The Laplace Transform: a powerful tool for solving differential equations

PLUS -- more info about the Machine Learning Course!

Before we begin, a quick announcement:

I’m starting a Machine Learning Course!

In the next few days, I’ll be launching a 6-week course teaching the fundamentals of machine learning.

In this course, you’ll work in a small group to build a machine learning model to solve a real-world problem of your choice.

It’s aimed at people who have no background in Machine Learning.

The course features:

• weekly video calls where I explain the content from that week

• an exclusive discord community where students can discuss concepts

• several presentations where you will present your progress on the project and I will give you detailed feedback.

You’ll hear more details about the course very shortly, so keep your eyes peeled!

With that, let’s get to the math:

Problem of the Week

Show that the equation

\(x^2 + y^2 = x^2 y^2\)

has no integer solutions other than x = y = 0.

Got a solution to the challenge problem? Submit it here.

Solution from last week

See here for the solution to last week’s problem. (Special thanks to Marc Caelles, whose solution is being featured this week!)

Topic of the week: the Laplace Transform

The Laplace transform is a powerful tool which is used to solve differential equations.

It takes as input a function of a real variable t, and it outputs a function of a complex variable s. It is defined by the following formula:

\(\mathcal{L}\{f \}(s) = \int_{0}^{\infty} f(t) e^{-st} \, dt.\)

Here are some properties of this transform:

1. The Laplace transform is linear. It satisfies the relations

   \(\mathcal{L}\{ f(t) + g(t) \} = \mathcal{L} \{ f(t) \} + \mathcal{L} \{ g(t) \}\)

   and

   \(\mathcal{L}\{ a f(t) \} = a\mathcal{L} \{ f(t) \} .\)

2. The Laplace transform converts multiplication by t into differentiation:

   \(\mathcal{L}\{ tf(t) \} = - \dfrac{d}{ds} \mathcal{L} \{f \}(s).\)

3. The Laplace transform converts convolution into multiplication:

   \(\mathcal{L} \{f \star g \} = \mathcal{L}\{ f\} \cdot \mathcal{L}\{ g\}.\)

What are the uses of this transform?

One application is in differential equations. In some special cases, it can transform a differential equation into an algebraic equation, which is in turn easier to solve.

This means that the Laplace transform comes up in all fields where linear differential equations are used. It comes up in electrical engineering, where you need to analyze circuit diagrams using differential equations. It also comes up in statistics in the form of moment generating functions (MGFs).

Here are some learning resources to learn about the Laplace transform in more detail:

1. Paul Lamar has some excellent notes about the Laplace transform. I used these notes quite heavily myself when I was learning this content during undergrad!

2. Gilbert Strang has two excellent videos explaining how to apply the Laplace transform to differential equations: see here and here. This gives a good sense of how to apply the Laplace transform to differential equations.

Thanks for reading and happy learning! Until next time,

Adithya