Differential Equations Notes

Math 243 - Fall 2025

• Examples
• Notes
• Schedule & Syllabus

Week 1 Notes

Mon, Aug 25

We talked about some examples of differential equations.

1. Exponential Growth/Decay. The rate of change in a variable $y$ with respect to time $t$ is proportional to $y$ itself. $$\dfrac{dy}{dt} = k y.$$
   1. Check that $y(t) = Ce^{kt}$ is a solution.
   2. Find the constant $C$ which satisfies the initial value problem with initial condition $y(0) = 1000$.

We talked about dependent and independent variables, the order of a differential equation and how to tell if a function is a solution of a differential equation. We also talked about initial conditions.

2. Spring-Mass Model. The force $F$ of a mass $m$ at the end of a spring can be modeled by Hooke’s Law which says $F = -k x$ where $x$ is the displacement of the spring from its rest position. $$m\dfrac{d^2 x}{dt^2} = -kx.$$

   1. Show that $x = \sin t$ and $x = \cos t$ are two different solutions of the spring equation when $m = k = 1$.
   2. How would the solutions change if $m$ and $k$ were not both equal to 1? How would the oscillation of the spring change if the mass was twice as heavy?
      
   3. How would the spring equation change if there was also a linear drag force $F = -b \frac{dx}{dt}$? What if there is a non-linear drag force $F = -b \left(\frac{dx}{dt}\right)^2$?

The last question led to a discussion of linear versus non-linear differential equations. It’s usually much harder to solve non-linear equations! We will also study systems of differential equations, like the following.

3. Rabbits and Foxes. Suppose there are $R$ rabbits and $F$ foxes in an ecosystem. The rabbit population would grow exponentially, except for the foxes which prey on the rabbits. The fox population would decay exponentially if there wasn’t food to eat, but as long as they can catch enough rabbits, it will grow. The number of rabbits that are eaten by foxes is proportional to the product $RF$.
   $$\begin{align*} \dfrac{dR}{dt} &= a R - b RF \\ \dfrac{dF}{dt} &= -c R + d RF \end{align*}$$

Here is a graph showing these equations as a vector field (with constants $a = 3, b = 4, c = 1, d= 2$).

4. Logistic Growth. $\dfrac{dP}{dt} = kP \left( 1 - \dfrac{P}{N} \right)$ where $k$ is a proportionality constant and $N$ is the carrying capacity.

   1. Under what circumstances does the population $P$ stop changing in this model?
   2. What are the units for the constants $k$ and $N$?
   3. Suppose that we use the logistic growth equation to model a population of rabbits in a region. What if we introduce a predator that always consumes $b$ rabbits per year. How would that change the differential equation above?

Wed, Aug 27

Today we talked about separable equations. We solved the following examples.

1. Solve $\dfrac{dy}{dx} = - x^2 y$.

2. Solve $xy^2 y' = x+1$. (https://youtu.be/1_Q4kndQrtk)

Not every differential equation is separable. For example: $$\frac{dy}{dx} = x-y$$ is not separable.

3. Which of the following differential equations are separable? (https://youtu.be/6vUjGgI8Dso)

   1. $xy' + y = 3$
      
   2. $2x + 2y + 2y' - 1 = 0$
   3. $y' = (x^2+x)(y^2+y)$
   4. $x \dfrac{dy}{dx} + y \dfrac{dy}{dx} = x$

We finished with this example:

4. Newton’s Law of Cooling. The temperature of a small object changes at a rate proportional to the difference between the object’s temperature and its surroundings.

   1. Express Newton’s Law of Cooling as a differential equation.
      
   2. Is that differential equation separable?

5. Mixing Problem. Salty water containing 0.02 kg of salt per liter is flowing into a mixing tank at a rate of 10 L/min. At the same time, water is draining from the tank at 10 L/min.

   1. Write a differential equation to model how the amount of salt in the tank changes with respect to time.
      
   2. Solve the differential equation if the amound of salt in the tank is initially 15 kg. (https://youtu.be/aFfAz9wnoyY)

We didn’t get to this last example in class, but it is a good practice problem.

6. $\dfrac{dy}{dx} = \dfrac{4 \sin x}{3 y^2}$ with initial condition $y(0) = 2$. (https://youtu.be/cc3qtMBdQlE)

Fri, Aug 29

Today we talked about slope fields.

