at$ ii) $t^2 e^t \sin 4t$
Laplace transform of periodic functions
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Definition. ==If $f(t+T)=f(t)$ for all $t$, then $L\{f(t)\} = \dfrac{1}{1-e^{-sT}}\displaystyle\int_0^T e^{-st}f(t)\,dt$.==
Key points.
- $T$ is the period, and only the integral over one period is needed.
- The factor $\frac{1}{1-e^{-sT}}$ comes from summing the geometric series of the shifted periods.
- Use it for square, sawtooth and rectified waves.
Finding inverse Laplace transform by different methods
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Definition. ==The inverse Laplace transform $L^{-1}\{F(s)\}=f(t)$ is the function whose transform is $F(s)$; it is linear, and $L^{-1}\{F(s+a)\} = e^{-at}L^{-1}\{F(s)\}$ (first shifting).==
Key points.
- Standard forms: $L^{-1}\{\frac1{s-a}\}=e^{at}$, $L^{-1}\{\frac{1}{s^2+a^2}\}=\frac{\sin at}{a}$, $L^{-1}\{\frac{s}{s^2+a^2}\}=\cos at$, $L^{-1}\{\frac{1}{s^n}\}=\frac{t^{n-1}}{(n-1)!}$.
- Completing the square: write the denominator as $(s+a)^2$ or $(s+a)^2+b^2$, rewrite the numerator in terms of $(s+a)$, and apply the first shifting theorem.
- Partial fractions: factor the denominator, split into simple fractions, find constants by substituting roots, and invert each term.
- Convolution theorem: use it when $F(s)$ is a product of two known transforms.
- Derivative rule: $L^{-1}\{F'(s)\} = -t\,f(t)$.
Example 1. $\dfrac{4s+12}{s^2+8s+16} = \dfrac{4(s+4)-4}{(s+4)^2} = \dfrac{4}{s+4}-\dfrac{4}{(s+4)^2}$, so $f(t) = 4e^{-4t} - 4te^{-4t} = 4e^{-4t}(1-t)$.
Example 2. $\dfrac{6s^2+22s+18}{(s+1)(s+2)(s+3)} = \dfrac{A}{s+1}+\dfrac{B}{s+2}+\dfrac{C}{s+3}$.
| Root | Numerator value | Divided by the other two factors | Constant |
|---|---|---|---|
| $s=-1$ | $6-22+18=2$ | $(1)(2)=2$ | $A=1$ |
| $s=-2$ | $24-44+18=-2$ | $(-1)(1)=-1$ | $B=2$ |
| $s=-3$ | $54-66+18=6$ | $(-2)(-1)=2$ | $C=3$ |
$f(t) = e^{-t}+2e^{-2t}+3e^{-3t}$
Answer frame. Open with "the denominator is a perfect square / has distinct linear factors, so I use shifting / partial fractions"; show the rewritten $F(s)$ line by line; invert each term using the standard table; close with the boxed $f(t)$.
Asked: [7 marks] (Dec 2025, Jun 2025) Find the inverse Laplace transform of $\frac{4s + 12}{s^2 + 8s + 16}$. / Evaluate $L^{-1}\left\{\frac{6s^{2}+22s+18}{s^{3}+6s^{2}+11s+6}\right\}$.
Convolution theorem
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Definition. ==If $L^{-1}\{F(s)\}=f(t)$ and $L^{-1}\{G(s)\}=g(t)$, then $L^{-1}\{F(s)G(s)\} = f * g = \displaystyle\int_0^t f(u)\,g(t-u)\,du$.==
Key points.
- The convolution of $f$ and $g$ is written $f*g$, and it is commutative: $f*g=g*f$.
- The theorem inverts a product of transforms, which ordinary partial fractions cannot handle when the factors repeat.
- Split $F(s)G(s)$ into two parts whose inverses are known, and write $f$ and $g$.
- Put the $f(u)$ and $g(t-u)$ into the integral, then use $2\cos A\cos B=\cos(A+B)+\cos(A-B)$ or $2\sin A\cos B=\sin(A+B)+\sin(A-B)$.
