Laplace Transform — Table, Properties & Inverse

The Laplace transform converts a time-domain function $f(t)$ into a new function $F(s)$ that is often easier to work with. In an introductory course, its main job is simple: turn differential equations with initial conditions into algebra problems, then use the inverse Laplace transform to return to $t$.

For the one-sided Laplace transform used in most differential-equations classes, the definition is

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

when the integral converges.

If $\operatorname{Re}(s)$ is large enough, the factor $e^{-st}$ suppresses large-$t$ behavior and the improper integral can stay finite. That convergence condition is part of the transform, not extra fine print.

What the Laplace transform helps you do

The transform does not change the problem's meaning. It repackages the problem in a form where differentiation becomes algebra.

That is why the method is especially useful for linear initial-value problems. You keep the initial condition, but the equation itself usually becomes easier to solve.

Laplace transform table: common pairs

These are the table entries students use most often. The condition in the right column matters because it tells you where the transform exists.

$f(t)$	$\mathcal\{L\}\{f(t)\}$	Valid when
$1$	$\frac\{1\}\{s\}$	$\operatorname\{Re\}(s) > 0$
$t$	$\frac\{1\}\{s^2\}$	$\operatorname\{Re\}(s) > 0$
$e^\{at\}$	$\frac\{1\}\{s-a\}$	$\operatorname\{Re\}(s) > a$
$\sin(bt)$	$\frac\{b\}\{s^2+b^2\}$	$\operatorname\{Re\}(s) > 0$
$\cos(bt)$	$\frac\{s\}\{s^2+b^2\}$	$\operatorname\{Re\}(s) > 0$

If you are working only with real-valued classroom examples, those conditions often appear as inequalities such as $s > 0$ or $s > a$. More generally, the transform is defined on a region of the complex $s$-plane.

Laplace transform properties that do most of the work

You do not need a long list. These three properties handle a large share of first-course problems.

Linearity

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

This lets you split a sum into simpler transforms.

Derivative rule

If $f$ is piecewise continuous on every finite interval and of exponential order, then

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

This is the key step in solving initial-value problems. The initial value appears automatically instead of being added later by hand.

Exponential shift

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

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

This is why many table entries are related by a simple shift in $s$.

Inverse Laplace transform: what it means

The inverse Laplace transform starts with $F(s)$ and recovers the time-domain function $f(t)$.

In theory there is a formal inversion formula. In most classroom problems, though, you do not evaluate that formula directly. You simplify $F(s)$ into known table forms, often with algebra or partial fractions, and then read the answer from the table.

Worked example: use the Laplace transform to solve an IVP

Consider

$$y'(t) + y(t) = 1, \qquad y(0) = 0$$

Let

$$Y(s) = \mathcal{L}\{y(t)\}$$

Take the Laplace transform of both sides:

$$\mathcal{L}\{y'(t)\} + \mathcal{L}\{y(t)\} = \mathcal{L}\{1\}$$

Use the derivative rule and the table entry for $1$:

$$sY(s) - y(0) + Y(s) = \frac{1}{s}$$

Since $y(0) = 0$,

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

So

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

Now split it into simpler fractions:

$$\frac{1}{s(s+1)} = \frac{1}{s} - \frac{1}{s+1}$$

Take the inverse Laplace transform term by term:

$$\mathcal{L}^{-1}\left\{\frac{1}{s}\right\} = 1, \qquad \mathcal{L}^{-1}\left\{\frac{1}{s+1}\right\} = e^{-t}$$

Therefore,

$$y(t) = 1 - e^{-t}$$

This is the full Laplace-transform workflow in one example: transform, solve in $s$, then invert. A differential equation became an algebra problem, and the initial condition was built into the calculation from the start.

Common Laplace transform mistakes

Forgetting the convergence condition

A table entry is only valid where the defining integral converges. If you ignore that condition, you are leaving out part of the answer.

Dropping the initial value in $\mathcal{L}\{f'(t)\}$

The term $-f(0)$ is easy to miss. If you leave it out, the transformed equation will usually solve the wrong problem.

Trying to invert too early

If $F(s)$ is a rational expression, it is often easier to simplify it first. Partial fractions are a common step before taking the inverse transform.

When the Laplace transform is used

The Laplace transform is especially useful for linear ordinary differential equations with initial conditions. That is the standard classroom use.

It also shows up in circuit analysis, control systems, signal modeling, and any setting where exponential responses and time-domain inputs need to be handled systematically.

Try your own version

Try the same workflow on

$$y'(t) + 2y(t) = 3, \qquad y(0) = 1$$

Transform the equation, solve for $Y(s)$, and then invert. If you want a quick check, compare your final $y(t)$ with the original initial condition and the differential equation at $t=0$.