1. Sketch the slope field for $\dfrac{dy}{dx} = x - y$. (https://youtu.be/24pxJ1DuWR8)

Here is a slope field grapher tool that I made a few years ago. You can also use Sage to plot slope fields. Here is an example from the book, with color added.

t, y = var('t, y')
f(t, y) = y^2/2 - t
plot_slope_field(f, (t, -1, 5), (y, -5, 10), headaxislength=3, headlength=3, 
    axes_labels=['$t$','$y(t)$'], color = "blue")

2. Consider the logistic equation with harvesting. $$\dfrac{dP}{dt} = k P \left(1 - \dfrac{P}{N} \right) - h$$ where $h$ is a number of rabbits that are harvested each year.

   1. If $k = 1$, $N = 8$, and $h = 1.5$, then what are the values of $P$ where $\dfrac{dP}{dt} = 0$? Slope field
   2. Graph the function $$y =k x \left(1 - \dfrac{x}{N} \right) - h.$$ Where does the graph cross the x-axis? Is the slope positive or negative at those crossing points?

The logistic equation (with or without harvesting) is autonomous which means that the rate of change $\dfrac{dP}{dt}$ does not depend on time, just on $P$. An equilibrium solution for an autonomous differential equation is a solution where $y'(t) = 0$ for all $t$.

3. In the logistic equation above, what happens to the equilibrium solutions when the rate of harvesting is increased to $h = 2$ and then to $h = 2.5$? What happens to the slope field? What does that mean about the population of rabbits?

Week 2 Notes

Wed, Sep 3

Last time we talked about equilibrium solutions of autonomous equations. An equilibrium $y_0$ for $y' = f(y)$ is stable (also known as a sink or attactor) if any solution with initial value close to $y_0$ converges to $y_0$ as $t \rightarrow \infty$. An equilibrium is unstable (also known as a source or repeller) if all solutions move away from $y_0$ as $t \rightarrow \infty$.

1. Consider the ODE $y' = y(y-2)(y+3)$. What are the equilibria for this ODE? Which are stable and which are unstable?

2. What would happen to the number of equilibrium solutions if we replaced $y(y-2)(y+3)$ by $y(y-2)(y+3) + 5$?

We talked about the phase line for an autonomous ODE.

2. Draw different phase lines for the logistic equation with harvesting parameters $h = 0, 1.5, 2, 2.5$ $$y' = y \left( 1 - \frac{y}{8} \right) - h$$

Suppose that $y' = f_\lambda(y)$ is a family of differential equations that depends on a parameter $\lambda$. A bifurcation point is a value of the parameter where the number of equilibrium solutions changes. A bifurcation diagram is a graph that shows how the phase lines change as the value of a parameter changes.

3. Draw a bifurcation diagram for the differential equation $y' = \lambda y - y^2$ showing the phase lines when $\lambda = -1, 0,$ and $1$.

You can use Desmos to help with the previous problem. Using $x$ to represent $\lambda$, you can graph the region where $dy/dt$ is positive in blue and the region where $dy/dt$ is negative in red. Then it is easier to draw the phase lines in the bifurcation diagram.

Fri, Sep 5

Today we talked about two important theorems in differential equations.

This theorem guarantees that in most circumstances, we are guarantee to have solutions to differential equations. But there are things to watch out for. Solutions might blow up in finite time, so they might not be defined on the whole interval $(a,b)$.

1. Solve the IVP $y' = y^2$ with initial condition $y(0) = 1$. Notice that the function $f(t,y) = y^2$ is continuous everywhere. But the solution is not.

If the partial derivative $f_y$ is not continuous, then we might not get unique solutions. Here is an example.

2. Solve the IVP $y' = y^{1/3}$, with $y(0) = 0$ using separation of variables. Then show that $y(t) = -(\tfrac{2}{3} t)^{3/2}$ and $y(t) = 0$ are also valid solutions of this IVP.

One very nice consequence of the uniqueness theorem is this important concept:

3. In our first homework we solved the IVP $xy' = \sqrt{1-y^2}$, with $y(1) = 0$. The solution was $y = \sin(\ln(t))$. But if you graph the solution with the slope field, there is something wrong! SageCell Plot

This illustrates that a formula for a solution to $y' = f(t,y)$ might not apply after we reach a point where $f_y$ is no longer continuous.

Week 3 Notes

Mon, Sep 8

Many ODEs cannot be solved analytically. That means there is no formula you can write down using standard functions for the solution. This is true even when the existence and uniqueness theorems apply. So there might be a solution that doesn’t have a solution you can write down. But you can still approximate the solution using numerical techniques.

