PAST PAPERS
Every question, solved step by step
Three MA2014 finals (Intakes 2021, 2022, 2023) with full solutions. Press Next step and try to write the next line yourself before you reveal it. Each question links back to the lesson that teaches it.
📊 Where the marks come from
Marks per topic in each paper. Brighter means more marks.
PatternLaplace and series solutions have appeared in every paper, at 25 marks each. Fourier series always shows up. The newest paper added Fourier transforms (22 marks), so expect them again. The two exact questions "\(xy''+y'+xy=0\) about \(x=0\)" appeared in both 2024 and 2025: learn that solution cold.
Q1 Laplace transform 25 marks
(a)(i) Prove \(\mathcal L\{f/t\}=\int_s^\infty F\,ds\) · (a)(ii) Prove \(\mathcal L\{\int_0^tf\}=F/s\) · 10 marks
\(\int_s^\infty F(p)dp=\int_s^\infty\int_0^\infty f(t)e^{-pt}dt\,dp=\int_0^\infty f(t)\left[\frac{-e^{-pt}}{t}\right]_{p=s}^{\infty}dt=\int_0^\infty\frac{f(t)}te^{-st}dt\) ∎
\(g=\int_0^tf\): \(g(0)=0\) and \(g'=f\). So \(F=\mathcal L\{g'\}=sG-g(0)=sG\), which gives \(G=F/s\) ∎
(b)(i) \(\mathcal L\{\int_0^{t/2}\frac{1-e^{-2x}}{x}dx\}=\frac{\ln(1+1/s)}s\) · (b)(ii) \(\int_0^\infty\int_0^te^{-t}\frac{\sin u}{u}du\,dt=\frac\pi4\) · 10 marks
\(\mathcal L\{\frac{1-e^{-2t}}{t}\}=\int_s^\infty(\frac1p-\frac1{p+2})dp=\ln\frac{s+2}{s}\).
\(g(t)=\int_0^t\): \(G=\frac1s\ln(1+\frac2s)\). Then \(g(t/2)\): \(\mathcal L=2G(2s)=\frac{\ln(1+1/s)}{s}\) ∎
The double integral is \(\mathcal L\{g\}(1)\) with \(g=\int_0^t\frac{\sin u}udu\). \(\mathcal L\{\frac{\sin t}t\}=\tan^{-1}\frac1s\), so \(\mathcal L\{g\}=\frac1s\tan^{-1}\frac1s\). At \(s=1\) this is \(\frac\pi4\) ∎
(c) LR circuit, \(L=1,R=1,V=\int_0^te^{2u}\sin3u\,du\), \(I(0)=0\) · 5 marks
\((s+1)I(s)=\frac{3}{s((s-2)^2+9)}\)
\(I(s)=\frac{3/13}{s}-\frac{1/6}{s+1}-\frac{5(s-2)+3}{78((s-2)^2+9)}\)
\(I(t)=\frac3{13}-\frac16e^{-t}-\frac{e^{2t}}{78}(5\cos3t+\sin3t)\)
Q2 Series solutions 25 marks
Lesson 4.1–4.3 → (the Frobenius engine has this equation)
(a) Explain ordinary and regular singular points · 5 marksFor \(P_0y''+P_1y'+P_2y=0\) with polynomial coefficients: \(x=a\) is ordinary if \(P_0(a)\ne0\). It is regular singular if \(P_0(a)=0\) but \((x-a)\frac{P_1}{P_0}\) and \((x-a)^2\frac{P_2}{P_0}\) are analytic (finite limits) at \(a\).
(b)(i) \(x(1-x)y''+(3x-1)y'+y=0\) near 0 · 10 marks
\(xP_1/P_0=\frac{3x-1}{1-x}\to-1\) and \(x^2P_2/P_0\to0\), so x = 0 is regular singular. Take \(y=\sum a_nx^{n+m}\).
Lowest power \(x^{m-1}\): \(a_0[m(m-1)-m]=0\), so \(m^2-2m=0\) and \(m=0,2\) (Case 3).
