Inverse Laplace Transform and Partial Fractions

How do you undo a Laplace transform once you have used it to crush a differential equation into algebra? This is the central practical question of this article. The Laplace Transform of Derivatives converts an initial value problem into a rational function Y(s)Y(s). The whole point of the method is wasted unless we can travel back from Y(s)Y(s) to y(t)y(t). That return trip is the inverse Laplace transform. For the rational functions that arise from linear ODEs with constant coefficients, the trip is always the same well-marked road. That road is partial fraction decomposition followed by table lookup. The big picture is that Laplace transform methods reduce differential equations to algebraic ones. But algebra is only useful if it is reversible. Uniqueness of the inverse transform (up to a technical equivalence) guarantees something important. Whichever algebraic path we take to simplify Y(s)Y(s), exactly one continuous function y(t)y(t) waits on the other side. Partial fractions is the mechanical bridge. It rewrites any proper rational function as a sum of simple pieces. Each piece is recognizable instantly from a short transform table. This article makes that bridge rigorous. We state and prove linearity and uniqueness properties of the inverse transform. We prove the partial fraction decomposition theorem itself. This covers both distinct linear factors and repeated/complex factors. We show in complete detail how to reconstruct y(t)y(t) from Y(s)Y(s). We cover three qualitatively different cases: real distinct roots, repeated real roots, and complex conjugate roots. We finish with two fully worked initial value problems. We also include a counterexample showing why properness of the rational function is essential. Finally, we look at where this machinery resurfaces. It reappears in circuit analysis, control theory, and the method of convolution.

Key takeaways
  • The inverse Laplace transform L−1\mathcal{L}^{-1} undoes L\mathcal{L}: if L{y}=Y\mathcal{L}\{y\}=Y, then L−1{Y}=y\mathcal{L}^{-1}\{Y\}=y, essentially uniquely.
  • Linearity of L−1\mathcal{L}^{-1} lets us invert term by term after splitting Y(s)Y(s) into simple pieces.
  • Every proper rational function P(s)/Q(s)P(s)/Q(s) decomposes uniquely into partial fractions over linear and irreducible quadratic factors of QQ.
  • Distinct real roots give exponentials, repeated roots give tkeatt^k e^{at} terms, complex roots give damped sines and cosines.
  • Two functions with the same Laplace transform can differ only on a set of measure zero (e.g., finitely many points) — this is why "uniqueness" needs a careful statement.
  • A rational function must be proper (degree of numerator less than degree of denominator) before partial fractions apply directly; otherwise divide first.
  • Key idea: partial fractions turn any transform coming from a linear ODE into a sum of table entries, making inversion a matching exercise instead of an integral.

Where this lives today

Partial fraction inversion of Laplace transforms is the computational backbone of electrical circuit analysis, where impedance networks produce rational transfer functions H(s)H(s) whose poles (from partial fractions) directly reveal transient response modes — decaying exponentials for resistive-dominated circuits, damped oscillations for underdamped RLCRLC circuits. In control theory, the poles of a closed-loop transfer function, found via the same factorization, determine stability: all poles must lie in the left half-plane, i.e. all real parts of aia_i and a±bia\pm bi must be negative, exactly the exponents appearing in the inverted time-domain response. In mechanical and structural engineering, forced vibration problems reduce to the same rational-function inversion, with resonance corresponding to poles approaching the imaginary axis. In pure mathematics, the technique generalizes to the method of residues in complex analysis (the Bromwich contour integral for L−1\mathcal{L}^{-1} is literally computed by summing residues at the poles, which is a global, complex-analytic restatement of the same partial-fraction bookkeeping). It also sets up the theory of Step Functions and Convolution, where discontinuous forcing and impulse responses require exactly the transform-and-invert cycle developed here, extended by the second shifting theorem and the convolution theorem.

Ask the tutor: How does the convolution theorem let us invert products of transforms without ever doing partial fractions?

DefinitionInverse Laplace Transform

Let f:[0,∞)→Rf:[0,\infty)\to\mathbb{R} be a piecewise continuous function of exponential order, meaning there exist constants M>0M>0, c∈Rc\in\mathbb{R}, and T≥0T\ge 0 such that

∣f(t)∣≤Mectfor all t≥T.|f(t)| \le M e^{ct} \quad \text{for all } t \ge T.

Its Laplace transform is

F(s)=L{f}(s)=∫0∞e−stf(t) dt,s>c.F(s) = \mathcal{L}\{f\}(s) = \int_0^\infty e^{-st} f(t)\, dt, \qquad s > c.

Given a function F(s)F(s) that arises as the Laplace transform of some piecewise continuous, exponential-order function ff, the inverse Laplace transform of FF, written L−1{F}\mathcal{L}^{-1}\{F\}, is defined to be that function ff (unique up to redefinition on a set of isolated points, see Uniqueness of the Laplace Transform discussed below):

L−1{F}(t)=f(t)  ⟺  L{f}(s)=F(s).\mathcal{L}^{-1}\{F\}(t) = f(t) \iff \mathcal{L}\{f\}(s) = F(s).

For rational functions F(s)=P(s)/Q(s)F(s) = P(s)/Q(s) with deg⁡P<deg⁡Q\deg P < \deg Q (a proper rational function) and QQ having real coefficients, the standard method of inversion is: 1. Factor Q(s)Q(s) over R\mathbb{R} into linear factors (s−a)m(s-a)^m and irreducible quadratic factors (s2+bs+c)n(s^2+bs+c)^n. 2. Decompose F(s)F(s) into a sum of partial fractions, one simple term (or sum of simple terms) per factor. 3. Invert each partial fraction individually using a table of elementary transforms, then sum by linearity.

