Laplace transform applied to IVP for second-order linear ODE with constant coefficients

At a Glance

• Frequency: 10 sub-parts across 9 of 13 years (2013, 2014, 2015, 2016, 2017, 2021, 2022, 2023, 2024)
• Priority tier: T1
• Marks (count): 10 (3), 15 (4), 17 (1), 20 (1), 6 (1)
• Average solve time: ~11 min
• Difficulty mix: medium 7, easy 2, hard 1
• Section: B | Dominant type: computation

Why This Chapter Matters

The Laplace transform appears in 9 of the last 13 years and is one of the most efficient exam techniques: ICs are baked in automatically, no complementary function is needed separately, and piecewise/impulse forcing that defeats other methods is handled mechanically. Six questions are standard constant-coefficient IVPs (the most procedural questions in Paper 1); two involve Heaviside step forcing; one has delta-function forcing; one is a convolution integral equation. The same five-step pipeline solves every variant.

Minimum Theory

Transform of derivatives. With $Y(s)=\mathcal L\{y(t)\}$: $\mathcal L\{y'\}=sY-y(0),\qquad\mathcal L\{y''\}=s^2Y-sy(0)-y'(0).$ These bake in the ICs at the start — no undetermined constants.

Standard Laplace pairs (must know):

First-shift theorem. $\mathcal L\{e^{at}f(t)\}=F(s-a)$. Used to invert $F(s+a)$.

Second-shift theorem. $\mathcal L^{-1}\{e^{-as}F(s)\}=u(t-a)f(t-a)$.

Convolution. $\mathcal L^{-1}\{F(s)G(s)\}=\int_0^t f(\tau)g(t-\tau)\,d\tau$.

Question Archetypes

Four patterns; the first covers 60% of questions.

laplace-ivp (6 question(s); 2013, 2014, 2015, 2016, 2021, 2023)

Recognition Cues

• Constant-coefficient second-order ODE with specified $y(0)$ and $y'(0)$.
• RHS may be $\sin$, $\cos$, $e^{at}\sin$, $e^{at}\cos$, polynomial, or $\sin(nt+\alpha)$ (requires addition formula).
• “Solve by using Laplace transform method.”

Solution Template

1. Apply $\mathcal L$ to both sides; substitute $\mathcal L\{y''\}=s^2Y-sy(0)-y'(0)$.
2. Solve for $Y(s)$.
3. Decompose $Y(s)$ into standard forms via partial fractions. If denominator has $(s^2+\omega^2)^2$, use resonance pairs; if $(s+a)^2+b^2$, complete the square.
4. Invert term by term. Apply first-shift theorem for $e^{at}$ multipliers.
5. Verify ICs.

Worked Example(s)

2016 Paper 1, 2016-P1-Q6d (10 marks)

Solve $y''-2y'-8y=0$, $y(0)=3$, $y'(0)=6$ via Laplace.

$\mathcal L$: $(s^2-2s-8)Y=sy(0)+y'(0)-2y(0)=3s+6-6=3s$.

$Y=\dfrac{3s}{(s-4)(s+2)}=\dfrac2{s-4}+\dfrac1{s+2}$.

$\boxed{\;y=2e^{4t}+e^{-2t}.\;}$

2015 Paper 1, 2015-P1-Q7a-ii (6 marks)

Solve $y''+y=t$, $y(0)=1$, $y'(0)=-2$.

$(s^2+1)Y=1/s^2+s-2$. Partial fractions: $1/[s^2(s^2+1)]=1/s^2-1/(s^2+1)$.

$Y=1/s^2-1/(s^2+1)+s/(s^2+1)-2/(s^2+1)=1/s^2+s/(s^2+1)-3/(s^2+1)$.

$\boxed{\;y=t+\cos t-3\sin t.\;}$

2013 Paper 1, 2013-P1-Q6d (15 marks)

Solve $(D^2+n^2)x=a\sin(nt+\alpha)$, $x(0)=x'(0)=0$.

