How to Find an Inverse Laplace Transform by Hand

Most newcomers assume how to find inverse laplace transform means matching F(s) to a table entry. That works for maybe three problems. The real work is algebraic: you must decompose F(s) into pieces the table can handle. A textbook F(s) is almost never in table-ready form. You will spend your time on partial fractions, completing the square, and handling exponential factors. A repeatable procedure inverts any rational function F(s) where the denominator polynomial has a higher degree than the numerator, using the case types from Boyce & DiPrima Elementary Differential Equations (11th ed., Chapter 6) as a reference. If the numerator degree equals or exceeds the denominator degree, you must perform polynomial division first. The method covers distinct real poles, repeated real poles, complex conjugate poles, and time delays introduced by e^−as factors. Each case maps to a specific step in the decision flow below.

Before you factor a single polynomial, check these conditions in order.

Is F(s) proper? If the numerator degree is greater than or equal to the denominator degree, perform polynomial division. This is step 1 below. Boyce & DiPrima Problem 31, §6.2, p. 316 gives an example of an improper rational function.

Does F(s) contain e^−as? If yes, treat the exponential as a time-shift factor. Invert the remaining F(s) first, then apply the second shifting theorem. Factor the denominator; the root types dictate the partial-fraction form: distinct real roots require the Heaviside cover-up method. Fastest path. Boyce & DiPrima Problem 1, §6.2, p. 315.

Repeated real roots: Use A/(s-p) + B/(s-p)² form. Do not use A/(s-p)² alone.

Complex conjugate roots: Keep the quadratic factor irreducible. Complete the square in the denominator. Use the s/( (s+b)² + ω² ) and ω/( (s+b)² + ω² ) table entries. Boyce & DiPrima Problem 13, §6.2, p. 315.

Repeated complex conjugate roots: Use the form for a repeated irreducible quadratic. Rare on exams, but Boyce & DiPrima Problem 19, §6.2, p. 315 covers it.

Is the denominator an irreducible quadratic with no real roots? Complete the square and apply the frequency-shift theorem (first shifting theorem). Boyce & DiPrima Problem 23, §6.2, p. 315.

This flow replaces guesswork. Match the root type, apply the correct algebraic form, and you are halfway to the time-domain answer.

If the degree of the numerator is greater than or equal to the degree of the denominator, you cannot decompose directly. Divide the numerator by the denominator. The quotient is a polynomial in s whose inverse is a combination of Dirac deltas δ(t) and its derivatives. The remainder is a proper rational function that goes through the rest of the procedure. Boyce & DiPrima Problem 31, §6.2, p. 316 shows this case. Skip this step if F(s) is already proper.

Factor the denominator of the proper rational function into linear factors (for real roots) and irreducible quadratics (for complex roots). The factors you get determine the partial-fraction form in the next step. Use the quadratic formula on any irreducible quadratic factor to find its complex roots, but keep the quadratic intact for the partial-fraction expansion. Do not factor it into complex linear factors unless you plan to use the Bromwich integral, which is not the method here.

Write the partial-fraction decomposition according to the root types from the decision flow.

Distinct real roots: Use the Heaviside cover-up method. For each root p, multiply F(s) by (s-p) and evaluate at s = p. That gives the residue directly. Example: for F(s) = 5/(s(s+2)), the residue at s=0 is 5/(0+2) = 2.5, and at s=-2 is 5/(-2) = -2.5. This is faster than solving a system of equations. Boyce & DiPrima Problem 1, §6.2, p. 315.

Repeated real roots: Write A/(s-p) + B/(s-p)². Solve for A and B by multiplying through by (s-p)² and equating coefficients or substituting values. Boyce & DiPrima Problem 7, §6.2, p. 315.

Complex conjugate roots: Keep the quadratic factor. Write (As + B)/( (s+b)² + ω² ). Complete the square if the quadratic is not already in the form (s+b)² + ω². Then match the numerator to the s and constant terms. Boyce & DiPrima Problem 13, §6.2, p. 315 and Problem 23, §6.2, p. 315.

Once F(s) is decomposed into a sum of terms of the form A/(s-p), A/(s-p)^k, (As+B)/((s+b)²+ω²), apply the standard inverse Laplace transform pairs. Use the frequency-shift theorem (first shifting theorem) for terms like A/((s+b)+ω²). The common table entries you need are: ℒ⁻¹{1/s} = 1, ℒ⁻¹{1/(s-a)} = e^at, ℒ⁻¹{ω/(s²+ω²)} = sin(ωt), ℒ⁻¹{s/(s²+ω²)} = cos(ωt), ℒ⁻¹{1/(s-a)^k} = t^k-1e^at/(k-1)!. A full inverse Laplace transform table is the single reference you need at this step.

