Worked examples — Inverse Laplace transform — partial fractions, tables

Before anything, a reminder of the two worlds we translate between. Read the picture left-to-right: a differential equation in real time $t$ becomes an algebra problem in the transform variable $s$, and the inverse Laplace transform ${\mathcal{L}}^{-1}$ is the return arrow.

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The first-shift theorem — state it once, use it everywhere

Several examples below lean on this single result, so let us pin it down cleanly before we need it.

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The scenario matrix

Every rational function $F(s)=\frac{P(s)}{Q(s)}$ we could meet is classified by what the denominator $Q(s)$ factors into and whether the fraction is proper. Here proper means the top degree is smaller than the bottom degree — that's the condition the table silently assumes.

Cell	Case class	Signature of $Q(s)$	Inverse looks like	Example
L1	Distinct real poles	$(s-a)(s-b)$, $a\ne b$	sum of $e^{at}$	Ex 1
L2	Real poles, one at $s=0$	$s(s-a)$	constant $+$ $e^{at}$	Ex 2
R1	Repeated real pole	$(s-a{)}^2$ or $(s-a{)}^3$	$e^{at}(c_0+c_{1}t+\dots )$	Ex 3
Q1	Irreducible quadratic, $a=0$	$s^2+{\omega }^2$	pure $\sin /\cos$	Ex 4
Q2	Irreducible quadratic, $a\ne 0$	$(s-a{)}^2+{\omega }^2$	damped/growing $e^{at}\sin$	Ex 5
H1	Real poles $\pm b$ (sign flip!)	$s^2-b^2$	$\sinh /\cosh$, not $\sin$	Ex 6
D1	Degenerate / improper	$\mathrm{deg}P\ge \mathrm{deg}Q$	$\delta (t)$ $+$ proper part	Ex 7
S1	Shifted in $t$ (Heaviside)	$e^{-cs}F(s)$	delayed switch-on	Ex 8
W1	Real-world word problem	comes from an ODE	physical curve	Ex 9
X1	Exam twist (mixed)	quadratic $\times$ linear	$e^{at}+$ damped osc.	Ex 10
RQ1	Repeated irreducible quadratic	$(s^2+{\omega }^2{)}^2$	$t\sin \omega t$, $t\cos \omega t$	Ex 11

The rest of the page walks one example per cell. Each is labelled with its cell so you can see the whole grid get covered.

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Ex 1 — Cell L1 · distinct real poles

Forecast: Two different real poles at $s=3$ and $s=-1$. Guess before reading on — the answer should be a mix of $e^{3t}$ (growing) and $e^{-t}$ (decaying), no wiggles at all.

1. Split into atoms. Write $\frac{2s-1}{(s-3)(s+1)}=\frac{A}{s-3}+\frac{B}{s+1}$. Why this step? Two distinct linear factors → the table only knows one pole at a time, so we must peel them apart.
2. Cover-up for $A$. Multiply by $(s-3)$ and set $s=3$: $A=\frac{2(3)-1}{3+1}=\frac{5}{4}$. Why this step? At $s=3$ the $B$-term still carries a factor $(s-3)$ so it dies; only $A$ survives.
3. Cover-up for $B$. Multiply by $(s+1)$ and set $s=-1$: $B=\frac{2(-1)-1}{-1-3}=\frac{-3}{-4}=\frac{3}{4}$. Why this step? Same trick at the other pole.
4. Look up each atom. ${\mathcal{L}}^{-1}\{1/(s-a)\}=e^{at}$, so $f(t)=\frac{5}{4}{e}^{3t}+\frac{3}{4}{e}^{-t}.$ Why this step? Linearity lets us invert the pieces separately and add.

Verify: Plug $t=0$: $f(0)=\frac{5}{4}+\frac{3}{4}=2$. Sanity: the initial-value theorem says $f(0^{+})={\lim }_{s\to \infty }sF(s)=\lim \frac{2s^2-s}{s^2-2s-3}=2$. ✓ Matches.

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Ex 2 — Cell L2 · a pole sitting at $s=0$

Forecast: One pole is at $s=0$. Recall ${\mathcal{L}}^{-1}\{1/s\}=1$, a constant. So expect "a constant plus $e^{2t}$" — a horizontal offset with a growing exponential.

