MGU Mathematics Series — Differential Equations
Part IV: Systems, Transforms, and Numerical Methods
Laplace Transform · $s$-Domain · Step Functions · IVPs · Circuits
Chapter 10 of 14
MGU Mathematics Series  ·  Differential Equations  ·  Library Textbook Edition

Chapter 10:
Laplace
Transforms

Converting differential equations into algebra — the transform idea, inverse transforms, step functions, discontinuous inputs, impulses, and circuit applications

■ Part IV: Systems, Transforms, and Numerical Methods ■

■ 14 Sections  •  Chapter 10 of 14 ■

Follows Chapter 9: Systems of Differential Equations  ·  Prepares students for Chapter 11: Numerical Methods

Chapter 10 — Front Matter

Purpose and Overview

Every method for solving differential equations studied so far has worked directly in the time domain. We identified the type of equation, applied a technique, integrated, determined constants, and read the result as a function of the independent variable. These methods are powerful, but they have limitations. When a differential equation involves discontinuous inputs, switching forces, impulse signals, or complicated initial conditions, direct methods become awkward and sometimes fail to produce clean solutions.

Chapter 10 introduces the Laplace transform, a different kind of tool. Rather than solving a differential equation directly, the Laplace transform converts the differential equation into an algebraic equation in a new variable. Algebra is then used to solve the transformed equation. An inverse transform converts the algebraic result back into the function of the original variable. The differential equation has been solved without integration in the usual sense, and without the repeated structure of undetermined coefficients or variation of parameters.

This is the essential idea behind transform methods: change the problem into a domain where it is easier to solve, solve it there, then translate the answer back. The Laplace transform is the most important example of this strategy in applied mathematics. It connects differential equations to algebra, complex analysis, control systems engineering, circuit analysis, signal processing, and mathematical physics.

This chapter builds the Laplace transform from the definition, develops a working table of transform pairs, introduces inverse transforms and partial fractions, solves initial value problems using the transform method, introduces the Heaviside step function and discontinuous inputs, previews impulse functions, and shows how circuits and control systems use transforms naturally.

Opening Question

A switch in an electrical circuit closes at time $t = 1$ second, applying a voltage that was not present before. You want to find the current in the circuit for all $t \geq 0$. The governing equation is a second-order linear differential equation, but the forcing function is not a smooth function of time — it is zero before $t = 1$ and nonzero after. The undetermined coefficients method cannot handle this kind of input cleanly. Can you still solve the problem systematically? The answer is yes, and the Laplace transform is how.

14Sections
13Objectives
15Key Terms
14+Worked Examples

Learning Objectives

  1. 1Explain the idea of a transform method and why it is useful for differential equations.
  2. 2Compute Laplace transforms from the definition for simple functions.
  3. 3Use a table of Laplace transforms to find transforms and inverse transforms of common functions.
  4. 4Apply linearity and shift properties of the Laplace transform.
  5. 5Find inverse Laplace transforms using partial fraction decomposition.
  6. 6Solve initial value problems using the Laplace transform method.
  7. 7Define and use the Heaviside step function.
  8. 8Express piecewise-defined forcing functions using step functions.
  9. 9Take the Laplace transform of equations with discontinuous inputs.
  10. 10Recognize the delta function as an idealized impulse and state its Laplace transform.
  11. 11Apply the Laplace transform to RC and RLC circuit models as an introduction.
  12. 12Interpret the transfer function concept in control systems.
  13. 13Identify and avoid common errors in Laplace transform work.
Chapter 10 — Contents

Sections at a Glance

Sections 10.1 – 10.5
The Transform Idea: Definition, Table, and Linearity
Why Transforms · The Idea · Definition · Basic Pairs · Linearity and Shifting
Section 10.1

Why Transform Methods Are Useful

In calculus, you encountered a useful idea: some problems become simpler when stated in different terms. Logarithms convert multiplication into addition. Completing the square converts a quadratic expression into a shifted square. A substitution can convert a complicated integral into a simpler one. The Laplace transform applies this same logic to differential equations.

Differentiation, which is the operation that makes differential equations hard to solve by algebra alone, is converted by the Laplace transform into multiplication by the transform variable $s$. A first derivative $y'(t)$ becomes $sY(s)$ minus an initial value. A second derivative $y''(t)$ becomes $s^2Y(s)$ minus combinations of initial values. This means a differential equation involving $y''(t)$, $y'(t)$, and $y(t)$ with initial conditions, once transformed, becomes an algebraic equation involving $Y(s)$. You solve the algebraic equation for $Y(s)$, then find the inverse transform to recover $y(t)$.

Three Situations Where the Laplace Transform Pays Off Most

Discontinuous forcing. Functions that switch on or off, like "voltage is zero for $t < 1$ then becomes 5 volts," do not fit into undetermined coefficients but fit naturally into the Laplace framework using the Heaviside step function.

Initial conditions at $t = 0$. The Laplace transform absorbs initial conditions directly into the transformed equation, so you never need to apply them separately after solving.

Standard engineering inputs. Electrical engineers and control systems engineers use the Laplace transform as their primary working language, because the transformed equation describes system behavior across all possible inputs at once.

Important Note on When to Use This Method

The Laplace transform is not always the fastest method. For a simple equation like $y' + 2y = 0$, $y(0) = 3$, the integrating factor method is faster. The Laplace transform pays off most clearly when the forcing function is discontinuous, when the system is complex, or when you want to analyze system behavior generally. Use the right tool for the problem.

Section 10.2

The Laplace Transform Idea

Before presenting the definition, it helps to understand the idea at a qualitative level. A function $f(t)$ defined for $t \geq 0$ contains information about how some quantity behaves over time. The Laplace transform takes that function and produces a new function $F(s)$, where $s$ is a new variable. The function $F(s)$ encodes the same information as $f(t)$, but in a form that turns differentiation into algebra.

Think of it as a dictionary. In one language, the differential equation says something complicated. In the other language, it says something algebraic. The transform is the dictionary. The inverse transform translates the answer back.

The new variable $s$ is often complex in theoretical work, but for most applied problems in this course, you can treat $s$ as a real positive variable and focus on the mechanics and the table of pairs. The function $F(s)$ is computed by multiplying $f(t)$ by the decaying exponential $e^{-st}$ and integrating over all $t \geq 0$. Choosing a large value of $s$ means the exponential decays fast, so $F(s)$ captures mainly what $f(t)$ does near $t = 0$. Choosing a smaller value means $F(s)$ captures behavior further into the future.

The resulting function $F(s)$ belongs to what is called the $s$-domain or the transform domain. The original $f(t)$ belongs to the time domain. The Laplace transform moves us from time to $s$, and the inverse transform moves us back.

