diff-eq-handbook/chapters/ch07_laplace_transforms.tex

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% ch07_laplace_transforms.tex
% Chapter 7: Laplace Transforms
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\section{Laplace Transforms}
\label{ch:laplace_transforms}

The Laplace transform is one of the most powerful tools for solving linear differential equations with constant coefficients. It converts differential equations in the time domain into algebraic equations in the \emph{frequency domain} (the ``$s$-domain''), where they can be manipulated with elementary algebra and then inverted back. The method is particularly valuable for problems involving initial conditions, discontinuous forcing functions, and impulse inputs --- situations where classical techniques become cumbersome.

\subsection{Definition and Existence}
\label{sec:ch07_definition_and_existence}

\begin{definition}
  \label{def:laplace_transform}
  Let $f(t)$ be a function defined for $t \ge 0$. The \textbf{Laplace transform} of $f(t)$ is the function
  \begin{equation}
    \label{eq:laplace_definition}
    \mathcal{L}\{f(t)\} = F(s) = \int_0^\infty e^{-st} f(t)\,\diff t,
  \end{equation}
  provided the integral converges.
\end{definition}

The variable $s$ is a complex parameter, though in most applications we treat $s$ as real and sufficiently large so that the improper integral converges. The function $F(s)$ is the \textbf{image} of $f(t)$ under the Laplace transform.

\begin{keyresult}
  \textbf{Existence conditions.} The Laplace transform $\mathcal{L}\{f(t)\}$ exists if $f(t)$ satisfies:
  \begin{enumerate}
    \item \textbf{Piecewise continuity} on $[0, \infty)$: $f(t)$ has at most a finite number of finite jump discontinuities on any finite subinterval of $[0, \infty)$.
    \item \textbf{Exponential order}: there exist constants $M > 0$, $a \in \R$, and $T > 0$ such that
    \[
      |f(t)| \le M e^{at} \quad \text{for all } t \ge T.
    \]
  \end{enumerate}
  Under these conditions, the integral \cref{eq:laplace_definition} converges for all $s > a$.
\end{keyresult}

\paragraph{Intuition.} The exponential factor $e^{-st}$ in the integrand acts as a ``damping'' factor. As long as $s$ is larger than the growth rate of $f(t)$, the product $e^{-st} f(t)$ decays and the integral converges.

We now compute a few Laplace transforms directly from the definition to build intuition.

\begin{workedexample}
  Compute $\mathcal{L}\{1\}$.

  \textbf{Solution.} By definition:
  \[
    \mathcal{L}\{1\} = \int_0^\infty e^{-st} \cdot 1 \,\diff t
    = \lim_{T \to \infty} \left[-\frac{1}{s} e^{-st}\right]_0^T
    = \lim_{T \to \infty} \left(-\frac{1}{s} e^{-sT} + \frac{1}{s}\right).
  \]
  For $s > 0$, $e^{-sT} \to 0$ as $T \to \infty$, so
  \[
    \mathcal{L}\{1\} = \frac{1}{s}, \qquad s > 0.
  \]
\end{workedexample}

\begin{workedexample}
  Compute $\mathcal{L}\{e^{at}\}$ for a constant $a$.

  \textbf{Solution.}
  \[
    \mathcal{L}\{e^{at}\} = \int_0^\infty e^{-st} e^{at}\,\diff t
    = \int_0^\infty e^{-(s-a)t}\,\diff t.
  \]
  This converges when $s > a$:
  \[
    \mathcal{L}\{e^{at}\} = \lim_{T \to \infty} \left[-\frac{1}{s-a} e^{-(s-a)t}\right]_0^T
    = \frac{1}{s-a}, \qquad s > a.
  \]
\end{workedexample}

\begin{workedexample}
  Compute $\mathcal{L}\{\sin(bt)\}$ for $b > 0$.

  \textbf{Solution.}
  \[
    \mathcal{L}\{\sin(bt)\} = \int_0^\infty e^{-st} \sin(bt)\,\diff t.
  \]
  We evaluate this using integration by parts twice. Let $I = \int e^{-st} \sin(bt)\,\diff t$. Set $u = \sin(bt)$, $\diff v = e^{-st}\,\diff t$:
  \[
    I = -\frac{1}{s} e^{-st} \sin(bt) + \frac{b}{s} \int e^{-st} \cos(bt)\,\diff t.
  \]
  Now integrate $\int e^{-st} \cos(bt)\,\diff t$ by parts with $u = \cos(bt)$, $\diff v = e^{-st}\,\diff t$:
  \[
    \int e^{-st} \cos(bt)\,\diff t
    = -\frac{1}{s} e^{-st} \cos(bt) + \frac{b}{s} \int e^{-st} \sin(bt)\,\diff t
    = -\frac{1}{s} e^{-st} \cos(bt) + \frac{b}{s} I.
  \]
  Substituting back:
  \[
    I = -\frac{1}{s} e^{-st} \sin(bt) + \frac{b}{s} \left(-\frac{1}{s} e^{-st} \cos(bt) + \frac{b}{s} I\right).
  \]
  Solving for $I$:
  \[
    I\left(1 - \frac{b^2}{s^2}\right) = -e^{-st}\left(\frac{1}{s} \sin(bt) + \frac{b}{s^2} \cos(bt)\right),
  \]
  \[
    I = -e^{-st}\frac{s \sin(bt) + b \cos(bt)}{s^2 + b^2}.
  \]
  Evaluating the definite integral from $0$ to $\infty$ (for $s > 0$):
  \[
    \mathcal{L}\{\sin(bt)\} = \left[0\right] - \left[-\frac{b}{s^2 + b^2}\right]
    = \frac{b}{s^2 + b^2}, \qquad s > 0.
  \]
\end{workedexample}

Direct computation from the definition is instructive but impractical for everyday use. In the next section, we compile a comprehensive table of transforms.

\subsection{Laplace Transform Table}
\label{sec:ch07_transform_table}

The following table collects the most commonly used Laplace transforms. Each entry includes the convergence condition on $s$.