- Integrate with respect to $u$, treating $t$ as a constant.
Example (each answer verified).
| Find | $f(t)$, $g(t)$ | Working | Result |
|---|---|---|---|
| $L^{-1}\frac{s^2}{(s^2+a^2)^2}$ | $\cos at,\ \cos at$ | $\frac12\int_0^t[\cos at+\cos(2au-at)]du = \frac12\left[t\cos at+\frac{\sin at}{a}\right]$ | $\frac{1}{2a}(\sin at + at\cos at)$ |
| $L^{-1}\frac{s}{(s^2+a^2)^2}$ | $\frac{\sin at}{a},\ \cos at$ | $\frac1{2a}\int_0^t[\sin at+\sin(2au-at)]du$, the second part integrates to 0 | $\frac{t\sin at}{2a}$ |
| $L^{-1}\frac{1}{s(s^2-a^2)}$ | $1,\ \frac{\sinh at}{a}$ | $\frac1a\int_0^t\sinh au\,du = \frac{1}{a^2}[\cosh au]_0^t$ | $\frac{\cosh at-1}{a^2}$ |
Answer frame. Open by stating the theorem with the integral; identify $F,G$ and write $f,g$; set up the integral, apply the product-to-sum identity, integrate and simplify; close with the result in bold.
Pitfall: In $g(t-u)$ replace $t$ by $t-u$ in $g$ only, not in $f$, and do not forget the limits $0$ to $t$.
Asked: [7 marks] (Nov 2022) Find $L^{-1}\left\{\frac{s^2}{(s^2+a^2)^2}\right\}$ Asked: [7 marks] (Jun 2025) State convolution theorem and hence evaluate $L^{-1}\left[\frac{1}{s(s^2 - a^2)}\right]$ Asked: [7 marks] (Dec 2025) State Convolution theorem and hence evaluate $L^{-1}\left\{\frac{s}{(s^{2}+a^{2})^{2}}\right\}$
Evaluation of integrals by Laplace transform
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Definition. ==If $L\{f(t)\}=F(s)$, then $\int_0^\infty e^{-st}\frac{f(t)}{t}\,dt = \int_s^\infty F(u)\,du$, and letting $s\to0$ gives $\int_0^\infty \frac{f(t)}{t}\,dt = \int_0^\infty F(u)\,du$.==
Key points.
- Put the given integral in the form $\int_0^\infty e^{-st}\,\phi(t)\,dt$ with $s\to 0$, and find $L\{\phi(t)\}$.
- When $\phi(t)=\frac{f(t)}{t}$, use the division-by-$t$ rule and integrate $F(u)$ from $s$ to $\infty$.
- Finally take the limit $s\to0$.
Example. $\int_0^\infty \frac{\cos at-\cos bt}{t}dt$: $L\{\cos at-\cos bt\}=\frac{s}{s^2+a^2}-\frac{s}{s^2+b^2}$, so $$L\left\{\frac{\cos at-\cos bt}{t}\right\}=\int_s^\infty\left(\frac{u}{u^2+a^2}-\frac{u}{u^2+b^2}\right)du=\frac12\ln\frac{s^2+b^2}{s^2+a^2}.$$ Let $s\to0$: value $=\ln\dfrac ba$.
Asked: [7 marks] (Nov 2022) Using Laplace Transform, Evaluate $\int_{0}^{\infty} \frac{\cos at - \cos bt}{t} \, dt$.
Solving ODEs by Laplace Transform method
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Definition. <mark>The Laplace method solves a linear ODE with constant coefficients and given initial conditions by transforming it to an algebraic equation for $Y(s)$, solving for $Y(s)$ and taking the inverse transform.</mark>
Formula. $L\{y'\}=sY-y(0)$, $\;L\{y''\}=s^2Y-s\,y(0)-y'(0)$.
Steps.
Step 1: Take the Laplace transform of both sides, using the derivative formulas.
Step 2: Put in y(0) and y'(0) and collect terms in Y(s).
Step 3: Solve for Y(s) = (polynomial in s) / (factored denominator).