Section 10.3

Definition of the Laplace Transform

Definition — The Laplace Transform

Let $f(t)$ be a function defined for $t \geq 0$. The Laplace transform of $f(t)$ is the function $F(s)$ defined by

$$\mathcal{L}\{f(t)\} = F(s) = \int_0^{\infty} e^{-st} f(t) \, dt$$

provided the integral converges. The notation $\mathcal{L}\{f(t)\}$ means "the Laplace transform of $f(t)$." The result is always written as $F(s)$, using a capital letter to match the lowercase $f$. This convention continues throughout: $\mathcal{L}\{y(t)\} = Y(s)$, $\mathcal{L}\{g(t)\} = G(s)$, and so on.

The integral $\int_0^{\infty} e^{-st} f(t) \, dt$ is an improper integral, interpreted as $\lim_{T \to \infty} \int_0^{T} e^{-st} f(t) \, dt$. For the transform to exist, the function $f(t)$ must not grow faster than some exponential as $t \to \infty$. Most functions encountered in differential equations, including polynomials, exponentials, sines, cosines, and products of these, satisfy this condition.

How You Will Actually Use This

You will compute a few transforms from the definition to build understanding. In practice, you will use a table of Laplace transform pairs. Memorizing the table is not required. Understanding how to use it and how to apply linearity, shifting, and partial fractions is what matters.

Example 10.3.1
The Laplace Transform of a Constant — $\mathcal{L}\{1\}$
  • ProblemFind $\mathcal{L}\{1\}$ directly from the definition.
  • ThinkApply the definition with $f(t) = 1$. The integral is a standard improper integral of a decaying exponential.
  • Solve $$F(s) = \int_0^{\infty} e^{-st} \cdot 1 \, dt = \lim_{T \to \infty} \left[ \frac{e^{-st}}{-s} \right]_0^{T} = \lim_{T \to \infty} \left( \frac{e^{-sT}}{-s} - \frac{1}{-s} \right)$$ For $s > 0$, we have $e^{-sT} \to 0$ as $T \to \infty$, so $F(s) = 0 + \frac{1}{s} = \frac{1}{s}$.
  • CheckThe integral converges for $s > 0$, which is the required condition. ✓
$$\mathcal{L}\{1\} = \frac{1}{s}, \quad s > 0$$
Example 10.3.2
The Laplace Transform of an Exponential — $\mathcal{L}\{e^{at}\}$
  • ProblemFind $\mathcal{L}\{e^{at}\}$ where $a$ is a constant.
  • ThinkApply the definition and combine the exponentials: $e^{-st} \cdot e^{at} = e^{-(s-a)t}$. This is the same integral as before, shifted by $a$.
  • Solve $$F(s) = \int_0^{\infty} e^{-st} e^{at} \, dt = \int_0^{\infty} e^{-(s-a)t} \, dt = \left[ \frac{e^{-(s-a)t}}{-(s-a)} \right]_0^{\infty}$$ For convergence we need $s > a$, so $e^{-(s-a)T} \to 0$. $$= 0 - \frac{1}{-(s-a)} = \frac{1}{s-a}$$
$$\mathcal{L}\{e^{at}\} = \frac{1}{s - a}, \quad s > a$$
Example 10.3.3
The Laplace Transform of $\sin(\omega t)$ and $\cos(\omega t)$
  • ProblemFind $\mathcal{L}\{\sin(\omega t)\}$ and $\mathcal{L}\{\cos(\omega t)\}$.
  • ThinkApply the definition. Integration by parts twice returns to a multiple of the original integral, which can be solved algebraically. The results are standard and appear in every transform table.
  • ResultThe integration by parts procedure gives the standard results:
$$\mathcal{L}\{\sin(\omega t)\} = \frac{\omega}{s^2 + \omega^2}, \qquad \mathcal{L}\{\cos(\omega t)\} = \frac{s}{s^2 + \omega^2}, \quad s > 0$$
Example 10.3.4
The Laplace Transform of $t^n$
  • ProblemFind $\mathcal{L}\{t^n\}$ for a nonnegative integer $n$.
  • ThinkThe integral $\int_0^{\infty} e^{-st} t^n \, dt$ is evaluated by repeated integration by parts or by the substitution $u = st$. The general result uses the factorial.
  • ResultFor $n = 0$: $\mathcal{L}\{1\} = \frac{1}{s}$. For $n = 1$: $\mathcal{L}\{t\} = \frac{1}{s^2}$. For $n = 2$: $\mathcal{L}\{t^2\} = \frac{2}{s^3}$.
$$\mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}}, \quad s > 0$$
Section 10.4

Basic Transform Pairs

Rather than computing every transform from scratch, the following table collects the most important Laplace transform pairs. These are your primary reference throughout the chapter. A more complete version appears in Appendix M.

$f(t)$ $\mathcal{L}\{f(t)\} = F(s)$ Condition
$1$$\dfrac{1}{s}$$s > 0$
$t^n$ ($n$ a nonneg integer)$\dfrac{n!}{s^{n+1}}$$s > 0$
$e^{at}$$\dfrac{1}{s-a}$$s > a$
$\sin(\omega t)$$\dfrac{\omega}{s^2 + \omega^2}$$s > 0$
$\cos(\omega t)$$\dfrac{s}{s^2 + \omega^2}$$s > 0$
$e^{at}\sin(\omega t)$$\dfrac{\omega}{(s-a)^2 + \omega^2}$$s > a$
$e^{at}\cos(\omega t)$$\dfrac{s-a}{(s-a)^2 + \omega^2}$$s > a$
$t^n e^{at}$$\dfrac{n!}{(s-a)^{n+1}}$$s > a$
$u_c(t)$ (unit step at $t=c$)$\dfrac{e^{-cs}}{s}$$s > 0$
$u_c(t)\, f(t-c)$$e^{-cs} F(s)$second shift theorem
$\delta(t-c)$ (impulse at $t=c$)$e^{-cs}$—
$f'(t)$$sF(s) - f(0)$transform of derivative
$f''(t)$$s^2 F(s) - sf(0) - f'(0)$transform of second derivative
The Core Theorem — Transforms of Derivatives

If $\mathcal{L}\{f(t)\} = F(s)$, then differentiation in the time domain becomes multiplication in the $s$-domain:

$$\mathcal{L}\{f'(t)\} = sF(s) - f(0)$$

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

These formulas follow from integration by parts. The boundary term contributes $-f(0)$ for the first derivative and $-sf(0) - f'(0)$ for the second. This is the precise mechanism by which a differential equation becomes an algebraic equation. Initial conditions are automatically included and need not be applied separately afterward.

Section 10.5

Linearity and the First Shift Theorem

Theorem — Linearity of the Laplace Transform

If $f$ and $g$ have Laplace transforms $F(s)$ and $G(s)$, and if $a$ and $b$ are constants, then

$$\mathcal{L}\{af(t) + bg(t)\} = aF(s) + bG(s)$$

This property allows us to take the transform of any linear combination by transforming each term separately, which is exactly how we handle the terms of a differential equation.

