Definition and Properties of Laplace Transforms

students, in this lesson you will learn one of the most useful tools in differential equations: the Laplace transform. 🚀 By the end, you should be able to explain what a Laplace transform is, why it matters, and how its main properties make hard problems easier to solve.

Why Laplace transforms matter

A Laplace transform turns a function of time, usually written as $f(t)$, into a new function of a variable $s$. This may sound abstract, but it is extremely powerful. The big idea is that a problem involving derivatives can often be converted into a simpler algebra problem.

For example, many differential equations in engineering and science involve sudden forces, switches turning on at a certain time, or initial conditions. Laplace transforms help because they handle these situations very well. Instead of solving directly in the time domain, we work in the $s$-domain, solve more easily, and then convert back.

Think of it like changing from one language to another. If a problem is hard to understand in one form, another form may make it much clearer. 📘

The Laplace transform is a central topic in Differential Equations because it connects directly to Midterm 1 material such as derivatives, initial conditions, and solving linear differential equations.

Definition of the Laplace transform

The Laplace transform of a function $f(t)$ is defined by

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

provided the integral converges.

Here is what the symbols mean:

$f(t)$ is the original function, usually given in terms of time $t$.
$F(s)$ is the transformed function.
$\mathcal{L}$ means “Laplace transform.”
$s$ is the new variable.
The exponential factor $e^{-st}$ helps the integral converge for many functions.

The interval starts at $0$, not at $-\infty$, which makes Laplace transforms especially useful for initial value problems. Most physical systems begin at a starting time, so this matches real-world situations well.

A simple example

Let $f(t)=1$. Then

$$\mathcal{L}\{1\}=\int_0^\infty e^{-st}\,dt.$$

If $s>0$, then

$$\int_0^\infty e^{-st}\,dt=\left[-\frac{1}{s}e^{-st}\right]_0^\infty=\frac{1}{s}.$$

$$\mathcal{L}\{1\}=\frac{1}{s}, \quad s>0.$$

This is one of the first formulas students memorize, and it becomes a building block for many others.

When the Laplace transform exists

Not every function has a Laplace transform, but many functions used in applications do. A common sufficient condition is that $f(t)$ is piecewise continuous on every finite interval and of exponential order.

A function is of exponential order if there exist constants $M$, $a$, and $T$ such that

$$|f(t)|\le Me^{at} \quad \text{for } t\ge T.$$

This condition means $f(t)$ does not grow too quickly. Functions like polynomials, exponentials, sines, and cosines usually work well.

For example:

$f(t)=t^2$ has a Laplace transform.
$f(t)=\sin t$ has a Laplace transform.
$f(t)=e^{3t}$ has a Laplace transform.

But a function that grows faster than any exponential may fail to have one.

The main properties you need to know

The real power of Laplace transforms comes from their properties. These rules let you transform complicated expressions one piece at a time. 🧩

1. Linearity

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

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

This means constants can be pulled out and sums can be transformed term by term.

Example

If $f(t)=3t+2$, then

$$\mathcal{L}\{3t+2\}=3\mathcal{L}\{t\}+2\mathcal{L}\{1\}.$$

Using known formulas,

$$\mathcal{L}\{t\}=\frac{1}{s^2}, \qquad \mathcal{L}\{1\}=\frac{1}{s},$$

$$\mathcal{L}\{3t+2\}=\frac{3}{s^2}+\frac{2}{s}.$$

2. Derivatives in the time domain

This property is one of the most important in differential equations. If $\mathcal{L}\{f(t)\}=F(s)$, then

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

For the second derivative,

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

More generally, derivatives become algebraic expressions in $s$, and initial conditions appear automatically.

Why this matters

A differential equation like

$$y''+3y'+2y=0$$

can become an algebra equation after transforming:

$$\mathcal{L}\{y''\}+3\mathcal{L}\{y'\}+2\mathcal{L}\{y\}=0.$$

If $\mathcal{L}\{y\}=Y(s)$, then the equation turns into something involving $Y(s)$ and the initial values $y(0)$ and $y'(0)$. This is much easier to solve than working with derivatives directly.

3. Exponential shift

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

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

This is called the first shifting theorem or exponential shift property.

Example

Since

$$\mathcal{L}\{\sin t\}=\frac{1}{s^2+1},$$

we get

$$\mathcal{L}\{e^{2t}\sin t\}=\frac{1}{(s-2)^2+1}.$$

The exponential factor $e^{2t}$ shifts $s$ to $s-2$.

4. Time shifting with the Heaviside function

Many real-world systems turn on at a specific time. The Heaviside step function is written as $u(t-a)$ and is defined by

$$u(t-a)=\begin{cases}0,& t<a,\\1,& t\ge a.\end{cases}$$

A key property is

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

This is called the second shifting theorem.

Real-world example

Suppose a heater turns on at $t=5$. A temperature model might include $u(t-5)$ to represent the switch. The Laplace transform handles this neatly because the turn-on time becomes the factor $e^{-5s}$.

