---
title: "The Laplace Transform: Definition, Properties, and Solving IVPs"
module: The Laplace Transform
moduleNumber: 5
lessonNumber: 1
order: 501
summary: >
  The Laplace transform sends a function of time to a function of a complex
  frequency by integrating it against the kernel e^{-st}. Differentiation in t
  becomes multiplication by s, so a linear constant-coefficient initial value
  problem turns into an algebraic equation. Existence rests on piecewise
  continuity and exponential order; the derivative rule folds in the initial
  data; and inversion runs through a transform table and partial fractions.
topics: [The Laplace Transform]
sources:
  - book: Boyce
    ref: "Ch. 6 — The Laplace Transform; §6.1 Definition of the Laplace Transform; §6.2 Solution of Initial Value Problems"
  - book: Simmons
    ref: "Ch. 9 — Laplace Transforms; §48 Introduction; §49 A Few Remarks on the Theory; §50 Applications to Differential Equations; §51 Derivatives and Integrals of Transforms"
---

A linear constant-coefficient equation can be solved by [guessing a form for the
particular solution](/differential-equations/second-order-linear/nonhomogeneous-undetermined-coefficients)
and fitting the arbitrary constants to the initial data at the end. The Laplace
transform reorganizes that work: it converts the whole initial value problem
into a single algebraic equation, folds the initial conditions in from the
start, and treats forced and unforced equations identically. It adds one
inversion step at the end, run against a table.

The idea is an **integral transform**: multiply the unknown function by a fixed
kernel $K(s,t)$ and integrate out the time variable, producing a new function of
$s$.[^sim-intro]

$$
F(s) = \int_\alpha^\beta K(s,t)\, f(t)\, \d t.
$$

Different kernels give different transforms. The Laplace transform takes
$K(s,t) = e^{-st}$ on $[0,\infty)$. That kernel suits linear
constant-coefficient differential equations because the derivatives of
$e^{-st}$ reproduce $e^{-st}$ up to a factor of $s$, so integration by parts
trades a derivative in $t$ for a multiplication by $s$.

## The definition

> **Definition (Laplace transform).** For a function $f$ defined on $t \ge 0$,
> the Laplace transform is
> $$
> \mathcal{L}\{f(t)\} = F(s) = \int_0^\infty e^{-st} f(t)\, \d t,
> $$
> defined for those $s$ for which the improper integral converges. The variable
> $s$ is the transform variable; $F$ lives in the **$s$-domain**, $f$ in the
> **$t$-domain**. Throughout, $s$ is real, though the full theory allows $s$
> complex.

The integral runs to infinity, so convergence is not automatic. Two examples fix
the boundary cases. For the constant function $f(t) = 1$,

$$
\mathcal{L}\{1\} = \int_0^\infty e^{-st}\, \d t
  = \lim_{A\to\infty} \left[ -\frac{e^{-st}}{s} \right]_0^A
  = \frac{1}{s}, \qquad s > 0,
$$

and the limit exists only when $s > 0$; for $s \le 0$ the integrand does not
decay and the integral diverges. For the exponential $f(t) = e^{at}$,

$$
\mathcal{L}\{e^{at}\} = \int_0^\infty e^{-(s-a)t}\, \d t = \frac{1}{s-a},
  \qquad s > a.
$$

Each transform comes with a **half-line of validity** $s > a$: the transform is
a function, but it is only defined where its defining integral converges.

## Existence: piecewise continuity and exponential order

Two structural conditions on $f$ guarantee the integral converges. The first
controls behavior on finite intervals; the second controls growth at infinity.

> **Definition (Piecewise continuous).** A function $f$ is piecewise continuous
> on $\alpha \le t \le \beta$ if the interval splits into finitely many
> subintervals on each of which $f$ is continuous and approaches finite limits
> at the endpoints. Equivalently, $f$ is continuous except for finitely many
> jump discontinuities. Its value at a jump does not affect any integral of $f$.