1. Split. $\frac{s+4}{s(s-2)}=\frac{A}{s}+\frac{B}{s-2}$. Why this step? $s=s-0$ is a perfectly ordinary linear factor; treat $0$ like any other pole.
2. Cover-up for $A$ (pole at $0$). Multiply by $s$, set $s=0$: $A=\frac{0+4}{0-2}=-2$. Why this step? Isolates the residue at the origin.
3. Cover-up for $B$. Multiply by $(s-2)$, set $s=2$: $B=\frac{2+4}{2}=3$. Why this step? Residue at the other pole.
4. Invert. ${\mathcal{L}}^{-1}\{1/s\}=1$ and ${\mathcal{L}}^{-1}\{1/(s-2)\}=e^{2t}$: $f(t)=-2+3e^{2t}.$

Verify: $f(0)=-2+3=1$. Initial-value check: ${\lim }_{s\to \infty }sF(s)=\lim \frac{s^2+4s}{s^2-2s}=1$. ✓

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Ex 3 — Cell R1 · repeated real pole (cubed)

Forecast: A triple pole. The parent note warned: a factor $(s-a{)}^k$ contributes $k$ terms, one per power. So expect $e^{-t}$ times a quadratic in $t$ — powers of $t$ appear precisely because of repetition.

1. Split into all three powers. $\frac{s^2}{(s+1{)}^3}=\frac{A}{s+1}+\frac{B}{(s+1{)}^2}+\frac{C}{(s+1{)}^3}$. Why this step? Underfitting with only the highest power throws away real information; the numerator can be any polynomial of degree $<3$.
2. Clear denominators. Multiply by $(s+1{)}^3$: $s^2=A(s+1{)}^2+B(s+1)+C$. Why this step? Turns fraction-matching into polynomial-matching, which we can compare coefficient by coefficient.
3. Substitute $s=-1$ for $C$. $(-1{)}^2=C\Rightarrow C=1$. Why this step? Cover-up on the highest power kills $A$ and $B$ instantly.
4. Expand and match. $A(s+1{)}^2+B(s+1)+C=A{s}^2+(2A+B)s+(A+B+C)$. Match $s^2$: $A=1$. Match $s^1$: $2A+B=0\Rightarrow B=-2$. (Check constant: $A+B+C=1-2+1=0=$ coeff of $s^0$ in $s^2$. ✓) Why this step? Comparing like powers gives a tidy system for the remaining coefficients.
5. Invert with the power rule + shift. Using ${\mathcal{L}}^{-1}\{1/(s+1{)}^{n+1}\}=\frac{t^n}{n!}{e}^{-t}$: $f(t)=e^{-t}\left(1-2t+\frac{1}{2}{t}^2\right).$ Why this step? $1/(s+1{)}^2\to t{e}^{-t}$ and $1/(s+1{)}^3\to \frac{t^2}{2}{e}^{-t}$ — the extra powers of $t$ are the signature of the repeated pole.

Verify: $f(0)=1$. Initial-value: ${\lim }_{s\to \infty }s\cdot \frac{s^2}{(s+1{)}^3}=1$. ✓

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Ex 4 — Cell Q1 · pure oscillation ($a=0$)

Forecast: Denominator $s^2+9=(s^2+{\omega }^2)$ with no $s$-shift, so no exponential envelope — pure undying $\sin$ and $\cos$ at frequency $\omega =\sqrt{9}=3$.

1. Read off $\omega$. $s^2+9\Rightarrow {\omega }^2=9\Rightarrow \omega =3$. Why this step? The table wants $\omega$, not ${\omega }^2$; a classic slip is using $9$.
2. Split by "what sits on top." $\frac{4s+6}{s^2+9}=4\cdot \frac{s}{s^2+9}+2\cdot \frac{3}{s^2+9}$. Why this step? $s$-on-top → $\cos$; the constant must be $\omega =3$-on-top → $\sin$. We wrote $6=2\cdot 3$ to manufacture that $\omega$.
3. Invert. $f(t)=4\cos 3t+2\sin 3t.$

Verify: $f(0)=4$; initial-value ${\lim }_{s\to \infty }s\frac{4s+6}{s^2+9}=4$. ✓ Also a second check with the final-value theorem is not valid here (poles on the imaginary axis) — good to notice.

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Ex 5 — Cell Q2 · damped oscillation (shifted quadratic)

This is the geometric heart of the topic: completing the square slides the poles off the imaginary axis, and that slide becomes an exponential envelope in time.

Forecast: The quadratic has no real roots (discriminant $16-52<0$). Completing the square will give $(s-2{)}^2+9$: the "$-2$" inside means poles at $s=2\pm 3i$, to the right of the imaginary axis → a growing $e^{2t}$ envelope times oscillation at $\omega =3$. Guess: $e^{2t}(\dots \cos 3t+\dots \sin 3t)$.