Today we introduced Euler’s method which is the simplest method to numerically approximate the solution of a first order differential equation. We used it to approximate the solution to $$\dfrac{dy}{dt} = \dfrac{y^2}{2} - t \text{ with initial condition } y(-1) = 0.$$

from numpy import *
import matplotlib.pyplot as plt

def EulersMethod(f,a,b,h,y0):
    ''' 
    Approximates the solution of y' = f(t, y) on the interval a < t < b with initial 
    condition y(a) = y0 and step size h. 
    Returns two lists, one of t-values and the other of y-values. 
    '''
    t, y = a, y0
    ts, ys = [a], [y0]
    while t < b:
        y = y + f(t,y)*h
        t = t + h
        ts.append(t)
        ys.append(y)
    return ts, ys

f = lambda t,y: y**2 / 2 - t

# h = 1
ts, ys = EulersMethod(f, -1, 5, 1, 0)
plt.plot(ts,ys)
# h = 0.1
ts, ys = EulersMethod(f, -1, 5, 0.1, 0)
plt.plot(ts,ys)
# h = 0.01
ts, ys = EulersMethod(f, -1, 5, 0.01, 0)
plt.plot(ts,ys)
plt.show()

Here’s the output for this code and here is a version with the slope field added.

After demonstrating how to implement Euler’s method in code, we talked about some simpler questions that we can answer with pencil & paper.

1. Suppose that we want to solve $\dfrac{dy}{dx} = x - y$ with initial condition $y(0) = 2$. Make a table showing the first three steps using Euler’s method with step size $h = 1$.

Euler’s method is only an approximation, so there is a gap between the actual y-value at $t = b$ and the Euler’s method approximation. That gap is the error in Euler’s method. There are two sources of error.

• Discretization Error - caused by using a discrete approximation to a continuously changing quantity.
• Rounding Error - caused by using a computer that can’t represent small decimal places accurately.

As $h$ gets smaller, the discretization error gets smaller, but the rounding error gets worse.
A worst case upper bound for the error is: $$\text{Error} \le \dfrac{e^{L(b-a)} - 1}{L} \left( \dfrac{Mh}{2} + \dfrac{\delta}{h} \right)$$ where $L = \max \left| \frac{\partial f(t,y)}{\partial y} \right|$, $M = \max |y''(t)|$, and $\delta$ is the smallest floating point number our computer can accurately represent. Using the standard base-64 floating point numbers, $\delta \approx 10^{-16}$. In practice, Euler’s method tends to get more accurate as $h$ gets smaller until around $h \approx 10^{-7}$. After that point the rounding error gets worse and there is no advantage to shrinking $h$ further.

Wed, Sep 10

Today I announced Project 1 which is due next Wednesday. I’ve been posting Python & Sage code examples, but if you would rather use Octave/Matlab, here are some Octave code examples.

Runge-Kutta methods are a family of methods to solve ODEs numerically. Euler’s method is a first order Runge-Kutta method, which means that the discretization error for Euler’s method is $O(h^1)$ which means that the error is less than a constant times $h$ to the first power.

Better Runge-Kutta methods have higher order error bounds. For example, RK4 is a popular method with fourth order error $O(h^4)$. Another Runge-Kutta method is the midpoint method also known as RK2 which has second order error.

In RK2 the slope used to calculate the next point from a point $P_1$ is the slope at the midpoint between $P_1$ and the Euler’s method next step. In RK4, the slope used is a weighted average of the slopes at $P_1$, $P_2$, $P_3$, and $P_4$ shown in the diagram above. Specifically, it is 1/6 of the slopes at $P_1$ and $P_4$ plus 1/3 of the slopes at $P_2$ and $P_3$.

There are even higher order Runge-Kutta methods, but there is a trade-off between increasing the order and increasing complexity.

After we talked about Runge-Kutta methods, we introduced the integrating factors method for solving first order linear ODEs $$y'(t) + f(t) y = g(t).$$ The key idea is that if $F(t)$ is an antiderivative of $f(t)$, then $e^{F(t)}$ is an integrating factor for the ODE. Since $$\dfrac{d}{dt} \left( e^{F(t)} y(t) \right) = e^{F(t)} y'(t) + e^{F(t)} f(t) y(t)$$ by the product rule, we can re-write the ODE as: $$\dfrac{d}{dt} \left( e^{F(t)} y(t) \right) = e^{F(t)} g(t).$$ Then just integrate both sides to find the solution.

1. $\dfrac{dy}{dt} + \dfrac{y}{t} = 3t$

2. $\dfrac{dy}{dx} + 2y = 3$

Fri, Sep 12

Today we looked at more examples of linear first order ODEs.

1. Suppose that a 200 gallon tank initially contains a 100 gallons of a saltwater solution at a concentration of 1 gram of salt per gallon. We start adding 5 gallons of saltwater per minute with a concentration of 2 grams per gallon. Meanwhile we let out 3 gallons per minute of well-mixed water from the tank.