> **Definition (Exponential order).** A function $f$ is of exponential order as
> $t \to \infty$ if there are constants $K > 0$, $a$, and $M > 0$ with
> $$
> |f(t)| \le K e^{at} \quad \text{for all } t \ge M.
> $$
> The bound says $f$ grows no faster than some exponential. The function
> $f(t) = e^{t^2}$ fails this: it outruns $K e^{at}$ for every $K$ and $a$.

$$
% caption: A function of exponential order stays under an envelope $K e^{at}$;
% multiplying by $e^{-st}$ with $s>a$ forces the integrand to decay, so the
% defining integral converges.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % axes
  \draw[->, black] (0,0) -- (7.4,0) node[right, black!70] {$t$};
  \draw[->, black] (0,-0.2) -- (0,4.0) node[above, black!70] {$y$};
  % exponential envelope K e^{a t}
  \draw[dashed, acc, thick] plot[smooth, domain=0:6.6, samples=40] (\x, {0.55*exp(0.30*\x)});
  \node[acc, anchor=south west] at (5.4,3.2) {$y = K e^{a t}$};
  % f(t): positive bounded oscillation staying under the envelope
  \draw[black, thick] plot[smooth, domain=0:6.6, samples=90]
    (\x, {0.45*exp(0.24*\x)*(0.6 + 0.4*sin(200*\x))});
  \node[black!75, anchor=west] at (3.1,0.35) {$y = f(t)$};
  % marker for M
  \draw[black, dotted] (1.0,0) -- (1.0,2.0);
  \node[black, anchor=north] at (1.0,-0.05) {$M$};
\end{tikzpicture}
$$

> **Theorem (Existence of the transform).** If $f$ is piecewise continuous on
> every interval $0 \le t \le A$ and of exponential order with bound
> $|f(t)| \le K e^{at}$ for $t \ge M$, then $\mathcal{L}\{f(t)\} = F(s)$ exists
> for every $s > a$.[^boyce-exist]

The proof splits $\int_0^\infty$ at $M$. The piece $\int_0^M$ exists because $f$
is piecewise continuous. On the tail, $|e^{-st}f(t)| \le K e^{(a-s)t}$, and
$\int_M^\infty e^{(a-s)t}\,\d t$ converges whenever $a - s < 0$, i.e. $s > a$. A
comparison with the decaying exponential does the rest. Every function treated
here except the [Dirac delta](/differential-equations/laplace/step-impulse-convolution)
satisfies these hypotheses.

## Linearity and the transform table

The integral is linear in its integrand, so the transform is a linear operator.

> **Theorem (Linearity).** If $\mathcal{L}\{f_1\}$ and $\mathcal{L}\{f_2\}$ exist
> for $s > a_1$ and $s > a_2$, then for constants $c_1, c_2$ and
> $s > \max(a_1, a_2)$,
> $$
> \mathcal{L}\{c_1 f_1(t) + c_2 f_2(t)\} = c_1 \mathcal{L}\{f_1(t)\}
>   + c_2 \mathcal{L}\{f_2(t)\}.
> $$

Linearity plus a few integrals fills a table. The transform of $t^n$ follows by
induction (or from the gamma function), and $\sinh, \cosh$ come from writing them
as sums of exponentials. The trigonometric entries take two integrations by
parts, which fold the transform back onto itself.[^sim-theory]

> **Worked example.** Find $\mathcal{L}\{\sin(at)\}$ from the definition.
>
> Set $F(s) = \int_0^\infty e^{-st}\sin(at)\,\d t$, $s > 0$. Integrate by parts
> with $u = e^{-st}$, $\d v = \sin(at)\,\d t$, so $v = -\cos(at)/a$:
> $$
> F(s) = \left[-\frac{e^{-st}\cos(at)}{a}\right]_0^\infty
>   - \frac{s}{a}\int_0^\infty e^{-st}\cos(at)\,\d t
>   = \frac{1}{a} - \frac{s}{a}\int_0^\infty e^{-st}\cos(at)\,\d t.
> $$
> Integrate the remaining integral by parts the same way, with $v = \sin(at)/a$:
> $$
> \int_0^\infty e^{-st}\cos(at)\,\d t
>   = \left[\frac{e^{-st}\sin(at)}{a}\right]_0^\infty
>   + \frac{s}{a}\int_0^\infty e^{-st}\sin(at)\,\d t
>   = \frac{s}{a}\,F(s).
> $$
> Substituting back, $F(s) = \tfrac{1}{a} - \tfrac{s^2}{a^2}F(s)$. Solving,
> $$
> F(s)\left(1 + \frac{s^2}{a^2}\right) = \frac{1}{a},
> \qquad
> \mathcal{L}\{\sin(at)\} = \frac{a}{s^2 + a^2}, \quad s > 0.
> $$