\begin{keyresult}
  \textbf{Laplace transform table.}
  \begin{center}
  \begin{tabular}{l l l}
    \toprule
    \textbf{$f(t)$} & \textbf{$F(s) = \mathcal{L}\{f(t)\}$} & \textbf{Convergence} \\
    \midrule
    $\delta(t - a)$ \quad ($a \ge 0$) & $e^{-as}$ & all $s$ \\[6pt]
    $1$ & $\dfrac{1}{s}$ & $s > 0$ \\[10pt]
    $t^n$ \quad ($n \in \N_0$) & $\dfrac{n!}{s^{n+1}}$ & $s > 0$ \\[10pt]
    $e^{at}$ & $\dfrac{1}{s - a}$ & $s > a$ \\[10pt]
    $t^n e^{at}$ & $\dfrac{n!}{(s - a)^{n+1}}$ & $s > a$ \\[10pt]
    $\sin(bt)$ & $\dfrac{b}{s^2 + b^2}$ & $s > 0$ \\[10pt]
    $\cos(bt)$ & $\dfrac{s}{s^2 + b^2}$ & $s > 0$ \\[10pt]
    $\sinh(bt)$ & $\dfrac{b}{s^2 - b^2}$ & $s > |b|$ \\[10pt]
    $\cosh(bt)$ & $\dfrac{s}{s^2 - b^2}$ & $s > |b|$ \\[10pt]
    $e^{at}\sin(bt)$ & $\dfrac{b}{(s-a)^2 + b^2}$ & $s > a$ \\[10pt]
    $e^{at}\cos(bt)$ & $\dfrac{s-a}{(s-a)^2 + b^2}$ & $s > a$ \\[10pt]
    $\sin^2(bt)$ & $\dfrac{2b^2}{s(s^2 + 4b^2)}$ & $s > 0$ \\[10pt]
    $\cos^2(bt)$ & $\dfrac{s^2+2b^2}{s(s^2 + 4b^2)}$ & $s > 0$ \\[10pt]
    $t\sin(bt)$ & $\dfrac{2bs}{(s^2 + b^2)^2}$ & $s > 0$ \\[10pt]
    $t\cos(bt)$ & $\dfrac{s^2 - b^2}{(s^2 + b^2)^2}$ & $s > 0$ \\[10pt]
    $u_c(t)$ \quad (Heaviside step) & $\dfrac{e^{-cs}}{s}$ & $s > 0$ \\[10pt]
    $J_0(bt)$ \quad (Bessel function) & $\dfrac{1}{\sqrt{s^2 + b^2}}$ & $s > 0$ \\[10pt]
    \bottomrule
  \end{tabular}
  \end{center}
\end{keyresult}

\begin{hintbox}
  \textbf{How to extend the table.} Most additional transforms can be generated from the entries above using the properties discussed in \cref{sec:ch07_properties}: linearity, shift theorems, and derivative transforms. For instance, $\mathcal{L}\{t^2 e^{3t}\}$ follows from the $t^n e^{at}$ entry with $n=2$, $a=3$.
\end{hintbox}

\subsection{Properties of the Laplace Transform}
\label{sec:ch07_properties}

The power of the Laplace transform comes from its algebraic properties, which allow us to compute transforms without reverting to the integral definition.

\subsubsection{Linearity}
\label{sec:ch07_linearity}

\begin{keyresult}
  \textbf{Linearity.} If $\mathcal{L}\{f(t)\} = F(s)$ and $\mathcal{L}\{g(t)\} = G(s)$, then for any constants $a, b \in \R$:
  \[
    \mathcal{L}\{a f(t) + b g(t)\} = a F(s) + b G(s).
  \]
\end{keyresult}

This follows immediately from the linearity of integration.

\begin{workedexample}
  Compute $\mathcal{L}\{3e^{2t} - 4\cos(5t) + 2t\}$.

  \textbf{Solution.} By linearity and the transform table:
  \[
    \begin{aligned}
      \mathcal{L}\{3e^{2t} - 4\cos(5t) + 2t\}
      &= 3\,\mathcal{L}\{e^{2t}\} - 4\,\mathcal{L}\{\cos(5t)\} + 2\,\mathcal{L}\{t\} \\[4pt]
      &= 3\cdot\frac{1}{s-2} - 4\cdot\frac{s}{s^2+25} + 2\cdot\frac{1}{s^2} \\[4pt]
      &= \frac{3}{s-2} - \frac{4s}{s^2+25} + \frac{2}{s^2}.
    \end{aligned}
  \]
\end{workedexample}

\subsubsection{First Shift Theorem (s-Shifting)}
\label{sec:ch07_first_shift}

\begin{keyresult}
  \textbf{First shift theorem.} If $\mathcal{L}\{f(t)\} = F(s)$, then
  \[
    \mathcal{L}\{e^{at} f(t)\} = F(s - a).
  \]
\end{keyresult}

\paragraph{Proof.}
\[
  \mathcal{L}\{e^{at} f(t)\} = \int_0^\infty e^{-st} e^{at} f(t)\,\diff t
  = \int_0^\infty e^{-(s-a)t} f(t)\,\diff t
  = F(s - a).
\]

\begin{workedexample}
  Find $\mathcal{L}\{e^{3t}\sin(2t)\}$.

  \textbf{Solution.} From the table, $\mathcal{L}\{\sin(2t)\} = \dfrac{2}{s^2 + 4}$. By the first shift theorem with $a = 3$:
  \[
    \mathcal{L}\{e^{3t}\sin(2t)\} = \frac{2}{(s-3)^2 + 4} = \frac{2}{(s-3)^2 + 2^2}.
  \]
\end{workedexample}

\begin{workedexample}
  Find $\mathcal{L}\{t^2 e^{-t}\}$.

  \textbf{Solution.} From the table, $\mathcal{L}\{t^2\} = \dfrac{2}{s^3}$. Applying the first shift with $a = -1$:
  \[
    \mathcal{L}\{t^2 e^{-t}\} = \frac{2}{(s+1)^3}.
  \]
\end{workedexample}

\subsubsection{Second Shift Theorem (t-Shifting)}
\label{sec:ch07_second_shift}

\begin{keyresult}
  \textbf{Second shift theorem.} Let $u_c(t)$ denote the Heaviside step function (\cref{sec:ch07_step_function}). If $\mathcal{L}\{f(t)\} = F(s)$, then
  \[
    \mathcal{L}\{u_c(t)\,f(t-c)\} = e^{-cs} F(s), \qquad c \ge 0.
  \]
  Equivalently,
  \[
    \mathcal{L}^{-1}\{e^{-cs} F(s)\} = u_c(t)\,f(t-c).
  \]
\end{keyresult}

\paragraph{Proof.}
\[
  \begin{aligned}
    \mathcal{L}\{u_c(t)\,f(t-c)\}
    &= \int_0^\infty e^{-st}\,u_c(t)\,f(t-c)\,\diff t \\
    &= \int_c^\infty e^{-st}\,f(t-c)\,\diff t.
  \end{aligned}
\]
Substitute $u = t - c$, so $\diff u = \diff t$, $t = u + c$:
\[
  \int_0^\infty e^{-s(u+c)} f(u)\,\diff u
  = e^{-cs} \int_0^\infty e^{-su} f(u)\,\diff u
  = e^{-cs} F(s).
\]

\begin{workedexample}
  Find $\mathcal{L}\{u_2(t)\,(t-2)^3\}$.

  \textbf{Solution.} Here $c = 2$ and $f(t) = t^3$, so $F(s) = \dfrac{6}{s^4}$. By the second shift theorem:
  \[
    \mathcal{L}\{u_2(t)\,(t-2)^3\} = e^{-2s} \cdot \frac{6}{s^4}.
  \]
\end{workedexample}

\begin{workedexample}
  Find $\mathcal{L}\{u_1(t)\,\sin(2(t-1))\}$.