1. Write down an IVP to model this situation using $y$ to represent the amount of salt in the tank.

2. Use integrating factors to solve the IVP. (https://youtu.be/b5QWC2DA5l4)

Sometimes it can be faster to use a guess-and-check method instead of integrating factors to solve linear ODEs. Here is an example. Consider the first order linear ODE: $$\dfrac{dy}{dx} + 4y = e^{-x}.$$ You might guess that there is a constant $A$ such that $y(t) = Ae^{-x}$ is a solution of this differential equation. This is true!

2. Find the constant $A$ by substituting $y = Ae^{-x}$ into the differential equation $y' + 4y = e^{-x}$.

So $y(t) = \tfrac{1}{3} e^{-x}$ is one particular solution for this ODE. To get all of the solutions, we need some theory:

3. Solve the homogeneous ODE $y' + 4y = 0$, then find the general solution of $y' + 4y = e^{-x}$ by combining the homogeneous solutions with the particular solution we found above.

Week 4 Notes

Mon, Sep 15

Consider the inhomogeneous linear ODE: $$\dfrac{dy}{dt} + 2y = \cos t.$$ If you know that waves can be modeled by equations of the form $A \sin t + B \cos t$, then you might guess that the solution $y(t)$ might have this form. Then substituting into the equation, we get $$\underbrace{A \cos t - B \sin t}_{y'} + \underbrace{2A \sin t + 2B \cos t}_{2y} = \cos t.$$
By combining like terms, we get a system of equations $$\begin{align*} A + 2B &= 1 \\ 2A - B &= 0. \end{align*}$$ The solution is $A = \tfrac{1}{5}$, $B = \tfrac{2}{5}$ which means that $y(t) = \tfrac{1}{5} \sin t + \tfrac{2}{5} \cos t$ is one solution to the ODE.

1. What is the corresponding homogeneous equation, and what is its solution?

2. What is the general solution to $y' + 2y = \cos t$?

3. Why is the method of integrating factors harder here?

After that, we introduced systems of differential equations. We started with this simple model of a predator-prey system with rabbits $R$ and foxes $F$:

$$\begin{align*} \dfrac{dR}{dt} &= 2R - RF \\ \dfrac{dF}{dt} &= -5F + RF \end{align*}$$

4. Find an equilibrium solution where both $dR/dt$ and $dF/dt$ are zero.

A graph of the vector field defined by a system of two differential equations is called a phase plane. Solution curves are parametric functions $R(t)$ and $F(t)$ that follow the vector field in the phase plane.

Converting a 2nd order equation to a system of equations

According to Hooke’s law the force of a spring is $$m \dfrac{d^2 x}{dt^2} = - k x$$ or equivalently $$\dfrac{d^2 x}{dt^2} + \dfrac{k}{m} x = 0.$$ This is a homogeneous 2nd order linear differential equation.

We can convert a second order ODE to a first order system of equations by using an extra variable equal to the first derivative $v = x'$. Then $x'' = v'$, so we get the system:

$$\begin{align*} \dfrac{dv}{dt} + \dfrac{k}{m} x &= 0 \\ \dfrac{dx}{dt} - v &= 0. \end{align*}$$

5. Convert the 2nd order equation $y'' + y' + 4y = \sin t$ into a 1st order system of equations.

Wed, Sep 17

Today we looked at more examples of systems of ODEs.

Suppose that we have two species that compete for resources and their populations $x$ and $y$ satisfy

$$\begin{align*} \dfrac{dx}{dt} = x(1-x) - \alpha xy \\ \dfrac{dy}{dt} = y(1-y) - \alpha xy \\ \end{align*}$$

1. Plot the phase plane for this system when $\alpha = 2$ and when $\alpha = \tfrac{1}{2}$ (Python). Describe the difference between the equilibrium solutions of the two systems.

Later in chapter 3 we will learn how to classify different types of equilibrium solutions on the phase plane using linear algebra. For now, here is a preview of some of the types of equilibria.

A simple model used to understand epidemics is the SIR-model, which stands for Susceptible-Infected-Recovered. The idea is that a disease will spread from people who are infected to people who are still susceptible. After infected people recover, they are usually immune to the disease, at least for a little while. In the system below, $S(t)$ is the percent of the population that is still susceptible, $I(t)$ is the percent that are currently infected, and $R(t)$ is the percent of the population that are recovered. The constants $\alpha$ and $\beta$ are the transmission rate and recovery rate, respectively.

$$\begin{align*} \dfrac{dS}{dt} &= -\alpha SI \\ \dfrac{dI}{dt} &= \alpha SI - \beta I \\ \dfrac{dR}{dt} &= \beta I \end{align*}$$

2. Under what circumstances is the number of infected people increasing?

3. If we introduce a vaccine, what effect might that have on the model?

4. What if the disease is fatal for some people? How would you change the model to account for that? Hint: You could have a constant $\gamma$ that represents the fatality rate, i.e., the proportion of the infected population that die each day.