Recurrence: \(a_{k+1}(k+m+1)(k+m-1)=a_k\big[(k+m)^2-4(k+m)-1\big]\).
\(m=2\): \(a_{k+1}=\frac{k^2-5}{(k+3)(k+1)}a_k\), giving \(y_1=x^2\left(1-\frac53x+\frac56x^2-\cdots\right)\).
\(m=0\): at \(k=1\) you get \(0\cdot a_2=-4a_1\) but \(a_1=a_0\ne0\). Contradiction, so a log term is needed:
\[y_2=-2y_1\ln x+1+x-\tfrac{40}9x^3+\cdots,\qquad y=c_1y_1+c_2y_2\]
(b)(ii) \((x+1)y''-(x+1)^2y=0\) near −1 · 10 marks
Divide by \(x+1\) and let \(t=x+1\): \(y''-ty=0\). Then \(t=0\) is an ordinary point.
\(\sum(n+2)(n+1)a_{n+2}t^n=\sum a_{n-1}t^n\Rightarrow a_2=0,\ a_{n+2}=\frac{a_{n-1}}{(n+2)(n+1)}\).
\(y=a_0\left[1+\frac{(x+1)^3}{6}+\frac{(x+1)^6}{180}+\cdots\right]+a_1\left[(x+1)+\frac{(x+1)^4}{12}+\frac{(x+1)^7}{504}+\cdots\right]\)
Q3 Heat equation 25 marks
Lesson 3.3 and 3.6 → (animated)
(a) General solution, all three cases · 12 marksSee the full derivation in lesson 3.3. Summary for \(u=XT\), \(\frac{T'}{c^2T}=\frac{X''}{X}=k\):
\(k=0\): \(u=(Ax+B)C\). · \(k=p^2\): \(u=(Ae^{px}+Be^{-px})Ce^{c^2p^2t}\). · \(k=-p^2\): \(u=(A\cos px+B\sin px)Ce^{-c^2p^2t}\), which is the physically relevant one.
\(k=0\): \(u=(Ax+B)C\). · \(k=p^2\): \(u=(Ae^{px}+Be^{-px})Ce^{c^2p^2t}\). · \(k=-p^2\): \(u=(A\cos px+B\sin px)Ce^{-c^2p^2t}\), which is the physically relevant one.
(b) Bar of length π at 100 °C, ends at 0 °C, c = 1 · 13 marks
\(u(0,t)=u(\pi,t)=0\) gives \(A=0\) and \(p=n\), so \(u=\sum B_n\sin nx\,e^{-n^2t}\).
\(u(x,0)=100\Rightarrow B_n=\frac2\pi\int_0^\pi100\sin nx\,dx=\frac{200(1-(-1)^n)}{n\pi}\).
\(u(x,t)=\frac{400}\pi\sum_{n\,odd}\frac{\sin nx}{n}e^{-n^2t}\)
Q4 Fourier series and Parseval 25 marks
Lesson 2.3, 2.6, 2.8 → (presets "1 + x" and "2x − 1")
\(f=1+x\) on \((-\pi,\pi)\). \(a_0=\frac1\pi\int(1+x)=2\). \(a_n=\frac1\pi\int(1+x)\cos nx=0\), since the \(x\cos nx\) part is odd and \(\int\cos nx=0\).
\(b_n=\frac1\pi\int x\sin nx\,dx=\frac{2(-1)^{n+1}}{n}\). So \(1+x=1+2\sum\frac{(-1)^{n+1}}{n}\sin nx\).
Parseval with \(l=\pi\): \(\frac1\pi\int_{-\pi}^{\pi}(1+x)^2dx=\frac1\pi\left(2\pi+\frac{2\pi^3}3\right)=2+\frac{2\pi^2}3\).
\(=\frac{a_0^2}2+\sum b_n^2=2+4\sum\frac1{n^2}\), so \(\sum\frac1{n^2}=\frac{\pi^2}6\) ∎
\(g=2x-1\) on \((0,1)\), cosine series with \(\ell=1\): \(a_0=2\int_0^1(2x-1)dx=0\).