  \textbf{Solution.} Here $c = 1$ and $f(t) = \sin(2t)$, so $F(s) = \dfrac{2}{s^2 + 4}$. By the second shift theorem:
  \[
    \mathcal{L}\{u_1(t)\,\sin(2(t-1))\} = e^{-s} \cdot \frac{2}{s^2 + 4}.
  \]
\end{workedexample}

\subsubsection{Derivative Transforms}
\label{sec:ch07_derivative_transforms}

Transforms of derivatives are the key to solving initial value problems via the Laplace transform. The initial conditions appear automatically in the transformed equation.

\begin{keyresult}
  \textbf{Transform of the first derivative.} If $f(t)$ is continuous and $f'(t)$ is piecewise continuous on $[0, \infty)$, then
  \[
    \mathcal{L}\{f'(t)\} = s F(s) - f(0).
  \]
\end{keyresult}

\paragraph{Proof.} Integrate by parts with $u = e^{-st}$ and $\diff v = f'(t)\,\diff t$:
\[
  \begin{aligned}
    \mathcal{L}\{f'(t)\} &= \int_0^\infty e^{-st} f'(t)\,\diff t \\
    &= \left[e^{-st} f(t)\right]_0^\infty - \int_0^\infty (-s) e^{-st} f(t)\,\diff t \\
    &= \left(\lim_{T \to \infty} e^{-sT} f(T) - f(0)\right) + s \int_0^\infty e^{-st} f(t)\,\diff t.
  \end{aligned}
\]
Since $f(t)$ is of exponential order, the limit $\lim_{T \to \infty} e^{-sT} f(T) = 0$ for $s > a$. Thus:
\[
  \mathcal{L}\{f'(t)\} = s F(s) - f(0).
\]

Higher derivatives follow from repeated application.

\begin{keyresult}
  \textbf{Higher derivative transforms.}
  \[
  \begin{aligned}
    \mathcal{L}\{f''(t)\} &= s^2 F(s) - s f(0) - f'(0), \\[4pt]
    \mathcal{L}\{f^{(n)}(t)\} &= s^n F(s) - s^{n-1} f(0) - s^{n-2} f'(0) - \cdots - f^{(n-1)}(0).
  \end{aligned}
  \]
\end{keyresult}

\begin{workedexample}
  Solve the IVP $y'' - 3y' + 2y = 0$, $y(0) = 1$, $y'(0) = 0$ using Laplace transforms.

  \textbf{Solution.} Let $Y(s) = \mathcal{L}\{y(t)\}$. Transforming each term:
  \[
  \begin{aligned}
    \mathcal{L}\{y''\} &= s^2 Y(s) - s y(0) - y'(0) = s^2 Y(s) - s, \\
    \mathcal{L}\{y'\}  &= s Y(s) - y(0) = s Y(s) - 1, \\
    \mathcal{L}\{y\}   &= Y(s).
  \end{aligned}
  \]
  Substituting into the ODE:
  \[
    (s^2 Y - s) - 3(s Y - 1) + 2Y = 0.
  \]
  Collect terms in $Y(s)$:
  \[
    (s^2 - 3s + 2) Y(s) - s + 3 = 0,
  \]
  \[
    Y(s) = \frac{s - 3}{s^2 - 3s + 2} = \frac{s - 3}{(s-1)(s-2)}.
  \]
  Partial fraction decomposition:
  \[
    \frac{s-3}{(s-1)(s-2)} = \frac{A}{s-1} + \frac{B}{s-2}.
  \]
  Multiplying through: $s - 3 = A(s-2) + B(s-1)$. Setting $s = 1$: $-2 = -A$, so $A = 2$. Setting $s = 2$: $-1 = B$, so $B = -1$.
  \[
    Y(s) = \frac{2}{s-1} - \frac{1}{s-2}.
  \]
  Inverting:
  \[
    y(t) = 2e^t - e^{2t}.
  \]
\end{workedexample}

\subsubsection{Integral Transform}
\label{sec:ch07_integral_transform}

\begin{keyresult}
  \textbf{Transform of an integral.} If $f(t)$ is of exponential order, then
  \[
    \mathcal{L}\left\{\int_0^t f(\tau)\,\diff\tau\right\} = \frac{F(s)}{s}.
  \]
\end{keyresult}

\paragraph{Proof.} Let $g(t) = \int_0^t f(\tau)\,\diff\tau$. Then $g'(t) = f(t)$ and $g(0) = 0$. By the derivative transform:
\[
  \mathcal{L}\{g'(t)\} = s \mathcal{L}\{g(t)\} - g(0) = s \mathcal{L}\{g(t)\}.
\]
But $\mathcal{L}\{g'(t)\} = \mathcal{L}\{f(t)\} = F(s)$. Hence $\mathcal{L}\{g(t)\} = F(s)/s$.

\begin{workedexample}
  Find $\mathcal{L}\left\{\int_0^t e^{3\tau}\sin(2\tau)\,\diff\tau\right\}$.

  \textbf{Solution.} From the table (or the first shift theorem):
  \[
    \mathcal{L}\{e^{3t}\sin(2t)\} = \frac{2}{(s-3)^2 + 4}.
  \]
  By the integral transform:
  \[
    \mathcal{L}\left\{\int_0^t e^{3\tau}\sin(2\tau)\,\diff\tau\right\}
    = \frac{2}{s\bigl((s-3)^2 + 4\bigr)}.
  \]
\end{workedexample}

\subsubsection{Multiplication by $t^n$}
\label{sec:ch07_multiplication_by_t}

\begin{keyresult}
  \textbf{Multiplication by $t^n$.} If $\mathcal{L}\{f(t)\} = F(s)$, then
  \[
    \mathcal{L}\{t^n f(t)\} = (-1)^n \frac{\diff^n}{\diff s^n} F(s).
  \]
  In particular, $\mathcal{L}\{t f(t)\} = -F'(s)$.
\end{keyresult}

This property is useful for deriving additional table entries. For example, $\mathcal{L}\{t \sin(bt)\} = -\dfrac{\diff}{\diff s}\left(\dfrac{b}{s^2+b^2}\right) = \dfrac{2bs}{(s^2+b^2)^2}$.

\subsection{Inverse Laplace Transforms}
\label{sec:ch07_inverse_transforms}

The \textbf{inverse Laplace transform}, denoted $\mathcal{L}^{-1}$, recovers the time-domain function from its $s$-domain image:
\[
  \text{if } F(s) = \mathcal{L}\{f(t)\}, \text{ then } f(t) = \mathcal{L}^{-1}\{F(s)\}.
\]

By linearity, $\mathcal{L}^{-1}\{a F(s) + b G(s)\} = a f(t) + b g(t)$. In practice, we compute inverse transforms using two complementary strategies: table lookup and partial fraction decomposition.

\begin{hintbox}
  \textbf{Inverse transform strategies.}
  \begin{enumerate}
    \item \textbf{Table lookup:} Match $F(s)$ directly to a table entry. Look for recognizable forms: $\dfrac{1}{s-a}$, $\dfrac{b}{s^2+b^2}$, $\dfrac{s}{s^2+b^2}$, etc.
    \item \textbf{Algebraic manipulation:} If $F(s)$ does not match a table entry exactly, try completing the square in the denominator, splitting into partial fractions, or using shift theorems.
    \item \textbf{Second shift theorem:} If $F(s)$ contains a factor $e^{-cs}$, the result will involve a Heaviside step $u_c(t)$.
  \end{enumerate}
\end{hintbox}

\paragraph{Completing the square.} Many denominators are of the form $s^2 + 2as + (a^2 + b^2)$, which can be rewritten as $(s+a)^2 + b^2$. This allows us to use the first shift theorem in reverse.

\begin{workedexample}
  Find $\mathcal{L}^{-1}\left\{\dfrac{1}{s^2 + 6s + 13}\right\}$.