5. If you divide $dI/dt$ by $dS/dt$, you get the differential equation $$\dfrac{dI}{dS} = -1 + \dfrac{\beta}{\alpha} \dfrac{1}{S}.$$ Solve this differential equation with initial condition $S = 1$ and $I = 0$.

Here is a plot showing the solution superimposed on the direction field (for $S$ and $I$ only).

Fri, Sep 19

Today we talked about decoupled systems and partially coupled systems.

A system of equations $$\begin{align*} \dfrac{dx}{dt} &= f(x) \\ \dfrac{dy}{dt} &= g(y) \\ \end{align*}$$ is called decoupled since the $x$-variable doesn’t depend on $y$, and the $y$-variable doesn’t depend on $x$. You can solve the differential equations in a decouple system separately.

A system of equations $$\begin{align*} \dfrac{dx}{dt} &= f(x, y) \\ \dfrac{dy}{dt} &= g(y) \\ \end{align*}$$ is partially coupled. You can solve for $y(t)$ first, and then substitute into the first equation to create a single variable differential equation for $x(t)$.

1. Solve the system $$\begin{align*} x' &= -x - y \\ y' &= -3y \\ \end{align*}$$

2. Solve the system $$\begin{align*} x' &= 2x + y^2 \\ y' &= -y \\ \end{align*}$$ with initial conditions $x(0) = 3$ and $y(0) = 2$. (https://youtu.be/sJ3CuM-QmOk)

Here is a Desmos graph showing the solutions to the last problem as different parametric curves.

Week 5 Notes

Mon, Sep 22

Today we introduced Euler’s method for systems of differential equations equations.

1. We started by implementing Euler’s method for the system of rabbits and foxes: $$\begin{align*} \dfrac{dR}{dt} &= 2R - RF \\ \dfrac{dF}{dt} &= -5F + RF \end{align*}$$

2. A more realistic model for the rabbits & foxes might be if the rabbits growth was constrained by a carrying capacity of 10 thousand rabbits (logistic growth), in the absence of foxes. How would this change the differential equation above?

3. Now use Euler’s method to investigate the long-run behavior of the rabbits & foxes with this new model. What changes?

Now talk about the equation for a pendulum: $$\dfrac{d^2 \theta}{d t^2} = - \dfrac{g}{L} \sin \theta.$$

4. Re-write this 2nd order ODE as a system of 1st order equations.

Typically in introductory physics, you find an approximate solution of this equation by assuming that the angle $\theta$ stays small and so $\sin \theta \approx \theta$. But we can use Euler’s method instead to generate solutions numerically (Python).

Wed, Sep 24

Today we introduced complex numbers and talked about how they can arise in differential equations.

1. Calculate $(-2 + 2i)(3+2i)$.

2. Show that for any complex number $z \cdot \overline{z} = |z|^2$.

A complex-valued function is a function $z(t) = x(t) + i y(t)$ where both $x(t)$ and $y(t)$ are real-valued functions. You can integrate and differentiate complex-valued functions by integrating/differentiating the real and imaginary parts.

3. Show that $z(t) = \cos t + i \sin t$ is a solution of the differential equation $\dfrac{dz}{dt} = i z$.

4. Convert $\sqrt{2} + \sqrt{2}i$ and $1 + \sqrt{3} i$ to polar form, then multiply them by applying the formula $$e^{i \alpha} e^{i \beta} = e^{i (\alpha + \beta)}.$$

5. Solve the differential equation $z' + z = i$.

6. Show that $e^{it}$ is a solution for the differential equation $y'' + y = 0$. Hint: The chain rule applies to complex-valued functions, so $\frac{d}{dt} e^{it} = i e^{it}$.

Fri, Sep 26

Today we talked about homogeneous second order linear differential equations with constant coefficients.

$$ay'' + b y' + c = 0.$$

These equations are used to model simple harmonic oscillators such as a spring where the total force depends on a spring force $-k x$ and a friction or damping force $-b x'$:

$$m \dfrac{d^2 x}{dt^2} = - b \dfrac{dx}{dt} - k x.$$

1. Show that $e^{\lambda t}$ is a solution of $ay'' + by' + cy$ if and only if $\lambda$ is a root of the characteristic polynomial $ax^2 + bx + c$.

