Inverse Laplace transform — partial fractions, tables: Exercises

The one table we lean on the whole way down — keep it visible:

Here $s$ is the frequency-domain variable, $t \geq 0$ is time, $a$ is a real growth/decay rate, $ω$ is an angular frequency, and $n$ is a whole-number power.

Level 1 — Recognition

Goal: match a single atom to the table on sight.

Problem 1.1

Find $L^{- 1} {\frac{5}{s}}$ .

Recall Solution 1.1

LOOK-UP. The atom $\frac{1}{s}$ inverts to the constant $1$ . Linearity pulls the $5$ straight through: $L^{- 1} {\frac{5}{s}} = 5 \cdot L^{- 1} {\frac{1}{s}} = 5 \cdot 1 = 5 .$

Problem 1.2

Find $L^{- 1} {\frac{1}{s + 4}}$ .

Recall Solution 1.2

LOOK-UP. The atom $\frac{1}{s - a}$ inverts to $e^{a t}$ . Here the denominator is $s - (- 4)$ , so $a = - 4$ : $L^{- 1} {\frac{1}{s + 4}} = e^{- 4 t} .$ The negative $a$ means decay — the function shrinks as $t$ grows.

Problem 1.3

Find $L^{- 1} {\frac{6}{s ^{2} + 9}}$ .

Recall Solution 1.3

LOOK-UP. The sine atom is $\frac{ω}{s ^{2} + ω ^{2}}$ . Here $ω^{2} = 9 \Rightarrow ω = 3$ , so the sine numerator should be $3$ . We have $6 = 2 \cdot 3$ , so factor a $2$ out: $L^{- 1} {\frac{6}{s ^{2} + 9}} = 2 \cdot L^{- 1} {\frac{3}{s ^{2} + 9}} = 2 sin 3 t .$

Level 2 — Application

Goal: one small preparation step (scale, shift, or split into two) before the table.

Problem 2.1

Find $L^{- 1} {\frac{s}{s ^{2} + 16}} + L^{- 1} {\frac{2}{s ^{2} + 16}}$ .

Recall Solution 2.1

LOOK-UP two atoms. Here $ω^{2} = 16 \Rightarrow ω = 4$ .

$s$ -on-top → cosine: $\frac{s}{s ^{2} + 16} \to cos 4 t$ .
constant-on-top → sine, but we need $ω = 4$ on top and only have $2 = \frac{1}{2} \cdot 4$ : $\frac{2}{s ^{2} + 16} = \frac{1}{2} \cdot \frac{4}{s ^{2} + 16} \to \frac{1}{2} sin 4 t$ . $cos 4 t + \frac{1}{2} sin 4 t .$

Problem 2.2

Find $L^{- 1} {\frac{1}{( s - 3 ) ^{2}}}$ .

Recall Solution 2.2

First-shift theorem. Start from the base atom $\frac{1}{s ^{2}} \to t$ (this is $\frac{1}{s ^{n + 1}} \to \frac{t ^{n}}{n !}$ with $n = 1$ ). Replacing $s \to s - 3$ multiplies the answer by $e^{3 t}$ : $L^{- 1} {\frac{1}{( s - 3 ) ^{2}}} = t e^{3 t} .$

Problem 2.3

Find $L^{- 1} {\frac{2 s + 5}{s ^{2} + 4}}$ .

Recall Solution 2.3

SPLIT, then LOOK-UP. $ω^{2} = 4 \Rightarrow ω = 2$ . Break by numerator pieces: $\frac{2 s + 5}{s ^{2} + 4} = 2 \cdot \frac{s}{s ^{2} + 4} + \frac{5}{s ^{2} + 4} .$

$2 \cdot \frac{s}{s ^{2} + 4} \to 2 cos 2 t$ .
$\frac{5}{s ^{2} + 4} = \frac{5}{2} \cdot \frac{2}{s ^{2} + 4} \to \frac{5}{2} sin 2 t$ . $2 cos 2 t + \frac{5}{2} sin 2 t .$

Level 3 — Analysis

Goal: real partial fractions, completing the square, and distinguishing $sin$ from $sinh$ .

Problem 3.1

Find $L^{- 1} {\frac{s - 1}{( s + 1 ) ( s - 3 )}}$ .