| $f(t)$ | $F(s) = \mathcal{L}\{f(t)\}$ | valid for |
| --- | --- | --- |
| $1$ | $1/s$ | $s > 0$ |
| $e^{at}$ | $1/(s-a)$ | $s > a$ |
| $t^n$, $n$ a positive integer | $n!/s^{n+1}$ | $s > 0$ |
| $t^p$, $p > -1$ | $\Gamma(p+1)/s^{p+1}$ | $s > 0$ |
| $\sin(at)$ | $a/(s^2 + a^2)$ | $s > 0$ |
| $\cos(at)$ | $s/(s^2 + a^2)$ | $s > 0$ |
| $\sinh(at)$ | $a/(s^2 - a^2)$ | $s > \lvert a\rvert$ |
| $\cosh(at)$ | $s/(s^2 - a^2)$ | $s > \lvert a\rvert$ |
| $e^{at}\sin(bt)$ | $b/((s-a)^2 + b^2)$ | $s > a$ |
| $e^{at}\cos(bt)$ | $(s-a)/((s-a)^2 + b^2)$ | $s > a$ |
| $t^n e^{at}$ | $n!/(s-a)^{n+1}$ | $s > a$ |

Because a transform corresponds to essentially one function, reading the table
right to left inverts a transform. Two piecewise-continuous functions can differ
at isolated points and share a transform, but among continuous functions the
correspondence is one-to-one, so the **inverse transform** $\mathcal{L}^{-1}$ is
well defined on continuous functions, and linear:
$\mathcal{L}^{-1}\{F_1 + \dots + F_n\} = \mathcal{L}^{-1}\{F_1\} + \dots + \mathcal{L}^{-1}\{F_n\}$.

## The frequency-shift rule

Multiplying $f$ by $e^{ct}$ in the $t$-domain slides its transform in the
$s$-domain. This one rule generates the last three table rows and handles
quadratic denominators.

> **Theorem (First shifting / $s$-shift).** If $F(s) = \mathcal{L}\{f(t)\}$
> exists for $s > a$ and $c$ is a constant, then
> $$
> \mathcal{L}\{e^{ct} f(t)\} = F(s - c), \qquad s > a + c.
> $$

The proof is one line: $\int_0^\infty e^{-st} e^{ct} f(t)\, \d t
= \int_0^\infty e^{-(s-c)t} f(t)\, \d t = F(s-c)$. Applied to
$\mathcal{L}\{\sin(bt)\} = b/(s^2+b^2)$, it gives
$\mathcal{L}\{e^{at}\sin(bt)\} = b/((s-a)^2 + b^2)$ directly.

$$
% caption: The first shifting rule slides the transform: multiplying $f$ by
% $e^{ct}$ moves the graph of $F(s)$ a distance $c$ to the right in the
% $s$-domain, taking $F(s)$ to $F(s-c)$.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (-0.3,0) -- (6.0,0) node[right, black!70] {$s$};
  \draw[->, black] (0,-0.2) -- (0,2.3) node[above, black!70] {};
  \draw[black, thick] plot[smooth, domain=-0.2:5.6, samples=90]
    (\x, {1.7/(1+(\x-0.9)*(\x-0.9))});
  \node[black!75, anchor=south] at (0.9,1.72) {$F(s)$};
  \draw[acc, thick] plot[smooth, domain=-0.2:5.8, samples=90]
    (\x, {1.7/(1+(\x-3.6)*(\x-3.6))});
  \node[acc, anchor=south] at (3.6,1.72) {$F(s$-$c)$};
  \draw[->, black] (1.7,1.05) -- (2.8,1.05) node[midway, above, black!70] {shift by $c$};
\end{tikzpicture}
$$

Read backwards, the rule inverts transforms whose denominators do not factor
over the reals: complete the square.

> **Worked example.** Invert $G(s) = 1/(s^2 - 4s + 5)$.
>
> The denominator has complex roots, so complete the square:
> $s^2 - 4s + 5 = (s-2)^2 + 1$. Then $G$ is a shifted copy of a table entry,
> $$
> G(s) = \frac{1}{(s-2)^2 + 1} = F(s-2), \qquad
> F(s) = \frac{1}{s^2+1} = \mathcal{L}\{\sin t\}.
> $$
> By the first shifting theorem, inverting $F(s-2)$ multiplies $\sin t$ by
> $e^{2t}$:
> $$
> \mathcal{L}^{-1}\{G(s)\} = e^{2t}\sin t.
> $$

## Transforms of derivatives

The property that turns calculus into algebra relates the transform of $f'$ to
the transform of $f$.