If F(s) includes a factor e^−as, remove it temporarily. Find the inverse Laplace transform of the remaining G(s) = F(s)/e^−as. Call the result g(t). Then apply the second shifting theorem: ℒ⁻¹{e^−asG(s)} = g(t-a)u(t-a), where u(t) is the unit step function. The factor u(t-a) makes the result zero for t < a. Forgetting this unit step function is a common mistake. Boyce & DiPrima Problem 1, §6.4, p. 331 shows this.

Find ℒ⁻¹{ (3s+2) / (s(s+1)) }.

Step 1: F(s) is proper (numerator degree 1, denominator degree 2). No division needed.

Step 2: Factor denominator: s(s+1).

Step 3: Partial fractions using Heaviside cover-up: (3s+2)/(s(s+1)) = A/s + B/(s+1). At s=0: A = (3*0+2)/(0+1) = 2. At s=-1: B = (3*(-1)+2)/(-1) = (-3+2)/(-1) = (-1)/(-1) = 1. So F(s) = 2/s + 1/(s+1).

Step 4: Inverse each term: ℒ⁻¹{2/s} = 2, ℒ⁻¹{1/(s+1)} = e^−t.

Result: f(t) = 2 + e^−t, for t ≥ 0. This matches the pattern of Boyce & DiPrima Problem 1, §6.2, p. 315.

Find ℒ⁻¹{ (s+5) / (s² + 4s + 13) }.

Step 1: Proper. No division needed.

Step 2: Denominator has complex roots. Complete the square: s² + 4s + 13 = (s+2)² + 3².

Step 3: Rewrite numerator in terms of (s+2): (s+5) = (s+2) + 3. So F(s) = (s+2)/((s+2)²+9) + 3/((s+2)²+9). This is now in the form (s+2)/((s+2)²+3²) + 3/((s+2)²+3²).

Step 4: Apply frequency-shift theorem: ℒ⁻¹{(s+2)/((s+2)²+9)} = e^−2tcos(3t). ℒ⁻¹{3/((s+2)²+9)} = e^−2tsin(3t).

Result: f(t) = e^−2t(cos(3t) + sin(3t)), for t ≥ 0. This follows the pattern of Boyce & DiPrima Problem 13, §6.2, p. 315.

Find ℒ⁻¹{ e^−3s / (s+1) }.

Step 1: Identify the delay: a = 3. G(s) = 1/(s+1).

Step 2: Invert G(s): ℒ⁻¹{1/(s+1)} = e^−t.

Step 3: Apply second shifting theorem: ℒ⁻¹{e^−3s * 1/(s+1)} = e^−(t-3) u(t-3).

Result: f(t) = u(t-3) e^3−t. The function is zero for t < 3, and equals e^3−t for t ≥ 3. This is the pattern from Boyce & DiPrima Problem 1, §6.4, p. 331.

The most frequent errors cluster in four areas.

Forgetting the Unit Step Function

When using the second shifting theorem, the factor u(t-a) is not optional. The inverse of e^−asF(s) is f(t-a)u(t-a). Without u(t-a), the function is defined for negative t, which violates causality.

Wrong Partial-Fraction Form for Repeated Poles

A repeated pole at s=p requires terms A/(s-p) + B/(s-p)². Using A/(s-p)² alone gives the wrong time-domain function. Boyce & DiPrima Problem 7, §6.2, p. 315 is the reference.

Confusing the Transforms of Sin and Cos

The pair for sin(ωt) is ω/(s²+ω²). The pair for cos(ωt) is s/(s²+ω²). Using the wrong numerator gives a 90° phase error.

Applying the Final Value Theorem to an Unstable F(s)

The final value theorem (lim_{t→∞} f(t) = lim_{s→0} sF(s)) requires all poles of F(s) to be in the left half-plane. If any pole is on the imaginary axis (like an undamped oscillator), the theorem gives a finite wrong answer when the true response oscillates forever. Check pole locations before applying it.

Incorrect Sign in the Frequency-Shift Theorem

If F(s) = G(s-a), then ℒ⁻¹{F(s)} = e^atg(t). If F(s) = G(s+a), the inverse is e^−atg(t). The sign of the shift in s maps to the opposite sign in the exponential in time.