\(a_n=2\int_0^1(2x-1)\cos n\pi x\,dx=\frac{4((-1)^n-1)}{n^2\pi^2}\). So \(2x-1=-\frac8{\pi^2}\sum_{n\,odd}\frac{\cos n\pi x}{n^2}\).
Put \(x=0\), where the even extension is continuous: \(-1=-\frac8{\pi^2}\sum_{n\ge0}\frac1{(2n+1)^2}\), so \(\sum\frac1{(2n+1)^2}=\frac{\pi^2}8\) ∎
Q1 Dirac delta and Laplace applications 25 marks
\(\int_0^\infty f\delta_n=n\int_{a-1/2n}^{a+1/2n}f\,dt=f(c_n)\) by the mean value theorem, with \(c_n\) in the interval. As \(n\to\infty\), \(c_n\to a\), so the limit is \(f(a)\) by continuity.
Put \(f=e^{-st}\): \(\mathcal L\{\delta(t-a)\}=e^{-as}\).
\(y''+y=\delta(t-3)\) with zero ICs: \((s^2+1)Y=e^{-3s}\), so \(y=\sin(t-3)\,u(t-3)\).
\(y''''=-\delta(x-2)\). Transform: \(s^4Y-s^3y(0)-s^2y'(0)-sy''(0)-y'''(0)=-e^{-2s}\).
With \(y(0)=y''(0)=0\) and \(y'(0)=y'''(0)=1\): \(Y=\frac1{s^2}+\frac1{s^4}-\frac{e^{-2s}}{s^4}\), so \(y=x+\frac{x^3}6-\frac{(x-2)^3}6u(x-2)\).
But that gives \(y(4)\ne0\), which contradicts \(y(L)=0\). The figure shows a simply supported beam. Treat \(y'(0)=A\) and \(y'''(0)=B\) as unknowns: \(y=Ax+\frac{Bx^3}6-\frac{(x-2)^3}6u(x-2)\). Then \(y''(4)=4B-2=0\Rightarrow B=\frac12\), and \(y(4)=4A+4=0\Rightarrow A=-1\).
\[\boxed{y=\frac{x^3}{12}-x-\frac{(x-2)^3}{6}u(x-2)}\]
Mid-span deflection \(y(2)=-\frac43=-\frac{FL^3}{48EI}\) ✔. Write both versions in the exam and state your interpretation.
Q2 Series solutions 25 marks
(a) Define all three point types · 6 marksOrdinary: \(P_0(a)\ne0\). Regular singular: \(P_0(a)=0\) and both \((x-a)P_1/P_0\) and \((x-a)^2P_2/P_0\) are analytic at \(a\). Irregular singular: a singular point that is not regular.
\(y''-xy=0\) about \(-2\). Let \(t=x+2\): \(y''-(t-2)y=0\), an ordinary point.
\((n+2)(n+1)a_{n+2}+2a_n-a_{n-1}=0\). So \(a_2=-a_0\), \(a_3=\frac{a_0-2a_1}6\), \(a_4=\frac{2a_0+a_1}{12}\).
\(y=a_0\left[1-(x+2)^2+\frac{(x+2)^3}6+\cdots\right]+a_1\left[(x+2)-\frac{(x+2)^3}3+\frac{(x+2)^4}{12}+\cdots\right]\)
\(xy''+y'+xy=0\): \(I(m)=m^2\), so \(m=0\) is a double root (Case 2). Recurrence: \(k^2a_k=-a_{k-2}\) and \(a_1=0\).
\(y_1=1-\frac{x^2}4+\frac{x^4}{64}-\cdots=J_0(x)\). \(y_2=y_1\ln x+\frac{x^2}4-\frac{3x^4}{128}+\frac{11x^6}{13824}-\cdots\)
Q3 Wave equation 25 marks
Lesson 3.7 → (string simulator has both cases)
\(y=XT\Rightarrow X''+p^2X=0\) and \(T''+c^2p^2T=0\). Ends fixed, so \(p=\frac{n\pi}l\): \(y=\sum\sin\frac{n\pi x}l\left(C_n\cos\frac{n\pi ct}l+D_n\sin\frac{n\pi ct}l\right)\).
