DIFFERENTIAL EQUATIONS • LAPLACE TRANSFORMS

Transforms of Common Functions

Learn how the Laplace transform converts everyday functions into a simpler algebraic world where differential equations become easy to solve.

Historical Context & Motivation

Solving differential equations — equations involving rates of change — has been a central challenge in mathematics for centuries. Engineers designing bridges, physicists studying circuits, and biologists modeling population growth all encounter these equations. The Laplace transform was developed as a powerful shortcut: instead of solving a complicated differential equation directly, you transform it into a simpler algebra problem, solve that, and then transform back. Think of it like translating a difficult sentence into a language where the grammar is easier, working out the meaning, and then translating the answer back.

1737
Euler's Integral Methods
Leonhard Euler explored integral transforms as tools for solving differential equations, laying the mathematical groundwork that later mathematicians would refine.
1782
Laplace's Probability Work
Pierre-Simon Laplace introduced a version of the transform in his studies of probability theory, giving the technique its lasting name.
1892
Heaviside's Engineering Calculus
Oliver Heaviside popularized operational methods for analyzing electrical circuits. His practical techniques were later justified rigorously using the Laplace transform.
1937
Modern Formalization
Gustav Doetsch published the first comprehensive textbook on the Laplace transform, establishing the standard table of transforms of common functions used in classrooms today.

The central question this lesson addresses is straightforward: when you apply the Laplace transform to familiar functions like constants, powers of t, exponentials, sines, and cosines, what do you get? Memorizing or understanding these standard transform pairs is the key that unlocks every Laplace-based technique you will use going forward.

Core Principles & Definitions

Before diving into specific transforms, you need to understand a few foundational ideas. The Laplace transform takes a function of time, f(t), and produces a new function of a complex variable s, written as F(s). The variable s lives in what we call the s-domain (or frequency domain), while t lives in the time domain.

1

The Definition Integral

The Laplace transform is defined as ℒ{f(t)} = ∫₀^∞ e−st f(t) dt. The exponential e−st acts as a "weight" that makes the integral converge for large t.
2

Linearity Property

The transform is linear: ℒ{af(t) + bg(t)} = aF(s) + bG(s). This means you can break complicated functions into simpler pieces, transform each one, and add the results.
3

Region of Convergence

Each transform is only valid when s is large enough. For example, the transform of eat requires s > a. This restriction is noted but rarely causes problems in practice.
4

One-to-One Pairing

Each time-domain function f(t) maps to exactly one s-domain function F(s), and vice versa. This uniqueness guarantees you can always "go back" — a process called the inverse Laplace transform.
KEY TAKEAWAY
Think of the Laplace transform like a universal translator app on your phone. You speak a sentence in one language (the time domain), the app converts it into another language (the s-domain) where the problem is easier to work with, you solve it, and then you translate the answer back. The "dictionary" of this translator is the table of transforms of common functions you'll learn in this lesson.

Visual Explanation — Time Domain to s-Domain

The diagram below illustrates the core idea: common functions in the time domain on the left are mapped by the Laplace transform into algebraic expressions in the s-domain on the right. The arrow labeled ℒ represents the transform, and the arrow labeled ℒ−1 represents the inverse. Notice how each time-domain function — whether it's a polynomial, an exponential, or a sinusoid — becomes a rational function of s (a fraction with polynomials on top and bottom).

Each row shows a common time-domain function (left) and its Laplace transform (right). The forward transform ℒ maps f(t) to F(s); the inverse ℒ⁻¹ maps back.

Study the pairs above carefully. Notice a pattern: simple time-domain functions produce simple algebraic fractions in the s-domain. A constant becomes 1/s, a linear ramp t becomes 1/s², and so on. Each higher power of t just adds another factor of s in the denominator. Exponentials shift the location of a pole, and trig functions create quadratic denominators involving s² + b².

Mathematical Framework

Let's now look at the formal definitions behind these transform pairs. Each result below is obtained by evaluating the defining integral ∫₀^∞ e−st f(t) dt for a specific choice of f(t). You are not expected to re-derive every one of these in a test setting, but understanding how the integral works for at least one or two cases helps you trust the table and use it correctly.