Example 10.5.1
Using Linearity
  • ProblemFind $\mathcal{L}\{3e^{2t} - 5\cos(4t) + 2t^3\}$.
  • ThinkApply the transform to each term using linearity and the table. Handle each piece independently.
  • Solve $$\mathcal{L}\{3e^{2t}\} = \frac{3}{s-2}, \qquad \mathcal{L}\{5\cos(4t)\} = \frac{5s}{s^2 + 16}, \qquad \mathcal{L}\{2t^3\} = 2 \cdot \frac{3!}{s^4} = \frac{12}{s^4}$$
$$F(s) = \frac{3}{s-2} - \frac{5s}{s^2 + 16} + \frac{12}{s^4}$$
Theorem — First Shift Theorem ($s$-Shift)

If $\mathcal{L}\{f(t)\} = F(s)$, then multiplying by $e^{at}$ in the time domain shifts the transform by $a$ in the $s$-domain:

$$\mathcal{L}\{e^{at} f(t)\} = F(s - a)$$

This explains the entries in the table for $e^{at}\sin(\omega t)$ and $e^{at}\cos(\omega t)$: their transforms are the standard sine and cosine transforms with $s$ replaced by $s - a$.

Example 10.5.2
Using the First Shift Theorem
  • ProblemFind $\mathcal{L}\{e^{-3t}\cos(2t)\}$.
  • ThinkThis is $e^{at}\cos(\omega t)$ with $a = -3$ and $\omega = 2$. Apply the first shift theorem: replace $s$ with $s - (-3) = s + 3$.
  • Solve$$\mathcal{L}\{e^{-3t}\cos(2t)\} = \frac{s - (-3)}{(s-(-3))^2 + 2^2} = \frac{s+3}{(s+3)^2 + 4}$$
$$\mathcal{L}\{e^{-3t}\cos(2t)\} = \frac{s+3}{(s+3)^2 + 4}$$
Sections 10.6 – 10.8
Inverse Transforms and Solving Initial Value Problems
Inverse Transforms · Partial Fractions · The Five-Step IVP Method
Section 10.6

Inverse Laplace Transforms

Solving a differential equation using the Laplace transform produces a function $Y(s)$ in the $s$-domain. The final step is to recover $y(t)$ from $Y(s)$. This is the inverse Laplace transform, written $\mathcal{L}^{-1}\{Y(s)\} = y(t)$.

The inverse transform is unique for continuous functions: if $F(s) = G(s)$ for all $s$ in a half-plane, then $f(t) = g(t)$ for all $t \geq 0$. In practice, the inverse transform is computed not by a formula but by recognizing patterns in $F(s)$ that match known transform pairs. The main tool is partial fraction decomposition combined with the transform table.

Approach to Inverse Transforms

Step 1: If $F(s)$ matches a table entry directly, read off $f(t)$.

Step 2: If $F(s)$ is a sum, use linearity: $\mathcal{L}^{-1}\{aF(s) + bG(s)\} = af(t) + bg(t)$.

Step 3: If $F(s)$ is a rational function, apply partial fraction decomposition to write it as a sum of simpler terms, then invert each term using the table.

Example 10.6.1
Direct Inverse Transform
  • ProblemFind $\mathcal{L}^{-1}\!\left\{ \dfrac{3}{s+4} \right\}$.
  • ThinkThe form $\frac{1}{s-a}$ corresponds to $e^{at}$. Here $s + 4 = s - (-4)$, so $a = -4$.
  • Solve$$\mathcal{L}^{-1}\!\left\{ \frac{3}{s+4} \right\} = 3 \cdot e^{-4t}$$
$y(t) = 3e^{-4t}$
Example 10.6.2
Inverse Transform Using the Shift Theorem
  • ProblemFind $\mathcal{L}^{-1}\!\left\{ \dfrac{5}{(s+2)^2 + 9} \right\}$.
  • ThinkThis resembles $\frac{\omega}{(s-a)^2 + \omega^2}$ with $a = -2$, $\omega = 3$. Write the numerator as a multiple of $\omega = 3$.
  • Solve $$\frac{5}{(s+2)^2 + 9} = \frac{5}{3} \cdot \frac{3}{(s+2)^2 + 9}$$ By the first shift theorem with $a = -2$, $\omega = 3$, this inverts to $\frac{5}{3} e^{-2t}\sin(3t)$.
$y(t) = \dfrac{5}{3} e^{-2t}\sin(3t)$
Section 10.7

Partial Fractions Review

When $F(s)$ is a rational function (ratio of polynomials), partial fraction decomposition breaks it into terms that can be found in the transform table. We assume the degree of the numerator is less than the degree of the denominator; if not, perform polynomial long division first.

Factor Type in $Q(s)$Partial Fraction FormInverts To
Distinct linear: $(s - r)$$\dfrac{A}{s - r}$$A e^{rt}$
Repeated linear: $(s - r)^k$$\dfrac{A_1}{s-r} + \dfrac{A_2}{(s-r)^2} + \cdots + \dfrac{A_k}{(s-r)^k}$$t^{n}e^{rt}$ terms
Irreducible quadratic: $(s-a)^2 + \omega^2$$\dfrac{A(s-a) + B\omega}{(s-a)^2 + \omega^2}$$e^{at}[\cos + \sin]$ terms
Example 10.7.1
Partial Fractions with Distinct Linear Factors
  • ProblemFind $\mathcal{L}^{-1}\!\left\{ \dfrac{5s + 3}{(s-1)(s+2)} \right\}$.
  • ThinkTwo distinct linear factors. Set up $\frac{A}{s-1} + \frac{B}{s+2}$ and find constants by cover-up method.
  • Solve Multiply both sides by $(s-1)(s+2)$: $5s + 3 = A(s+2) + B(s-1)$.
    Set $s = 1$: $8 = 3A$, so $A = \frac{8}{3}$.
    Set $s = -2$: $-7 = -3B$, so $B = \frac{7}{3}$. $$f(t) = \frac{8}{3} e^{t} + \frac{7}{3} e^{-2t}$$
  • CheckSubstitute $s = 0$ into the original: $\frac{3}{(-1)(2)} = -\frac{3}{2}$. From partial fractions: $-\frac{8}{3} + \frac{7}{6} = -\frac{16}{6} + \frac{7}{6} = -\frac{9}{6} = -\frac{3}{2}$. ✓
$f(t) = \dfrac{8}{3} e^{t} + \dfrac{7}{3} e^{-2t}$
Example 10.7.2
Partial Fractions with an Irreducible Quadratic
  • ProblemFind $\mathcal{L}^{-1}\!\left\{ \dfrac{2s + 7}{s^2 + 4s + 13} \right\}$.
  • ThinkDiscriminant: $16 - 52 < 0$, so the quadratic is irreducible. Complete the square, then rewrite the numerator in terms of $(s+2)$.
  • Solve Complete the square: $s^2 + 4s + 13 = (s+2)^2 + 9$.
    Rewrite numerator: $2s + 7 = 2(s+2) + 3$. $$\frac{2(s+2)}{(s+2)^2 + 9} + \frac{3}{(s+2)^2 + 9}$$ First term inverts to $2e^{-2t}\cos(3t)$. Second: $\frac{3}{9} \cdot \frac{3}{(s+2)^2+9}$ inverts to $e^{-2t}\sin(3t)$.
$f(t) = 2e^{-2t}\cos(3t) + e^{-2t}\sin(3t)$
Section 10.8

Solving Initial Value Problems

The Five-Step IVP Method Using Laplace Transforms

Step 1: Take the Laplace transform of both sides of the differential equation.

