In the previous lesson, we studied the Laplace transform and highlighted different ideas through step-by-step examples. In this lesson, we will present the **inverse Laplace transform definition** and how it can be used to solve related problems!

## The inverse Laplace Transform

When we apply the Laplace transform method to solve the initial-value problem below

\left\{\begin{array}{l} ay''+by'+cy=f(t), \\ y(0)=y_{0} \\ y'(0)=y_{1} \end{array}\right.

we do not determine y(t) directly. Instead, an equation envolving the function Y(s) (the Laplace Transform of y(t) ) comes up.

However, as we are interested to find the function y(t) and since we have its Laplace Transform, the main idea is to make the inverse process, i.e., given \mathscr{L}\{y(t)\} , we look for y(t) .

To compute the inverse Laplace transform, we identify the parameters a,b and n in F(s) to be able to read off \mathscr{L}^{-1}\{f(t)\}.

## Illustrative examples about inverse Laplace transform

### Example 01:

Let:

\displaystyle F(s)=\frac{1}{s-5}

be a function in the variable s . To evaluate its Laplace inverse transform, we need to discover f(t) such that:

f(t)=\mathscr{L}^{-1}\{F(s)\}\Rightarrow f(t)=\mathscr{L}^{-1}\left(\frac{1}{s-5}\right)

i.e., f is the function which Laplace Transform is:

\displaystyle F(s)=\frac{1}{s-5}

From the Example 2 in Lesson 1, if f(t)=e^{at}, then

\displaystyle F(s)=\mathscr{L}\{f(t)\}=\frac{1}{s-a}

This shows us that it just requires to identify the parameter a. Certainly a=5 in F(s) given. Therefore \mathscr{L}^{-1}\{F(s)\} is the function

f(t)=e^{at}\xRightarrow{a=5} f(t)=e^{5t}

### Example 02:

If f(t)=\cos(at) , the Laplace transform of f is the function

\displaystyle F(s)=\frac{s}{s^{2}+a^{2}}

As we did before, a is the only parameter we need to identify.

So, let F be a function in the variable s such that:

\displaystyle F(s)=\frac{s}{s^{2}+9}

We can rewrite F as follows:

\displaystyle F(s)=\frac{s}{s^{2}+9}=\frac{s^{2}}{s^{2}+3^{2}}

Therefore, a=3. Hence:

f(t)=\mathscr{L}^{-1}\{F(s)\}=\cos(3t)

### Example 03:

To find the inverse Laplace transform of:

\displaystyle F(s)=\frac{3}{s^{2}-4s+13}

it may be not so simple like the last examples, due to the fact it does not match exactly to any Laplace transform. At first, since (check this post):

\displaystyle \mathscr{L}\{e^{at}\sin(bt)\}=\frac{b}{(s-a)^{2}+b^{2}}\Rightarrow\mathscr{L}^{-1}\left\{\frac{b}{(s-a)^{2}+b^{2}}\right\}=e^{at}\sin(bt)

then we have to rewrite F in a way it comes up a square of a difference. As:

(x+y)^{2}=x^{2}+2\cdot x\cdot y+y^{2}

and

s^{2}-4s=s^{2}-2\cdot s\cdot 2

so the second term of the difference is 2, i.e.,

s^{2}-4s+2^{2}=(s-2)^{2}

Furthermore,

13=4+9\Rightarrow13=2^{2}+3^{2}

Hence,

\begin{aligned} F(s) & = \frac{3}{s^{2}-4s+13} \\ & = \frac{3}{s^{2}-2\cdot s\cdot2+2^{2}+3^{2}} \\ & = \frac{3}{(s-2)^{2}+3^{2}} \end{aligned}

which gives us:

a=2\quad\text{and}\qquad b=3

Thus,

\mathscr{L}^{-1}\{F(s)\}=f(t)=e^{2t}\sin(3t)

## Conclusion

In this lesson, we presented the** inverse Laplace transform definition**. To illustrate ideas, we applied the definition to three different examples.