Key step. Cannot use shift theorem for $\sin(nt+\alpha)$ — expand via addition formula: $\sin(nt+\alpha)=\sin nt\cos\alpha+\cos nt\sin\alpha$.

$X(s)=\dfrac{a(n\cos\alpha+s\sin\alpha)}{(s^2+n^2)^2}$.

Use resonance pairs ($\mathcal L\{t\sin nt\}=2ns/(s^2+n^2)^2$, $\mathcal L\{\sin nt-nt\cos nt\}=2n^3/(s^2+n^2)^2$):

$\boxed{\;x=\frac{a}{2n^2}\bigl[\cos\alpha\sin nt-nt\cos(nt+\alpha)\bigr].\;}$ The secular term $-nt\cos(nt+\alpha)$ grows unboundedly — this is resonance (driving frequency $=$ natural frequency).

2014 Paper 1, 2014-P1-Q8c (20 marks)

Solve $y''+y=8e^{-2t}\sin t$, $y(0)=y'(0)=0$.

$\mathcal L\{e^{-2t}\sin t\}=1/(s^2+4s+5)$ (shift theorem, $\sin t\to F(s+2)$).

$(s^2+1)Y=8/(s^2+4s+5)$. Partial fractions: $Y=(-s+1)/(s^2+1)+(s+3)/(s^2+4s+5)$.

Invert: first term $\to -\cos t+\sin t$; second: complete square $(s+2)^2+1$, write $s+3=(s+2)+1$, shift back $\to e^{-2t}(\cos t+\sin t)$.

$\boxed{\;y=\sin t-\cos t+e^{-2t}(\sin t+\cos t).\;}$

2021 Paper 1, 2021-P1-Q5b (10 marks)

Solve $y''+4y=e^{-2x}\sin2x$, $y(0)=y'(0)=0$.

$Y=2/[(s^2+4)(s^2+4s+8)]$. Four-unknown partial fractions: $A=-1/10$, $B=1/10$, $C=1/10$, $D=3/10$.

$y=\frac{1}{10}\!\left[\tfrac{\sin2x}{2}-\cos2x+e^{-2x}\!\left(\cos2x+\tfrac{\sin2x}{2}\right)\right].$

$\boxed{\;y=\frac{\sin2x(1+e^{-2x})}{20}+\frac{\cos2x(e^{-2x}-1)}{10}.\;}$

2023 Paper 1, 2023-P1-Q8a (15 marks)

Solve $y''-4y'+3y=f(t)$, $y(0)=1$, $y'(0)=0$ for general $f(t)$.

$Y(s^2-4s+3)=F(s)+s-4$. Partial fractions: $\dfrac{s-4}{(s-1)(s-3)}=\dfrac{3/2}{s-1}-\dfrac{1/2}{s-3}$.

IC-driven part: $\frac32e^t-\frac12e^{3t}$. $F$-driven part via convolution: $\frac12\int_0^t f(\tau)[e^{3(t-\tau)}-e^{t-\tau}]\,d\tau$.

$\boxed{\;y=\frac32e^t-\frac12e^{3t}+\frac12\int_0^t f(\tau)\!\left[e^{3(t-\tau)}-e^{t-\tau}\right]d\tau.\;}$

Common Traps

• $\mathcal L\{y''\}=s^2Y-sy(0)-y'(0)$: the $y'(0)$ term is subtracted, not added. With $y'(0)=-2$ (2015), this contributes $+2$.
• For $\sin(nt+\alpha)$: expand via addition formula first — there is no Laplace shift theorem for argument shifts.
• Resonance pair: $\mathcal L\{t\sin nt\}=\dfrac{2ns}{(s^2+n^2)^2}$, not $\dfrac{t}{(s^2+n^2)}$ — students frequently misremember this.
• Partial fractions for two irreducible quadratics (2014, 2021): 4 unknowns, 4 equations from matching $s^3,s^2,s^1,s^0$. Must complete the square in the second denominator for the inverse transform.
• The constant $GMm/(2a)$ in the 2023 planet problem contributes zero to derivatives — never skip it conceptually, but it vanishes in all Hamilton’s equations.

