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Inverse Laplace Transform – Definition, Formulas, and Examples

The inverse Laplace transform is important when using Laplace transformation in differential equations. Knowing how to reverse the process of Laplace transformation leads to simpler processes when working on linear differential equations, since applying the inverse Laplace transform would be the last step.

The inverse Laplace transform allows us to reverse the process of Laplace transformation. The easiest way to find the inverse Laplace transform of functions is by having a table of transformations ready!

In this article, we’ll show you how an inverse Laplace transform operator works, and the essential properties defining this relationship. We’ll also make sure that you have enough examples to work on to know how the inverse Laplace transform works. For now, let’s do a recap on what Laplace transform represents and establish the rules for inverse Laplace transformation.

What Is an Inverse Laplace Transform?

From its name alone, the inverse Laplace transformation represents the inverse operation of the Laplace transform operator. Now, let’s define the inverse Laplace transform mathematically. Suppose that $F(s)$ is the Laplace transform of $f(t)$, we define $f(t)$ as the inverse Laplace transform of $F(s)$.

\begin{aligned}\boldsymbol{f(t) = \mathcal{L}^{-1}\{F(s)\}} &\boldsymbol{\Leftrightarrow }\boldsymbol{F(s) = \mathcal{L}\{f(t)\}}\end{aligned}

Keep in mind that the definition of the inverse Laplace transform remains true only when it satisfies the following condition: that the function, $f(t)$ is continuous on the interval, $t \in [0, \infty)$. Let us show you the general integral form of the inverse Laplace transform first:

\begin{aligned}\mathcal{L}^{-1} \{F(s)\} &= f(t)\\&= \dfrac{1}{2\pi k} \int_{c – k\infty}^{ c + k\infty} F(s) e^{st} \phantom{x} ds\end{aligned}

Don’t worry, thanks to our knowledge of Laplace transforms, we’ll have a simpler process of evaluating the inverse Laplace transforms.

How To Find Inverse Laplace Transform?

We can find the inverse Laplace transform of $F(s)$ by intuitively thinking of the function, $f(t)$, whose Laplace transform is equal to $F(s)$. It’s similar to when we have just started learning about integrals. We work backwards by thinking of a function, $f(x)$, that returns a derivative of $f^{\prime}(x)$.

\begin{aligned}\textbf{Inverse Laplace Transform}: \boldsymbol{f(t) = \mathcal{L}^{-1}\{F(s)\}}\end{aligned}

This means that it is important that you have the table of transformations from your notes on Laplace transform if you want to understand that. We’re simply reversing the process of Laplace transforms and now, instead of $F(s)$, we’re now finding the expression for $f(t)$.

Inverse Laplace Transform Formula of Common Functions

Let us rewrite the transformation table to highlight the inverse Laplace transform operator instead. To summarize the table, we can find the inverse Laplace transform of $F(s)$ by finding the function, $f(t)$, that has $F(s)$ as its Laplace transform.

\begin{aligned}\boldsymbol{F(s)} \end{aligned}

\begin{aligned}\boldsymbol{f(t) = \mathcal{L}^{-1} \{F(s)\}}\end{aligned}

\begin{aligned} \dfrac{1}{s}, \phantom{x}s > 0\end{aligned}

\begin{aligned} 1\end{aligned}

\begin{aligned} \dfrac{n!}{s^{n +1}},\phantom{x}s > 0\end{aligned}

\begin{aligned} t^n\end{aligned}

\begin{aligned} \dfrac{1}{s – k}, \phantom{x}s > k\end{aligned}

\begin{aligned} e^{kt}\end{aligned}

\begin{aligned} \dfrac{c}{s^2 + c^2},\phantom{x}s > 0\end{aligned}

\begin{aligned} \sin (ct)\end{aligned}

\begin{aligned} \dfrac{s}{s^2 + c^2}, \phantom{x}s > 0\end{aligned}

\begin{aligned} \cos (ct)\end{aligned}

\begin{aligned} \dfrac{c}{s^2 – c^2},\phantom{x}s >\pm c\end{aligned}

\begin{aligned} \sinh (ct)\end{aligned}

\begin{aligned} \dfrac{s}{s^2 – c^2}, \phantom{x}s > \pm c\end{aligned}

\begin{aligned} \cosh (ct)\end{aligned}

Here are some other properties of the inverse Laplace transform that will be helpful to keep in mind:

\begin{aligned}&(1) \phantom{x}\mathcal{L}^{-1} \{F(s) + G(s)\} = \mathcal{L}^{-1}\{F(s)\} + \mathcal{L}^{-1}\{G(s)\}\\\\&(2) \phantom{x}\mathcal{L}^{-1} \{cF(s)\} = c\mathcal{L}^{-1}\{F(s)\}\\\\ &(3)\phantom{x}\text{If }\mathcal{L}^{-1}\{G(s)\} = g(t) \text{ then }\mathcal{L}^{-1}\{G(s – k)\} = e^{kt}g(t)\\\\ &(4)\phantom{x}\text{If }\mathcal{L}^{-1}\{G(s)\} = g(t) \text{ then }\mathcal{L}^{-1}\left\{\dfrac{G(s)}{s}\right\} = \int_{0}^{t} g(t) \phantom{x}dt  \end{aligned}

Using Inverse Laplace Transform Formula

Now, let’s use the transformation table shown above to work on simpler and more complex functions. Here is some guideline to help you start applying the inverse Laplace transform formulas:

  • Rewrite the function, $F(s)$, so that it is broken down into expressions that have general forms found in a Laplace transform table.
  • Apply partial fraction decomposition when necessary to break down more complex expressions.
  • Use the transformation table and the properties of inverse Laplace transform to find the expression for $f(t)$.

Let’s begin with a simple example first by finding the inverse Laplace transform of $F(s) = \dfrac{6}{s^4}$. When given a rational function with $s^n$ in the denominator, try to rewrite the expression so that it is of the form, $\dfrac{n!}{s^{n +1}}$.

\begin{aligned} F(s) &= \dfrac{6}{s^4}\\&= \dfrac{3!}{s^{3 + 1}}\end{aligned}

Apply the inverse Laplace transform, $f(t) =\mathcal{L}^{-1} \left\{\dfrac{n!}{s^{n +1}}\right\}  = t^n$.

\begin{aligned}\mathcal{L}^{-1}\{F(s)\} &=\mathcal{L}^{-1}\left\{\dfrac{3!}{s^{3 + 1}}\right\} \\&= t^3\end{aligned}

This means that the inverse Laplace transform of $F(s) = \dfrac{6}{s^4}$ is equal to $f(t) = t^3$.

Now, let’s work on a more complex function, $F(s) = \dfrac{24}{s^5} + \dfrac{12}{s – 6}$, and find its inverse Laplace transform. Let’s first work on the inverse Laplace transform of each of the terms.

\begin{aligned}\boldsymbol{\dfrac{24}{s^5}}\end{aligned}

\begin{aligned}\boldsymbol{\dfrac{12}{s -6}}\end{aligned}

\begin{aligned}\mathcal{L}^{-1}\left\{\dfrac{24}{s^5}\right\} &= \mathcal{L}^{-1}\left\{\dfrac{4!}{s^{4 + 1}}\right\}\\&= t^4, \phantom{xx} \color{Teal}\mathcal{L}^{-1}\left\{\dfrac{n!}{s^{n +1}} \right\} = t^n\end{aligned}

\begin{aligned}\mathcal{L}^{-1}\left\{\dfrac{12}{s -6}\right\} &= 12 \cdot\mathcal{L}^{-1}\left\{\dfrac{1}{s – 6}\right\}\\&= 12e^{6t}, \phantom{xx} \color{Teal}\mathcal{L}^{-1}\left\{\dfrac{1}{s – k} \right\} = e^{kt}\end{aligned}

Use these resulting functions to find the inverse Laplace transform of $F(s)$.

\begin{aligned}f(t) &= \mathcal{L}^{-1}\left\{F(s)\right\}\\&= \mathcal{L}^{-1}\left\{\dfrac{24}{s^5} + \dfrac{12}{s – 6}\right\}\\&= \mathcal{L}^{-1}\left\{\dfrac{24}{s^5} \right\} + \mathcal{L}^{-1}\left\{\dfrac{12}{s – 6}\right\}\\&= t^4 + 12e^{6t} \end{aligned}

This means that the inverse Laplace transform of $F(s)$ is equal to $f(t) = t^4 + 12e^{6t}$.

Example 1

Find the inverse Laplace transform of the function, $F(s) = \dfrac{s + 2}{s^2 + 4}$.