Recall Solution 3.1

SPLIT (distinct linear poles). Write $\frac{s - 1}{( s + 1 ) ( s - 3 )} = \frac{A}{s + 1} + \frac{B}{s - 3}$ . Cover-up at $s = - 1$ : $A = \frac{- 1 - 1}{- 1 - 3} = \frac{- 2}{- 4} = \frac{1}{2}$ . Cover-up at $s = 3$ : $B = \frac{3 - 1}{3 + 1} = \frac{2}{4} = \frac{1}{2}$ . LOOK-UP each $\frac{1}{s - a} \to e^{a t}$ : $f (t) = \frac{1}{2} e^{- t} + \frac{1}{2} e^{3 t} = \frac{1}{2} (e^{- t} + e^{3 t}) .$

Problem 3.2

Predict the form first, then find $L^{- 1} {\frac{3}{s ^{2} - 1}}$ . Look at the figure: real poles → exponential/hyperbolic, not oscillation.

Figure — Inverse Laplace transform — partial fractions, tables

Recall Solution 3.2

Predict: $s^{2} - 1 = (s - 1) (s + 1)$ has real roots $\pm 1$ . Real poles mean growing/decaying exponentials, which combine into $sinh / cosh$ — no wiggling sine. Compare with $s^{2} + 1$ (roots off the real axis) which would give $sin / cos$ . The figure shows both. SPLIT: $\frac{3}{( s - 1 ) ( s + 1 )} = \frac{A}{s - 1} + \frac{B}{s + 1}$ . Cover-up: $A = \frac{3}{1 + 1} = \frac{3}{2}$ , $B = \frac{3}{- 1 - 1} = - \frac{3}{2}$ . LOOK-UP: $f (t) = \frac{3}{2} e^{t} - \frac{3}{2} e^{- t} = 3 \cdot \frac{e ^{t} - e ^{- t}}{2} = 3 sinh t .$ Confirmed: hyperbolic sine, exactly as predicted.

Problem 3.3

Find $L^{- 1} {\frac{s + 7}{s ^{2} + 4 s + 13}}$ .

Recall Solution 3.3

COMPLETE the square: $s^{2} + 4 s + 13 = (s + 2)^{2} + 9$ , so $a = - 2$ , $ω^{2} = 9 \Rightarrow ω = 3$ . SPLIT numerator around $(s + 2)$ : we want an $(s + 2)$ (for cosine) plus a leftover (for sine): $\frac{s + 7}{( s + 2 ) ^{2} + 9} = \frac{( s + 2 ) + 5}{( s + 2 ) ^{2} + 9} = \frac{s + 2}{( s + 2 ) ^{2} + 9} + \frac{5}{( s + 2 ) ^{2} + 9} .$ LOOK-UP with the shift atoms:

$\frac{s + 2}{( s + 2 ) ^{2} + 9} \to e^{- 2 t} cos 3 t$ .
$\frac{5}{( s + 2 ) ^{2} + 9} = \frac{5}{3} \cdot \frac{3}{( s + 2 ) ^{2} + 9} \to \frac{5}{3} e^{- 2 t} sin 3 t$ . $f (t) = e^{- 2 t} (cos 3 t + \frac{5}{3} sin 3 t) .$

Level 4 — Synthesis

Goal: combine repeated poles, quadratics, and improper-fraction handling in one problem.

Problem 4.1

Find $L^{- 1} {\frac{2 s + 3}{( s + 1 ) ^{2} ( s + 2 )}}$ .

Recall Solution 4.1

SPLIT. The double pole $(s + 1)^{2}$ needs two terms; the simple pole $(s + 2)$ needs one: $\frac{2 s + 3}{( s + 1 ) ^{2} ( s + 2 )} = \frac{A}{s + 1} + \frac{B}{( s + 1 ) ^{2}} + \frac{C}{s + 2} .$ Multiply through by $(s + 1)^{2} (s + 2)$ : $2 s + 3 = A (s + 1) (s + 2) + B (s + 2) + C (s + 1)^{2} .$