> **Theorem (Transform of a derivative).** If $f$ is continuous and $f'$ is
> piecewise continuous on every $0 \le t \le A$, and $f$ is of exponential order
> with $|f(t)| \le K e^{at}$, then for $s > a$,
> $$
> \mathcal{L}\{f'(t)\} = s\, \mathcal{L}\{f(t)\} - f(0).
> $$

Integration by parts gives it:
$\int_0^A e^{-st} f'(t)\,\d t = e^{-sA}f(A) - f(0) + s\int_0^A e^{-st}f(t)\,\d t$,
and $e^{-sA}f(A) \to 0$ as $A \to \infty$ because $s > a$.[^boyce-deriv] Applying
the rule to $f'$ in place of $f$ compounds it:

$$
\mathcal{L}\{f''(t)\} = s\, \mathcal{L}\{f'(t)\} - f'(0)
  = s^2\, \mathcal{L}\{f(t)\} - s f(0) - f'(0).
$$

Each derivative pulls out one more power of $s$ and one initial value.

> **Theorem (Transform of the $n$th derivative).** If $f, f', \dots, f^{(n-1)}$
> are continuous and $f^{(n)}$ is piecewise continuous, all of exponential order,
> then
> $$
> \mathcal{L}\{f^{(n)}(t)\} = s^n F(s) - s^{n-1} f(0) - s^{n-2} f'(0)
>   - \cdots - f^{(n-1)}(0).
> $$

The polynomial in $s$ that multiplies $F(s)$ reproduces the characteristic
polynomial of the differential operator, and the trailing terms carry the
initial data.

## Solving initial value problems

Transforming a constant-coefficient linear equation replaces each $\mathcal{L}\{y^{(k)}\}$
by its expression in $Y(s) = \mathcal{L}\{y\}$ and the initial values. The
differential equation becomes an algebraic equation for $Y(s)$; solving it,
decomposing into table entries, and inverting recovers $y(t)$.

$$
% caption: The transform method routes around the differential equation:
% transform to the s-domain, solve algebraically for $Y(s)$, and invert. The
% direct route (dashed) is the classical method.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize,
  box/.style={draw, minimum width=34mm, minimum height=13mm, align=center}]
  \definecolor{acc}{HTML}{4A6FA5}
  \node[box, draw=acc, fill=acc!10] (ivp) at (0,2.6) {IVP in $t$\\ (linear, order $n$)};
  \node[box] (alg) at (6.6,2.6) {algebraic equation\\ in $Y(s)$};
  \node[box] (Y)   at (6.6,0)   {$Y(s)$};
  \node[box, draw=acc, fill=acc!10] (sol) at (0,0) {solution $y(t)$};
  \draw[->, acc, thick] (ivp) -- (alg) node[midway, above] {transform $\mathcal{L}$};
  \draw[->, acc, thick] (alg) -- (Y)   node[midway, right] {solve};
  \draw[->, acc, thick] (Y)   -- (sol) node[midway, above] {invert};
  \draw[->, dashed, black] (ivp) -- (sol) node[midway, left, black!70, align=center] {direct\\ methods};
\end{tikzpicture}
$$

> **Worked example.** Solve $y'' - y' - 2y = 0$ with $y(0) = 1$, $y'(0) = 0$.
>
> Transform both sides and use the derivative rules:
> $$
> \bigl(s^2 Y - s y(0) - y'(0)\bigr) - \bigl(s Y - y(0)\bigr) - 2Y = 0.
> $$
> Insert $y(0) = 1$, $y'(0) = 0$ and collect, $(s^2 - s - 2) Y - (s - 1) = 0$, so
> $$
> Y(s) = \frac{s - 1}{s^2 - s - 2} = \frac{s-1}{(s-2)(s+1)}.
> $$
> Factoring $s^2 - s - 2$ is the same factoring the characteristic equation
> requires. Expand in partial fractions, $Y(s) = a/(s-2) + b/(s+1)$, and clear
> denominators:
> $$
> s - 1 = a(s+1) + b(s-2).
> $$
> Setting $s = 2$ gives $a = 1/3$; setting $s = -1$ gives $b = 2/3$. Each simple
> fraction inverts by the $e^{at}$ row, so
> $$
> y(t) = \tfrac{1}{3} e^{2t} + \tfrac{2}{3} e^{-t},
> $$
> the same solution the characteristic-equation method gives, with the constants
> already correct.