1. Complete the square. $s^2-4s+13=(s-2{)}^2+9$, so $a=2$, $\omega =3$. Why this step? The shift table needs the shape $(s-a{)}^2+{\omega }^2$; only then can the first-shift theorem give the $e^{at}$ factor.
2. Rewrite the numerator around $(s-2)$. $3s-2=3(s-2)+4$. Why this step? We must match $\frac{s-a}{(s-a{)}^2+{\omega }^2}$ (gives $e^{at}\cos$). Split the numerator so the "shifted $s$" appears explicitly; $3(s-2)=3s-6$, and $-2-(-6)=+4$ is the leftover constant.
3. Group into the two table atoms. $\frac{3(s-2)}{(s-2{)}^2+9}+\frac{4}{(s-2{)}^2+9}=3\cdot \frac{s-2}{(s-2{)}^2+9}+\frac{4}{3}\cdot \frac{3}{(s-2{)}^2+9}.$ Why this step? The sine atom needs $\omega =3$ on top, so we factor $4=\frac{4}{3}\cdot 3$.
4. Invert (first-shift theorem). $f(t)=e^{2t}\left(3\cos 3t+\frac{4}{3}\sin 3t\right).$ Why this step? By the first-shift theorem stated above, $(s-2)$ in the denominator becomes the $e^{2t}$ envelope; the picture above shows the poles sitting at $s=2\pm 3i$ and the resulting growing wave.

Verify: $f(0)=e^0(3+0)=3$; initial-value ${\lim }_{s\to \infty }s\frac{3s-2}{s^2-4s+13}=3$. ✓

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Ex 6 — Cell H1 · the sign trap ($s^2-b^2$)

Forecast: Look hard — the sign is minus: $s^2-1$, not $s^2+1$. Real poles at $\pm 1$, so this is $\sinh /\cosh$ (hyperbolic), never $\sin /\cos$. If you wrote $\sin t$ here you fell into the classic trap.

1. Factor into real linear poles. $s^2-1=(s-1)(s+1)$. Why this step? The minus sign means the roots are real, so we split like distinct linear factors — no completing-the-square, no imaginary poles.
2. Split. $\frac{2s+1}{(s-1)(s+1)}=\frac{A}{s-1}+\frac{B}{s+1}$. Why this step? Two distinct real linear factors → two simple-pole atoms, one per factor, so each can be inverted by the ${\mathcal{L}}^{-1}\{1/(s-a)\}=e^{at}$ entry.
3. Cover-up. $A=\frac{2(1)+1}{1+1}=\frac{3}{2}$; $B=\frac{2(-1)+1}{-1-1}=\frac{-1}{-2}=\frac{1}{2}$. Why this step? Residues at the two real poles.
4. Invert (exponential form — always safe). $f(t)=\frac{3}{2}{e}^t+\frac{1}{2}{e}^{-t}.$ Why this step? ${\mathcal{L}}^{-1}\{1/(s-a)\}=e^{at}$ applied to each pole. This exponential form is the final answer.

Verify: $f(0)=\frac{3}{2}+\frac{1}{2}=2$; initial-value ${\lim }_{s\to \infty }s\frac{2s+1}{s^2-1}=2$. ✓

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Ex 7 — Cell D1 · degenerate / improper fraction

Forecast: Top degree $=$ bottom degree $=2$. This is improper — the table doesn't apply directly. Expect the leftover to invert to a $\delta (t)$ (the Dirac spike from Heaviside Step & Dirac Delta) plus a proper oscillation.

1. Long-divide first. $\frac{s^2+3s+5}{s^2+4}=1+\frac{3s+1}{s^2+4}$. Why this step? When $\mathrm{deg}P\ge \mathrm{deg}Q$, you must extract the polynomial part; the table only inverts proper fractions.
2. Invert the constant part. ${\mathcal{L}}^{-1}\{1\}=\delta (t)$. Why this step? A pure constant in $s$ corresponds to the impulse at $t=0$ — the transform of $\delta (t)$ is $1$.
3. Split the proper remainder. With $\omega =2$: $\frac{3s+1}{s^2+4}=3\cdot \frac{s}{s^2+4}+\frac{1}{2}\cdot \frac{2}{s^2+4}$. Why this step? $s$-on-top → $\cos 2t$; constant $1=\frac{1}{2}\cdot 2$ → $\sin 2t$ at $\omega =2$.
4. Assemble. $f(t)=\delta (t)+3\cos 2t+\frac{1}{2}\sin 2t.$ Why this step? Linearity of ${\mathcal{L}}^{-1}$ lets us invert each piece separately and simply add the results — the impulse from step 2 and the two oscillation atoms from step 3 combine with no cross-terms, because inverting a sum is the sum of the inverses.

