Inverse Laplace transform

At a Glance

• Frequency: 2 sub-parts across 2 of 14 years (2015, 2018)
• Priority tier: T3
• Marks (count): 5 (1), 6 (1)
• Average solve time: ~6 min
• Difficulty mix: medium 1, easy 1
• Section: B | Dominant type: computation

Why This Chapter Matters

Both UPSC questions on inverse Laplace transforms are short (5–6 marks) and follow a fixed recipe: decompose the expression into recognisable pieces, apply the matching inversion rule to each piece, combine. The 2018 question is a straightforward partial-fractions inversion. The 2015 question combines two harder rules — the log-derivative rule for $\ln(1+1/s^2)$ and the second-shift theorem for the $e^{-\pi s}$ factor — but each sub-step is mechanical. Knowing four rules cold (partial fractions, first-shift, second-shift, log-derivative via $\mathcal{L}\{tf\}=-F'(s)$) covers everything in this atom and also supports the IVP-solving questions in P1-OD-15.

Minimum Theory

Linearity. $\mathcal{L}^{-1}\{aF+bG\}=a\mathcal{L}^{-1}\{F\}+b\mathcal{L}^{-1}\{G\}$.

Standard inverse pairs.

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

Second shift theorem. If $\mathcal{L}^{-1}\{F(s)\}=f(t)$, then $\mathcal{L}^{-1}\{e^{-as}F(s)\}=f(t-a)\,u(t-a)$, where $u$ is the Heaviside step function.

Log-derivative rule. If $\mathcal{L}\{tf(t)\}=-F'(s)$, then $\mathcal{L}^{-1}\{F(s)\}=f(t)$ where $tf(t)=\mathcal{L}^{-1}\{-F'(s)\}$. Concretely: differentiate $F(s)$, recognise $-F'(s)$ as the transform of some $g(t)$, then $f(t)=g(t)/t$.

Partial fractions. For a proper rational function $N(s)/D(s)$ (degree $N <$ degree $D$) with distinct simple poles at $s=a_k$: cover-up rule gives residue $A_k = N(a_k)/D'(a_k)$, so $F(s)=\sum A_k/(s-a_k)$ and $f(t)=\sum A_k e^{a_k t}$.

Question Archetypes

inverse-laplace (2 question(s); 2015, 2018)

Recognition Cues

• “Find the inverse Laplace transform of $F(s)$.”
• $F(s)$ is a rational function with distinct poles: use partial fractions.
• $F(s)$ contains $e^{-as}$ as a factor: use the second-shift theorem.
• $F(s)$ is a logarithm: use the log-derivative rule $\mathcal{L}\{tf\}=-F'(s)$.
• $F(s)$ is a sum: split and invert each piece separately.

Solution Template

1. Split $F(s)$ into pieces, one per tool.
2. For rational pieces: check degree (proper = numerator degree $<$ denominator degree). Factor denominator; apply cover-up rule for residues; invert.
3. For $e^{-as}G(s)$ pieces: identify $g(t)=\mathcal{L}^{-1}\{G(s)\}$; answer is $g(t-a)u(t-a)$.
4. For logarithm pieces: differentiate $F(s)$; recognise $-F'(s)$ as a standard transform; divide by $t$.
5. Combine and box the answer.

Worked Example

2018 Paper 1, 2018-P1-Q5d-ii (5 marks)

Find the inverse Laplace transform of $\dfrac{5s^2+3s-16}{(s-1)(s-2)(s+3)}$.

Step 1 — Partial fractions. Degree of numerator (2) $<$ degree of denominator (3): no polynomial part.

$\frac{5s^2+3s-16}{(s-1)(s-2)(s+3)} = \frac{A}{s-1}+\frac{B}{s-2}+\frac{C}{s+3}.$

Cover-up rule:

$A = \frac{5(1)^2+3(1)-16}{(1-2)(1+3)} = \frac{-8}{-4} = 2.$

$B = \frac{5(4)+3(2)-16}{(2-1)(2+3)} = \frac{10}{5} = 2.$

$C = \frac{5(9)+3(-3)-16}{(-4)(-5)} = \frac{20}{20} = 1.$

Step 2 — Invert. Using $\mathcal{L}^{-1}\{1/(s-a)\}=e^{at}$:

$\boxed{\;\mathcal{L}^{-1}\!\left\{\frac{5s^2+3s-16}{(s-1)(s-2)(s+3)}\right\} = 2e^{t}+2e^{2t}+e^{-3t}.\;}$

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

Obtain the inverse Laplace transform of $\ln\!\left(1+\dfrac{1}{s^2}\right)+\dfrac{s}{s^2+25}\,e^{-\pi s}$.