Intuition

Think of the Laplace transform as a language-translation device: it turns "differentiate with respect to tt" into "multiply by ss," turning calculus into algebra. Once you have solved the algebra problem for Y(s)Y(s), you are holding a sentence in the ss-language, and you need to translate it back into the tt-language to actually read off the solution y(t)y(t). The trouble is that Y(s)Y(s), fresh out of the algebra, is usually a single complicated fraction — a ratio of two polynomials in ss — and there is no dictionary entry for "complicated fraction." There are dictionary entries for pieces like 1s−a\frac{1}{s-a}, 1(s−a)2\frac{1}{(s-a)^2}, and b(s−a)2+b2\frac{b}{(s-a)^2+b^2}. Partial fractions is the grammatical rule that lets you split one complicated sentence into several simple ones, each with its own dictionary entry. Geometrically, the roots of the denominator Q(s)Q(s) — the poles of Y(s)Y(s) — encode the qualitative behavior of y(t)y(t) before you even do the algebra. A real pole at s=as=a signals a term eate^{at}; a repeated pole at s=as=a (multiplicity mm) signals terms tkeatt^k e^{at} for k=0,…,m−1k=0,\dots,m-1; and a complex-conjugate pair of poles at s=a±bis = a\pm bi signals oscillation eatcos⁡(bt)e^{at}\cos(bt) and eatsin⁡(bt)e^{at}\sin(bt). The partial fraction decomposition is precisely the bookkeeping device that assigns the correct coefficient to each of these building blocks.

Figure
Poles of Y(s)Y(s) in the complex ss-plane and their time-domain signatures
Re(s) Im(s) s=-2 → e^{-2t} s=3 → e^{3t} s=1+2i s=1-2i → e^{t}cos(2t), e^{t}sin(2t) s=-1 (double) → e^{-t}, t e^{-t}

The formal theory

TheoremLinearity of the Inverse Laplace Transform

Let F(s)F(s) and G(s)G(s) be Laplace transforms of piecewise continuous, exponential-order functions f(t)f(t) and g(t)g(t) respectively, both convergent for s>cs > c for some common cc. Let α,β\alpha,\beta be real (or complex) constants. Then

L−1{αF(s)+βG(s)}(t)=αf(t)+βg(t).\mathcal{L}^{-1}\{\alpha F(s) + \beta G(s)\}(t) = \alpha f(t) + \beta g(t).
Proof

This is a direct consequence of linearity of the (forward) Laplace transform, which is itself immediate from linearity of the integral. For s>cs>c,

L{αf+βg}(s)=∫0∞e−st(αf(t)+βg(t)) dt=α∫0∞e−stf(t) dt+β∫0∞e−stg(t) dt,\mathcal{L}\{\alpha f + \beta g\}(s) = \int_0^\infty e^{-st}\big(\alpha f(t) + \beta g(t)\big)\,dt = \alpha \int_0^\infty e^{-st} f(t)\,dt + \beta \int_0^\infty e^{-st} g(t)\,dt,

where the splitting of the integral is valid because both integrals converge absolutely for s>cs > c (by the exponential order hypothesis, ∣e−stf(t)∣≤Me(c−s)t|e^{-st}f(t)| \le M e^{(c-s)t} is integrable on [0,∞)[0,\infty) when s>cs>c, and similarly for gg). Hence

L{αf+βg}(s)=αF(s)+βG(s).\mathcal{L}\{\alpha f + \beta g\}(s) = \alpha F(s) + \beta G(s).

By definition of the inverse transform (matching a function to the transform it produces), the function αf+βg\alpha f + \beta g is a function whose Laplace transform is αF+βG\alpha F + \beta G. By the uniqueness result below, it is the function meant by L−1{αF+βG}\mathcal{L}^{-1}\{\alpha F+\beta G\}, up to the stated equivalence. ■\blacksquare

∎

The linearity theorem is what licenses splitting Y(s)Y(s) into a sum of partial fractions and inverting term by term. We still owe two debts: a precise uniqueness statement (so that "L−1\mathcal{L}^{-1}" is well-defined as used above), and the partial fraction decomposition theorem itself.

TheoremLerch's Uniqueness Theorem

Suppose ff and gg are piecewise continuous functions on [0,∞)[0,\infty) of exponential order, and suppose L{f}(s)=L{g}(s)\mathcal{L}\{f\}(s) = \mathcal{L}\{g\}(s) for all ss in some common half-line s>cs > c. Then f(t)=g(t)f(t) = g(t) for every t≥0t \ge 0 at which both ff and gg are continuous. In particular, if f,gf,g are continuous on [0,∞)[0,\infty), then f≡gf \equiv g.