Set $s = - 1$ : $2 (- 1) + 3 = 1 = B (1) \Rightarrow B = 1$ .
Set $s = - 2$ : $2 (- 2) + 3 = - 1 = C (1) \Rightarrow C = - 1$ .
Compare $s^{2}$ -coefficients: $0 = A + C \Rightarrow A = 1$ . LOOK-UP: $\frac{1}{s + 1} \to e^{- t}$ , $\frac{1}{( s + 1 ) ^{2}} \to t e^{- t}$ (shift of $\frac{1}{s ^{2}} \to t$ ), $\frac{1}{s + 2} \to e^{- 2 t}$ : $f (t) = e^{- t} + t e^{- t} - e^{- 2 t} = (1 + t) e^{- t} - e^{- 2 t} .$

Problem 4.2

Find $L^{- 1} {\frac{s ^{2} + 2 s + 3}{s ^{2} + 1}}$ .

Recall Solution 4.2

CHECK proper first. Degree of numerator (2) $=$ degree of denominator (2): this is improper. Do polynomial long division: $\frac{s ^{2} + 2 s + 3}{s ^{2} + 1} = 1 + \frac{2 s + 2}{s ^{2} + 1} .$ LOOK-UP:

$L^{- 1} {1} = δ (t)$ (the Dirac delta — the inverse of a constant).
$\frac{2 s}{s ^{2} + 1} \to 2 cos t$ ( $ω = 1$ ).
$\frac{2}{s ^{2} + 1} \to 2 sin t$ . $f (t) = δ (t) + 2 cos t + 2 sin t .$

Level 5 — Mastery

Goal: full ODE round-trip, and the convolution route when the table won't split cleanly.

Problem 5.1

Solve $y^{''} + 4 y = 0$ with $y (0) = 1, y^{'} (0) = 6$ using Laplace transforms. Report $y (t)$ .

Recall Solution 5.1

Transform the ODE (recall $L {y^{''}} = s^{2} Y - s y (0) - y^{'} (0)$ , from Solving ODEs with Laplace Transforms): $s^{2} Y - s (1) - 6 + 4 Y = 0 \Rightarrow (s^{2} + 4) Y = s + 6 \Rightarrow Y = \frac{s + 6}{s ^{2} + 4} .$ SPLIT / LOOK-UP with $ω = 2$ : $Y = \frac{s}{s ^{2} + 4} + \frac{6}{s ^{2} + 4} = \frac{s}{s ^{2} + 4} + 3 \cdot \frac{2}{s ^{2} + 4} .$ $y (t) = cos 2 t + 3 sin 2 t .$ Sanity check: $y (0) = cos 0 + 3 sin 0 = 1$ ✓, and $y^{'} (t) = - 2 sin 2 t + 6 cos 2 t$ gives $y^{'} (0) = 6$ ✓.

Problem 5.2

Find $L^{- 1} {\frac{1}{s ^{2} ( s ^{2} + 1 )}}$ by partial fractions, then confirm the $\frac{1}{s ^{2}} \cdot \frac{1}{s ^{2} + 1}$ structure agrees with the Convolution Theorem.

Recall Solution 5.2

SPLIT. With only even powers, write $\frac{1}{s ^{2} ( s ^{2} + 1 )} = \frac{A}{s} + \frac{B}{s ^{2}} + \frac{C s + D}{s ^{2} + 1}$ . Multiply out: $1 = A s (s^{2} + 1) + B (s^{2} + 1) + (C s + D) s^{2} .$ Match coefficients: $s^{0} : B = 1$ ; $s^{1} : A = 0$ ; $s^{2} : B + D = 0 \Rightarrow D = - 1$ ; $s^{3} : A + C = 0 \Rightarrow C = 0$ . So $\frac{1}{s ^{2} ( s ^{2} + 1 )} = \frac{1}{s ^{2}} - \frac{1}{s ^{2} + 1}$ . LOOK-UP: $\frac{1}{s ^{2}} \to t$ , $\frac{1}{s ^{2} + 1} \to sin t$ : $f (t) = t - sin t .$ Convolution cross-check. $L^{- 1} {\frac{1}{s ^{2}}} = t$ and $L^{- 1} {\frac{1}{s ^{2} + 1}} = sin t$ , so the product transform inverts to the convolution $(t) * (sin t) = \int_{0}^{t} τ sin (t - τ) d τ = t - sin t,$ which matches. Two roads, one answer — that's the reassurance the theorem gives.