  \textbf{Solution.} Complete the square in the denominator:
  \[
    s^2 + 6s + 13 = (s+3)^2 + 4 = (s+3)^2 + 2^2.
  \]
  Thus:
  \[
    \frac{1}{s^2 + 6s + 13} = \frac{1}{(s+3)^2 + 2^2} = \frac{1}{2}\cdot\frac{2}{(s+3)^2 + 2^2}.
  \]
  From the table, $\mathcal{L}^{-1}\left\{\dfrac{2}{s^2+2^2}\right\} = \sin(2t)$. By the first shift theorem (with $a = -3$):
  \[
    \mathcal{L}^{-1}\left\{\frac{1}{2}\cdot\frac{2}{(s+3)^2+2^2}\right\}
    = \frac{1}{2} e^{-3t} \sin(2t).
  \]
\end{workedexample}

\subsection{Partial Fraction Decomposition}
\label{sec:ch07_partial_fractions}

When $F(s)$ is a rational function (ratio of polynomials) and the denominator degree exceeds the numerator degree, we decompose it into simpler fractions that match table entries.

\begin{hintbox}
  \textbf{Partial fraction decomposition guide.}
  \begin{itemize}
    \item \textbf{Distinct linear factors:} $\dfrac{P(s)}{(s-a)(s-b)} = \dfrac{A}{s-a} + \dfrac{B}{s-b}$.
    \item \textbf{Repeated linear factor:} $\dfrac{P(s)}{(s-a)^n} = \dfrac{A_1}{s-a} + \dfrac{A_2}{(s-a)^2} + \cdots + \dfrac{A_n}{(s-a)^n}$.
    \item \textbf{Irreducible quadratic:} $\dfrac{P(s)}{(s^2+bs+c)} = \dfrac{As+B}{s^2+bs+c}$. Complete the square in the denominator, then split into $\dfrac{A(s+b/2)}{(s+b/2)^2+(c-b^2/4)} + \dfrac{B'}{\text{denom}}$.
    \item \textbf{Combined:} Handle each factor type separately and combine.
  \end{itemize}
\end{hintbox}

\begin{workedexample}
  Find $\mathcal{L}^{-1}\left\{\dfrac{3s-5}{(s-1)(s+2)}\right\}$.

  \textbf{Solution.} Decompose:
  \[
    \frac{3s-5}{(s-1)(s+2)} = \frac{A}{s-1} + \frac{B}{s+2}.
  \]
  Multiplying both sides by $(s-1)(s+2)$:
  \[
    3s - 5 = A(s+2) + B(s-1).
  \]
  Set $s = 1$: $3(1) - 5 = A(3) \Rightarrow -2 = 3A \Rightarrow A = -\dfrac{2}{3}$.

  Set $s = -2$: $3(-2) - 5 = B(-3) \Rightarrow -11 = -3B \Rightarrow B = \dfrac{11}{3}$.

  Therefore:
  \[
    \frac{3s-5}{(s-1)(s+2)} = -\frac{2}{3}\cdot\frac{1}{s-1} + \frac{11}{3}\cdot\frac{1}{s+2}.
  \]
  Inverting:
  \[
    \mathcal{L}^{-1}\left\{\frac{3s-5}{(s-1)(s+2)}\right\} = -\frac{2}{3}e^t + \frac{11}{3}e^{-2t}.
  \]
\end{workedexample}

\begin{workedexample}
  Find $\mathcal{L}^{-1}\left\{\dfrac{2s+1}{(s-1)^2(s+3)}\right\}$.

  \textbf{Solution.} The repeated factor $(s-1)^2$ requires two terms:
  \[
    \frac{2s+1}{(s-1)^2(s+3)} = \frac{A}{s-1} + \frac{B}{(s-1)^2} + \frac{C}{s+3}.
  \]
  Multiply through:
  \[
    2s + 1 = A(s-1)(s+3) + B(s+3) + C(s-1)^2.
  \]
  Set $s = 1$: $2(1) + 1 = B(4) \Rightarrow 3 = 4B \Rightarrow B = \dfrac{3}{4}$.

  Set $s = -3$: $2(-3) + 1 = C(-4)^2 \Rightarrow -5 = 16C \Rightarrow C = -\dfrac{5}{16}$.

  To find $A$, expand and equate coefficients of $s^2$:
  \[
    A(s^2 + 2s - 3) + B(s+3) + C(s^2 - 2s + 1).
  \]
  The $s^2$ coefficient is $A + C$. The left side has no $s^2$ term, so $A + C = 0 \Rightarrow A = -C = \dfrac{5}{16}$.

  Thus:
  \[
    \frac{2s+1}{(s-1)^2(s+3)} = \frac{5/16}{s-1} + \frac{3/4}{(s-1)^2} - \frac{5/16}{s+3}.
  \]
  Inverting each term:
  \[
    \mathcal{L}^{-1}\left\{\frac{2s+1}{(s-1)^2(s+3)}\right\}
    = \frac{5}{16}e^t + \frac{3}{4}t\,e^t - \frac{5}{16}e^{-3t}.
  \]
\end{workedexample}

\subsection{Solving Initial Value Problems}
\label{sec:ch07_solving_ivps}

The Laplace transform method provides a systematic algorithm for solving linear IVPs with constant coefficients. Unlike classical methods (undetermined coefficients, variation of parameters), the Laplace approach incorporates initial conditions directly into the algebraic step, so there is no need to determine constants \emph{after} finding the general solution.

\begin{keyresult}
  \textbf{IVP algorithm via Laplace transforms.}
  \begin{enumerate}
    \item \textbf{Transform both sides.} Apply $\mathcal{L}$ to the differential equation. Use the derivative transforms to express $\mathcal{L}\{y'\}$, $\mathcal{L}\{y''\}$, etc.\ in terms of $Y(s)$ and the initial conditions.
    \item \textbf{Substitute initial conditions.} Replace $y(0)$, $y'(0)$, etc.\ with their given values.
    \item \textbf{Solve algebraically for $Y(s)$.} Collect all $Y(s)$ terms on one side and solve for $Y(s)$.
    \item \textbf{Partial fraction decomposition.} If $Y(s)$ is a rational function, decompose it into simpler fractions.
    \item \textbf{Invert.} Apply $\mathcal{L}^{-1}$ to obtain $y(t)$.
  \end{enumerate}
\end{keyresult}

\paragraph{Example 1: First-order IVP.}

\begin{workedexample}
  Solve $y' + 4y = e^{-2t}$, $y(0) = 3$.