Proof

(Stated with proof sketch beyond full analytic detail, as the complete proof requires the Weierstrass approximation theorem or complex-analytic uniqueness of analytic continuation; we give the standard real-variable argument.) Let h=f−gh = f - g; then hh is piecewise continuous, of exponential order, and L{h}(s)=0\mathcal{L}\{h\}(s) = 0 for all s>cs > c. We must show h(t)=0h(t) = 0 at every point of continuity. Fix s0>max⁡(c,0)s_0 > \max(c,0) and substitute s=s0+ns = s_0 + n for n=1,2,3,…n = 1, 2, 3,\dots, and further substitute u=e−tu = e^{-t} (so t=−ln⁡ut = -\ln u, dt=−du/udt = -du/u, and t∈[0,∞)t\in[0,\infty) corresponds to u∈(0,1]u \in (0,1]). Define

ϕ(u)=e−s0th(t)=us0h(−ln⁡u),u∈(0,1],\phi(u) = e^{-s_0 t} h(t) = u^{s_0} h(-\ln u), \qquad u \in (0,1],

extended by ϕ(0)=0\phi(0)=0; exponential order of hh guarantees ϕ\phi is bounded and piecewise continuous on [0,1][0,1]. Then

0=L{h}(s0+n)=∫0∞e−(s0+n)th(t) dt=∫01unϕ(u) du,n=0,1,2,…0 = \mathcal{L}\{h\}(s_0+n) = \int_0^\infty e^{-(s_0+n)t} h(t)\, dt = \int_0^1 u^{n} \phi(u)\, du, \qquad n = 0,1,2,\dots

(the case n=0n=0 recovers L{h}(s0)=0\mathcal{L}\{h\}(s_0)=0, and the change of variables is a standard exponential substitution). Thus ∫01unϕ(u) du=0\int_0^1 u^n \phi(u)\,du = 0 for every non-negative integer nn, which means ∫01p(u)ϕ(u) du=0\int_0^1 p(u)\phi(u)\,du = 0 for every polynomial pp. By the Weierstrass approximation theorem, polynomials are dense in C[0,1]C[0,1] under the sup norm, so this forces ∫01ϕ(u)2 du=0\int_0^1 \phi(u)^2\, du = 0 at every point where ϕ\phi is continuous (approximate ϕ\phi itself, or more carefully, approximate a continuous function agreeing with ϕ\phi off a small neighborhood of its jump points, and let the neighborhood shrink). Since ϕ\phi is continuous except at finitely many points on any compact interval, and ∫01ϕ2=0\int_0^1 \phi^2 = 0 with ϕ\phi piecewise continuous forces ϕ=0\phi = 0 wherever it is continuous. Undoing the substitution u=e−tu=e^{-t}, we conclude h(t)=0h(t) = 0 at every point of continuity of hh, i.e., f(t)=g(t)f(t)=g(t) there. ■\blacksquare

∎

This theorem is exactly why we may speak of "the" inverse Laplace transform: two candidate functions with the same transform agree everywhere except possibly at finitely many isolated jump discontinuities, which never affects the value of any integral or the physical interpretation of the solution to a differential equation. In practice, when solving IVPs, the reconstructed y(t)y(t) is continuous, so this ambiguity never arises.

TheoremPartial Fraction Decomposition

Let P(s)P(s) and Q(s)Q(s) be polynomials with real coefficients, deg⁡P<deg⁡Q\deg P < \deg Q, and suppose QQ has no repeated complex factors in common with PP (i.e., P/QP/Q is already in lowest terms). Factor QQ over R\mathbb{R} as

Q(s)=∏i=1r(s−ai)mi∏j=1k(s2+bjs+cj)nj,Q(s) = \prod_{i=1}^{r} (s-a_i)^{m_i} \prod_{j=1}^{k} (s^2+b_js+c_j)^{n_j},

where the aia_i are the distinct real roots with multiplicities mim_i, and the quadratics s2+bjs+cjs^2+b_js+c_j are distinct and irreducible over R\mathbb{R} (i.e. bj2−4cj<0b_j^2 - 4c_j < 0) with multiplicities njn_j. Then there exist unique real constants Ai,ℓA_{i,\ell} (1≤i≤r1\le i\le r, 1≤ℓ≤mi1\le \ell \le m_i) and Bj,p,Cj,pB_{j,p}, C_{j,p} (1≤j≤k1\le j \le k, 1≤p≤nj1\le p\le n_j) such that

P(s)Q(s)=∑i=1r∑ℓ=1miAi,ℓ(s−ai)ℓ  +  ∑j=1k∑p=1njBj,p s+Cj,p(s2+bjs+cj)p.\frac{P(s)}{Q(s)} = \sum_{i=1}^r \sum_{\ell=1}^{m_i} \frac{A_{i,\ell}}{(s-a_i)^{\ell}} \;+\; \sum_{j=1}^k \sum_{p=1}^{n_j} \frac{B_{j,p}\,s + C_{j,p}}{(s^2+b_js+c_j)^{p}}.
Proof

We argue by strong induction on deg⁡Q\deg Q, using the technique of reduction via a single factor.

Base case. If deg⁡Q=1\deg Q = 1, then Q(s)=s−aQ(s) = s-a for some real aa (up to an overall constant absorbed into PP), and P(s)=AP(s) = A is a constant since deg⁡P<1\deg P < 1. The formula holds trivially with A1,1=AA_{1,1}=A.

Inductive step, real root case. Suppose Q(s)=(s−a)mR(s)Q(s) = (s-a)^m R(s) where R(a)≠0R(a) \ne 0 and deg⁡R=deg⁡Q−m\deg R = \deg Q - m. Consider the constant

Am:=P(a)R(a),A_{m} := \frac{P(a)}{R(a)},

which is well-defined since R(a)≠0R(a) \ne 0. We claim

P(s)Q(s)−Am(s−a)m=P(s)−AmR(s)(s−a)mR(s)\frac{P(s)}{Q(s)} - \frac{A_m}{(s-a)^m} = \frac{P(s) - A_m R(s)}{(s-a)^m R(s)}

has the property that its numerator P(s)−AmR(s)P(s) - A_m R(s) vanishes at s=as=a. Indeed, P(a)−AmR(a)=P(a)−P(a)=0P(a) - A_m R(a) = P(a) - P(a) = 0 by construction. Hence (s−a)(s-a) divides P(s)−AmR(s)P(s)-A_mR(s), say P(s)−AmR(s)=(s−a)P~(s)P(s) - A_mR(s) = (s-a)\widetilde P(s) for a polynomial P~\widetilde P. Substituting,

P(s)Q(s)−Am(s−a)m=P~(s)(s−a)m−1R(s).\frac{P(s)}{Q(s)} - \frac{A_m}{(s-a)^m} = \frac{\widetilde P(s)}{(s-a)^{m-1}R(s)}.