$$
% caption: The transform method as a fixed procedure: four steps from the
% initial value problem to the solution, with the table used at both the
% transform and the inversion step.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize,
  st/.style={draw, minimum width=58mm, minimum height=11mm, align=center}]
  \definecolor{acc}{HTML}{4A6FA5}
  \node[st] (a) at (0,3.9)  {transform both sides; insert the initial data};
  \node[st] (b) at (0,2.6)  {solve the algebraic equation for $Y(s)$};
  \node[st] (c) at (0,1.3)  {split $Y(s)$ into partial fractions};
  \node[st, draw=acc, fill=acc!10] (d) at (0,0) {invert term by term to get $y(t)$};
  \draw[->, acc, thick] (a) -- (b);
  \draw[->, acc, thick] (b) -- (c);
  \draw[->, acc, thick] (c) -- (d);
\end{tikzpicture}
$$

Written as a procedure, the method never varies:

```algorithm
caption: $\textsc{SolveIVP}$ — solve $a y'' + b y' + c y = g(t)$ with $y(0)=y_0,\ y'(0)=y_0'$
transform both sides: $a\,\mathcal{L}\{y''\} + b\,\mathcal{L}\{y'\} + c\,\mathcal{L}\{y\} = G(s)$
substitute $\mathcal{L}\{y'\} = sY - y_0$ and $\mathcal{L}\{y''\} = s^2 Y - s y_0 - y_0'$
collect terms and solve the algebraic equation for $Y(s)$
decompose $Y(s)$ into partial fractions matching table entries
invert term by term: $y(t) \gets \mathcal{L}^{-1}\{Y(s)\}$
return $y(t)$
```

The transform $Y(s)$ is found by solving
an **algebraic** equation, not a differential one. The initial conditions enter
at the transform step, so no arbitrary constants remain to fit at the end. A
nonhomogeneous equation is handled exactly like a homogeneous one, with $G(s)$
on the right rather than zero. And the method extends to any order $n$ unchanged,
using the $n$th-derivative rule. The one genuine difficulty is the last line:
recovering $y(t)$ from $Y(s)$, the **inversion problem**.

## Partial fractions and inversion

Inversion works by decomposing $Y(s)$ into a sum of terms that appear in the
table. When $Y(s) = P(s)/Q(s)$ is a rational function with $\deg P < \deg Q$ and
$Q$ has distinct real roots, the decomposition is a sum of simple fractions.

$$
% caption: Partial-fraction inversion: a proper rational $Y(s)=P(s)/Q(s)$ with
% simple real roots splits into terms $c_k/(s+p_k)$, each inverting to an
% exponential $c_k e^{r_k t}$ by the table.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize,
  nd/.style={draw, minimum width=26mm, minimum height=12mm, align=center},
  lf/.style={draw, minimum width=26mm, minimum height=10mm, align=center}]
  \definecolor{acc}{HTML}{4A6FA5}
  \node[nd, draw=acc, fill=acc!10] (root) at (0,3.0) {$\dfrac{P(s)}{Q(s)}$};
  \node[nd] (l) at (-3.2,1.1) {$\dfrac{c_1}{s+p_1}$};
  \node[nd] (r) at (3.2,1.1)  {$\dfrac{c_2}{s+p_2}$};
  \node[lf] (li) at (-3.2,-1.0) {$c_1 e^{r_1 t}$};
  \node[lf] (ri) at (3.2,-1.0)  {$c_2 e^{r_2 t}$};
  \draw[->, acc, thick] (root) -- (l) node[midway, above left, black!70] {split};
  \draw[->, acc, thick] (root) -- (r);
  \draw[->, black] (l) -- (li) node[midway, right, black!70] {invert};
  \draw[->, black] (r) -- (ri) node[midway, right, black!70] {invert};
\end{tikzpicture}
$$

The homogeneous worked example above ran exactly this split, with
$Q(s) = (s-2)(s+1)$ and residues $1/3$ and $2/3$. When the denominator carries an
irreducible quadratic factor, the corresponding numerator is linear.

> **Worked example.** Solve $y'' + y = \sin(2t)$ with $y(0) = 2$, $y'(0) = 1$.
>
> Transform, using $\mathcal{L}\{\sin 2t\} = 2/(s^2+4)$:
> $$
> s^2 Y - 2s - 1 + Y = \frac{2}{s^2+4},
> \qquad
> Y(s) = \frac{2s^3 + s^2 + 8s + 6}{(s^2+1)(s^2+4)}.
> $$
> Both quadratic factors are irreducible, so the decomposition uses linear
> numerators, $Y(s) = (as+b)/(s^2+1) + (cs+d)/(s^2+4)$. Matching coefficients
> gives $a = 2$, $b = 5/3$, $c = 0$, $d = -2/3$, hence
> $$
> Y(s) = \frac{2s}{s^2+1} + \frac{5/3}{s^2+1} - \frac{2/3}{s^2+4},
> \qquad
> y(t) = 2\cos t + \tfrac{5}{3}\sin t - \tfrac{1}{3}\sin(2t).
> $$

For a denominator $Q(s)$ of degree $n$ with distinct roots $r_1, \dots, r_n$,
the coefficient of $1/(s - r_k)$ is $P(r_k)/Q'(r_k)$, and the inverse transform is
$\sum_k \tfrac{P(r_k)}{Q'(r_k)} e^{r_k t}$ — a compact residue formula.

