Chapter 1 #

What are we studying? #

Describe the function \[ f(x) = 2x \]

If you said: “A line where for every step to the right, one goes two steps up”. You would be exactly right! Of course, there are other ways of saying this. Something like “take the value of x and f(x) is twice that” would be correct.

Let’s go deeper. Using the power rule and that \(x = x^1\), \[ \frac{d}{dx}\bigl(x^1\bigr) = 1\cdot x^{1-1} = 1\cdot 1 = 1\\ f'(x) = \frac{d}{dx}(2x) = 2\cdot\frac{d}{dx}(x^1) = 2\cdot1 = 2 \]

Now, does the derivative being 2 imply that f(x) = 2x?

Try taking the derivative of f(x) = 2x + 1. Still 2?

Interesting…

Ordinary Differential Equations (ODE): #

If you’ve been following along, you’ve already seen an ODE! \[ \frac{dy}{dx} = y' = 2 \]

Definition: Ordinary Differential Equation (ODE): An equation that contains one or more derivatives of an unknown function.

We also have the same notion for partial derivatives. \[ \frac{\partial^2\phi}{\partial x^2} + \frac{\partial^2\phi}{\partial y^2} = 0 \] The Laplace Equation.

Definition: Partial Differential Equation (PDE): An equation that contains one or more partial derivatives of an unknown function of two or more variables.

Practice: #

\[ \text{Is }\space y''' + 3y'' + 3y' + y = 0 \space \text{a differential equation?} \]

Yes!

\[ \text{Give an example of a differential equation about something in the real world.} \]

Examples in the book.

Order of Differential Equations: #

The order of an ODE is simply the highest derivative.

\[ \text{Find the order of: } y'' + y''' = 7 \]

3 (from y''')

Definition: Explicit form of a differential equation: y’ = f(x,y)

Solutions of ODE #

A solution is simply the function that satisfies the differential equation.

The “best” example is \[ y' = 0.2y. \]

Plug in \(y = e^{0.2x}\).
Plug in \(y = 2e^{0.2x}\).
Plug in \(y = ce^{0.2x}\) where c is a constant.
Definition: General Solution: Solutions to differential equations with arbitrary constant(s) c.

Definition: Particular Solution: A solution to a differential equation without arbitrary constants.

Practice 2: #

Using the solution to the differential equation above, try to find a solution to: \[ y' = 0.5y \]

\(y = ce^{0.5y}\)

Hint

Look at the example and replace 0.2 with the appropriate value.

Initial Value Problems #

Definition: Initial Value Problem An ODE with an initial value.

With a line:

Take a point, say (0,0) and a slope, say 2. The only function with this property is f(x) = 2x.

From the book, \[ y' = \frac{dy}{dx} = 3y \quad y(0) = 5.7 \]

First, you should SOLVE the ODE.
Then, PLUG IN using the initial value condition(s).

Chapter 1.2 #

Definition: Direction Field A description of the rate of change on the possible values of a function.

See the book.

Euler’s Method #

Given h (time step) and f’ (derivative), simply take the initial x and y values \((x_0,y_0)\). Now, using the slope at \((x_0,y_0)\), take the new y \((y_1)\) to be the time step (h) times the derivative \(f’(x_0,y_0)\).

Chapter 1.3 #

Separable ODEs.

Definition: Separable ODE Equations that are of the form or can be manipulated into the form \(g(y)y’ = f(x)\).

Homogeneous ODE #

\(y’ = f(\frac{y}{x})\) f is differentiable with respect to \(\frac{y}{x}\).

Let \(u = \frac{y}{x} \implies y = ux \implies y’ = u’x + u\)

1.4 #

Definition: Differential: Given a function has continuous partial derivatives, the differential is \(du = \frac{\partial u}{\partial x}dx + \frac{\partial u}{\partial y}dy\)

Exact ODE #

Definition: Exact Differential Equation:
M(x,y) + N(x,y)y’ = 0

This is equivalent to \[ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \] .

Step 1: Test for exactness: #

Essentially, “complete” the partials. \[ \frac{\partial M}{\partial y} = \frac{\partial^2 u}{\partial y\, \partial x}, \quad \frac{\partial N}{\partial x} = \frac{\partial^2 u}{\partial x\, \partial y}. \] .

Step 2: Integrate in either way #

Calculate:

Calculate: \[ u = \int M dx + k(y) \\ or \\ u = \int N dy + l(x) \]

Depending on which is easier.

Find \(k\)(y) or \(l\)(x) by taking the partial derivate of u with respect to the varible of the \(k\) or \(l\) function.
The result \(\frac{\partial u}{\partial (x \space or \space y)}\) is either M or N.
This gives you the derivative of the \(k\) or \(l\) function.
Integrate \(k\) or \(l\) to get u(x,y).

Step 3: Check u(x,y) is correct #

Plug in \(du = \frac{\partial u}{\partial x}dx + \frac{\partial u}{\partial y}dy\)

Chapter 1