Laplace transforms can be used to turn a differential equation into an complex algebra equation.

The Laplace Transform is defined by the following equation.

Note that we have an improper integral here, so you should remember how to solve an improper integral.

It is usually defined as an operator $L$ that transforms the function $f(t)$ into the Laplace Transform $F(s)$

Example 1: Find the Laplace transform of $f(t)=1$

I just skipped the steps for solving improper integrals. If you want to see the full steps, Click here.

So, we have $F(s)=\frac{1}{s}$, where $x>0$

However, rather than solving a improper integral every time, we usually just use a table of Laplace Transforms. For a table of Laplace transforms, click here.

I will list a few common transforms below.

Inverse Laplace Transforms

We can actually take the inverse of a Laplace transform, which is basically going from $F(s)$ to $f(t)$, rather than $f(t)$ to $F(s)$. This is usually defined by the operator $L^{-1}$.

To find the inverse Laplace transform, just use the table above.

Example 2: Find $L^{-1}\{F(s)\}$, where $F(s)=\frac{6}{s} + \frac{1}{s-8} + \frac{4}{s-3}$.

Solution: One property of Laplace Transforms (and inverse Laplace transforms) is that there is no need to worry about the sums or differences of Laplace transforms, just take the transform of each individual part and sum the rest.

In other words

Take the individual inverse Laplace transforms

• $L^{-1}\{\frac{6}{s}\}=6$
• $L^{-1}\{\frac{1}{s-8}\}=e^{8t}$
• $L^{-1}\{\frac{4}{s-3}\}=4e^{^{3t}}$

I found these inverses by using the table that I showed above.

Substitute these inverse transforms in to get a final answer.