step-forcing-laplace (2 question(s); 2017, 2022)

Recognition Cues

• Piecewise forcing: ”$r(x)=8\sin x$ for $x<\pi$, $0$ for $x\ge\pi$” or “switch off at $t=4$”.
• Write $r(t)=f_1(t)-f_1(t)u(t-a)$ first (Heaviside decomposition).
• Apply the second-shift theorem: $\mathcal L\{g(t-a)u(t-a)\}=e^{-as}G(s)$.

Solution Template

1. Express $r(t)$ as a Heaviside combination: $r(t)=f(t)[1-u(t-a)]=f(t)-f(t)u(t-a)$.
2. Rewrite the shifted piece as $g(t-a)u(t-a)$: need to express $f(t)$ as a function of $(t-a)$.
3. Transform; solve for $Y(s)$; partial fractions.
4. Invert — the $e^{-as}$ terms give shifted functions times $u(t-a)$.
5. Write the solution piecewise; verify continuity of $y$ and $y'$ at $t=a$.

Worked Example(s)

2022 Paper 1, 2022-P1-Q7b (15 marks)

$y''-3y'+2y=h(t)$, $h=2$ on $(0,4)$, $h=0$ for $t>4$; $y(0)=y'(0)=0$.

$h(t)=2-2u(t-4)$. Transform: $\mathcal L\{h\}=\frac{2(1-e^{-4s})}{s}$.

$(s^2-3s+2)Y=\frac{2(1-e^{-4s})}{s}$. Partial fractions: $\frac1{s(s-1)(s-2)}=\frac{1/2}s-\frac1{s-1}+\frac{1/2}{s-2}$.

Invert both pieces ($f(t)=1-2e^t+e^{2t}$): $y(t)=f(t)-f(t-4)u(t-4)=(1-2e^t+e^{2t})-[1-2e^{t-4}+e^{2(t-4)}]u(t-4).$

For $0<t<4$: $y=1-2e^t+e^{2t}$. For $t>4$: add the shifted piece. At $t=0$: $y=1-2+1=0$ ✓; $y'=0-2+2=0$ ✓.

2017 Paper 1, 2017-P1-Q8b (17 marks)

$y''+9y=r(x)$, $y(0)=0$, $y'(0)=4$; $r=8\sin x$ for $x<\pi$, $0$ for $x\ge\pi$.

$r(x)=8\sin x-8\sin x\,u(x-\pi)$. Key identity: $\sin x=-\sin(x-\pi)$, so the shifted piece is $8\sin(x-\pi)\,u(x-\pi)$.

$(s^2+9)Y-4=\frac8{s^2+1}+\frac{8e^{-\pi s}}{s^2+1}$.

Partial fractions: $\frac8{(s^2+1)(s^2+9)}=\frac1{s^2+1}-\frac1{s^2+9}$.

Invert (unshifted part = $\sin x+\sin 3x$; shifted part applies second-shift theorem):

$y=\begin{cases}\sin x+\sin 3x,&0\le x\le\pi,\\\tfrac43\sin3x,&x>\pi.\end{cases}$

After forcing switches off, only the natural frequency $\omega=3$ survives.

Common Traps

• Write the Heaviside form first, before touching the Laplace transform.
• For $\sin x\,u(x-\pi)$: must rewrite in terms of $(x-\pi)$. The identity $\sin x=-\sin(x-\pi)$ is essential — without it, the second-shift theorem cannot be applied.
• $\mathcal L^{-1}\{e^{-as}F(s)\}=f(t-a)u(t-a)$ — the unit step factor $u(t-a)$ is mandatory; omitting it gives a solution that’s nonzero before the forcing switches.
• Check continuity of $y$ and $y'$ at the switching time.

impulse-forcing-laplace (1 question(s); 2024)

Recognition Cues

• "$\delta(t-a)$" on the right-hand side.
• $\mathcal L\{\delta(t-a)\}=e^{-as}$ (the Laplace transform of an impulse is a pure exponential).