  \textbf{Solution.} \textbf{Step 1.} Transform both sides:
  \[
    \mathcal{L}\{y'\} + 4\mathcal{L}\{y\} = \mathcal{L}\{e^{-2t}\}.
  \]
  \textbf{Step 2.} Use the derivative transform and substitute $y(0) = 3$:
  \[
    (s Y(s) - 3) + 4Y(s) = \frac{1}{s+2}.
  \]
  \textbf{Step 3.} Solve for $Y(s)$:
  \[
    (s + 4) Y(s) = \frac{1}{s+2} + 3 = \frac{1 + 3(s+2)}{s+2} = \frac{3s + 7}{s+2},
  \]
  \[
    Y(s) = \frac{3s + 7}{(s+2)(s+4)}.
  \]
  \textbf{Step 4.} Partial fractions:
  \[
    \frac{3s+7}{(s+2)(s+4)} = \frac{A}{s+2} + \frac{B}{s+4}.
  \]
  Multiply: $3s + 7 = A(s+4) + B(s+2)$.
  Set $s = -2$: $3(-2) + 7 = A(2) \Rightarrow 1 = 2A \Rightarrow A = \dfrac{1}{2}$.
  Set $s = -4$: $3(-4) + 7 = B(-2) \Rightarrow -5 = -2B \Rightarrow B = \dfrac{5}{2}$.
  \textbf{Step 5.} Invert:
  \[
    y(t) = \frac{1}{2}e^{-2t} + \frac{5}{2}e^{-4t}.
  \]
\end{workedexample}

\paragraph{Example 2: Second-order IVP.}

\begin{workedexample}
  Solve $y'' + 2y' + 5y = \sin(3t)$, with $y(0) = 1$ and $y'(0) = 0$.

  \textbf{Solution.} \textbf{Step 1.} Transform:
  \[
    \mathcal{L}\{y''\} + 2\mathcal{L}\{y'\} + 5\mathcal{L}\{y\} = \mathcal{L}\{\sin(3t)\}.
  \]
  \textbf{Step 2.} Use derivative transforms and substitute ICs:
  \[
    (s^2 Y - s\cdot 1 - 0) + 2(s Y - 1) + 5Y = \frac{3}{s^2 + 9}.
  \]
  Simplify:
  \[
    (s^2 + 2s + 5) Y(s) - s - 2 = \frac{3}{s^2 + 9}.
  \]
  \textbf{Step 3.} Solve for $Y(s)$:
  \[
    Y(s) = \frac{s+2}{s^2 + 2s + 5} + \frac{3}{(s^2+9)(s^2+2s+5)}.
  \]
  Complete the square in the first denominator: $s^2 + 2s + 5 = (s+1)^2 + 4$. Rewrite the first term:
  \[
    \frac{s+2}{(s+1)^2 + 4} = \frac{s+1}{(s+1)^2+4} + \frac{1}{(s+1)^2+4}.
  \]
  Inverting the first part: $\mathcal{L}^{-1}\left\{\dfrac{s+1}{(s+1)^2+4}\right\} = e^{-t}\cos(2t)$ and $\mathcal{L}^{-1}\left\{\dfrac{1}{(s+1)^2+4}\right\} = \dfrac{1}{2}e^{-t}\sin(2t)$.

  \textbf{Step 4.} For the second term, partial fractions on $s^2$:
  \[
    \frac{3}{(s^2+9)(s^2+2s+5)} = \frac{As+B}{s^2+9} + \frac{Cs+D}{s^2+2s+5}.
  \]
  Multiply: $3 = (As+B)(s^2+2s+5) + (Cs+D)(s^2+9)$. Expanding and collecting powers of $s$:
  \[
    3 = (A+C)s^3 + (2A+B+D)s^2 + (5A+2B+9C)s + (5B+9D).
  \]
  Equating coefficients:
  \[
  \begin{cases}
    A + C = 0 \\
    2A + B + D = 0 \\
    5A + 2B + 9C = 0 \\
    5B + 9D = 3
  \end{cases}
  \]
  From the first: $C = -A$. Substitute into the third: $5A + 2B - 9A = 0 \Rightarrow 2B = 4A \Rightarrow B = 2A$. From the second: $2A + 2A + D = 0 \Rightarrow D = -4A$. From the fourth: $5(2A) + 9(-4A) = 3 \Rightarrow 10A - 36A = 3 \Rightarrow -26A = 3 \Rightarrow A = -\dfrac{3}{26}$.

  So $C = \dfrac{3}{26}$, $B = -\dfrac{6}{26} = -\dfrac{3}{13}$, $D = \dfrac{12}{26} = \dfrac{6}{13}$.

  The second term becomes:
  \[
    \frac{-\frac{3}{26}s - \frac{3}{13}}{s^2+9} + \frac{\frac{3}{26}s + \frac{6}{13}}{s^2+2s+5}.
  \]
  Rewrite for inversion. The first fraction:
  \[
    -\frac{3}{26}\cdot\frac{s}{s^2+9} - \frac{3}{13}\cdot\frac{1}{s^2+9}
    = -\frac{3}{26}\cdot\frac{s}{s^2+9} - \frac{1}{13}\cdot\frac{3}{s^2+9}.
  \]
  Inverting: $-\dfrac{3}{26}\cos(3t) - \dfrac{1}{13}\sin(3t)$.

  The second fraction: complete the square in $s^2+2s+5 = (s+1)^2+4$:
  \[
    \frac{\frac{3}{26}s + \frac{6}{13}}{(s+1)^2+4}
    = \frac{3}{26}\cdot\frac{s+1}{(s+1)^2+4} + \frac{9}{26}\cdot\frac{1}{(s+1)^2+4}.
  \]
  Inverting: $\dfrac{3}{26}e^{-t}\cos(2t) + \dfrac{9}{52}e^{-t}\sin(2t)$.

  \textbf{Step 5.} Combining all terms:
  \[
    y(t) = e^{-t}\cos(2t) + \frac{1}{2}e^{-t}\sin(2t)
    - \frac{3}{26}\cos(3t) - \frac{1}{13}\sin(3t)
    + \frac{3}{26}e^{-t}\cos(2t) + \frac{9}{52}e^{-t}\sin(2t).
  \]
  Grouping like terms:
  \[
    y(t) = \frac{29}{26}e^{-t}\cos(2t) + \frac{11}{52}e^{-t}\sin(2t)
    - \frac{3}{26}\cos(3t) - \frac{1}{13}\sin(3t).
  \]
\end{workedexample}

\paragraph{Example 3: System of first-order IVPs.}

\begin{workedexample}
  Solve the system
  \[
  \begin{cases}
    x' = 3x - 2y, \quad x(0) = 1, \\
    y' = x + y, \quad y(0) = 0.
  \end{cases}
  \]

  \textbf{Solution.} Let $X(s) = \mathcal{L}\{x(t)\}$ and $Y(s) = \mathcal{L}\{y(t)\}$. Transform both equations:
  \[
  \begin{cases}
    sX - x(0) = 3X - 2Y, \\
    sY - y(0) = X + Y.
  \end{cases}
  \]
  Substitute initial conditions:
  \[
  \begin{cases}
    (s-3)X + 2Y = 1, \\
    -X + (s-1)Y = 0.
  \end{cases}
  \]
  From the second equation: $X = (s-1)Y$. Substitute into the first:
  \[
    (s-3)(s-1)Y + 2Y = 1,
  \]
  \[
    \bigl((s-3)(s-1) + 2\bigr)Y = 1,
  \]
  \[
    (s^2 - 4s + 5)Y = 1 \quad\Longrightarrow\quad Y(s) = \frac{1}{s^2 - 4s + 5}.
  \]
  Complete the square: $s^2 - 4s + 5 = (s-2)^2 + 1$. Thus:
  \[
    Y(s) = \frac{1}{(s-2)^2 + 1}.
  \]
  Inverting: $y(t) = e^{2t}\sin(t)$.

