Lecture notes from the first lecture of [[MATH 325]], taught by Anthony Humphries. In this lecture: information on the course (grading scheme, expectations, etc) and an introduction to ODEs, complete with a solved example.

Under construction.

[TOC]

Course information

Announcements and postings

The course outline is available through MyCourses. Announcements will be made either in class or via the MyCourses email system or both. Assignments will be posted on MyCourses.

Course material and textbooks

This course will focus on the theory of ODEs, with minimal coverage of applications. There is no required textbook, although there are two recommended textbooks: Elementary Differential Equations, by W.E. Boyce and R.C. DiPrima, and a textbook of the same name by C.H. Edwards and D.E. Penny. The latest versions are the 10th and 6th edition, respectively, though earlier versions are fine as well. According to the profesosr, neither book is substantial enough for this course, which is why they are only supplementary; some of the underlying theory is glossed over in the texts. Both textbooks should be available in the library. If you want to buy the textbooks, you don't need the "boundary problems" versions.

Since there is no required textbook, it is important to attend class in order to avoid missing material. I will try to post the lectures on wikinotes as soon as possible after class, but these notes should not be considered a substitute for regularly attending lectures.

Grading scheme

Final mark: 20% written assigments (6 total), 20% in-class midterm (on Thursday, February 28), 60% final exam. There will be no other grading scheme options. Assignments must be handed in on time, as late assignments will not be accepted (unless other prior arrangements have been made with the professor).

There will be WebWork questions, but they will be not be for credit and are merely intended to serve as practice problems.

Introduction to ODEs

Ordinary: one independent variable Differential: derivatives will be involved Equations: note the presence of an equals sign

ODES differ from PDEs, which involve multiple independent variables (and thus partial derivatives, hence the name),

Trivial examples

$\displaystyle \frac{dy}{dx} = 4x$. To solve it: integrate both sides with respect to $x$. Solution: $y = 2x^2 + C$ lol.
$\displaystyle \frac{d^2y}{dx^2} = 0$. Solution: $y = C_1x + C_2$.
$\displaystyle \frac{dy}{dx} = y$. Solution: $y = e^{x+C}$ (or, more generally: $y = ke^x$ where $k$ is some constant; this is more general because it includes the valid solution $y=0$, while the other does not). The question of where to put the constant is a bit more tricky in this situation, since if we integrate both sides of the ODE we get that $y = \int y,dx$ and so we can't just have $y = e^x + C$.

ODEs with constraints

Often, we'll try to solve ODEs subject to constarints. Of these types of problems, we will focus on initial value problems, in which we are given $y'$, as well as the value of as many derivatives as needed at some point $x_0$ (so $y(x_0)$, possibly, $y'(x_0)$, and maybe $y''(x_0)$, etc).

In example 3 above, with $y = ke^x$, we only need the value of $y(x_0)$ to be able to solve it, since there's only one constant. In example 2, however, we would need $y(x_0)$ and $y'(x_0)$, since there are two constants. In general, if the ODE has a derivative of degree $n$ (i.e., $\displaystyle\frac{d^ny}{dx^n}$), we would need $n$ derivatives (including $y(x_0)$) to get a unique solution.

Some notes on notation

In this class, we will usually use $y'$ instead of $\frac{dy}{dx}$, and occasionally $\dot y$ instead of $\frac{dy}{dt}$, where $t$ is time (this was Newton's notation). Less frequently, we'll use $Dy$ (Leibniz's notation), which indicates the derivative with respect to the argument.

A detailed example

Here's an example that Newton came up with in 1671:

$$\frac{\dot y}{\dot x} = 1 - 3x + y + xx + xy$$

Translated into modern notation:

$$y' = 1-3x+y+x^2+xy$$

Equivalently, we can write

$$y' - (1+x)y = 1-3x+x^2$$

(This is a specific example of the general class of problems $y' + p(x)y = g(x)$, which we will consider in a later lecture.)

First, some remarks on continuity. Clearly $y$ is continuous and differentiable, since we have $y'$ up there. Also, since $y'=g(x)-p(x)y$, $y'$ is also continuous. Furthermore, if $p$ and $g$ are infinitely differentiable, so is $y$. Not sure why this is relevant or even if I wrote this down accurately.

Now, let's look at how Newton solved this. His method will seem rather clunky compared to how we might approach this today, but then again, his technique was state-of-the-art at the time and in any case it's quite instructive. He formulated this as an initial value problem, with $y(0) = 0$. Then, $y'(0) = 1-3(0) + 0 + 0^2 + 0 = 1$, so $y' = 1 + \ldots$. If we integrate this, $y = x + \ldots$.

We continue in the same fashion to improve the approximation. $y' = x + (1 -3x +\ldots) = 1 - 2x + \ldots$ and so $y = x-x^2 + \ldots$. Then $y' = (1+x)(x-x^2+\ldots) + (1-3x+x^2+\ldots) = 1 - 2x + x^2 + \ldots$, so $y = x-x^2+\frac{1}{3}x^3 + \ldots$. Eventually, we'll get

$$y = x-x^2+\frac{1}{3}x^3-\frac{1}{6}x^4 + \frac{1}{30}x^5 - \frac{1}{45}x^7 + \ldots$$

So this technique reduces the solving of the ODE to a simple algorithnm, but one that requires infinitely many integrations[^newton]. Another technique that involves fewer (infinitely fewer, in fact) integrations makes use of Taylor series. If we ignore the issue of whether or not the series converges, we can write

$$y(x) = \sum_{n=0}^{\infty} \frac{1}{n!} y^{(n)}(x_0) (x-x_0)^n$$

(where $y^{(n)}$ is the $n$th derivative). Since $y(0) = 0$, we have that

$$y(x) = \sum_{n=1}^{\infty} \frac{1}{n!} y^{(n)}(0)x^n$$

But $y'(0) = 1$, as we know from earlier. We can obtain the second derivative by differentiating $y'(x)$: $y''(x) = -3 + y' + 2x + y + y'x$, and so $y''(0) = -3 + 1 + 2(0) + 0 + 1(0) = -2$. Likewise for the third derivative: $y'''(x) = y'' + 2 + y' + y''x + y'$, so $y'''(0) = (-2) + 2 + 1 + (-2)(0) + 1 = 2$. Then, if we substitute these into the summation, we get:

$$y(x) = y(0) + y'(0)x + \frac{1}{2} y''(0)x^2 + \frac{1}{6} y'''(0)x^3 + \ldots = 0 + 1 - x^2 + \frac{1}{3}x^3 + \ldots$$

which is the same as what we would have gotten using Newton's method. And we didn't even have to integrate anything!

[newton]: I'm not really sure how this works to be honest. To be revisited.

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