Verify: For $t>0$ (ignoring the impulse), $f(0^{+})=3$. Initial-value applied to the proper part $\frac{3s+1}{s^2+4}$: ${\lim }_{s\to \infty }s\cdot \frac{3s+1}{s^2+4}=3$. ✓

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Ex 8 — Cell S1 · time-delayed switch-on (Heaviside)

Forecast: The factor $e^{-2s}$ is the second-shift theorem signature — it says "nothing happens until $t=2$, then the answer starts." Expect a Heaviside step $u(t-2)$ multiplying whatever $\frac{1}{s^2+s}$ inverts to, with $t$ replaced by $t-2$.

1. Handle the un-shifted core first. $\frac{1}{s^2+s}=\frac{1}{s(s+1)}=\frac{1}{s}-\frac{1}{s+1}$. Why this step? Peel the $e^{-2s}$ off mentally; find $g(t)={\mathcal{L}}^{-1}\{1/(s^2+s)\}$ first, because the second-shift theorem needs that base function $g$.
2. Invert the core. $g(t)=1-e^{-t}$. Why this step? ${\mathcal{L}}^{-1}\{1/s\}=1$ and ${\mathcal{L}}^{-1}\{1/(s+1)\}=e^{-t}$, applied term-by-term via linearity.
3. Apply the second-shift theorem with $c=2$. Replace $t$ by $t-2$ and switch on at $t=2$: $f(t)=(1-e^{-(t-2)})u(t-2).$ Why this step? The $e^{-2s}$ factor delays $g(t)$ by $2$ and multiplies by the Heaviside step $u(t-2)$ to enforce "off before $t=2$" — this is precisely the second-shift theorem stated above.

Verify: At $t=2^{+}$: $f=1-e^0=0$, so it starts continuously from $0$ (no jump). As $t\to \infty$: $f\to 1$. Both sensible for a switched-on rising exponential.

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Ex 9 — Cell W1 · real-world word problem

Forecast: Quadratic denominator with no real roots → damped oscillation. Complete the square to see how fast it decays and at what frequency it wobbles.

1. Complete the square. $s^2+2s+10=(s+1{)}^2+9$, so $a=-1$, $\omega =3$. Why this step? The "$+1$" inside gives poles at $s=-1\pm 3i$ — left of the imaginary axis → a decaying $e^{-t}$ envelope (a stable, settling system).
2. Match the sine atom. $\frac{10}{(s+1{)}^2+9}=\frac{10}{3}\cdot \frac{3}{(s+1{)}^2+9}$. Why this step? The sine atom needs $\omega =3$ on top; factor $10=\frac{10}{3}\cdot 3$.
3. Invert (first-shift theorem). $x(t)=\frac{10}{3}{e}^{-t}\sin 3t.$ Why this step? By the first-shift theorem, the shifted denominator $(s+1{)}^2+9$ → an $e^{-t}$ envelope times $\sin 3t$.

Verify (units & physics): $x(0)=0$ (mass starts at rest position — matches an impulse nudge). Envelope $\frac{10}{3}{e}^{-t}$ shrinks → oscillations die out → stable system, exactly what a real damper does. Frequency $\omega =3$ rad/s. Initial-value: ${\lim }_{s\to \infty }sX(s)=0$. ✓

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Ex 10 — Cell X1 · exam twist (linear × quadratic mix)

Forecast: One real pole ($s=-1$) and an irreducible quadratic ($s^2+4$). Expect a blend: an $e^{-t}$ term plus an undamped $\cos 2t/\sin 2t$ pair.