Step 2: Use the derivative formulas to replace $y'$ with $sY - y(0)$ and $y''$ with $s^2Y - sy(0) - y'(0)$.

Step 3: Substitute the known initial conditions.

Step 4: Solve algebraically for $Y(s)$.

Step 5: Use partial fractions and the inverse transform table to find $y(t) = \mathcal{L}^{-1}\{Y(s)\}$.

Example 10.8.1
First-Order IVP
  • ProblemSolve $y' + 3y = 6$, $y(0) = 2$.
  • Step 1–3 Transform: $[sY - y(0)] + 3Y = \frac{6}{s}$. Substitute $y(0) = 2$: $$sY - 2 + 3Y = \frac{6}{s} \implies (s+3)Y = \frac{6}{s} + 2 = \frac{2s+6}{s}$$
  • Step 4$$Y(s) = \frac{2s+6}{s(s+3)} = \frac{2(s+3)}{s(s+3)} = \frac{2}{s}$$
  • Step 5$$y(t) = \mathcal{L}^{-1}\!\left\{\frac{2}{s}\right\} = 2$$
  • Check$y' = 0$, so $0 + 3(2) = 6$ ✓ and $y(0) = 2$ ✓.
  • InterpretThe equilibrium is $y = 2$. The initial condition starts exactly at equilibrium, so the solution stays constant forever.
$y(t) = 2$
Example 10.8.2
Second-Order Homogeneous IVP
  • ProblemSolve $y'' - 5y' + 6y = 0$, $y(0) = 2$, $y'(0) = 3$.
  • Step 1–3 $$[s^2Y - 2s - 3] - 5[sY - 2] + 6Y = 0$$ $$s^2Y - 2s - 3 - 5sY + 10 + 6Y = 0 \implies (s^2 - 5s + 6)Y = 2s - 7$$
  • Step 4$$Y(s) = \frac{2s-7}{(s-2)(s-3)}$$
  • Partial fractions $2s - 7 = A(s-3) + B(s-2)$. Set $s=2$: $A = 3$. Set $s=3$: $B = -1$. $$Y(s) = \frac{3}{s-2} - \frac{1}{s-3}$$
  • Step 5$$y(t) = 3e^{2t} - e^{3t}$$
  • Check$y(0) = 3 - 1 = 2$ ✓, $y'(0) = 6 - 3 = 3$ ✓. Verify: $y'' - 5y' + 6y = (12-30+18)e^{2t} + (-9+15-6)e^{3t} = 0$ ✓.
  • InterpretBoth roots $r = 2$ and $r = 3$ are positive, so the solution grows without bound. This reflects unstable exponential behavior.
$y(t) = 3e^{2t} - e^{3t}$
Example 10.8.3
Second-Order IVP with Nonhomogeneous Forcing
  • ProblemSolve $y'' + 4y = \sin(2t)$, $y(0) = 0$, $y'(0) = 1$.
  • Step 1–3 $$[s^2Y - 0 - 1] + 4Y = \frac{2}{s^2+4} \implies (s^2+4)Y = 1 + \frac{2}{s^2+4} = \frac{s^2+6}{s^2+4}$$
  • Step 4$$Y(s) = \frac{s^2+6}{(s^2+4)^2} = \frac{s^2+4+2}{(s^2+4)^2} = \frac{1}{s^2+4} + \frac{2}{(s^2+4)^2}$$
  • Step 5 First term: $\mathcal{L}^{-1}\!\left\{\frac{1}{s^2+4}\right\} = \frac{1}{2}\sin(2t)$.
    Second term (using extended table with $\omega = 2$): $\mathcal{L}^{-1}\!\left\{\frac{2}{(s^2+4)^2}\right\} = \frac{1}{8}[\sin(2t) - 2t\cos(2t)]$. $$y(t) = \frac{5}{8}\sin(2t) - \frac{1}{4}t\cos(2t)$$
  • InterpretThe term $-\frac{1}{4}t\cos(2t)$ is a resonance term: the forcing frequency $\omega = 2$ matches the natural frequency of $y'' + 4y = 0$. The solution grows without bound over time.
$y(t) = \dfrac{5}{8}\sin(2t) - \dfrac{1}{4}t\cos(2t)$
Sections 10.9 – 10.11
Step Functions, Discontinuous Inputs, and Impulses
Heaviside Function · Second Shift Theorem · Piecewise Forcing · Dirac Delta
Section 10.9

Step Functions

One of the most powerful features of the Laplace transform is its ability to handle functions that are not smooth everywhere. The unit step function (also called the Heaviside step function) is the standard way to represent switching behavior.

Definition — The Unit Step Function

$$u_c(t) = \begin{cases} 0 & t < c \\ 1 & t \geq c \end{cases}$$

where $c \geq 0$. The function $u_c(t)$ is zero before $t = c$ and jumps to $1$ at $t = c$. It models a switch being turned on at time $t = c$. Some texts use the notation $\mathcal{H}(t-c)$ or $H(t-c)$ for the same function.

Its Laplace transform follows directly from the definition: $\int_c^{\infty} e^{-st} dt = \frac{e^{-cs}}{s}$.

$$\mathcal{L}\{u_c(t)\} = \frac{e^{-cs}}{s}, \quad s > 0$$

Theorem — Second Shift Theorem (Time Shift)

If $\mathcal{L}\{f(t)\} = F(s)$, then

$$\mathcal{L}\{u_c(t)\, f(t-c)\} = e^{-cs} F(s)$$

Multiplying a function by $u_c(t)$ and delaying it by $c$ in the time domain multiplies the transform by $e^{-cs}$ in the $s$-domain. Equivalently: $\mathcal{L}^{-1}\{e^{-cs} F(s)\} = u_c(t) f(t-c)$.

Critical: The Delay Must Be Built In

The second shift theorem requires the function to be evaluated at $t - c$, not $t$. Writing $\mathcal{L}\{u_c(t)\, f(t)\} = e^{-cs} F(s)$ is wrong. The function inside must be the delayed version $f(t-c)$. Before applying the theorem in either direction, confirm the function is written correctly as $f(t-c)$ multiplied by $u_c(t)$.