Step 4: Split by partial fractions or shifting.
Step 5: Take the inverse transform term by term to get y(t).
Key points.
- The initial conditions are built into the transform, so no arbitrary constants appear and the answer is the particular solution.
- The right-hand side must be transformed with the standard results and the shifting theorem: $L\{e^{-t}\sin t\}=\frac{1}{(s+1)^2+1}$.
- Write $Y(s)$ with the factored denominator, since the factors decide the partial fractions.
- Check the answer with $y(0)$ and $y'(0)$.
Example. Solve $(D^2+2D+5)y=e^{-t}\sin t$, $y(0)=0$, $y'(0)=1$.
Transform: $(s^2Y-1)+2sY+5Y=\frac{1}{(s+1)^2+1}$, so with $p=(s+1)^2$, $Y(s^2+2s+5) = Y(p+4)$: $$Y=\frac{1}{p+4}+\frac{1}{(p+1)(p+4)}=\frac{1}{p+4}+\frac13\left[\frac1{p+1}-\frac1{p+4}\right]=\frac{2/3}{(s+1)^2+4}+\frac{1/3}{(s+1)^2+1}$$ Inverse by first shifting: $y(t)=\dfrac{e^{-t}}{3}\,(\sin t+\sin 2t)$; check: $y(0)=0$ and $y'(0)=\frac13(1+2)=1$.
Other two asked equations (results checked numerically).
| Equation | $Y(s)$ | Answer |
|---|---|---|
| $(D^2+1)y=t\cos2t$, $y(0)=y'(0)=0$ | $\frac{s^2-4}{(s^2+1)(s^2+4)^2}=-\frac{5/9}{s^2+1}+\frac{5/9}{s^2+4}+\frac{8/3}{(s^2+4)^2}$ | $y=-\frac59\sin t+\frac49\sin2t-\frac t3\cos2t$ |
| $(D^2+6D+9)y=\sin x$, $y(0)=1$, $y'(0)=0$ | $\frac{s+6}{(s+3)^2}+\frac{1}{(s^2+1)(s+3)^2}$ | $y=e^{-3x}\left(\frac{53}{50}+\frac{31x}{10}\right)+\frac{4\sin x-3\cos x}{50}$ |
For the first row use $L\{t\cos2t\}=\frac{s^2-4}{(s^2+4)^2}$ and $L^{-1}\frac{1}{(s^2+4)^2}=\frac{\sin2t-2t\cos2t}{16}$. In the second row, read $y=1$, $y'=0$ at $x=0$ (the paper's second "$y=0$" must be $y'=0$); the fractions are $\frac{-3s/50+2/25}{s^2+1}+\frac{3/50}{s+3}+\frac{1/10}{(s+3)^2}$.
Answer frame. Open with "taking the Laplace transform of both sides"; write the transformed equation with the initial conditions substituted; solve for $Y(s)$ and show the partial fractions; invert term by term; close with $y(t)$ and the check of the initial conditions.
Pitfall: Write $L\{y''\}=s^2Y-sy(0)-y'(0)$ with the right signs; a wrong sign on $y'(0)$ ruins the whole 14 marks.
Asked: [14 marks] (Nov 2022, Jun 2023, Dec 2025) Using Laplace Transform, Solve $(D^2 + 2D + 5)y = e^{-t} \sin t$, given that $y(0) = 0, y'(0) = 1$. / Solve $(D^2 + 1) y = t \cos 2t, y(0) = 0 = y'(0), t > 0$. / Solve $(D^{2}+6D+9)y = \sin x$ given that $y=1,y=0$ when $x=0$.
Fourier transforms
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Definition. ==The Fourier cosine transform of $f(x)$ is $F_c\{f\}=\sqrt{\frac2\pi}\int_0^\infty f(x)\cos sx\,dx$, the sine transform is $F_s\{f\}=\sqrt{\frac2\pi}\int_0^\infty f(x)\sin sx\,dx$, and the complex transform is $F(s)=\int_{-\infty}^\infty f(x)e^{isx}dx$.==
Key points.