The new denominator (s−a)m−1R(s)(s-a)^{m-1}R(s) has strictly smaller degree than QQ, and one checks deg⁡P~<deg⁡((s−a)m−1R(s))\deg \widetilde P < \deg\big((s-a)^{m-1}R(s)\big) (properness is preserved: subtracting a proper fraction from a proper fraction and clearing denominators cannot increase the degree of the numerator past that of the new, smaller-degree denominator, since both fractions individually tend to 00 as s→∞s\to\infty, so their difference does too, forcing numerator degree less than denominator degree). By the inductive hypothesis applied to P~(s)/[(s−a)m−1R(s)]\widetilde P(s)/\big[(s-a)^{m-1}R(s)\big], this remainder decomposes into partial fractions of the required form (now with (s−a)(s-a) appearing to powers 1,…,m−11,\dots,m-1 and RR's factorization contributing the rest). Adding back Am/(s−a)mA_m/(s-a)^m completes the decomposition for QQ.

Inductive step, irreducible quadratic case. Suppose instead Q(s)=(s2+bs+c)nS(s)Q(s) = (s^2+bs+c)^n S(s) with s2+bs+cs^2+bs+c irreducible (b2−4c<0b^2-4c<0) and coprime to SS. Let α=−b+i4c−b22\alpha = \frac{-b+i\sqrt{4c-b^2}}{2} be a root of s2+bs+cs^2+bs+c in C\mathbb{C}. Since S(α)≠0S(\alpha)\ne 0 (coprimality), define the complex number

w:=P(α)S(α),w := \frac{P(\alpha)}{S(\alpha)},

and set Bns+CnB_n s + C_n to be the unique real-coefficient linear polynomial with Bnα+Cn=wB_n\alpha + C_n = w (solve the 2×22\times2 real linear system obtained by separating real and imaginary parts of w=Bnα+Cnw = B_n\alpha+C_n; this system is nonsingular because α∉R\alpha \notin \mathbb{R}, so 1,α1,\alpha are linearly independent over R\mathbb{R}). Because s2+bs+cs^2+bs+c has real coefficients, its other root is αˉ\bar\alpha, and one checks directly that Bnαˉ+Cn=wˉ=P(αˉ)/S(αˉ)B_n\bar\alpha+C_n = \bar w = P(\bar\alpha)/S(\bar\alpha) as well (complex conjugation commutes with evaluation of real polynomials at conjugate points). Hence the polynomial P(s)−(Bns+Cn)S(s)P(s) - (B_ns+C_n)S(s) vanishes at both s=αs=\alpha and s=αˉs=\bar\alpha, so it is divisible by (s−α)(s−αˉ)=s2+bs+c(s-\alpha)(s-\bar\alpha) = s^2+bs+c, say

P(s)−(Bns+Cn)S(s)=(s2+bs+c) P~(s).P(s) - (B_ns+C_n)S(s) = (s^2+bs+c)\,\widetilde P(s).

Then, exactly as before,

P(s)Q(s)−Bns+Cn(s2+bs+c)n=P~(s)(s2+bs+c)n−1S(s),\frac{P(s)}{Q(s)} - \frac{B_ns+C_n}{(s^2+bs+c)^n} = \frac{\widetilde P(s)}{(s^2+bs+c)^{n-1}S(s)},

a proper rational function with strictly smaller denominator degree, to which the inductive hypothesis applies.

Termination. Each application of the reduction strictly lowers deg⁡Q\deg Q (by 11 in the real-root case, by 22 in the quadratic case), so after finitely many steps we reach the base case, having produced a term for every factor (s−ai)ℓ(s-a_i)^\ell, ℓ=1,…,mi\ell=1,\dots,m_i, and every factor (s2+bjs+cj)p(s^2+b_js+c_j)^p, p=1,…,njp=1,\dots,n_j. This proves existence.

Uniqueness. Suppose two decompositions of the stated form both equal P(s)/Q(s)P(s)/Q(s). Subtracting, we get an identity

0=∑i,ℓAi,ℓ′(s−ai)ℓ+∑j,pBj,p′s+Cj,p′(s2+bjs+cj)p0 = \sum_{i,\ell} \frac{A_{i,\ell}'}{(s-a_i)^\ell} + \sum_{j,p}\frac{B_{j,p}'s+C_{j,p}'}{(s^2+b_js+c_j)^p}

for all ss outside the (finite) root set of QQ, where primed coefficients are differences of the two decompositions' coefficients. Multiply through by (s−a1)m1(s-a_1)^{m_1} and let s→a1s\to a_1: every term except the one with ℓ=m1, i=1\ell=m_1,\, i=1 has a factor (s−a1)(s-a_1) remaining (or higher), which vanishes in the limit, forcing A1,m1′=0A_{1,m_1}' = 0. Subtract that term and repeat with (s−a1)m1−1(s-a_1)^{m_1-1}, obtaining A1,m1−1′=0A_{1,m_1-1}'=0, and so on down to A1,1′=0A_{1,1}'=0; repeat for each real root aia_i. A symmetric argument — multiplying by (s2+bjs+cj)nj(s^2+b_js+c_j)^{n_j} and evaluating at the complex root s=αjs=\alpha_j — forces Bj,nj′αj+Cj,nj′=0B_{j,n_j}'\alpha_j + C_{j,n_j}' = 0, and since αj∉R\alpha_j\notin\mathbb R this forces Bj,nj′=Cj,nj′=0B_{j,n_j}'=C_{j,n_j}'=0 (as 1,αj1,\alpha_j are R\mathbb{R}-linearly independent); descend inductively as before. Hence all primed coefficients vanish, i.e. the two decompositions coincide. ■\blacksquare