Example 10.9.1
Transform of a Switched Function
  • ProblemFind $\mathcal{L}\{u_2(t)(t-2)^3\}$.
  • ThinkThis is $u_c(t) f(t-c)$ with $c = 2$ and $f(t) = t^3$. So $F(s) = \mathcal{L}\{t^3\} = \frac{6}{s^4}$.
  • SolveBy the second shift theorem: $e^{-2s} \cdot \frac{6}{s^4}$.
$\dfrac{6e^{-2s}}{s^4}$
Example 10.9.2
Inverse Transform with $e^{-cs}$
  • ProblemFind $\mathcal{L}^{-1}\!\left\{ \dfrac{e^{-3s}}{s-2} \right\}$.
  • ThinkThis is $e^{-cs} F(s)$ with $c = 3$ and $F(s) = \frac{1}{s-2}$, so $f(t) = e^{2t}$.
  • SolveBy the second shift theorem: $u_3(t) \cdot e^{2(t-3)}$.
  • InterpretThis is zero for $t < 3$ and equals $e^{2(t-3)}$ for $t \geq 3$ — a growing exponential that switches on at $t = 3$.
$f(t) = u_3(t)\, e^{2(t-3)}$
Section 10.10

Discontinuous Inputs

The step function allows us to write piecewise-defined functions in a single expression, which the Laplace transform can then process. A forcing function switching between $g_1(t)$ and $g_2(t)$ at time $t = c$ can be written as:

$$g(t) = g_1(t) + u_c(t)[g_2(t) - g_1(t)]$$

For $t < c$: $u_c = 0$, giving $g_1(t)$. For $t \geq c$: $u_c = 1$, giving $g_2(t)$. This single expression is ready to be transformed.

Example 10.10.1
IVP with a Discontinuous Forcing Function
  • ProblemSolve $y' + 2y = g(t)$, $y(0) = 0$, where $g(t) = 0$ for $0 \leq t < 1$ and $g(t) = 1$ for $t \geq 1$.
  • ThinkWrite $g(t) = u_1(t)$. Take the Laplace transform of both sides.
  • Transform $$[sY - 0] + 2Y = \frac{e^{-s}}{s} \implies (s+2)Y = \frac{e^{-s}}{s} \implies Y(s) = \frac{e^{-s}}{s(s+2)}$$
  • Partial fractions $\frac{1}{s(s+2)} = \frac{1/2}{s} - \frac{1/2}{s+2}$, so $Y(s) = e^{-s}\!\left[\frac{1/2}{s} - \frac{1/2}{s+2}\right]$.
  • InvertBy the second shift theorem with $c = 1$: $$y(t) = \frac{1}{2} u_1(t)\left[1 - e^{-2(t-1)}\right]$$
  • InterpretFor $t < 1$: $y(t) = 0$. Nothing happens because no input is applied. For $t \geq 1$: the system responds to the suddenly applied constant input, rising from $0$ and approaching $\frac{1}{2}$ asymptotically. This is the classic first-order step response.
$y(t) = \dfrac{1}{2}\, u_1(t)\left[1 - e^{-2(t-1)}\right]$
Example 10.10.2
A Rectangular Pulse — Two-Switch Forcing
  • ProblemA forcing function is $0$ for $0 \leq t < 1$, equals $3$ for $1 \leq t < 3$, and returns to $0$ for $t \geq 3$. Write it using step functions and find its Laplace transform.
  • ThinkTurning off is accomplished by subtracting the same value with a second step function at the switch-off time.
  • Solve $$g(t) = 3u_1(t) - 3u_3(t) = 3[u_1(t) - u_3(t)]$$ Check: $t < 1$: both zero, $g = 0$ ✓. $1 \leq t < 3$: $u_1 = 1, u_3 = 0$, $g = 3$ ✓. $t \geq 3$: both one, $g = 0$ ✓. $$\mathcal{L}\{g\} = \frac{3(e^{-s} - e^{-3s})}{s}$$
$G(s) = \dfrac{3(e^{-s} - e^{-3s})}{s}$
Modeling Note — Rectangular Pulses

A rectangular pulse (on for a finite interval, then off) arises constantly in engineering: a voltage applied for a fixed time, a force applied briefly, a drug infused over a period, a heater running on a schedule. The step function difference $u_a(t) - u_b(t)$ is the standard representation for any such on-off input.

Section 10.11

Impulse Functions as an Introduction

Some physical inputs are extreme: a hammer blow, an explosion, a sudden spike of current. These inputs deliver a large effect over an extremely short time. The idealized version of such an input is the Dirac delta function, also called the impulse function.

The Dirac Delta Function $\delta(t-c)$

Defined informally by two properties: $\delta(t - c) = 0$ for $t \neq c$, and $\int_{-\infty}^{\infty} \delta(t - c)\, dt = 1$.

The delta function is zero everywhere except at $t = c$, yet its integral is $1$. It is the idealization of a unit impulse: infinite height, zero width, area exactly $1$.

Its key property is the sifting property: $\int_{-\infty}^{\infty} f(t)\, \delta(t-c)\, dt = f(c)$. The delta function picks out the value of $f$ at the single point $t = c$.

Laplace transform: $\mathcal{L}\{\delta(t-c)\} = e^{-cs}$. For $c = 0$: $\mathcal{L}\{\delta(t)\} = 1$.

Example 10.11.1
IVP with an Impulse Input
  • ProblemSolve $y'' + 4y = \delta(t - \pi)$, $y(0) = 0$, $y'(0) = 0$.
  • ThinkThe system starts at rest and receives a unit impulse at time $t = \pi$. Find the response for all $t$.
  • Transform $$(s^2 + 4)Y = e^{-\pi s} \implies Y(s) = \frac{e^{-\pi s}}{s^2 + 4} = e^{-\pi s} \cdot \frac{1}{s^2+4}$$
  • Invert $\mathcal{L}^{-1}\!\left\{\frac{1}{s^2+4}\right\} = \frac{1}{2}\sin(2t)$. By the second shift theorem: $$y(t) = u_{\pi}(t)\cdot\frac{1}{2}\sin(2(t-\pi)) = \frac{1}{2}\, u_{\pi}(t)\sin(2t)$$ since $\sin(2(t-\pi)) = \sin(2t)$.
  • InterpretFor $t < \pi$: $y(t) = 0$. The system is completely undisturbed. For $t \geq \pi$: the impulse sets the system into free oscillation with amplitude $\frac{1}{2}$ and angular frequency $2$ — as if a string had been plucked or a pendulum had been struck.
$y(t) = \dfrac{1}{2}\, u_{\pi}(t)\sin(2t)$
Looking Ahead — Impulse Response and Convolution

The impulse response of a system is its output when the input is $\delta(t)$. Once the impulse response is known, the response to any input can in principle be computed using an operation called convolution, which is fundamental in signal processing and introduced in advanced courses. For now, the key facts are: $\mathcal{L}\{\delta(t-c)\} = e^{-cs}$, and an impulse gives a system a sudden kick that sets it into free motion.

Sections 10.12 – 10.14
Applications, Common Mistakes, and Preparing for Numerical Methods
Circuits and Control Systems · Eight Common Errors · Bridge to Chapter 11
Section 10.12

Laplace Transforms in Circuits and Control Systems

The Laplace transform is not a mathematical curiosity. It is the primary working language of electrical engineering, control systems, and signal processing. This section introduces the connection through a concrete circuit example and the transfer function concept.