\(y_t(x,0)=0\Rightarrow D_n=0\). \(y(x,0)=\sum C_n\sin\frac{n\pi x}l=200\sin\frac{4\pi x}l\Rightarrow C_4=200\), all others 0. So \(y=200\sin\frac{4\pi x}l\cos\frac{4\pi ct}l\) ∎
\(y(x,0)=0\Rightarrow C_n=0\). \(y_t(x,0)=\sum D_n\frac{n\pi c}l\sin\frac{n\pi x}l=100\sin\frac{\pi x}l\Rightarrow D_1=\frac{100l}{\pi c}\). So \(y=\frac{100l}{\pi c}\sin\frac{\pi x}l\sin\frac{\pi ct}l\).
Q4 Fourier series 25 marks
Half-range cosine of \(x(\pi-x)\) on \((0,\pi)\): \(a_0=\frac2\pi\int_0^\pi x(\pi-x)dx=\frac{\pi^2}3\).
\(a_n=\frac2\pi\int_0^\pi x(\pi-x)\cos nx\,dx=-\frac{2(1+(-1)^n)}{n^2}\). That is \(-\frac{4}{n^2}\) for even \(n\) and 0 for odd. Writing \(n=2k\): \(-\frac1{k^2}\cos2kx\). So \(x(\pi-x)=\frac{\pi^2}6-\sum\frac{\cos2kx}{k^2}\) ∎
Parseval (half-range cosine): \(\frac2\pi\int_0^\pi x^2(\pi-x)^2dx=\frac2\pi\cdot\frac{\pi^5}{30}=\frac{\pi^4}{15}=\frac{\pi^4}{18}+\sum\frac1{k^4}\), so \(\sum\frac1{n^4}=\frac{\pi^4}{90}\).
\(f=\cos x\) on \((-\frac\pi2,0)\). A sine series needs an odd function on \((-\frac\pi2,\frac\pi2)\), so \(f=-\cos x\) on \((0,\frac\pi2)\) with \(\ell=\frac\pi2\).
\(b_n=\frac{2}{\pi/2}\int_0^{\pi/2}(-\cos x)\sin2nx\,dx=-\frac4\pi\cdot\frac{2n}{4n^2-1}\). So \(f(x)=-\frac8\pi\sum\frac{n\sin2nx}{4n^2-1}\).
The series has period \(\pi\). \(f(\pi)=f(0)\), and at 0 there is a jump between \(1\) and \(-1\), so the value is \(0\). \(f(\pi/4)=-\cos\frac\pi4=-\frac1{\sqrt2}\).
\(c_n=\frac12\int_{-1}^1e^{-x}e^{-in\pi x}dx=\frac{(-1)^n\sinh1}{1+in\pi}\), so \(e^{-x}=\sum_{n=-\infty}^{\infty}\frac{(-1)^n\sinh 1\,(1-in\pi)}{1+n^2\pi^2}e^{in\pi x}\).
Q1 Laplace transform 25 marks
Lessons 1.1, 1.3 and 1.10 → (two-mass animation)
\(\mathcal L\{e^{iat}\}=\frac1{s-ia}=\frac{s+ia}{s^2+a^2}\) for \(\operatorname{Re}s>0\). The real part gives \(\mathcal L\{\cos at\}=\frac s{s^2+a^2}\) and the imaginary part gives \(\mathcal L\{\sin at\}=\frac a{s^2+a^2}\).
By parts: \(\mathcal L\{f'\}=[e^{-st}f]_0^\infty+sF=sF-f(0)\), because exponential order makes the upper limit vanish. Apply twice: \(\mathcal L\{f''\}=s^2F-sf(0)-f'(0)\).
\((s^2+4)Y_1-2Y_2=1\) and \(-2Y_1+(s^2+4)Y_2=1\). Subtracting gives \(Y_1=Y_2\), so \((s^2+2)Y_1=1\).