∎
CorollaryTable-based inversion

If Y(s)=P(s)/Q(s)Y(s) = P(s)/Q(s) is proper with real coefficients and QQ factors as above, then

y(t)=L−1{Y}(t)=∑i,ℓAi,ℓ L−1{1(s−ai)ℓ}(t)+∑j,pL−1{Bj,ps+Cj,p(s2+bjs+cj)p}(t),y(t) = \mathcal{L}^{-1}\{Y\}(t) = \sum_{i,\ell} A_{i,\ell}\, \mathcal{L}^{-1}\left\{\frac{1}{(s-a_i)^\ell}\right\}(t) + \sum_{j,p} \mathcal{L}^{-1}\left\{\frac{B_{j,p}s+C_{j,p}}{(s^2+b_js+c_j)^p}\right\}(t),

by linearity (Linearity of the Inverse Laplace Transform), and each term on the right is a standard table entry:

  • L−1{1/(s−a)}=eat\mathcal{L}^{-1}\{1/(s-a)\} = e^{at}
  • L−1{1/(s−a)ℓ}=tℓ−1(ℓ−1)!eat\mathcal{L}^{-1}\{1/(s-a)^\ell\} = \dfrac{t^{\ell-1}}{(\ell-1)!}e^{at}
  • L−1{s−a(s−a)2+b2}=eatcos⁡(bt)\mathcal{L}^{-1}\left\{\dfrac{s-a}{(s-a)^2+b^2}\right\} = e^{at}\cos(bt), and L−1{b(s−a)2+b2}=eatsin⁡(bt)\mathcal{L}^{-1}\left\{\dfrac{b}{(s-a)^2+b^2}\right\} = e^{at}\sin(bt) (obtained after completing the square in s2+bjs+cj=(s−a)2+b2s^2+b_js+c_j = (s-a)^2+b^2).

Worked examples

Example 1: Distinct real and complex poles. Solve the IVP

y′′−y′−2y=0,y(0)=1,  y′(0)=0,y'' - y' - 2y = 0, \qquad y(0) = 1,\; y'(0) = 0,

by Laplace transform — but let us instead illustrate the partial fraction machinery on a forced problem where the poles are genuinely mixed. Consider

y′′+4y=8cos⁡(2t)⋅0+δ-free forcing:y′′+y′−6y=et,y(0)=0, y′(0)=1.y'' + 4y = 8\cos(2t)\cdot 0 + \delta\text{-free forcing}: \quad y'' + y' - 6y = e^{t}, \qquad y(0)=0,\ y'(0)=1.

Taking the Laplace transform of both sides and using L{y′′}=s2Y−sy(0)−y′(0)\mathcal{L}\{y''\} = s^2Y - sy(0) - y'(0), L{y′}=sY−y(0)\mathcal{L}\{y'\} = sY - y(0), and L{et}=1/(s−1)\mathcal{L}\{e^t\} = 1/(s-1):

(s2Y−1)+(sY)−6Y=1s−1.\big(s^2Y - 1\big) + \big(sY\big) - 6Y = \frac{1}{s-1}.

Solve for YY:

(s2+s−6)Y=1+1s−1=ss−1,(s^2+s-6)Y = 1 + \frac{1}{s-1} = \frac{s}{s-1},
Y(s)=s(s−1)(s2+s−6)=s(s−1)(s+3)(s−2).Y(s) = \frac{s}{(s-1)(s^2+s-6)} = \frac{s}{(s-1)(s+3)(s-2)}.

The denominator has three distinct real roots: s=1,−3,2s=1,-3,2. Write

Y(s)=As−1+Bs+3+Cs−2.Y(s) = \frac{A}{s-1} + \frac{B}{s+3} + \frac{C}{s-2}.

Compute each coefficient by the cover-up rule (multiply by the factor and evaluate at its root, valid because that is exactly the limiting argument used in the uniqueness proof above):

A=s(s+3)(s−2)∣s=1=1(4)(−1)=−14,A = \frac{s}{(s+3)(s-2)}\Big|_{s=1} = \frac{1}{(4)(-1)} = -\frac14,
B=s(s−1)(s−2)∣s=−3=−3(−4)(−5)=−320,B = \frac{s}{(s-1)(s-2)}\Big|_{s=-3} = \frac{-3}{(-4)(-5)} = \frac{-3}{20},
C=s(s−1)(s+3)∣s=2=2(1)(5)=25.C = \frac{s}{(s-1)(s+3)}\Big|_{s=2} = \frac{2}{(1)(5)} = \frac{2}{5}.

Check the partial fraction identity at a convenient test value, say s=0s=0: left side is Y(0)=0Y(0)=0; right side is A/(−1)+B/(3)+C/(−2)=14−120−15=520−120−420=0A/(-1)+B/(3)+C/(-2) = \frac14 - \frac{1}{20} - \frac15 = \frac{5}{20}-\frac1{20}-\frac4{20}=0. ✓. By the table,

y(t)=−14et−320e−3t+25e2t.y(t) = -\frac14 e^{t} - \frac{3}{20} e^{-3t} + \frac{2}{5} e^{2t}.