Split into two pieces.

Piece 1: $\mathcal{L}^{-1}\!\left\{\dfrac{s}{s^2+25}e^{-\pi s}\right\}$.

Standard pair: $\mathcal{L}^{-1}\!\left\{\dfrac{s}{s^2+25}\right\}=\cos 5t$.

Second-shift theorem with $a=\pi$:

$\mathcal{L}^{-1}\!\left\{\frac{s}{s^2+25}e^{-\pi s}\right\} = \cos(5(t-\pi))\,u(t-\pi).$

Simplify: $\cos(5t-5\pi)=\cos 5t\cos 5\pi=-\cos 5t$ (since $\cos 5\pi=\cos\pi=-1$).

So Piece 1 $= -\cos(5t)\,u(t-\pi)$.

Piece 2: $\mathcal{L}^{-1}\!\left\{\ln\!\left(1+\dfrac{1}{s^2}\right)\right\} = \mathcal{L}^{-1}\!\left\{\ln\dfrac{s^2+1}{s^2}\right\}$.

Let $F(s)=\ln(s^2+1)-\ln(s^2)$. Differentiate:

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

So $-F'(s)=\dfrac{2}{s}-\dfrac{2s}{s^2+1}=\mathcal{L}\{2\}-\mathcal{L}\{2\cos t\}=\mathcal{L}\{2-2\cos t\}$.

Since $\mathcal{L}\{tf(t)\}=-F'(s)$, we have $tf(t)=2-2\cos t$, so $f(t)=\dfrac{2(1-\cos t)}{t}$.

$\mathcal{L}^{-1}\!\left\{\ln\!\left(1+\frac{1}{s^2}\right)\right\} = \frac{2(1-\cos t)}{t}.$

Combine:

$\boxed{\;\mathcal{L}^{-1}\!\left\{\ln\!\left(1+\frac{1}{s^2}\right)+\frac{s}{s^2+25}e^{-\pi s}\right\} = \frac{2(1-\cos t)}{t} - \cos(5t)\,u(t-\pi).\;}$

Common Traps

• Partial fractions (2018): Always check that the numerator degree is strictly less than the denominator degree before writing the partial-fraction form. If not, do polynomial long division first.
• Cover-up at $s=-3$ (2018): Sign errors are common. $5(9)+3(-3)-16=45-9-16=20$ and $(-3-1)(-3-2)=(-4)(-5)=20$, giving $C=1$ — verify the negatives carefully.
• Second shift (2015): The formula is $\mathcal{L}^{-1}\{e^{-as}F(s)\}=f(t-a)u(t-a)$, not $f(t)\cdot u(t-a)$. Write out $\cos(5(t-\pi))$ before simplifying.
• Log inversion (2015): The log-derivative rule proceeds as: differentiate $F(s)$, recognise $-F'(s)$ as a known transform $\mathcal{L}\{g(t)\}$, then $f(t)=g(t)/t$. Do not try to invert $F(s)$ directly as if it were a ratio.
• The simplification $\cos(5t-5\pi)=-\cos 5t$ (2015): $5\pi$ is an odd multiple of $\pi$, so $\cos(5t-5\pi)=\cos(5t)\cos(5\pi)+\sin(5t)\sin(5\pi)=-\cos(5t)$.

Marks-Aware Writing

5-mark question (2018): Three residues (1 mark each) plus inversion and boxed answer (2 marks). Each cover-up computation should show the numerator evaluated at the pole and the denominator product; do not just write the answer.

6-mark question (2015): Two pieces of roughly equal weight. For the second-shift piece: state the standard pair, apply the theorem, simplify the cosine expression (2–3 marks). For the log piece: differentiate, recognise $-F'(s)$, divide by $t$ (2–3 marks). Combine clearly.

Practice Set