Solution

Recall that we can break down $F(s)$ as a sum of simpler rational functions and the inverse Laplace transform of $F(s)$ will still be the same.

\begin{aligned}F(s) &= \dfrac{s + 2}{s^2 + 4}\\&= \dfrac{s}{s^2 + 4} + \dfrac{2}{s^2 + 4}\end{aligned}

These two rational functions look familiar because they have the general forms we can find in our table of transformations:  $\mathcal{L}^{-1}\left\{\dfrac{s}{s^2 + c^2}\right\} = \cos ct$ and $\mathcal{L}^{-1}\left\{\dfrac{c}{s^2 + c^2}\right\} = \sin ct$.

\begin{aligned}\mathcal{L}^{-1}\{F(s)\} &= \mathcal{L}^{-1} \left\{\dfrac{s}{s^2 + 4} + \dfrac{2}{s^2 + 4}\right\}\\&=\mathcal{L}^{-1} \left\{\dfrac{s}{s^2 + 4}\right\} + \mathcal{L}^{-1} \left\{\dfrac{2}{s^2 + 4}\right\}\\&=\mathcal{L}^{-1} \left\{\dfrac{s}{s^2 + 2^2}\right\} + \mathcal{L}^{-1} \left\{\dfrac{2}{s^2 + 2^2}\right\}\\&= \sin (2t) + \cos (2t) \end{aligned}

This means that the inverse Laplace transform, $F(s) = \dfrac{s + 2}{s^2 + 4}$, is equal to $f(t) = \sin (2t) + \cos (2t)$.

Example 2

Find the inverse Laplace transform of the function, $F(s) = \dfrac{12}{s^2 – 4s -12}$.

Solution

First, rewrite $F(s)$ so that it is a rational function . In this way, it contains factors or terms that look similar from the expressions we have from the table of transformations.

\begin{aligned}F(s) &= \dfrac{12}{s^2 – 4s -12}\\&=\dfrac{12}{s^2 – 4s + 4 – 16}\\&=\dfrac{12}{(s – 2)^2 – 16}\\&=\dfrac{3 \cdot 4}{(s – 2)^2 – 4^2} \\&=3 \cdot\dfrac{ 4}{(s – 2)^2 – 4^2} \end{aligned}

Our rational expression looks like $F(s) = \dfrac{c}{s^2 – c^2}$ and, as we know, it has an inverse Laplace transform of $f(t) = \sinh ct$. However, rational expression has $(s- 2)^2$ in its denominator, so we’ll need to apply the third property of inverse Laplace transform:

\begin{aligned (3)\phantom{x}\text{If }\mathcal{L}^{-1}\{G(s)\} = g(t) \text{ then }\mathcal{L}^{-1}\{G(s – k)\} = e^{kt}g(t) \end{aligned}

Let’s now evaluate $\mathcal{L}^{-1} \left\{\dfrac{12}{s^2 – 4s -12}\right\}$ using the inverse Laplace transform formula and the property as shown below.

\begin{aligned}\mathcal{L}^{-1}\{F(s – 2) \}&= \mathcal{L}^{-1}\left\{3 \cdot\dfrac{ 4}{(s – 2)^2 – 4^2}\right\} \\&= 3 \cdot\mathcal{L}^{-1}\left\{\dfrac{ 4}{(s – 2)^2 – 4^2}\right\}\\&= {\color{Teal}3e^{4t} }\mathcal{L}^{-1}\left\{\dfrac{ 4}{s^2 – 4^2}\right\}, \phantom{x}\color{Teal}\mathcal{L}^{-1}\{G(s – k)\} = e^{kt}g(t)\\&= 3e^{4t} {\color{Teal}\mathcal{L}^{-1}\{\sinh (4t)\}}, \phantom{x}\color{Teal}\mathcal{L}^{-1}\left\{\dfrac{c}{s^2 – c^2}\right\} = \sinh (ct)\\&= 3e^{4t}\sinh (4t)  \end{aligned}

This means that the inverse Laplace transform of $F(s) = \dfrac{12}{s^2 – 4s -12}$ is equal to $f(t) = 3e^{4t}\sinh (4t)$.

Example 3

Find the inverse Laplace transform of the function, $F(s) = \dfrac{s + 6}{s^2 – 6s – 16}$.

Solution

We can first rewrite $F(s)$ by decomposing it into sums of partial fractions.  As a refresher, the denominator of the rational function gives us an idea of how we can decompose it into partial fractions.