Using the language of linear algebra, we can describe the result above several ways:

• $f$ and $g$ are a basis for the set of general solutions.
• $f$ and $g$ span the set of all general solutions.
• Every solution is a linear combination of $f$ and $g$.

We applied the theorem above to the following two examples:

2. Find the general solution to $y'' + 3y' + 2y = 0$. (https://youtu.be/Pxc7VIgr5kc?t=241)

3. Find the general solution of $y'' + 2 y' + 2 y = 0$. Hint: Use the quadratic formula.

Week 6 Notes

Mon, Sep 29

We talked about the midterm 1 review problems. We also looked at this example:

1. Find the general solution of the differential equation $y' + 3y = t + 1$ using the guess & check method. Hint: A good guess for the particular solution is that $y$ is a linear function, so $y(t) = A t + B$ for some constants $A$ and $B$.

Fri, Oct 3

Today we talked about homogeneous linear systems of differential equations. These can be expressed using a matrix. For example, if $\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$, then the system of differential equations $$\dfrac{dx_1}{dt} = a x_1 + b x_2$$ $$\dfrac{dx_2}{dt} = c x_1 + d x_2$$ can be re-written as $$\dfrac{d \mathbf{x}}{dt} = A \mathbf{x} \text{ where } A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}.$$ It turns out that the eigenvectors and eigenvalues of $A$ tell you a lot about the solutions of the system. We did these exercises in class.

1. Find the characteristic polynomial and eigenvalues of the matrix $A = \begin{bmatrix} 3 & 5 \\ 2 & 6 \end{bmatrix}.$

2. Show that $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ is an eigenvector for $A$. What is its eigenvalue?

3. Find an eigenvector for $A$ corresponding to the eigenvalue $\lambda = 1$ by finding the null space of $A - \lambda I$.

After those examples, we did a workshop.

• Workshop: Linear Algebra Review

We also talked about how to calculate the eigenvectors of a matrix using a computer. In Python, the sympy library lets you calculate the eigenvectors of a matrix exactly when possible. You can also do this in Octave if you load the symbolic package.

from sympy import *

A = Matrix([[3,5],[2,6]])
'''
The .eigenvects() method returns a list of tuples containing: 
1. an eigenvalue, 
2. its multiplicity (how many times it is a root), and
3. a list of corresponding eigenvectors. 
'''
pretty_print(A.eigenvects())

Week 7 Notes

Mon, Oct 6

Today we talked about how to solve a homogeneous linear system $\dfrac{dx}{dt} = Ax$ using the eigenvectors and eigenvalues of $A$ when the eigenvalues are all real with no repeats. We did the following examples:

1. Show that $x(t) = e^{8t} \begin{bmatrix} 1 \\ 1 \end{bmatrix}$ is a solution to the linear system $\dfrac{dx}{dt} = \begin{bmatrix} 3 & 5 \\ 2 & 6 \end{bmatrix} x$.

We used these facts to find the general solutions for the following systems.

2. $\dfrac{dx}{dt} = \begin{bmatrix} 3 & 5 \\ 2 & 6 \end{bmatrix} x$.

3. $\dfrac{dx}{dt} = \begin{bmatrix} 1 & 2 \\ 4 & 3 \end{bmatrix} x$. (https://youtu.be/DWzq_jMPRgc)

We also talked about how to graph the solutions.

We finished with the following question:

4. The zero vector is always an equilibrium solution of $x' = Ax$. Under what conditions will there be other equilibrium solutions?

Wed, Oct 8

Last time we talked about how to find general solutions for linear systems. Today we talked about how to find specific solutions that satisfy an initial condition. Before that, we answered this question:

1. Under what circumstances does $x' = A x$ have more than one equilibrium solution?

   Any time that zero is an eigenvalue of $A$ (i.e., the null space of $A$ is non-trivial), all of the eigenvectors corresponding to zero (i.e., the vectors in the null space) will be equilibrium points. It helps to know following theorem:

   Invertible Matrix Theorem. For an n-by-n matrix $A$, the following are equivalent.

   1. $A$ is invertible.
   2. $\det A \ne 0$.
   3. The columns of $A$ are linearly independent.
   4. The rows of $A$ are linearly independent.
   5. The null space of $A$ is $\{0\}$.
   6. Zero is not an eigenvalue of $A$.

After that, we talked about the classification of equilibria for linear systems.

Then we solved the following initial value problems.