1. Set up the mixed decomposition. $\frac{2s^2+s+3}{(s+1)(s^2+4)}=\frac{A}{s+1}+\frac{Bs+C}{s^2+4}$. Why this step? Linear factor → constant top; irreducible quadratic → a full linear top $Bs+C$ (it may need both a $\cos$ and a $\sin$ piece).
2. Cover-up for $A$. Multiply by $(s+1)$, set $s=-1$: $A=\frac{2(1)+(-1)+3}{1+4}=\frac{4}{5}$. Why this step? Isolates the real-pole residue cleanly.
3. Clear denominators & match. $2s^2+s+3=A(s^2+4)+(Bs+C)(s+1)$. Match $s^2$: $2=A+B\Rightarrow B=2-\frac{4}{5}=\frac{6}{5}$. Match constant: $3=4A+C\Rightarrow C=3-\frac{16}{5}=-\frac{1}{5}$. (Check $s^1$: $1=B+C=\frac{6}{5}-\frac{1}{5}=1$ ✓.) Why this step? Comparing powers pins down $B,C$ once $A$ is known.
4. Split the quadratic atom & invert. With $\omega =2$: $\frac{\frac{6}{5}s-\frac{1}{5}}{s^2+4}=\frac{6}{5}\cdot \frac{s}{s^2+4}-\frac{1}{10}\cdot \frac{2}{s^2+4}.$ So $f(t)=\frac{4}{5}{e}^{-t}+\frac{6}{5}\cos 2t-\frac{1}{10}\sin 2t.$ Why this step? $s$-on-top → $\cos 2t$; constant $-\frac{1}{5}=-\frac{1}{10}\cdot 2$ → $\sin 2t$.

Verify: $f(0)=\frac{4}{5}+\frac{6}{5}+0=\frac{10}{5}=2$. Initial-value: ${\lim }_{s\to \infty }s\frac{2s^2+s+3}{(s+1)(s^2+4)}=2$. ✓

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Ex 11 — Cell RQ1 · repeated irreducible quadratic

Forecast: The quadratic $s^2+4$ is squared — this is the complex-pole analogue of a repeated real pole. Just as a repeated real pole gave a factor of $t$ (Ex 3), a repeated complex pole gives a factor of $t$ times an oscillation. So expect something with $t\sin 2t$ or $t\cos 2t$ in it.

1. Identify $\omega$ and pick the right table atom. Here $\omega =2$. The standard result for repeated complex poles is ${\mathcal{L}}^{-1}\left\{\frac{2{\omega }^3}{(s^2+{\omega }^2{)}^2}\right\}=\sin \omega t-\omega t\cos \omega t,{\mathcal{L}}^{-1}\left\{\frac{2\omega s}{(s^2+{\omega }^2{)}^2}\right\}=t\sin \omega t.$ Why this step? These two are the atoms the $(s^2+{\omega }^2{)}^2$ shape decomposes into; the $t$-factor is the fingerprint of the repeated (squared) pole, exactly parallel to Ex 3.
2. Match our fraction to the first atom. We want $\frac{1}{(s^2+4{)}^2}$. With $\omega =2$, $2{\omega }^3=2\cdot 8=16$, so $\frac{1}{(s^2+4{)}^2}=\frac{1}{16}\cdot \frac{16}{(s^2+4{)}^2}=\frac{1}{16}\cdot \frac{2{\omega }^3}{(s^2+{\omega }^2{)}^2}.$ Why this step? Scale our numerator $1$ up to the atom's numerator $2{\omega }^3=16$ by pulling out $\frac{1}{16}$.
3. Invert. $f(t)=\frac{1}{16}(\sin 2t-2t\cos 2t).$ Why this step? Direct read-off of the first atom with $\omega =2$, then multiply by our $\frac{1}{16}$.

Verify: $f(0)=\frac{1}{16}(0-0)=0$. Initial-value: ${\lim }_{s\to \infty }s\frac{1}{(s^2+4{)}^2}=0$ ✓. The $-2t\cos 2t$ term is the tell-tale growing-in-amplitude oscillation that repeated complex poles always produce (resonance!).

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Recall Self-test — which cell does each shape belong to?

$\frac{1}{s^2-9}$ is which cell? ::: H1 — real poles $\pm 3$, so $\sinh /\cosh$, not $\sin$. $\frac{5}{(s-2{)}^3}$ is which cell? ::: R1 — repeated pole, gives $\frac{5}{2}{t}^2{e}^{2t}$. $\frac{s^3}{s^2+1}$ is which cell? ::: D1 — improper, long-divide first (yields a derivative-of-$\delta$ term). $\frac{e^{-3s}}{s}$ is which cell? ::: S1 — delayed step $u(t-3)$. $\frac{s+1}{s^2+2s+2}$ is which cell? ::: Q2 — complete square to $(s+1{)}^2+1$, gives $e^{-t}\cos t$. $\frac{s}{(s^2+1{)}^2}$ is which cell? ::: RQ1 — repeated complex pole, gives $\frac{1}{2}t\sin t$.

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Back to the parent: Inverse Laplace transform — partial fractions & tables. Prerequisites worth a look: Partial Fraction Decomposition (Algebra), Laplace Transform — definition and existence, Convolution Theorem.