  Now find $X(s) = (s-1)Y(s)$:
  \[
    X(s) = \frac{s-1}{(s-2)^2 + 1} = \frac{s-2+1}{(s-2)^2+1}
    = \frac{s-2}{(s-2)^2+1} + \frac{1}{(s-2)^2+1}.
  \]
  Inverting:
  \[
    x(t) = e^{2t}\cos(t) + e^{2t}\sin(t) = e^{2t}\bigl(\cos(t) + \sin(t)\bigr).
  \]
\end{workedexample}

\subsection{Step and Delta Functions}
\label{sec:ch07_step_and_delta}

\subsubsection{Heaviside Step Function}
\label{sec:ch07_step_function}

The \textbf{Heaviside step function} (or simply \textbf{unit step}) is defined by
\[
  u_c(t) = \begin{cases}
    0, & t < c, \\
    1, & t \ge c,
  \end{cases}
  \qquad c \ge 0.
\]
When $c = 0$, we write $u(t)$ instead of $u_0(t)$.

\begin{keyresult}
  \textbf{Laplace transform of the unit step.}
  \[
    \mathcal{L}\{u_c(t)\} = \frac{e^{-cs}}{s}, \qquad s > 0.
  \]
  This follows directly from the definition:
  \[
    \int_0^\infty e^{-st} u_c(t)\,\diff t = \int_c^\infty e^{-st}\,\diff t = \frac{e^{-cs}}{s}.
  \]
\end{keyresult}

The step function is the building block for modeling \textbf{piecewise continuous} forcing functions. Any piecewise-defined function $g(t)$ can be expressed as a sum of step functions.

\begin{workedexample}
  Express $g(t) = \begin{cases} t, & 0 \le t < 2 \\ 2, & t \ge 2 \end{cases}$ using Heaviside step functions, and find its Laplace transform.

  \textbf{Solution.} Write $g(t)$ as:
  \[
    g(t) = t + (2 - t)\,u_2(t).
  \]
  To see this: for $t < 2$, $u_2(t) = 0$, so $g(t) = t$. For $t \ge 2$, $g(t) = t + 2 - t = 2$.

  Now transform:
  \[
    \mathcal{L}\{g(t)\} = \mathcal{L}\{t\} + \mathcal{L}\{2u_2(t) - t\,u_2(t)\}.
  \]
  We have $\mathcal{L}\{t\} = \dfrac{1}{s^2}$. For the second part, use the second shift theorem:
  \[
    \mathcal{L}\{u_2(t)\cdot 2\} = \mathcal{L}\{u_2(t)\cdot 2\cdot 1\} = 2\cdot\frac{e^{-2s}}{s}.
  \]
  For $\mathcal{L}\{t\,u_2(t)\}$, rewrite as $u_2(t)\cdot t = u_2(t)\cdot((t-2)+2)$:
  \[
    \mathcal{L}\{u_2(t)\cdot(t-2)\} = e^{-2s}\cdot\frac{1}{s^2}, \quad
    \mathcal{L}\{u_2(t)\cdot 2\} = 2\cdot\frac{e^{-2s}}{s}.
  \]
  Therefore:
  \[
    \mathcal{L}\{t\,u_2(t)\} = \frac{e^{-2s}}{s^2} + \frac{2e^{-2s}}{s}.
  \]
  Putting it all together:
  \[
    \mathcal{L}\{g(t)\} = \frac{1}{s^2} + \frac{2e^{-2s}}{s} - \left(\frac{e^{-2s}}{s^2} + \frac{2e^{-2s}}{s}\right)
    = \frac{1 - e^{-2s}}{s^2}.
  \]
\end{workedexample}

\subsubsection{Dirac Delta Function}
\label{sec:ch07_delta_function}

The \textbf{Dirac delta function} $\delta(t-a)$ is a generalized function (distribution) characterized by its \textbf{sifting property}:
\[
  \int_{-\infty}^\infty \delta(t-a)\,f(t)\,\diff t = f(a).
\]
Intuitively, $\delta(t-a)$ models an instantaneous impulse of unit magnitude occurring at time $t = a$.

\begin{keyresult}
  \textbf{Laplace transform of the delta function.} For $a \ge 0$:
  \[
    \mathcal{L}\{\delta(t-a)\} = e^{-as}.
  \]
\end{keyresult}

\paragraph{Relationship to the step function.} The delta function can be viewed as the ``derivative'' of the step function:
\[
  \frac{\diff}{\diff t} u_c(t) = \delta(t-c).
\]
Consistently, differentiating $\mathcal{L}\{u_c(t)\} = e^{-cs}/s$ with respect to $t$ should yield the transform of $\delta(t-c)$. Since $\mathcal{L}\{u_c'\} = s\,\mathcal{L}\{u_c\} - u_c(0) = s\cdot\dfrac{e^{-cs}}{s} - 0 = e^{-cs}$, we recover the correct result.

\begin{workedexample}
  Solve $y'' + 4y = \delta(t-\pi)$, $y(0) = 0$, $y'(0) = 0$.

  \textbf{Solution.} Transform both sides:
  \[
    (s^2 Y(s) - s\cdot 0 - 0) + 4Y(s) = e^{-\pi s}.
  \]
  Solving for $Y(s)$:
  \[
    Y(s) = \frac{e^{-\pi s}}{s^2 + 4} = \frac{1}{2}\,e^{-\pi s}\cdot\frac{2}{s^2+4}.
  \]
  Invert using the second shift theorem. Since $\mathcal{L}^{-1}\left\{\dfrac{2}{s^2+4}\right\} = \sin(2t)$:
  \[
    y(t) = \frac{1}{2}\,u_\pi(t)\,\sin\bigl(2(t-\pi)\bigr).
  \]
  Note that $\sin(2(t-\pi)) = \sin(2t - 2\pi) = \sin(2t)$. Therefore:
  \[
    y(t) = \begin{cases}
      0, & 0 \le t < \pi, \\[4pt]
      \dfrac{1}{2}\sin(2t), & t \ge \pi.
    \end{cases}
  \]
  The system remains at rest until the impulse at $t = \pi$, after which it begins oscillating.
\end{workedexample}

\begin{workedexample}
  Find the impulse response of the system $y'' + 3y' + 2y = \delta(t)$, with $y(0) = 0$ and $y'(0) = 0$.