RC Circuits

An RC circuit consists of a resistor $R$ (ohms) and a capacitor $C$ (farads) in series with a voltage source $V(t)$ (volts). The governing equation for the voltage $v_C(t)$ across the capacitor is

RC Circuit Governing Equation

$$RC\, v_C'(t) + v_C(t) = V(t)$$

This is a first-order linear differential equation with time constant $\tau = RC$. Taking the Laplace transform with initial condition $v_C(0)$:

$$V_C(s) = \frac{\mathcal{L}\{V(t)\} + RC\, v_C(0)}{RCs + 1}$$

The denominator $RCs + 1$ characterizes the circuit. For any input $V(t)$, the capacitor voltage is found by computing this expression and inverting.

Example 10.12.1
RC Circuit Step Response
  • ProblemAn RC circuit with $\tau = RC = 1\,\text{s}$, $v_C(0) = 0$, has a constant 5 V applied for $t \geq 0$. Find $v_C(t)$.
  • ThinkThe equation simplifies to $v_C' + v_C = 5$, $v_C(0) = 0$. Apply the Laplace method.
  • Solve $(s+1)V_C = \frac{5}{s} \implies V_C(s) = \frac{5}{s(s+1)} = \frac{5}{s} - \frac{5}{s+1}$ $$v_C(t) = 5 - 5e^{-t} = 5(1 - e^{-t})$$
  • InterpretThe capacitor starts uncharged and asymptotically charges toward the applied voltage $5\,\text{V}$. At $t = 1\,\text{s}$ (one time constant), $v_C \approx 3.16\,\text{V}$, which is about $63\%$ of the final value. After five time constants ($t = 5\,\text{s}$), the capacitor is $> 99\%$ charged.
$v_C(t) = 5(1 - e^{-t})$ V

Transfer Functions in Control Systems

In control systems, the Laplace transform is used to represent the input-output relationship of a system in terms of the transfer function. If a system is described by a linear differential equation relating input $u(t)$ to output $y(t)$, and if all initial conditions are zero, then the transfer function is the ratio

Transfer Function

$$H(s) = \frac{Y(s)}{U(s)} = \frac{\mathcal{L}\{y(t)\}}{\mathcal{L}\{u(t)\}}$$

This ratio depends only on the system's structure (its coefficients), not on the specific input. The first-order system $y' + ay = bu(t)$ has transfer function $H(s) = \frac{b}{s + a}$.

The poles of $H(s)$ (roots of the denominator) determine stability: poles with negative real parts give stable systems; poles with positive real parts give unstable systems. This connection between algebra and stability is the foundation of control systems design.

MGU Library Connection
  • RC and RLC circuit analysis — MGU Library: Electrical Engineering Foundations
  • Bode plots and frequency response — MGU Library: Control Systems I
  • Transfer functions and feedback control design — MGU Library: Control Systems II
  • Stability analysis using poles — MGU Library: Linear Systems and Signals
  • Fourier transforms as a related tool — MGU Library: Signals and Systems
Section 10.13

Common Laplace Transform Mistakes

Mistake 1 — Omitting initial condition terms
$\mathcal{L}\{y''\} = s^2 Y$

Correct: $\mathcal{L}\{y''\} = s^2Y(s) - sy(0) - y'(0)$. Students who drop the initial condition terms produce a wrong transformed equation. The error is invisible until the solution fails the initial conditions. Always write the full formula before substituting.

Mistake 2 — Applying the second shift theorem to $f(t)$ instead of $f(t-c)$
$\mathcal{L}\{u_c(t) f(t)\} = e^{-cs} F(s)$

Correct: The theorem requires $f(t-c)$, not $f(t)$. The function must be the delayed version evaluated at $t-c$. Before applying the theorem in either direction, confirm the argument is $(t-c)$.

Mistake 3 — Errors in partial fraction constants
Constants found without verification

After computing the constants $A$, $B$, etc., substitute a specific value of $s$ back into the original expression and verify that both sides agree. Use $s = 0$ or another convenient value. This check costs almost no time and catches most errors.

Mistake 4 — Confusing $Y(s)$ with $y(t)$
"The solution is $Y(s) = \frac{3}{s+2}$."

Correct: $Y(s) = \frac{3}{s+2}$ is the transform. The solution is $y(t) = \mathcal{L}^{-1}\{Y(s)\} = 3e^{-2t}$. The inverse transform step is never optional. The work is not done until $y(t)$ is written explicitly.

Mistake 5 — Wrong polynomial transform formula
$\mathcal{L}\{t^2\} = \frac{2}{s^2}$

Correct: $\mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}}$, so $\mathcal{L}\{t^2\} = \frac{2!}{s^3} = \frac{2}{s^3}$. The denominator power is $n+1$, not $n$. Verify: $n=0 \to \frac{1}{s}$; $n=1 \to \frac{1}{s^2}$; $n=2 \to \frac{2}{s^3}$.

Mistake 6 — Missing the numerator adjustment for irreducible quadratics
Writing $\frac{2s+7}{(s+2)^2+9}$ as $\frac{2}{(s+2)^2+9} + \frac{7}{(s+2)^2+9}$

Correct: The numerator must be split into a multiple of $(s+2)$ and a multiple of $3$ (the $\omega$ value). Write $2s+7 = 2(s+2) + 3$, giving $\frac{2(s+2)}{(s+2)^2+9} + \frac{3}{(s+2)^2+9}$. The first inverts with cosine, the second with sine.

Mistake 7 — Writing a piecewise function with wrong step function signs
"On for $1 \leq t < 3$" represented as $u_1(t) + u_3(t)$

Correct: To switch on at $t=1$ and off at $t=3$, use $u_1(t) - u_3(t)$, not $+$. Adding two step functions keeps the value on permanently. Subtracting the second one switches it back off. Always test your formula at specific $t$ values in each region.

Mistake 8 — Stopping after finding $Y(s)$ without inverting
Solution presented as $Y(s) = \frac{2s-7}{(s-2)(s-3)}$

The function $Y(s)$ is not the answer. It is the transform of the answer. The solution to a differential equation must be a function of $t$. Complete the partial fractions and inverse transform to write $y(t)$ explicitly.

Section 10.14

Preparing for Numerical Methods

Chapter 10 closes the part of the book devoted to exact solution methods. You now have a substantial toolkit: separable equations, integrating factors, second-order homogeneous equations with constant coefficients, undetermined coefficients, variation of parameters, systems with eigenvalue methods, and Laplace transforms for equations with initial conditions and discontinuous forcing.

These methods are powerful, but they share a common limitation. They require the differential equation to have a form that is analytically tractable. Real-world differential equations are not always so cooperative. A population model with seasonal variation may involve time-varying coefficients. A mechanical system with nonlinear damping cannot be solved by undetermined coefficients. A fluid dynamics equation may have no known closed-form solution at all. In these situations, exact methods fail, not because the student is missing a technique, but because no such technique exists.