LAPLACE TRANSFORM DEFINITION
ℒ{f(t)} = F(s) = ∫₀^∞ e^(−st) f(t) dt
f(t) is the original time-domain function, F(s) is the transformed function, s is a complex variable (s > 0 for convergence), and the integral runs from 0 to ∞.
TRANSFORM OF A CONSTANT
ℒ{1} = 1/s, s > 0
Setting f(t) = 1, the integral becomes ∫₀^∞ e−st dt = [−e−st/s]₀^∞ = 1/s. This is the simplest and most fundamental transform pair.
TRANSFORM OF tⁿ
ℒ{tⁿ} = n! / s^(n+1), s > 0, n = 0, 1, 2, …
Here n! means n factorial (for example, 3! = 6). When n = 1 you get ℒ{t} = 1/s². When n = 2 you get ℒ{t²} = 2/s³. Each higher power of t raises the power of s in the denominator by one.
TRANSFORM OF eᵃᵗ
ℒ{eᵃᵗ} = 1 / (s − a), s > a
The exponential function shifts the pole from s = 0 to s = a. If a is positive the function grows; if a is negative it decays. The convergence condition s > a ensures the integral doesn't blow up.
TRANSFORMS OF SINE AND COSINE
ℒ{sin(bt)} = b / (s² + b²); ℒ{cos(bt)} = s / (s² + b²), s > 0
Both sine and cosine produce the same denominator s² + b², but sine puts the frequency b in the numerator while cosine puts s in the numerator. This is a common source of mix-ups, so be careful!

Complete Reference Table & Graph

The table below collects all the standard transform pairs you'll need. Keep this table handy — most textbooks include it on a formula sheet for exams. Each row shows the time-domain function, its Laplace transform, and the condition on s that ensures convergence.

Standard Laplace Transform Pairs
f(t)F(s) = ℒ{f(t)}Convergence
11/ss > 0
t1/s²s > 0
tⁿ (n = 0, 1, 2, …)n! / sn+1s > 0
eᵃᵗ1 / (s − a)s > a
sin(bt)b / (s² + b²)s > 0
cos(bt)s / (s² + b²)s > 0
eᵃᵗ sin(bt)b / ((s − a)² + b²)s > a
eᵃᵗ cos(bt)(s − a) / ((s − a)² + b²)s > a
tⁿ eᵃᵗn! / (s − a)n+1s > a
Four panels showing the time-domain behavior of common functions. Top-left: constant, linear, and quadratic functions. Top-right: growing and decaying exponentials. Bottom-left: sine and cosine oscillations. Bottom-right: a damped sinusoid with its exponential envelope shown as dashed lines.
💡 Pattern to Remember
Notice that exponential shifts in the time domain correspond to replacing s with (s − a) in the s-domain. So ℒ{eᵃᵗ sin(bt)} looks just like ℒ{sin(bt)} = b/(s² + b²), except every s is replaced by (s − a), giving b/((s − a)² + b²). This is called the First Shifting Theorem.

Worked Example

Let's find the Laplace transform of the function f(t) = 3t² + 5e−2t − 4 sin(7t). This is the kind of problem you'll see on homework and exams, where you need to combine several entries from the transform table using the linearity property.

Find ℒ{3t² + 5e⁻²ᵗ − 4 sin(7t)}
1
Step 1 — Apply LinearitySince the Laplace transform is linear, we can transform each term separately and combine the results: ℒ{3t² + 5e−2t − 4 sin(7t)} = 3 · ℒ{t²} + 5 · ℒ{e−2t} − 4 · ℒ{sin(7t)}.
2
Step 2 — Transform t²Using the formula ℒ{tⁿ} = n!/sn+1 with n = 2, we get ℒ{t²} = 2!/s³ = 2/s³.
3 · ℒ{t²} = 3 × (2/s³) = 6/s³
3
Step 3 — Transform e⁻²ᵗUsing the formula ℒ{eᵃᵗ} = 1/(s − a) with a = −2, we get ℒ{e−2t} = 1/(s − (−2)) = 1/(s + 2).
5 · ℒ{e⁻²ᵗ} = 5 × 1/(s + 2) = 5/(s + 2)
4
Step 4 — Transform sin(7t)Using the formula ℒ{sin(bt)} = b/(s² + b²) with b = 7, we get ℒ{sin(7t)} = 7/(s² + 49).
−4 · ℒ{sin(7t)} = −4 × 7/(s² + 49) = −28/(s² + 49)
5
Step 5 — Combine ResultsAdding the three transformed terms together gives the final answer.
F(s) = 6/s³ + 5/(s + 2) − 28/(s² + 49), s > 0
⚠️ Watch the Signs
In Step 3, notice that when a = −2, the denominator becomes s − (−2) = s + 2. This sign detail trips up many students. Always substitute the value of a carefully, including its sign.