  \textbf{Solution.} Transform:
  \[
    s^2 Y(s) + 3s Y(s) + 2Y(s) = 1,
  \]
  \[
    Y(s) = \frac{1}{s^2 + 3s + 2} = \frac{1}{(s+1)(s+2)}.
  \]
  Partial fractions:
  \[
    \frac{1}{(s+1)(s+2)} = \frac{1}{s+1} - \frac{1}{s+2}.
  \]
  Inverting:
  \[
    y(t) = e^{-t} - e^{-2t}.
  \]
  This function, often denoted $h(t)$, is the \textbf{impulse response}: the system's output when driven by a unit impulse at $t=0$. For any forcing function $f(t)$, the solution is given by the convolution $y(t) = (h*f)(t)$, as discussed in the next section.
\end{workedexample}

\subsection{Convolution}
\label{sec:ch07_convolution}

\begin{definition}
  \label{def:convolution}
  The \textbf{convolution} of two functions $f(t)$ and $g(t)$ is defined by
  \begin{equation}
    \label{eq:convolution_definition}
    (f * g)(t) = \int_0^t f(\tau)\,g(t-\tau)\,\diff\tau.
  \end{equation}
  Convolution is commutative: $f*g = g*f$.
\end{definition}

The fundamental theorem connecting convolution and the Laplace transform is the following.

\begin{theorem}[Convolution Theorem]
  \label{thm:convolution_theorem}
  If $\mathcal{L}\{f(t)\} = F(s)$ and $\mathcal{L}\{g(t)\} = G(s)$, then
  \[
    \mathcal{L}\{(f*g)(t)\} = F(s)\,G(s).
  \]
  Equivalently,
  \[
    \mathcal{L}^{-1}\{F(s)\,G(s)\} = (f*g)(t).
  \]
\end{theorem}

\paragraph{Proof sketch.} Consider $\mathcal{L}\{(f*g)(t)\}$:
\[
  \int_0^\infty e^{-st} \left(\int_0^t f(\tau)\,g(t-\tau)\,\diff\tau\right)\diff t.
\]
The region of integration is $0 \le \tau \le t < \infty$. Changing the order of integration (Fubini's theorem) and substituting $u = t - \tau$:
\[
  \int_0^\infty f(\tau) \left(\int_\tau^\infty e^{-st} g(t-\tau)\,\diff t\right)\diff\tau
  = \int_0^\infty f(\tau) e^{-s\tau} \left(\int_0^\infty e^{-su} g(u)\,\diff u\right)\diff\tau.
\]
The inner integral is $G(s)$, and the remaining integral is $F(s)$. Thus the result is $F(s)G(s)$.

\begin{workedexample}
  Use convolution to find $\mathcal{L}^{-1}\left\{\dfrac{1}{s^2(s^2+1)}\right\}$.

  \textbf{Solution.} Write $F(s)G(s)$ with $F(s) = \dfrac{1}{s^2}$ and $G(s) = \dfrac{1}{s^2+1}$. Then $f(t) = t$ and $g(t) = \sin(t)$. By the convolution theorem:
  \[
    \mathcal{L}^{-1}\left\{\frac{1}{s^2(s^2+1)}\right\} = (t * \sin t)(t) = \int_0^t \tau\,\sin(t-\tau)\,\diff\tau.
  \]
  Integrate by parts with $u = \tau$, $\diff v = \sin(t-\tau)\,\diff\tau$:
  \[
    \begin{aligned}
      \int_0^t \tau\,\sin(t-\tau)\,\diff\tau
      &= \Bigl[\tau\cos(t-\tau)\Bigr]_0^t - \int_0^t \cos(t-\tau)\,\diff\tau \\
      &= \bigl(t\cos(0) - 0\bigr) - \Bigl[-\sin(t-\tau)\Bigr]_0^t \\
      &= t - \bigl(-\sin(0) + \sin(t)\bigr) \\
      &= t - \sin(t).
    \end{aligned}
  \]
  Therefore:
  \[
    \mathcal{L}^{-1}\left\{\frac{1}{s^2(s^2+1)}\right\} = t - \sin(t).
  \]
\end{workedexample}

\begin{workedexample}
  Use convolution to solve $y'' + y = f(t)$, with $y(0) = 0$ and $y'(0) = 0$, where $f(t)$ is an arbitrary forcing function.

  \textbf{Solution.} Transform:
  \[
    s^2 Y(s) + Y(s) = F(s) \quad\Longrightarrow\quad Y(s) = \frac{F(s)}{s^2 + 1}
    = F(s) \cdot \frac{1}{s^2 + 1}.
  \]
  Since $\mathcal{L}^{-1}\left\{\dfrac{1}{s^2+1}\right\} = \sin(t)$, the convolution theorem gives:
  \[
    y(t) = (\sin * f)(t) = \int_0^t \sin(\tau)\,f(t-\tau)\,\diff\tau.
  \]
  This formula provides the solution for \emph{any} $f(t)$ without needing to perform partial fractions or guess particular solutions. The function $h(t) = \sin(t)$ is the impulse response of this system.
\end{workedexample}

\subsection{Transfer Functions}
\label{sec:ch07_transfer_functions}

The \textbf{transfer function} provides a compact description of a linear system's input-output behavior in the frequency domain.

\begin{definition}
  Consider a linear differential equation with constant coefficients:
  \[
    a_n y^{(n)} + a_{n-1} y^{(n-1)} + \cdots + a_1 y' + a_0 y = f(t),
  \]
  with all initial conditions equal to zero. The \textbf{transfer function} is the ratio
  \[
    H(s) = \frac{Y(s)}{F(s)},
  \]
  where $Y(s) = \mathcal{L}\{y(t)\}$ and $F(s) = \mathcal{L}\{f(t)\}$.
\end{definition}

\begin{keyresult}
  \textbf{Transfer function from ODE.} Taking the Laplace transform of the ODE with zero initial conditions:
  \[
    (a_n s^n + a_{n-1} s^{n-1} + \cdots + a_1 s + a_0) Y(s) = F(s),
  \]
  so the transfer function is
  \[
    H(s) = \frac{1}{a_n s^n + a_{n-1} s^{n-1} + \cdots + a_1 s + a_0}.
  \]
  The output in the $s$-domain is simply $Y(s) = H(s) F(s)$, and in the time domain $y(t) = (h * f)(t)$ where $h(t) = \mathcal{L}^{-1}\{H(s)\}$ is the impulse response.
\end{keyresult}

\subsubsection{Pole Analysis and Stability}
\label{sec:ch07_pole_analysis}

The \textbf{poles} of $H(s)$ are the roots of the denominator polynomial. They determine the long-term behavior (stability) of the system:

\begin{itemize}
  \item All poles have negative real parts (left half-plane, LHP): the system is \textbf{stable} and the output decays to zero for bounded inputs.
  \item Any pole has a positive real part (right half-plane, RHP): the system is \textbf{unstable} and the output grows without bound.
  \item Poles on the imaginary axis (purely imaginary, no right-half-plane poles): the system is \textbf{marginally stable}; the output oscillates persistently but does not grow.
\end{itemize}

\begin{workedexample}
  Find the transfer function and analyze the stability of the RLC circuit modeled by
  \[
    L\,q'' + R\,q' + \frac{1}{C} q = E(t),
  \]
  where $q(t)$ is the charge on the capacitor, $L$ is the inductance, $R$ is the resistance, $C$ is the capacitance, and $E(t)$ is the applied voltage.