2. Find the solution to $\dfrac{dx}{dt} = \begin{bmatrix} 3 & 5 \\ 2 & 6 \end{bmatrix} x$ that satisfies $x(0) = \begin{bmatrix} 7 \\ 0 \end{bmatrix}.$

3. Find the solution to $\dfrac{dx}{dt} = \begin{bmatrix} 1 & 2 \\ 4 & 3 \end{bmatrix} x$ that satisfies $x(0) = \begin{bmatrix} 5 \\ 4 \end{bmatrix}.$

We used row reduction to solve these initial value problems. Here’s a video with a similar exercise (https://youtu.be/Mnz-S-RvpDw) if you want a different take. Octave makes this kind of matrix computation very easy.

pkg load symbolic
A = sym([3 5; 2 6])

# The columns of V are the eigenvectors and the diagonal entries of D are the eigenvalues.
[V, D] = eig(A)

# The backslash operator A \ b solves the matrix equation Ax = b. 
V \ [7;0]

Fri, Oct 10

Today we talked about systems with complex eigenvalues. We started by looking at three different direction fields for the following three matrices:

$$A= \begin{bmatrix} 0 & 2 \\ -3 & 2 \end{bmatrix} ~~~~~~ B = \begin{bmatrix} -2 & 3 \\ -1 & -2 \end{bmatrix} ~~~~~~ C = \begin{bmatrix} 2 & 5 \\ -4 & -2 \end{bmatrix}$$

We talked about why this is true using Euler’s formula $$e^{\alpha t \pm i \beta t} = e^{\alpha t} ( \cos (\beta t) \pm i \sin (\beta t) ).$$

It is still true that the general solution is $$x(t) = C_1 e^{\lambda_1 t} v_1 + C_2 e^{\lambda_2 t} v_2$$ but the coefficients are typically complex numbers and we only want real solutions. Next time we’ll discuss a better way to find the real-valued solutions.

Week 8 Notes

Wed, Oct 15

Last time we talked about planar systems with complex eigenvalues. We haven’t seen how to find nice solutions for those systems yet. Here is the key idea we need:

Here’s how to use these facts to find the general real-valued solution:

• Step 1. Find an eigenvector $v$ and its corresponding eigenvalue $\lambda = \alpha + i \beta$. Then $$x(t) = e^{\lambda t} v$$ is one complex solution for the system.

• Step 2. Use Euler’s formula to convert $$e^{\lambda t} = e^{\alpha t} (\cos (\beta t) + i \sin (\beta t)).$$

• Step 3. Expand $x(t)$ to find the real and imaginary parts.

We used this approach to find the general (real) solutions for the following planar systems.

1. $\dfrac{dx}{dt} = \begin{bmatrix} -2 & 3 \\ -1 & -2 \end{bmatrix} x$

2. Suppose a planar system $x' = Ax$ has an eigenvector $\begin{bmatrix} 1 \\ 4 - 3i \end{bmatrix}$ with corresponding eigenvalue $1 - 4i$. What is the general solution for the system?

3. Find a solution for the previous example with initial condition $x(0) = \begin{bmatrix} 2 \\ -6 \end{bmatrix}$.

4. $\dfrac{dx}{dt} = \begin{bmatrix} 1 & -1 \\ 4 & 1 \end{bmatrix} x$ (https://youtu.be/j-qvdT8nSnw)

Fri, Oct 17

If you change the parameters of a system of differential equations, a bifurcation happens when the type or number of equilibria changes. For planar systems, that happens when $\det A = 0$, $\operatorname{tr}A = 0$, or when $(\operatorname{tr}A)^2 = 4 \det A$.

1. Consider the 1-parameter family $\dfrac{d\mathbf{x}}{dt} = \begin{bmatrix} a & a^2 - a \\ 1 & a \end{bmatrix} \mathbf{x}$. For what values of $a$ do you get bifurcations?

2. Consider the harmonic oscillator $m \dfrac{d^2 x}{dt^2} + b \dfrac{dx}{dt} + k x = 0$.

   1. Re-write this as a linear system.
   2. Express the trace and determinant of the system as functions of $m, b,$ and $k$.
   3. Describe how the type of the equilibrium in phase space changes depending on $m, b, k$.

Here are two more examples that we didn’t have time for in class.

3. Consider the 1-parameter family $\dfrac{d\mathbf{Y}}{dt} = \begin{bmatrix} a & a \\ 1 & a \end{bmatrix} \mathbf{Y}$. For what values of $a$ do you get bifurcations? (https://youtu.be/F2ew7_dD5Zg)

4. Consider the 2-parameter family $\dfrac{d\mathbf{y}}{dt} = \begin{bmatrix} 0 & a \\ b & 0 \end{bmatrix} \mathbf{y}$. Describe how the type of equilibrium depends on the parameters $a$ and $b$.

