Lesson 21 - Definition of Laplace Transforms

We have seen roughly $8$ methods for solving differential equations. Let's take a break from this to discuss a powerful tool in solving differential equations, which involves transforming a function, typically some function in terms of time, into the $s$ domain, meaning we will not have a function in terms of $s$ .

f (t) \to f (s)

In doing so, we will uncover a few Laplace transforms that often come up in differential equations, and in the end we will construct a table of common Laplace transforms that we will utilize. The lesson will be the derivation of those Laplace transforms.

To start, let's introduce a definition of the Laplace transform.

Let $f (t)$ be a function on $[0, \infty)$ . The Laplace transform of the function $f (t)$ is the function $F (s)$ defined by the integral

L [f (t)] = F (s) := \int_{0}^{\infty} e^{- s t} f (t) d t

In order to evaluate this improper integral, we can evaluate as such:

\int_{0}^{\infty} e^{- s t} f (t) d t = lim_{b \to \infty} \int_{0}^{b} e^{- s t} f (t) d t

The property of linearity applied to Laplace transforms as well:

L [a f_{1} (t) + b f_{2} (t)] = a L [f_{1} (t)] + b L [f_{2} (t)]

Example

Find the following Laplace transforms:

$L [1]$
$L [e^{a t}]$

For $L [1]$

L [1] = \int_{0}^{\infty} (e^{- s t} \cdot 1) d t

Using our definition of Laplace transforms:

lim_{b \to \infty} \int_{0}^{b} (e^{- s t}) d t

Recall that the derivative of $e^{t}$ $=$ $e^{t}$ .

Now, recall that the derivative $e^{k t}$ $=$ $k e^{k t}$ .

We want to find the integral of $e^{k t}$ . So, we use the inverse of $k$ . $\int e^{k t} d t$ $=$ $\frac{1}{k} e^{k t} + c$ . If we take the derivative of $\frac{1}{k} e^{k t} + c$ , we end up with the original integral $e^{k t}$ . This will be useful for us going forward.

= lim_{b \to \infty} {[- \frac{1}{s} e^{- s t}]}_{0}^{b}

= lim_{b \to \infty} [- \frac{1}{s e^{s b}} + \frac{1}{s}]

If we analyze the limit of this function, as $- \frac{1}{b e^{s t}}$ approaches infinity, the function goes to zero.

= [0 + \frac{1}{s}] = \frac{1}{s}

So,

L [1] = \frac{1}{s}

For $L [e^{a t}]$ , where $a$ is a constant.

L [e^{a t}] = \int_{0}^{\infty} (e^{- s t} \cdot e^{a t}) d t

= lim_{b \to \infty} \int_{0}^{b} e^{- s t + a t} d t

Factor out $- t$ in the exponent.

= lim_{b \to \infty} \int_{0}^{b} e^{- t (s - a)} d t

= lim_{b \to \infty} {[- \frac{1}{s - a} e^{- t (s - a)}]}_{0}^{b}

= [- \frac{1}{s - a} e^{- b (s - a)} - \frac{- 1}{s - a} e^{- (s - a) 0}]

Similarly as before, as $- \frac{1}{s - a} e^{- b (s - a)}$ approaches to infinity, the function goes to 0.

= 0 + \frac{1}{s - a} = \frac{1}{s - a}

So,

L [e^{a t}] = \frac{1}{s - a}

This next try for yourself contains the remaining Laplace transforms. I encourage you to try these and compare to the Laplace transform table in the next mini lesson.

Try for Yourself
3. $L [\sin (α t)]$
4. $L [\cos (α t)]$
5. $L [t]$
6. $L [t^{n}]$
7. $L [e^{a t} f (t)]$
8. $L [e^{a t} t^{n}]$

Next Lesson: Lesson 21.1 - Table of Laplace Transforms

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