Strengths and Limitations

Like any mathematical tool, the Laplace transform has situations where it shines and situations where other methods might be easier. Understanding these trade-offs helps you choose the right strategy for each problem.

Strengths vs. Limitations of the Laplace Transform Method
StrengthsLimitations
Converts differential equations into algebraic equations — no guessing neededRequires the function to grow no faster than an exponential (must satisfy an "exponential order" condition)
Handles initial conditions automatically — they get "baked in" during the transformFinding the inverse transform can require partial fractions, which is algebraically messy
Works beautifully with piecewise and discontinuous functions (e.g., on/off switches)Not well-suited for nonlinear differential equations — it's mainly a linear tool
Standard tables make the process almost mechanical for common functionsRequires memorizing or having access to the transform table
KEY TAKEAWAY
Think of the Laplace transform like a GPS for solving differential equations. When the road is a straight highway (a linear equation with constant coefficients), the GPS gives you perfect directions every time. But if you go off-road into rough terrain (nonlinear or variable-coefficient equations), the GPS may not have a map. Use this tool when the problem fits its strengths, and you'll save yourself a lot of work.

Connection to Advanced Theory

The transform pairs you've learned in this lesson are the starting blocks for much deeper ideas. As you progress, you'll encounter additional theorems that extend the basic table to cover an enormous variety of functions. The table below previews some of these advanced connections.

What You Know NowWhere It Leads
ℒ{eᵃᵗ f(t)} replaces s with (s − a)First Shifting Theorem — used to handle exponentially modified signals in engineering
Transform of sin and cosTransfer functions — describes how electrical circuits or mechanical systems respond to oscillating inputs
Linearity of the transformConvolution theorem — multiplication in the s-domain corresponds to a special integral operation in the time domain
Basic table of common functionsInverse Laplace via partial fractions — break a complicated F(s) into pieces you recognize from the table, then invert each piece

In future lessons you'll learn how to handle step functions (Heaviside functions), impulse functions (Dirac delta), and the Second Shifting Theorem for time-delayed signals. All of these build directly on the basic transform pairs from this lesson. Master the table now, and those advanced topics will feel like natural extensions rather than new material.

Practice Problems

PROBLEM 1CONCEPTUAL
Explain in your own words why the Laplace transform of sin(bt) and cos(bt) share the same denominator s² + b² but have different numerators. What does the numerator tell you about which trig function you started with?
PROBLEM 2BASIC CALCULATION
Find the Laplace transform of f(t) = 4t³.
PROBLEM 3INTERMEDIATE
Find the Laplace transform of f(t) = 2e3t + 7 cos(5t) − t.
PROBLEM 4APPLIED
An electrical circuit has a voltage source that produces V(t) = 10 sin(60t) volts (where t is in seconds and 60 is the angular frequency in rad/s). Write the Laplace transform V(s) that an engineer would use as the input when analyzing this circuit in the s-domain.
PROBLEM 5CRITICAL THINKING
Suppose you know that ℒ{sin(bt)} = b/(s² + b²). Use the First Shifting Theorem (if ℒ{f(t)} = F(s), then ℒ{eᵃᵗ f(t)} = F(s − a)) to derive the Laplace transform of eᵃᵗ sin(bt) without looking at the table. Verify your result matches the table entry.

Lesson Summary

The Laplace transform converts time-domain functions into s-domain expressions using the integral ℒ{f(t)} = ∫₀^∞ e−st f(t) dt. The essential transform pairs include: ℒ{1} = 1/s, ℒ{tⁿ} = n!/sⁿ⁺¹, ℒ{eᵃᵗ} = 1/(s − a), ℒ{sin(bt)} = b/(s² + b²), and ℒ{cos(bt)} = s/(s² + b²). Because the transform is linear, you can break any sum of these functions into individual terms, transform each one separately, and combine the results.

The First Shifting Theorem extends the table further: multiplying a function by eᵃᵗ in the time domain replaces every s with (s − a) in the s-domain, producing transforms like ℒ{eᵃᵗ sin(bt)} = b/((s−a)²+b²). Mastering these standard pairs and the linearity property gives you the essential toolkit for solving linear differential equations using Laplace methods. Keep the transform table close by — it is your most important reference as you move into more advanced topics like inverse transforms and solving initial value problems.

Varsity Tutors • Differential Equations • Transforms of Common Functions