  \textbf{Solution.} Taking the Laplace transform with zero initial conditions:
  \[
    (L s^2 + R s + \tfrac{1}{C}) Q(s) = E(s),
  \]
  so the transfer function is
  \[
    H(s) = \frac{Q(s)}{E(s)} = \frac{1}{L s^2 + R s + \tfrac{1}{C}}.
  \]
  The poles are the roots of $L s^2 + R s + \dfrac{1}{C} = 0$:
  \[
    s = \frac{-R \pm \sqrt{R^2 - \tfrac{4L}{C}}}{2L}.
  \]
  Since $L, R, C > 0$, we have $-R < 0$. If $R^2 \ge \dfrac{4L}{C}$ (overdamped or critically damped), both roots are real and negative. If $R^2 < \dfrac{4L}{C}$ (underdamped), the roots are complex conjugates with real part $-\dfrac{R}{2L} < 0$. In all cases, the real parts are negative, so the system is stable. The physical intuition is that the resistor dissipates energy, ensuring that oscillations decay.
\end{workedexample}

\begin{workedexample}
  Consider the spring-mass-damper system
  \[
    m\,x'' + c\,x' + k\,x = F(t),
  \]
  where $m$ is mass, $c$ is damping, and $k$ is spring stiffness. Find the transfer function $H(s) = X(s)/F(s)$ and analyze the stability.

  \textbf{Solution.} With zero initial conditions:
  \[
    (ms^2 + cs + k)X(s) = F(s),
  \]
  so
  \[
    H(s) = \frac{1}{m s^2 + c s + k}.
  \]
  The poles are
  \[
    s = \frac{-c \pm \sqrt{c^2 - 4mk}}{2m}.
  \]
  Since $m, c, k > 0$, the real parts of the poles are always negative (the same analysis as the RLC circuit above). The system is therefore stable. If $c = 0$ (no damping), the poles lie on the imaginary axis: $s = \pm i\sqrt{k/m}$, giving undamped oscillations (marginal stability).
\end{workedexample}

\subsubsection{Initial and Final Value Theorems}
\label{sec:ch07_value_theorems}

Two important theorems allow us to extract information about the time-domain behavior of $f(t)$ directly from $F(s)$ without computing the full inverse transform.

\begin{theorem}[Initial Value Theorem]
  \label{thm:initial_value}
  If $\mathcal{L}\{f(t)\} = F(s)$ and the limit exists, then
  \[
    \lim_{t \to 0^+} f(t) = \lim_{s \to \infty} s F(s).
  \]
\end{theorem}

\begin{theorem}[Final Value Theorem]
  \label{thm:final_value}
  If $\mathcal{L}\{f(t)\} = F(s)$ and all poles of $sF(s)$ lie in the open left half-plane, then
  \[
    \lim_{t \to \infty} f(t) = \lim_{s \to 0} s F(s).
  \]
\end{theorem}

\begin{hintbox}
  \textbf{Caveat for the final value theorem.} The condition that all poles of $sF(s)$ lie in the open LHP is essential. If $F(s)$ has poles on the imaginary axis (e.g., $\dfrac{1}{s^2+1}$ corresponding to $\sin t$), the limit $\lim_{t\to\infty} f(t)$ does not exist, and the final value theorem gives a misleading result. Always check pole locations before applying.
\end{hintbox}

\begin{workedexample}
  Verify the initial and final value theorems for $y(t) = \dfrac{1}{2}e^{-2t} + \dfrac{5}{2}e^{-4t}$ (the solution to the first-order IVP in \cref{sec:ch07_solving_ivps}).

  \textbf{Solution.} From the solution, $Y(s) = \dfrac{3s+7}{(s+2)(s+4)}$.

  \textbf{Initial value.} By the theorem:
  \[
    \lim_{s \to \infty} s Y(s) = \lim_{s \to \infty} \frac{s(3s+7)}{(s+2)(s+4)}
    = \lim_{s \to \infty} \frac{3s^2 + 7s}{s^2 + 6s + 8}
    = 3.
  \]
  Directly: $y(0) = \dfrac{1}{2} + \dfrac{5}{2} = 3$. $\checkmark$

  \textbf{Final value.} All poles are at $s=-2$ and $s=-4$ (both in the LHP), so the theorem applies:
  \[
    \lim_{s \to 0} s Y(s) = \lim_{s \to 0} \frac{s(3s+7)}{(s+2)(s+4)} = \frac{0\cdot 7}{8} = 0.
  \]
  Directly: $\lim_{t \to \infty} y(t) = 0$ since both exponentials decay. $\checkmark$
\end{workedexample}

\subsection{Summary}
\label{sec:ch07_summary}

The Laplace transform is a powerful method for solving linear differential equations. The key ideas are:

\begin{itemize}
  \item The transform $\mathcal{L}\{f(t)\} = F(s)$ converts functions of time into functions of frequency, turning differentiation into multiplication.
  \item A comprehensive transform table allows us to quickly look up common transforms (\cref{sec:ch07_transform_table}).
  \item Linearity, shift theorems, and derivative transforms form the algebraic backbone of the method (\cref{sec:ch07_properties}).
  \item Inverse transforms are obtained via table lookup and partial fraction decomposition (\cref{sec:ch07_inverse_transforms}, \cref{sec:ch07_partial_fractions}).
  \item The method naturally handles initial conditions, making it ideal for IVPs (\cref{sec:ch07_solving_ivps}).
  \item Step and delta functions allow modeling of discontinuous and impulsive inputs (\cref{sec:ch07_step_and_delta}).
  \item The convolution theorem connects multiplication in the $s$-domain with convolution in the time domain (\cref{sec:ch07_convolution}).
  \item Transfer functions provide a compact input-output description and enable stability analysis via pole locations (\cref{sec:ch07_transfer_functions}).
\end{itemize}

\begin{table}[htbp]
\centering
\caption{Key Laplace transforms and properties}
\label{tab:ch07_summary}
\begin{tabular}{l l}
\toprule
\textbf{Concept} & \textbf{Key Formula/Method} \\
\midrule
Definition & $\mathcal{L}\{f(t)\} = \int_0^\infty e^{-st} f(t)\,\diff t$ \\
Linearity & $\mathcal{L}\{af+bg\} = aF(s) + bG(s)$ \\
First shift & $\mathcal{L}\{e^{at}f(t)\} = F(s-a)$ \\
Second shift & $\mathcal{L}\{u_c(t)f(t-c)\} = e^{-cs}F(s)$ \\
First derivative & $\mathcal{L}\{f'\} = sF(s) - f(0)$ \\
Second derivative & $\mathcal{L}\{f''\} = s^2F(s) - sf(0) - f'(0)$ \\
Integral & $\mathcal{L}\left\{\int_0^t f(\tau)\,\diff\tau\right\} = F(s)/s$ \\
Step function & $\mathcal{L}\{u_c(t)\} = e^{-cs}/s$ \\
Delta function & $\mathcal{L}\{\delta(t-a)\} = e^{-as}$ \\
Convolution theorem & $\mathcal{L}\{f*g\} = F(s)G(s)$ \\
Transfer function & $H(s) = Y(s)/F(s)$ (zero ICs) \\
Initial value theorem & $\lim_{t\to 0^+}f(t) = \lim_{s\to\infty} sF(s)$ \\
Final value theorem & $\lim_{t\to\infty}f(t) = \lim_{s\to 0} sF(s)$ \\
\bottomrule
\end{tabular}
\end{table}

The Laplace transform method will be used throughout the remainder of this handbook, particularly in the analysis of systems of equations (\cref{ch:systems}) and in the construction of Fourier series (\cref{ch:fourier_series}).