Chapter 11 introduces numerical methods for differential equations. Rather than finding an exact formula, numerical methods build an approximate table of values step by step. The method begins at the initial condition, uses the differential equation to compute the slope at each point, takes a small step, updates the position, and repeats. Euler's method, the improved Euler method, and the Runge-Kutta method are all based on this core idea, with increasing levels of accuracy.

Numerical methods are how most real differential equations in science, engineering, biology, economics, and data science are actually solved. They are implemented in Python, MATLAB, Julia, R, and specialized scientific computing environments. Chapter 11 will build these methods from scratch, analyze their errors, and connect them to modern computational practice.

Chapter 10 — Back Matter

Chapter Summary

Key Terms Review

Laplace transform
The integral $F(s) = \int_0^{\infty} e^{-st} f(t)\,dt$ that converts a time-domain function into an $s$-domain function, turning differentiation into multiplication.
$s$-domain
The domain of the variable $s$ in which the transformed equation is algebraic. Also called the transform domain or frequency domain.
time domain
The domain of the independent variable $t$ in which the original differential equation lives.
transform pair
A matched pair $(f(t), F(s))$ where $F(s) = \mathcal{L}\{f(t)\}$. Tables of transform pairs are the primary computational tool.
linearity
The property that $\mathcal{L}\{af + bg\} = aF + bG$. Allows the transform of any linear combination to be computed term by term.
first shift theorem
The identity $\mathcal{L}\{e^{at}f(t)\} = F(s-a)$. Multiplication by $e^{at}$ in time shifts the transform by $a$ in the $s$-domain.
inverse Laplace transform
The operation $\mathcal{L}^{-1}\{F(s)\} = f(t)$ that recovers the time-domain function from its transform. Computed via partial fractions and table lookup.
partial fraction decomposition
The technique of writing a rational function as a sum of simpler fractions with denominator factors matching the transform table. The essential tool for inverse transforms.
Heaviside step function
The function $u_c(t)$ that is $0$ for $t < c$ and $1$ for $t \geq c$. Models a switch turning on at time $c$. Transform: $\mathcal{L}\{u_c\} = e^{-cs}/s$.
second shift theorem
The identity $\mathcal{L}\{u_c(t)f(t-c)\} = e^{-cs}F(s)$. Handles switching inputs by connecting time delay in the time domain to multiplication by $e^{-cs}$ in the $s$-domain.
piecewise forcing
A forcing function defined by different expressions on different time intervals. Written as a combination of Heaviside step functions before taking the Laplace transform.
Dirac delta function
An idealized unit impulse $\delta(t-c)$ concentrated at $t = c$. Defined by $\delta = 0$ for $t \neq c$ and $\int \delta\, dt = 1$. Transform: $\mathcal{L}\{\delta(t-c)\} = e^{-cs}$.
impulse response
The system's output when the input is a unit impulse $\delta(t)$. The fundamental characterization of a linear system's behavior.
transfer function
The ratio $H(s) = Y(s)/U(s)$ describing a system's input-output relationship with zero initial conditions. The poles of $H(s)$ determine the system's stability.
resonance (via transform)
In the $s$-domain, resonance appears as a repeated factor $(s^2 + \omega^2)^2$ in the denominator of $Y(s)$. Its inverse transform contains the secular term $t\sin(\omega t)$ or $t\cos(\omega t)$, growing without bound.

Skill Practice

Part ATransform Table Practice
  1. Verify $\mathcal{L}\{e^{-2t}\} = \frac{1}{s+2}$ from the definition.
  2. Find $\mathcal{L}\{4 - 3t + 2t^2\}$.
  3. Find $\mathcal{L}\{7\cos(5t) - 2\sin(5t)\}$.
  4. Find $\mathcal{L}\{e^{3t}\cos(4t)\}$.
  5. Find $\mathcal{L}\{t^2 e^{-t}\}$.
  6. Find $\mathcal{L}\{6e^{-2t}\sin(3t)\}$.
  7. Find $\mathcal{L}\{3t^3 - 4e^{2t} + \cos(t)\}$.
  8. Verify the first shift theorem for $\mathcal{L}\{e^{2t}t^3\}$.
Part BInverse Transform Practice
  1. $\mathcal{L}^{-1}\!\left\{\frac{4}{s+5}\right\}$
  2. $\mathcal{L}^{-1}\!\left\{\frac{3}{s^2+9}\right\}$
  3. $\mathcal{L}^{-1}\!\left\{\frac{2s+1}{s^2+4s+8}\right\}$
  4. $\mathcal{L}^{-1}\!\left\{\frac{s}{s^2-4}\right\}$
  5. $\mathcal{L}^{-1}\!\left\{\frac{3s-5}{(s-1)(s+3)}\right\}$
  6. $\mathcal{L}^{-1}\!\left\{\frac{2}{s^2(s+1)}\right\}$
  7. $\mathcal{L}^{-1}\!\left\{\frac{s+3}{(s+1)(s^2+4)}\right\}$
  8. $\mathcal{L}^{-1}\!\left\{\frac{1}{(s+2)^3}\right\}$
Part CInitial Value Problems
  1. $y' - 4y = 0$, $y(0) = 3$
  2. $y' + 5y = 10$, $y(0) = 1$
  3. $y'' + 9y = 0$, $y(0) = 2$, $y'(0) = 0$
  4. $y'' - 3y' + 2y = 0$, $y(0) = 1$, $y'(0) = 0$
  5. $y'' + 2y' + 5y = 0$, $y(0) = 1$, $y'(0) = 0$
  6. $y'' + y = \cos(t)$, $y(0) = 0$, $y'(0) = 0$
  7. $y'' - y' - 6y = e^{3t}$, $y(0) = 0$, $y'(0) = 1$
  8. $y'' + 4y' + 4y = 0$, $y(0) = 2$, $y'(0) = -1$
Part DStep Function Practice
  1. Write the following function using step functions and find its Laplace transform. $$g(t) = \begin{cases} 2 & 0 \leq t < 3 \\ 0 & t \geq 3 \end{cases}$$
  2. Find $\mathcal{L}\{u_2(t)(t-2)^2\}$.
  3. Find $\mathcal{L}^{-1}\!\left\{\dfrac{e^{-2s}}{s-3}\right\}$.
  4. Find $\mathcal{L}^{-1}\!\left\{\dfrac{e^{-s}(s+1)}{s^2+4}\right\}$.
  5. Solve $y' + y = u_2(t)$, $y(0) = 0$.
  6. Write the following function using step functions: $$g(t) = \begin{cases} t & 0 \leq t < 1 \\ 1 & 1 \leq t < 3 \\ 4-t & t \geq 3 \end{cases}$$

Application Problems

For each application: set up the governing differential equation, solve using the Laplace transform, and interpret the result in physical terms.