\(y_1(t)=y_2(t)=\frac{1}{\sqrt2}\sin(\sqrt2\,t)\)
Q2 Fourier series and Fourier transform 25 marks
Builder preset "−1 for |x| < 1" → · Lesson 5.4 →
(a)(i) Sketch on [−2, 6]
The function is even with \(\ell=2\), so \(b_n=0\). \(a_0=\frac22\int_0^2f=\int_0^1(-1)dx=-1\).
\(a_n=\int_0^1-\cos\frac{n\pi x}2dx=-\frac{2}{n\pi}\sin\frac{n\pi}2\). So \(f=-\frac12-\frac2\pi\left(\cos\frac{\pi x}2-\frac13\cos\frac{3\pi x}2+\frac15\cos\frac{5\pi x}2-\cdots\right)\).
Parseval: \(\frac12\int_{-2}^2f^2=\frac12\cdot2=1=\frac{1}{2}+\frac{4}{\pi^2}\sum_{odd}\frac1{n^2}\), so \(\sum_{n\ge0}\frac{1}{(2n+1)^2}=\frac{\pi^2}8\).
By parts: \(\mathcal F\{f'\}=[fe^{-i\omega t}]_{-\infty}^{\infty}+i\omega\int fe^{-i\omega t}=i\omega F\), since \(f\to0\) at \(\pm\infty\).
\(te^{-(t-1)^2}-e^{-(t-1)^2}=g(t-1)\) with \(g=te^{-t^2}=-\frac12(e^{-t^2})'\). So \(G=-\frac{i\omega}{2}\sqrt\pi e^{-\omega^2/4}\), and the answer is \(-\frac{i\sqrt\pi}{2}\omega e^{-\omega^2/4-i\omega}\).
Q3 Laplace's equation 25 marks
Lesson 3.4 heat map → · Lesson 5.7 strip →
\(X''+k^2X=0\) and \(Y''-k^2Y=0\). \(U(0,y)=U(1,y)=0\Rightarrow X=\sin n\pi x\). \(U(x,1)=0\Rightarrow Y=\sinh n\pi(1-y)\).
\(U(x,0)=\sum A_n\sinh(n\pi)\sin n\pi x=\sinh\pi\sin\pi x\Rightarrow A_1=1\). So \(U=\sin\pi x\,\sinh\pi(1-y)\).
Transform in x: \(\hat U_{yy}=\omega^2\hat U\). Then \(\hat U(\omega,0)=0\) and \(\hat U(\omega,1)=\sqrt\pi e^{-\omega^2/4}\).
\(\hat U=\sqrt\pi e^{-\omega^2/4}\frac{\sinh\omega y}{\sinh\omega}\), so \(U=\frac1{\sqrt\pi}\int_0^\infty e^{-\omega^2/4}\frac{\sinh\omega y}{\sinh\omega}\cos\omega x\,d\omega\).
Q4 Series solutions 25 marks
Lesson 4.2 preset "2025 Q4(a)" →
\(xy''+y=0\) about 1: \(P_0(1)=1\ne0\), so it is ordinary. With \(t=x-1\): \((t+1)y''+y=0\).
\((n+2)(n+1)a_{n+2}+(n+1)na_{n+1}+a_n=0\). So \(a_2=-\frac{a_0}2\), \(a_3=\frac{a_0-a_1}6\), \(a_4=\frac{2a_1-a_0}{24}\).
\(y=a_0\left[1-\frac{(x-1)^2}2+\frac{(x-1)^3}6-\cdots\right]+a_1\left[(x-1)-\frac{(x-1)^3}6+\frac{(x-1)^4}{12}-\cdots\right]\)
Identical to 2022 Q2(b)(ii): \(m=0\) is a double root. \(y_1=J_0=1-\frac{x^2}4+\frac{x^4}{64}\), and \(y_2=J_0\ln x+\frac{x^2}4-\frac{3x^4}{128}+\cdots\)