As a sanity check, y(0)=−14−320+25=−520−320+820=0y(0) = -\frac14 - \frac{3}{20} + \frac25 = -\frac{5}{20} - \frac{3}{20} + \frac{8}{20} = 0 ✓, and differentiating, y′(t)=−14et+920e−3t+45e2ty'(t) = -\frac14 e^t + \frac{9}{20}e^{-3t} + \frac45 e^{2t}, so y′(0)=−14+920+45=−520+920+1620=1y'(0) = -\frac14+\frac{9}{20}+\frac45 = -\frac5{20}+\frac9{20}+\frac{16}{20}=1 ✓, matching both initial conditions.

Example 2: A repeated real root and an irreducible quadratic. Solve

y′′+2y′+y=4e−t+2sin⁡(2t),y(0)=0, y′(0)=0.y'' + 2y' + y = 4e^{-t} + 2\sin(2t), \qquad y(0)=0,\ y'(0)=0.

Note s2+2s+1=(s+1)2s^2+2s+1 = (s+1)^2, a repeated real root. Transform both sides:

L{y′′}=s2Y,L{y′}=sY,L{4e−t}=4s+1,L{2sin⁡2t}=4s2+4.\mathcal{L}\{y''\} = s^2Y,\qquad \mathcal{L}\{y'\}=sY, \qquad \mathcal{L}\{4e^{-t}\} = \frac{4}{s+1},\qquad \mathcal{L}\{2\sin2t\} = \frac{4}{s^2+4}.

So

(s2+2s+1)Y=4s+1+4s2+4,(s^2+2s+1)Y = \frac{4}{s+1} + \frac{4}{s^2+4},
Y(s)=4(s+1)3+4(s+1)2(s2+4).Y(s) = \frac{4}{(s+1)^3} + \frac{4}{(s+1)^2(s^2+4)}.

Handle the second term by partial fractions:

4(s+1)2(s2+4)=As+1+B(s+1)2+Cs+Ds2+4.\frac{4}{(s+1)^2(s^2+4)} = \frac{A}{s+1} + \frac{B}{(s+1)^2} + \frac{Cs+D}{s^2+4}.

Multiply both sides by (s+1)2(s2+4)(s+1)^2(s^2+4):

4=A(s+1)(s2+4)+B(s2+4)+(Cs+D)(s+1)2.4 = A(s+1)(s^2+4) + B(s^2+4) + (Cs+D)(s+1)^2.

Set s=−1s=-1 (kills the AA and C,DC,D terms except BB): 4=B(1+4)=5B4 = B(1+4) = 5B, so B=45B = \frac45. To find the rest, expand and match coefficients. Expand each piece:

A(s+1)(s2+4)=A(s3+s2+4s+4),A(s+1)(s^2+4) = A(s^3+s^2+4s+4),
B(s2+4)=45s2+165,B(s^2+4) = \frac45 s^2 + \frac{16}{5},
(Cs+D)(s+1)2=(Cs+D)(s2+2s+1)=Cs3+(2C+D)s2+(C+2D)s+D.(Cs+D)(s+1)^2 = (Cs+D)(s^2+2s+1) = Cs^3 + (2C+D)s^2 + (C+2D)s + D.

Sum and match to 44 (i.e. coefficient 44 on constant term, 00 on s,s2,s3s,s^2,s^3):

  • s3s^3: A+C=0A + C = 0
  • s2s^2: A+45+2C+D=0A + \frac45 + 2C+D = 0
  • s1s^1: 4A+C+2D=04A + C+2D = 0
  • s0s^0: 4A+165+D=44A + \frac{16}{5} + D = 4

From the last equation: 4A+D=4−165=454A + D = 4 - \frac{16}{5} = \frac{4}{5}, so D=45−4AD = \frac45 - 4A. From s3s^3: C=−AC = -A. Substitute into s1s^1: 4A−A+2D=0⇒3A+2D=0⇒D=−3A24A - A + 2D = 0 \Rightarrow 3A + 2D = 0 \Rightarrow D = -\frac{3A}{2}. Set the two expressions for DD equal: 45−4A=−3A2\frac45 - 4A = -\frac{3A}{2}. Multiply by 1010: 8−40A=−15A8 - 40A = -15A, so 8=25A8 = 25A, giving

A=825.A = \frac{8}{25}.

Then C=−825C = -\frac{8}{25}, and D=−32⋅825=−1225D = -\frac{3}{2}\cdot\frac{8}{25} = -\frac{12}{25}. Check with the s2s^2 equation: A+45+2C+D=825+2025+2(−825)−1225=8+20−16−1225=0A + \frac45 + 2C + D = \frac{8}{25} + \frac{20}{25} + 2\left(-\frac{8}{25}\right) - \frac{12}{25} = \frac{8+20-16-12}{25} = 0 ✓. So

4(s+1)2(s2+4)=8/25s+1+4/5(s+1)2+−825s−1225s2+4.\frac{4}{(s+1)^2(s^2+4)} = \frac{8/25}{s+1} + \frac{4/5}{(s+1)^2} + \frac{-\frac{8}{25}s - \frac{12}{25}}{s^2+4}.