5. Multiplication by $t$ and differentiation in $s$

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

$$\mathcal{L}\{tf(t)\}=-\frac{d}{ds}F(s).$$

More generally,

$$\mathcal{L}\{t^n f(t)\}=(-1)^n\frac{d^n}{ds^n}F(s).$$

This property is useful when a function includes factors like $t\sin t$ or $t^2e^{at}$.

Common transform formulas

Knowing a small set of transform pairs is very helpful. Here are several important ones:

$$\mathcal{L}\{1\}=\frac{1}{s}$$

$$\mathcal{L}\{t\}=\frac{1}{s^2}$$

$$\mathcal{L}\{t^n\}=\frac{n!}{s^{n+1}}$$

$$\mathcal{L}\{e^{at}\}=\frac{1}{s-a}$$

$$\mathcal{L}\{\sin bt\}=\frac{b}{s^2+b^2}$$

$$\mathcal{L}\{\cos bt\}=\frac{s}{s^2+b^2}$$

These formulas are often used together with linearity.

Example with trig functions

Find $\mathcal{L}\{2\cos(3t)-5\sin(3t)\}$.

Using linearity,

$$\mathcal{L}\{2\cos(3t)-5\sin(3t)\}=2\mathcal{L}\{\cos(3t)\}-5\mathcal{L}\{\sin(3t)\}.$$

Then

$$=2\left(\frac{s}{s^2+9}\right)-5\left(\frac{3}{s^2+9}\right)=\frac{2s-15}{s^2+9}.$$

How these properties connect to solving differential equations

Laplace transforms are not just for definitions and formulas. They are a method for solving differential equations.

A typical workflow is:

Take the Laplace transform of both sides of the differential equation.
Use derivative properties to rewrite the equation in terms of $Y(s)$.
Solve for $Y(s)$ algebraically.
Use partial fractions or transform tables.
Apply the inverse Laplace transform to get $y(t)$.

For example, if

$$y'+2y=4, \quad y(0)=1,$$

taking Laplace transforms gives

$$\mathcal{L}\{y'\}+2\mathcal{L}\{y\}=\mathcal{L}\{4\}.$$

$$sY(s)-y(0)+2Y(s)=\frac{4}{s}.$$

Using $y(0)=1$,

$$sY(s)-1+2Y(s)=\frac{4}{s}.$$

Then

$$Y(s)=\frac{\frac{4}{s}+1}{s+2}.$$

At that point, algebra and inverse transforms finish the problem.

This is why Laplace transforms are so important in Midterm 1 and the rest of the unit: they bridge calculus, algebra, and differential equations.

A quick check of understanding

students, ask yourself these questions:

Can I state the definition of $\mathcal{L}\{f(t)\}$?
Do I know why $\mathcal{L}\{f'(t)\}=sF(s)-f(0)$ is so useful?
Can I use linearity to transform sums quickly?
Do I recognize when a shift property is being used?
Can I explain how Laplace transforms simplify initial value problems?

If you can answer yes to most of these, you are building a strong foundation. ✅

Conclusion

The Laplace transform changes a function $f(t)$ into a new function $F(s)$ using the integral

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

Its strength comes from the properties that make differential equations easier: linearity, derivative rules, shifting rules, and formulas for common functions. These tools are especially useful when initial conditions and piecewise behavior appear, which is common in science and engineering.

For Midterm 1 and Laplace Transforms I, the main goal is not just memorizing formulas. It is understanding how the transform works, why it exists, and how its properties turn difficult differential equations into manageable algebra. That is the core idea you will use again and again in this course. 💡

Study Notes

The Laplace transform is defined by $\mathcal{L}\{f(t)\}=\int_0^\infty e^{-st}f(t)\,dt$.
The transform changes a time function $f(t)$ into an $s$-domain function $F(s)$.
Linearity means $\mathcal{L}\{af(t)+bg(t)\}=aF(s)+bG(s)$.
Derivatives become algebraic expressions, such as $\mathcal{L}\{f'(t)\}=sF(s)-f(0)$.
The first shifting theorem says $\mathcal{L}\{e^{at}f(t)\}=F(s-a)$.
The second shifting theorem uses the Heaviside function $u(t-a)$ and gives $\mathcal{L}\{u(t-a)f(t-a)\}=e^{-as}F(s)$.
Common transforms to know include $\mathcal{L}\{1\}=\frac{1}{s}$, $\mathcal{L}\{t\}=\frac{1}{s^2}$, $\mathcal{L}\{e^{at}\}=\frac{1}{s-a}$, $\mathcal{L}\{\sin bt\}=\frac{b}{s^2+b^2}$, and $\mathcal{L}\{\cos bt\}=\frac{s}{s^2+b^2}$.
Laplace transforms are especially useful for initial value problems because they include initial conditions naturally.
The standard solving process is transform, algebra, simplify, and inverse transform.
Laplace transforms connect directly to the main goals of Differential Equations by turning calculus problems into algebraic ones.