Worked Example(s)

2024 Paper 1, 2024-P1-Q8a (15 marks)

Solve $y''+2y'+5y=\delta(t-2)$, $y(0)=y'(0)=0$.

$(s^2+2s+5)Y=e^{-2s}\Rightarrow Y=\dfrac{e^{-2s}}{(s+1)^2+4}$.

Invert: $\mathcal L^{-1}\{1/(s^2+2s+5)\}=\frac12e^{-t}\sin2t$. Apply second-shift:

$\boxed{\;y(t)=u(t-2)\cdot\tfrac12e^{-(t-2)}\sin2(t-2).\;}$ For $t<2$: $y=0$ (impulse hasn’t fired). For $t\ge2$: damped oscillation triggered by the kick.

Common Traps

• $\mathcal L\{\delta(t-2)\}=e^{-2s}$, not $1$. The exponential factor survives.
• Complete the square: $(s+1)^2+4$, so $\omega=2$ and the inversion needs the $1/\omega$ factor: $\frac12e^{-t}\sin2t$.
• The unit step $u(t-2)$ is essential — the system is at rest until the impulse fires.

integral-equation-laplace (1 question(s); 2024)

Recognition Cues

• $y(t)=g(t)+\int_0^t y(\tau)k(t-\tau)\,d\tau$: the integral is a convolution $y\ast k$.
• Apply Laplace: $Y=G+Y\cdot K$; solve for $Y=G/(1-K)$.

Worked Example(s)

2024 Paper 1, 2024-P1-Q5b (10 marks)

Solve $y(t)=\cos t+\int_0^t y(x)\cos(t-x)\,dx$.

$Y=\frac s{s^2+1}+Y\cdot\frac s{s^2+1}$. Solve: $Y\cdot\frac{s^2-s+1}{s^2+1}=\frac s{s^2+1}$, so $Y=\frac s{s^2-s+1}$.

Complete the square: $s^2-s+1=(s-\frac12)^2+\frac34$. Write $s=(s-\frac12)+\frac12$:

$Y=\frac{s-1/2}{(s-1/2)^2+3/4}+\frac{1/\sqrt3\cdot\sqrt3/2}{(s-1/2)^2+3/4}.$

$\boxed{\;y(t)=e^{t/2}\!\left[\cos\!\left(\frac{\sqrt3\,t}{2}\right)+\frac{1}{\sqrt3}\sin\!\left(\frac{\sqrt3\,t}{2}\right)\right].\;}$

Common Traps

• Recognise the convolution immediately — missing the structure costs significant time.
• $Y=G/(1-K)$, not $Y=G\cdot K$ or $Y=G+K$.
• $\omega=\sqrt{3/4}=\sqrt3/2$, not $\sqrt3$ — the factor $\frac12$ in $\omega$ appears in the $1/\omega$ denominator.

Marks-Aware Writing

6-mark questions (2015): Transform, partial fractions (one line), invert, state answer. No verification needed for 6 marks.

10-mark questions (2016, 2021, 2024-Q5b): Show the transformed equation, partial fraction decomposition, and inversion term by term. Verify ICs in one line.

15-mark questions (2013, 2014, 2021, 2022, 2023, 2024-Q8a): Write the full partial fraction derivation (show 4 equations, 4 unknowns if applicable). For resonance (2013): clearly state the secular term. For convolution (2023): write the integral formula explicitly.

17-20-mark questions (2017, 2014): For Heaviside (2017): show the Heaviside decomposition step and the $\sin x=-\sin(x-\pi)$ identity. For the 4-unknown partial fraction (2014): match coefficients of all four powers of $s$ explicitly. Verify both ICs and the ODE residual.

Practice Set