Rewrite the last term to match table form ss2+4→cos⁡2t\frac{s}{s^2+4}\to\cos2t, 2s2+4→sin⁡2t\frac{2}{s^2+4}\to\sin2t:

−825⋅ss2+4−1225⋅1s2+4=−825⋅ss2+4−625⋅2s2+4.-\frac{8}{25}\cdot\frac{s}{s^2+4} - \frac{12}{25}\cdot\frac{1}{s^2+4} = -\frac{8}{25}\cdot\frac{s}{s^2+4} - \frac{6}{25}\cdot\frac{2}{s^2+4}.

Now assemble the full Y(s)Y(s), including the original 4(s+1)3\frac{4}{(s+1)^3} term (table entry ℓ=3\ell=3: L−1{1/(s+1)3}=t22e−t\mathcal{L}^{-1}\{1/(s+1)^3\} = \frac{t^2}{2}e^{-t}):

Y(s)=4(s+1)3+8/25s+1+4/5(s+1)2−825⋅ss2+4−625⋅2s2+4.Y(s) = \frac{4}{(s+1)^3} + \frac{8/25}{s+1} + \frac{4/5}{(s+1)^2} - \frac{8}{25}\cdot\frac{s}{s^2+4} - \frac{6}{25}\cdot\frac{2}{s^2+4}.

Inverting term by term:

y(t)=4⋅t22e−t+825e−t+45t e−t−825cos⁡(2t)−625sin⁡(2t),y(t) = 4\cdot\frac{t^2}{2}e^{-t} + \frac{8}{25}e^{-t} + \frac45 t\,e^{-t} - \frac{8}{25}\cos(2t) - \frac{6}{25}\sin(2t),
y(t)=2t2e−t+45te−t+825e−t−825cos⁡(2t)−625sin⁡(2t).\boxed{y(t) = 2t^2 e^{-t} + \frac45 t e^{-t} + \frac{8}{25}e^{-t} - \frac{8}{25}\cos(2t) - \frac{6}{25}\sin(2t).}

As a check, y(0)=0+0+825−825−0=0y(0) = 0 + 0 + \frac{8}{25} - \frac{8}{25} - 0 = 0 ✓. Differentiating and evaluating at t=0t=0 is routine and (omitting the arithmetic here for brevity) confirms y′(0)=0y'(0)=0, consistent with the given initial conditions and with Y(s)→0Y(s)\to 0 appropriately as s→∞s\to\infty (since YY is proper, lim⁡s→∞sY(s)=y(0)=0\lim_{s\to\infty} sY(s) = y(0) = 0, which the reader can verify directly from the original expression for Y(s)Y(s)).

CounterexampleWhy properness (deg⁡P<deg⁡Q\deg P < \deg Q) is essential

Drop the hypothesis deg⁡P<deg⁡Q\deg P < \deg Q and consider

F(s)=s2s2−1=s2(s−1)(s+1).F(s) = \frac{s^2}{s^2-1} = \frac{s^2}{(s-1)(s+1)}.

Here deg⁡P=deg⁡Q=2\deg P = \deg Q = 2, so FF is not proper. If we naively apply the partial fraction ansatz for distinct linear factors,

F(s)As−1+Bs+1,F(s) \overset{?}{=} \frac{A}{s-1} + \frac{B}{s+1},

we immediately get a contradiction on degree grounds: the right side is a proper rational function (it vanishes as s→∞s\to\infty), while the left side satisfies lim⁡s→∞F(s)=1≠0\lim_{s\to\infty} F(s) = 1 \ne 0. So no choice of A,BA,B can work, and the decomposition theorem's conclusion (existence of such A,BA,B) simply fails — because its hypothesis deg⁡P<deg⁡Q\deg P<\deg Q was violated. The correct procedure is to first perform polynomial long division:

s2s2−1=1+1s2−1=1+1/2s−1−1/2s+1.\frac{s^2}{s^2-1} = 1 + \frac{1}{s^2-1} = 1 + \frac{1/2}{s-1} - \frac{1/2}{s+1}.

Now inverting requires care: L−1{1}\mathcal{L}^{-1}\{1\} is not an ordinary function at all (the constant function 11 has Laplace transform 1/s1/s, not 11; the "function" whose transform is identically 11 is the Dirac delta δ(t)\delta(t), a generalized function, not covered by the piecewise-continuous exponential-order theory in the definition above). This shows concretely why the properness hypothesis cannot be dropped: violating it can push you entirely outside the class of ordinary functions that L−1\mathcal{L}^{-1} was defined to produce. In the context of solving ODEs, an improper Y(s)Y(s) signals either an algebra error or (legitimately) a forcing term itself involving derivatives of δ\delta, which lies beyond the present scope and belongs to the theory of Step Functions and Convolution.

Scope, edge cases & common traps

SituationWhat to doCommon error
deg⁡P≥deg⁡Q\deg P \ge \deg QLong-divide first; invert polynomial part separately (often signals δ\delta-forcing)Applying partial fractions directly to an improper fraction
Repeated real root (s−a)m(s-a)^mInclude A1/(s−a)+⋯+Am/(s−a)mA_1/(s-a) + \cdots + A_m/(s-a)^m, all powers up to mmOnly including the top power Am/(s−a)mA_m/(s-a)^m
Irreducible quadratic (s2+bs+c)n(s^2+bs+c)^nUse linear numerator Bps+CpB_ps+C_p at each power, not a constantUsing constant numerators, or factoring over C\mathbb{C} inconsistently
Complex roots after completing the squareMatch to s−a(s−a)2+b2\frac{s-a}{(s-a)^2+b^2} and b(s−a)2+b2\frac{b}{(s-a)^2+b^2} exactlyOff-by-factor errors in the sin⁡\sin coefficient (forgetting to force numerator to be exactly bb)
Checking workEvaluate decomposition identity at a convenient extra point, or match a coefficient, or check y(0),y′(0)y(0),y'(0)Trusting algebra without any independent check

The mathematician behind it

Origins · 1 minMathias Lerch and how this was discovered
ML
Mathias Lerch
1860–1922 · Czech mathematician, Charles University, Prague

Lerch worked broadly in analysis and number theory, including elliptic and gamma functions, but is honored here for the 1903 uniqueness theorem that bears his name — the rigorous foundation making "the" inverse Laplace transform meaningful as a well-defined operation, decades before engineers popularized Laplace-transform methods for solving differential equations.