Denominator

Partial Fraction Decomposition

\begin{aligned}Ax + B\end{aligned}

\begin{aligned}\dfrac{A}{Ax + B}\end{aligned}

\begin{aligned}(Ax + B)^n\end{aligned}

\begin{aligned}\dfrac{A_1}{Ax + B} + \dfrac{A_2}{Ax + B}^2+ …+ \dfrac{A_n}{Ax + B}^n\end{aligned}

\begin{aligned} Ax^2 + Bx + C\end{aligned}

\begin{aligned}\dfrac{Ax + B}{Ax^2 + Bx + C} \end{aligned}

\begin{aligned} (Ax^2 + Bx + C)^n\end{aligned}

\begin{aligned}\dfrac{A_1x + B_1}{Ax^2 + Bx + C} +\dfrac{A_2x + B_2}{(Ax^2 + Bx + C)^2} + …+ +\dfrac{A_nx + B_n}{(Ax^2 + Bx + C)^n} \end{aligned}

Let’s use these general forms to write down the partial fraction decomposition for $F(s)$.

\begin{aligned}F(s)&= \dfrac{s + 6}{s^2 – 6s – 16}\\&=\dfrac{A}{s – 8} +\dfrac{B}{s + 2}\\&=\dfrac{A(s + 2) +B(s -8)}{(s – 8)(s + 2)} \\\\s + 6 &= (A+B)s + (2A – 8B)\\A = \dfrac{7}{5}&, B = – \dfrac{2}{5} \end{aligned}

This means that $F(s)$ can be written as $\dfrac{7/5}{s – 8} +\dfrac{-2/5}{s + 2}$.These two rational functions has a similar form to $\dfrac{1}{s – k}$, so it will now be easier for us to evaluate the inverse Laplace transform of $F(s)$.

\begin{aligned}F(s)&= \dfrac{7/5}{s – 8} +\dfrac{-2/5}{s + 2}\\&= \dfrac{7}{5} \cdot\dfrac{1}{s – 8} – \dfrac{2}{5} \cdot \dfrac{1}{s + 2}\\\\\mathcal{L}^{-1}\{F(s)\} &=\mathcal{L}^{-1}\left\{\dfrac{7}{5} \cdot\dfrac{1}{s – 8} – \dfrac{2}{5} \cdot \dfrac{1}{s + 2}\right\} \\&= \mathcal{L}^{-1}\left\{\dfrac{7}{5} \cdot\dfrac{1}{s – 8}\right\} – \mathcal{L}^{-1}\left\{\dfrac{2}{5} \cdot \dfrac{1}{s + 2}\right\} \\&= \dfrac{7}{5} \cdot\mathcal{L}^{-1}\left\{\dfrac{1}{s – 8}\right\} – \dfrac{2}{5} \cdot\mathcal{L}^{-1}\left\{ \dfrac{1}{s + 2}\right\}\\&= \dfrac{7}{5}e^{8t} – \dfrac{2}{5}e^{-2t}, \phantom{x}\color{Teal}\mathcal{L}^{-1} \left\{\dfrac{1}{s – k}\right\} = e^{kt}\end{aligned}

This means that the inverse Laplace transform of the function, $F(s) = \dfrac{s + 6}{s^2 – 6s – 16}$, is equal to $f(t) = \dfrac{7}{5}e^{8t} – \dfrac{2}{5}e^{-2t}$.

Practice Questions

1. Find the inverse Laplace transform of the function, $F(s) = \dfrac{5}{s – 5} + \dfrac{24}{s^4} – \dfrac{4}{s}$.
2. Find the inverse Laplace transform of the function, $G(s) = \dfrac{s + 3}{s^2 + 9}$.
3. Find the inverse Laplace transform of the function, $H(s) = \dfrac{s }{s^2 + 4s + 5}$.
4. Find the inverse Laplace transform of the function, $I(s) = \dfrac{s + 3}{s^2 – 7s – 18}$.
5. Find the inverse Laplace transform of the function, $J(s) = \dfrac{6s – 15}{(s – 1)(s + 2)(2s – 3)}$.

Answer Key

1. $\mathcal{L}^{-1}\{F(s)\} = f(t) = 5e^{5t} + 4t^3 – 4$
2. $\mathcal{L}^{-1}\{G(s)\} = g(t) = \sin (3t) + \cos (3t)$
3. $\mathcal{L}^{-1}\{H(s)\} = h(t) = e^{-2t}\cos t – 2e^{-2t}\sin t$
4. $\mathcal{L}^{-1}\{I(s)\} = i(t) = -\dfrac{1}{11}e^{-2t} + \dfrac{12}{11}e^{9t}$
5. $\mathcal{L}^{-1}\{I(s)\} = j(t) = 3e^t- \dfrac{9}{7}e^{-2t} – \dfrac{12}{7}e^{\frac{3t}{2}}$

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