- Use the cosine transform for even extensions and the sine transform for odd extensions on $(0,\infty)$.
- The standard integral is $\int_0^\infty e^{-ax}\cos bx\,dx=\frac{a}{a^2+b^2}$, and $\int_0^\infty e^{-ax}\sin bx\,dx=\frac{b}{a^2+b^2}$.
- Some books drop the factor $\sqrt{2/\pi}$; state the definition you use.
Example. $f(x)=e^{-x}$: $F_c=\sqrt{\frac2\pi}\int_0^\infty e^{-x}\cos sx\,dx = \sqrt{\dfrac2\pi}\cdot\dfrac{1}{1+s^2}$ (using $a=1$, $b=s$).
Asked: [7 marks] (Jun 2023) Find the Fourier Cosine Transform of $e^{-x}$
Last-minute revision
- $L\{f\}=\int_0^\infty e^{-st}f(t)\,dt$.
- $L\{t^n\}=\frac{n!}{s^{n+1}}$, $L\{e^{at}\}=\frac1{s-a}$, $L\{\sin at\}=\frac{a}{s^2+a^2}$, $L\{\cos at\}=\frac{s}{s^2+a^2}$.
- First shifting: $L\{e^{at}f\}=F(s-a)$; $L\{t^nf\}=(-1)^nF^{(n)}(s)$; $L\{f/t\}=\int_s^\infty F$.
- $L\{y'\}=sY-y(0)$, $L\{y''\}=s^2Y-sy(0)-y'(0)$.
- Periodic: $L\{f\}=\frac{1}{1-e^{-sT}}\int_0^Te^{-st}f\,dt$.
- Convolution: $L^{-1}\{FG\}=\int_0^tf(u)g(t-u)\,du$.
- $L^{-1}\frac{s^2}{(s^2+a^2)^2}=\frac{\sin at+at\cos at}{2a}$; $L^{-1}\frac{s}{(s^2+a^2)^2}=\frac{t\sin at}{2a}$; $L^{-1}\frac{1}{s(s^2-a^2)}=\frac{\cosh at-1}{a^2}$.
- $\int_0^\infty\frac{\cos at-\cos bt}{t}dt=\ln\frac ba$.
- $L^{-1}\frac{4s+12}{(s+4)^2}=4e^{-4t}(1-t)$.
- $y''+2y'+5y=e^{-t}\sin t$ gives $y=\frac{e^{-t}}3(\sin t+\sin2t)$.
- $F_c\{e^{-x}\}=\sqrt{2/\pi}\,/(1+s^2)$.
Memory hooks
- Shift in $t$-multiplier: $e^{at}$ moves $s$ (to $s-a$), $t^n$ differentiates.
- ODE recipe: transform, insert initial values, solve for $Y$, split, invert.
- Convolution: "$F$ times $G$ in $s$ is $f$ folded with $g$ in $t$", limits $0$ to $t$.
- $s^2/(s^2+a^2)^2$ has a plus, $s/(s^2+a^2)^2$ has only $t\sin$.
- Division by $t$ integrates from $s$ to $\infty$, then $s\to0$ gives the integral.
Coverage checklist
- Laplace Transform: Dec 2025 proof of $L\{\cos\sqrt t/\sqrt t\}$.
- Properties of Laplace Transform: Jun 2023 $t^2\cos at$ and $t^2e^t\sin4t$.
- Laplace transform of periodic functions: formula and key points (not asked recently).
- Finding inverse Laplace transform by different methods: Jun 2025 and Dec 2025 $\frac{4s+12}{s^2+8s+16}$ and $\frac{6s^2+22s+18}{s^3+6s^2+11s+6}$.
- convolution theorem: Nov 2022, Jun 2025 and Dec 2025 inverse transforms.
- Evaluation of integrals by Laplace transform: Nov 2022 $\int_0^\infty\frac{\cos at-\cos bt}{t}dt$.
- solving ODEs by Laplace Transform method: Nov 2022, Jun 2023 and Dec 2025 (three ODEs).
- Fourier transforms: Jun 2023 cosine transform of $e^{-x}$.