History
- 1782: Pierre-Simon Laplace studies the integral transform that will bear his name, in the context of probability theory.
- 1903: Mathias Lerch proves the uniqueness theorem showing the inverse transform is essentially well-defined.
- 1910s–1920s: Oliver Heaviside's operational calculus popularizes transform methods for circuits and ODEs, without full rigor.
- 1920s–1930s: Bromwich, Carson, and Van der Pol give the operational calculus a rigorous complex-analytic (Laplace) foundation.
- Mid-20th century: partial-fraction-based Laplace methods become a standard textbook technique for linear ODEs and control systems.
What exactly does "inverse Laplace transform" mean if the Laplace transform is an integral?

It means finding a function f(t)f(t) whose Laplace transform integral evaluates to a given F(s)F(s). Formally, there is a contour integral formula (the Bromwich/Mellin inversion integral) that computes f(t)f(t) directly from F(s)F(s) using complex analysis, but for the rational functions arising from linear ODEs with constant coefficients, that contour integral always reduces to summing residues at poles — which is exactly what partial fractions computes without ever invoking contour integration explicitly. So in this course, "inverse Laplace transform" is a table-lookup operation justified by linearity and uniqueness, not literally an integral you compute by hand.

Why does partial fractions even work — why is every proper rational function a sum of such simple pieces?

This is the content of the Partial Fraction Decomposition Theorem proved above. The short intuition is induction: peel off one root of the denominator at a time. At each step, subtracting the "correct" simple fraction (the cover-up value) from the whole expression cancels the singularity at that root, leaving a new proper rational function with one fewer power of denominator. Repeating this until the denominator is exhausted produces exactly the sum of simple pieces, and a separate limiting argument shows the coefficients you get this way are the only ones that work.

Do I need to know complex numbers to invert transforms with quadratic denominators?

Not for the final answer, but complex numbers streamline the proof and one common computational shortcut. You can always complete the square and match real-numerator coefficients purely with real algebra, as done in Example 2 and in the theorem's real-variable proof. Some instructors also present a complex-partial-fractions shortcut, factoring s2+bs+cs^2+bs+c into (s−(a+bi))(s−(a−bi))(s - (a+bi))(s-(a-bi)) over C\mathbb{C} and using the cover-up rule with complex arithmetic, then combining conjugate terms back into a real sine/cosine pair at the end.

What if the denominator doesn't factor nicely, e.g., a cubic with no rational roots?

The theorem still guarantees a decomposition exists, since every real polynomial factors (in principle) into real linear and irreducible quadratic factors — this is the Fundamental Theorem of Algebra combined with pairing complex conjugate roots. In practice, if you cannot factor the cubic by hand, you either use numerical root-finding, or in a course setting the denominator is deliberately chosen to factor cleanly, since the point of the topic is the inversion procedure, not root-finding algorithms.

Is partial fraction decomposition the only way to invert a Laplace transform?

No. Other tools include the convolution theorem (expressing y(t)y(t) as ∫0tg(t−τ)r(τ) dτ\int_0^t g(t-\tau)r(\tau)\,d\tau for the impulse response gg and forcing rr, useful when Y(s)Y(s) is a product of two transforms you already recognize), the second shifting theorem for transforms with e−cse^{-cs} factors (discontinuous/delayed forcing), and direct contour integration. Partial fractions is preferred for rational Y(s)Y(s) because it is purely algebraic and always terminates in finitely many algebra steps.

How do I know my partial fraction coefficients are correct before inverting?

Three checks, used throughout the worked examples: (1) evaluate both sides of the partial fraction identity at a convenient value of ss not equal to a pole (e.g. s=0s=0) and confirm they match numerically; (2) match polynomial coefficients degree by degree after clearing denominators, and confirm all equations are simultaneously consistent; (3) after inverting, plug t=0t=0 and (via differentiation) t=0t=0 again into y(t)y(t) and y′(t)y'(t) and confirm they reproduce the original initial conditions of the ODE.

Why do repeated roots give tkeatt^k e^{at} terms instead of just more copies of eate^{at}?

This is dictated by the transform table entry L−1{1/(s−a)ℓ}=tℓ−1(ℓ−1)!eat\mathcal{L}^{-1}\{1/(s-a)^\ell\} = \dfrac{t^{\ell-1}}{(\ell-1)!}e^{at}, which itself follows from differentiating L{eat}=1/(s−a)\mathcal{L}\{e^{at}\}=1/(s-a) repeatedly with respect to ss (using the transform-of-tnf(t)t^n f(t) rule from Transforms of Derivatives-adjacent identities). A repeated pole at s=as=a means the "resonance" of the exponential mode with itself, analogous to repeated-root behavior in the characteristic equation of the homogeneous ODE, which independently forces tkeatt^k e^{at} solutions there too — the two theories are consistent by design.