  1. RC Circuit. An RC circuit satisfies $v_C' + 2v_C = 4$, $v_C(0) = 0$. Solve using the Laplace transform. Find the time constant and the long-term voltage. At what time does $v_C = 1.5\,\text{V}$?
  2. Mechanical Impulse. A mass-spring system satisfies $y'' + 9y = \delta(t-1)$, $y(0) = 0$, $y'(0) = 0$. Solve and describe the motion for $t < 1$ and $t \geq 1$. What does the impulse represent physically?
  3. Switching Load. An electrical system satisfies $y'' + y' + y = g(t)$, $y(0) = 0$, $y'(0) = 0$, where $g(t) = u_2(t)$. Solve and describe the response.
  4. Delayed Deposits. A bank account satisfies $A' - 0.05A = 100\, u_2(t)$, $A(0) = 1000$. Deposits begin at $t = 2$ years at $\$100$/year. Solve for $A(t)$ and find the balance at $t = 5$ years.
  5. Harvested Population. A harvested population satisfies $P' - 0.2P = -50\, u_3(t)$, $P(0) = 500$. Harvesting of 50 individuals per year begins at $t = 3$. Solve for $P(t)$ and discuss long-term behavior.
  6. Two-Step Forcing. A first-order system satisfies $y' + 3y = g(t)$, $y(0) = 0$, where $g(t)$ is $0$ for $0 \leq t < 1$, equals $6$ for $1 \leq t < 4$, and is $0$ for $t \geq 4$. Write $g(t)$ using step functions, solve using the Laplace transform, and sketch the response.

Error Analysis

Each problem shows a student's attempted solution. Identify the error and provide the correct solution with full explanation.

Problem 1 — Missing Initial Condition Terms

A student solves $y'' + 4y' + 3y = 0$, $y(0) = 1$, $y'(0) = 0$.

$(s^2 + 4s + 3)Y = 0 \implies Y = 0 \implies y(t) = 0$

What is the error? What is the correct solution?

Problem 2 — Second Shift Theorem Misapplication

A student is asked to find $\mathcal{L}^{-1}\!\left\{\frac{e^{-2s}}{s+3}\right\}$.

$f(t) = u_2(t)\, e^{-3t}$

Is this correct? If not, what is the right answer and why?

Problem 3 — Wrong Shifted Laplace Transform

A student finds $\mathcal{L}\{u_3(t) \cdot t^2\}$ and writes:

$\mathcal{L}\{u_3(t) \cdot t^2\} = e^{-3s} \cdot \dfrac{2}{s^3}$

What is wrong with this application of the second shift theorem? What is the correct approach?

Problem 4 — Partial Fraction Shortcut Without Verification

A student solves $y' + y = e^t$, $y(0) = 0$, correctly arrives at $Y(s) = \frac{1}{(s-1)(s+1)}$, then writes without working:

$y(t) = e^t - e^{-t}$ by pattern matching.

Verify whether the answer is correct by performing the partial fractions properly, and identify whether the student's reasoning is reliable.

Chapter 10 Checkpoint

Chapter 10 Checkpoint
Complete without assistance — assesses all key skills and concepts from Chapter 10
Part A — Transforms and Inverse Transforms
  1. Find $\mathcal{L}\{3t^2 - 5e^{-4t} + 2\sin(3t)\}$.
  2. Find $\mathcal{L}\{e^{-2t}\cos(5t)\}$.
  3. Find $\mathcal{L}^{-1}\!\left\{\dfrac{5s - 1}{s^2 + 6s + 10}\right\}$. (Complete the square.)
  4. Find $\mathcal{L}^{-1}\!\left\{\dfrac{3}{s(s^2 + 9)}\right\}$. (Partial fractions.)
Part B — Initial Value Problems
  1. Solve $y' + 6y = 12$, $y(0) = 0$, using the Laplace transform. Show all five steps.
  2. Solve $y'' + 5y' + 6y = 0$, $y(0) = 1$, $y'(0) = -1$. Verify the initial conditions.
  3. Solve $y'' + 4y = 8\cos(2t)$, $y(0) = 0$, $y'(0) = 0$. Identify resonance if present and explain what it means.
Part C — Step Functions and Discontinuous Inputs
  1. Write the following function using step functions and find $\mathcal{L}\{g(t)\}$: $$g(t) = \begin{cases} 0 & 0 \leq t < 2 \\ 5 & 2 \leq t < 4 \\ 0 & t \geq 4 \end{cases}$$
  2. Find $\mathcal{L}^{-1}\!\left\{\dfrac{e^{-3s}}{s^2+4}\right\}$. Interpret the result.
  3. Solve $y' + 2y = u_1(t)$, $y(0) = 1$. Describe the behavior for $t < 1$ and $t \geq 1$.
Part D — Applications
  1. A circuit satisfies $v' + 4v = 8\, u_2(t)$, $v(0) = 2$. Solve for $v(t)$. Describe the response: what happens before $t = 2$, what happens after, and what is the long-term voltage?
  2. A system at rest receives an impulse at $t = \pi/2$: $y'' + y = \delta(t - \pi/2)$, $y(0) = 0$, $y'(0) = 0$. Solve for $y(t)$ and interpret the physical meaning of the impulse and the subsequent motion.
Part E — Error Identification
  1. A student transforms $y'' - 4y = 0$, $y(0) = 3$, $y'(0) = 2$, and writes $(s^2 - 4)Y = 0$. What is wrong? What should the right-hand side of the transformed equation be?
  2. A student writes $g(t) = 3[u_2(t) + u_5(t)]$ for the function that equals $3$ on $2 \leq t < 5$ and $0$ otherwise. Identify the error and write the correct expression.
Looking Ahead
Chapter 11: Numerical Methods for Differential Equations

Chapter 10 closes the analytical portion of the book. You now possess a complete set of symbolic tools: first-order methods, second-order methods, systems, and Laplace transforms. These tools are powerful, but they require specific equation structures. Most real differential equations in science, engineering, biology, and computing do not have analytically tractable forms — not because a technique is missing, but because none exists.

Chapter 11 introduces numerical methods, a fundamentally different category of tool. Rather than finding a formula, a numerical method builds an approximate table of values step by step using the differential equation itself to supply the direction of change at each point. Euler's method, the improved Euler method, and the Runge-Kutta method (RK4) achieve progressively higher accuracy using the same basic idea. These are the methods that actually run every simulation in physics, climate science, epidemiology, spacecraft navigation, and financial modeling. Chapter 11 will build them from scratch and connect them to modern computational practice.

MGU Library Connection
  • Laplace Transform Table — Appendix M of this textbook
  • Integration Techniques Review — Appendix B (partial fractions, integration by parts)
  • Systems of Differential Equations, Chapter 9 — prereq for transfer function understanding
  • MGU Library: Electrical Engineering Foundations — RC and RLC circuits in full
  • MGU Library: Control Systems I — transfer functions, Bode plots, stability margins
  • MGU Library: Signals and Systems — Fourier transforms, convolution, frequency response
  • MGU Dictionary entries: Laplace transform, transfer function, Heaviside function, Dirac delta