## Differentiating and integrating the transform

Differentiating $F(s)$ under the integral sign multiplies the original function
by $-t$, which produces transforms of $t^n f(t)$ without new integration.

> **Theorem (Derivative of the transform).** If $F(s) = \mathcal{L}\{f(t)\}$,
> then differentiation in $s$ corresponds to multiplication by $-t$ in $t$:
> $$
> F'(s) = \mathcal{L}\{-t f(t)\}, \qquad
> F^{(n)}(s) = \mathcal{L}\{(-t)^n f(t)\}.
> $$

For example, from $\mathcal{L}\{\sin(at)\} = a/(s^2+a^2)$,

$$
\mathcal{L}\{t\sin(at)\} = -\frac{\d}{\d s}\!\left(\frac{a}{s^2+a^2}\right)
  = \frac{2as}{(s^2+a^2)^2}.
$$

Because $t$ appears as a first-power coefficient in some variable-coefficient
equations, this rule occasionally converts such an equation into a first-order
equation for $Y(s)$; that is how the transform of the Bessel function
[$J_0$](/differential-equations/series-solutions/bessel-and-special-functions)
can be found.[^sim-derint] Integrating the transform runs the other way: if
$f(t)/t$ has a limit at $0$, then $\mathcal{L}\{f(t)/t\} = \int_s^\infty F(\sigma)\,\d\sigma$,
and letting $s \to 0$ evaluates $\int_0^\infty f(t)/t\,\d t$. For $f = \sin t$ this
gives the value $\pi/2$ for the Dirichlet integral.

## Benefits and costs of the method

The transform is a change of representation, with definite benefits and costs.

- **Differentiation becomes multiplication.** Each $\d/\d t$ becomes a factor of
  $s$, and the differential equation becomes algebraic.
- **Initial data is built in.** The values $y(0), y'(0), \dots$ enter at the
  transform step, so the general-solution-then-fit-constants detour disappears.
- **Forcing is uniform.** Nonhomogeneous and homogeneous equations transform the
  same way; [discontinuous and impulsive forcing](/differential-equations/laplace/step-impulse-convolution)
  are no harder in the $s$-domain.
- **Roots still matter.** The denominator of $Y(s)$ is the characteristic
  polynomial, so factoring it — the same obstruction as before — is unavoidable.
- **Inversion is the real work.** Going from $Y(s)$ back to $y(t)$ relies on the
  table, partial fractions, the shift rules, and the convolution theorem.

[^sim-intro]: **Simmons**, _Differential Equations with Applications and Historical Notes_, §48 — Introduction: the Laplace transform as an integral transform with kernel $e^{-pt}$, converting differentiation into multiplication.
[^boyce-exist]: **Boyce**, _Elementary Differential Equations and Boundary Value Problems_, §6.1, Theorems 6.1.1–6.1.2 — the comparison test for improper integrals and existence of the transform for functions piecewise continuous and of exponential order, valid for $s > a$.
[^sim-theory]: **Simmons**, §49 — A Few Remarks on the Theory; and **Boyce**, §6.1, Examples 4–8 and Table 6.2.1 — the elementary transforms of $1$, $e^{at}$, $t^n$, $\sin(at)$, $\cos(at)$, and their combinations, assembled by linearity.
[^boyce-deriv]: **Boyce**, §6.2, Theorem 6.2.1 and Corollary 6.2.2 — $\mathcal{L}\{f'\} = sF(s) - f(0)$ and its $n$-fold extension, the identity that reduces a constant-coefficient IVP to an algebraic equation whose denominator is the characteristic polynomial.
[^sim-derint]: **Simmons**, §51 — Derivatives and Integrals of Laplace Transforms: $\mathcal{L}\{-t f(t)\} = F'(s)$ and $\mathcal{L}\{f(t)/t\} = \int_s^\infty F(\sigma)\,\d\sigma$, with the Bessel-equation and Dirichlet-integral applications.