Week 9 Notes

Mon, Oct 20

Today we talked about the matrix exponential and how it solves any homogeneous linear system!

To calculate a matrix exponential, start by diagonalizing the matrix, if possible.

1. Diagonalize the matrix $A = \begin{bmatrix} 1 & -1 \\ 2 & 4 \end{bmatrix}$ and then use the diagonalization to compute $A^{100}$. (https://www.youtube.com/watch?v=uHW2zThZDEw)

To solve the last problem, it helps to know the following formula.

2. Use the diagonalization of $A = \begin{bmatrix} 1 & -1 \\ 2 & 4 \end{bmatrix}$ to solve the initial value problem $$\frac{d \mathbf{x}}{dt} = \begin{bmatrix} 1 & -1 \\ 2 & 4 \end{bmatrix} \mathbf{x}, ~ \mathbf{x}(0) = \begin{bmatrix} 1 \\ 2 \end{bmatrix}.$$

3. Calculate $e^{tA}$ for $A = \begin{bmatrix} 3 & 4 \\ -4 & 3 \end{bmatrix}$ and use it to find the general solution to $\mathbf{x}' = A \mathbf{x}$. Hint: the eigenvalues of $A$ are $3 \pm 4i$ with corresponding eigenvectors $\begin{bmatrix} 1 \\ \pm i \end{bmatrix}$.

Since calculating a matrix exponential by hand is so tedious, we’ll usually use a computer.

• Symbolic matrix exponentials (Python, Octave)
• Numerical matrix exponentials (Python, Octave)

Be careful computing symbolic matrix exponentials. It only works for very simple matrices.

Wed, Oct 22

Today we talked about homogeneous linear systems in 3-dimensions. We started with these examples.

1. Use the matrix exponential function to find the solution to the initial value problem $$\dfrac{dX}{dt} = \begin{bmatrix} 1 & 10 & 0 \\ -10 & 1 & 0 \\ -4 & 6 & -1 \end{bmatrix}X, \text{ with } X(0) = \begin{bmatrix} 1 \\ 0 \\ 100 \end{bmatrix}.$$

2. Use Desmos 3D to graph the solution above.

3. Find the straight-line solutions and the general solution for the system $$\dfrac{dY}{dt} = \begin{bmatrix} -3 & 1 & 2 \\ 4 & 0 & -4 \\ -4 & 1 & 3 \end{bmatrix} Y.$$ Hint: The matrix for this system has eigenpairs $$\lambda_1 = -1, \mathbf{v}_1 = \begin{bmatrix}1 \\ 0 \\ 1 \end{bmatrix}, \lambda_2 = 0, \mathbf{v}_2 = \begin{bmatrix} 1 \\ 1\\ 1\end{bmatrix}, \lambda_3 = 1, \mathbf{v}_3 = \begin{bmatrix} 1 \\ 4 \\ 0 \end{bmatrix}.$$

4. Consider the system $\dfrac{dZ}{dt} = \begin{bmatrix} 0 & 2 & 0 \\ -2 & 0 & 0 \\ 0 & 0 & -1 \end{bmatrix} Z$.

   1. The vector $\begin{bmatrix} 1 \\ i \\ 0 \end{bmatrix}$ is an eigenvector with eigenvalue $2i$. Find the real and imaginary parts of the corresponding solution.
      
   2. The system above decouples since $Z_3$ does not depend on $Z_1$ or $Z_2$. Find a solution for the third component of $Z$. Express that solution as a straight line solution in vector form assuming that $Z_1 = Z_2 = 0$.

Fri, Oct 24

There is one other case of linear systems that we haven’t considered yet. What happens when there are repeated eigenvalues?

1. The matrix $A = \begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix}$ has the eigenvalue $\lambda = 2$ repeated twice, but it only has one eigenvector (up to scaling).

We applied this theorem to the following examples:

2. Consider the system $\dfrac{d\mathbf{x}}{dt} = \begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix} \mathbf{x}$.
   1. Find the straight-line solutions.
   2. Find the solution with initial condition $\mathbf{x}_0 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}.$

This is an example of what is called a degenerate source equilibrium at the origin. You can also have a degenerate sink if the repeated eigenvalue is negative.

3. We also talked about Homework 8, exercise 9.

We finished by talking about what happens when both eigenvalues are zero. In that case you get uniform motion in a straight line parallel to the eigenvector, with a velocity proportional to the distance from the eigenvector.

Week 10 Notes

Mon, Oct 27

Today we talked about the midterm 2 review problems. We also went over the problems on the quiz about the trace-determinant plane.

Week 11 Notes

Week 12 Notes

Week 13 Notes

Week 14 Notes