15.4 Exercises

Exercise 15.1 Write the following systems of equations in matrix notation ($\displaystyle \frac{ d \vec{x} }{dt} = A \vec{x}$):

\[\begin{equation} \begin{split} \frac{dx}{dt} &= 3x-4y \\ \frac{dy}{dt} &= 2x-y, \end{split} \end{equation}\]
\[\begin{equation} \begin{split} \frac{dx}{dt} &= x+y \\ \frac{dy}{dt} &= y-x, \end{split} \end{equation}\]
\[\begin{equation} \begin{split} \frac{dx}{dt} &= 5x-4y + z \\ \frac{dy}{dt} &= y - 9z, \\ \frac{dz}{dt} &= 7x-z, \\ \end{split} \end{equation}\]
\[\begin{equation} \begin{split} \frac{dx}{dt} &= -cx \\ \frac{dy}{dt} &= rcx-y, \end{split} \end{equation}\]

Exercise 15.2 Verify that $x=0$ and $y =e^{t}$ and $x=e^{2t}$ and $y=e^{2t}$ are solutions for the differential equation \[\begin{equation} \begin{split} \frac{dx}{dt} &= 2x \\ \frac{dy}{dt} &= x+y \end{split} \end{equation}\]

Exercise 15.3 Verify that $x=0$ and $y=0$ are solutions to the differential equation \[\begin{equation} \begin{split} \frac{dx}{dt} &= -3x \\ \frac{dy}{dt} &= .2x-y, \end{split} \end{equation}\]

Exercise 15.4 Verify that $x=0$, $y=0$, and $z=0$ are solutions to the differential equation \[\begin{equation} \begin{split} \frac{dx}{dt} &= 5x-4y + z \\ \frac{dy}{dt} &= y - 9z, \\ \frac{dz}{dt} &= 7x-z, \\ \end{split} \end{equation}\]

Exercise 15.5 Consider the differential equation $\displaystyle \frac{dx}{dt} = -3x$. This exercise will help you work through the details of creating a two dimensional system of equations by re-parameterizing $s=t$.

Define the variable $t = s$. For this case, what is $\displaystyle \frac{dt}{ds}$?
Explain if $ x = f (t (s))$ ($x$ is a composition between $t$ and $s$), explain why the chain rule has $\displaystyle \frac{dx}{ds} = \frac{dx}{dt} \cdot \frac{dt}{ds}$.
Use the fact that $\displaystyle \frac{dx}{ds} = \frac{dx}{dt} \cdot \frac{dt}{ds}$ to explain why $\displaystyle \frac{dx}{ds} = -3x$.
Finally use your previous work to determine the system of equations for $\displaystyle \frac{dx}{ds}$ and $\displaystyle \frac{dt}{ds}$.

Exercise 15.6 This problem considers the differential equation

\[\begin{equation} \begin{split} \frac{dx}{dt} &= x+y \\ \frac{dy}{dt} &= y-x, \end{split} \end{equation}\]

Use the command phaseplane to create a phaseplane of this differential equation.
Change the number of arrows shown to 5 and 20 (2 different plots). What do you notice about the updated phaseplane?
Change the viewing window from the default to -10 to 10 in both axes. Now change the number of arrows shown to 5 and 20 (2 different plots). What do you notice about the updated phaseplane?

Exercise 15.7 Explain why we call $x=0$ and $y=0$ equilibrium solutions to the general linear differential equation $\displaystyle \frac{ d \vec{x} }{dt} = A \vec{x}$. In other words, why is the word equilibrium important? (Hint: Think about the graph of $x$ and $y$ for the solution in this case.)

Exercise 15.8 (Inspired by logan_mathematical_2009) Consider the following differential equation:

\[\begin{equation} \begin{split} \frac{dx}{dt} &= -ax-y \\ \frac{dy}{dt} &= x-ay \end{split} \end{equation}\]

Write this system in the form $\displaystyle \frac{d\vec{x}}{dt}=A \vec{x}$.
Let $a= -2, \; -1, \; -0.5, \; 0, \; 0.5, \; 1, \; 2$. Generate a phase plane for each of these values of $a$ and characterize the behavior of the equilibrium solution.

Exercise 15.9 Generate a phaseplane for the following differential equations and using your result, classify if the equilibrium solution is stable or unstable.

$\displaystyle \frac{dx}{dt} = -x, \; \frac{dy}{dt} = -2y$
$\displaystyle \frac{dx}{dt} = 3x+y, \; \frac{dy}{dt} = 2x+4y$
$\displaystyle \frac{dx}{dt} = 8x-11y, \; \frac{dy}{dt} = 6x-9y$
$\displaystyle \frac{dx}{dt}= 3x-y, \; \frac{dy}{dt}=3y$
$\displaystyle \frac{dx}{dt} = -2x-3y, \; \frac{dy}{dt} = 3x-2y$

Exercise 15.10 Consider the following differential equation:

\[\begin{equation} \begin{split} \frac{dx}{dt} &= -y \\ \frac{dy}{dt} &= x \end{split} \end{equation}\]

Generate a phase plane diagram of this system. What do you notice?
Verify that $x(t)=A \cos(t)$ and $y(t)=A \sin(t)$ is a solution to this differential equation.
An equation of a circle of radius $R$ is $x^{2}+y^{2}=R^{2}$. Use implicit differentiation to differentiate this equation. Remember you are differentiating with respect to $t$, and $x$ and $y$ are functions of time $t$.
Substitute the differential equation into your implicit derivative to verify $x^{2}+y^{2}=R^{2}$ is a solution to the differential equation.
Verify that $x(t)=A \cos(t) + B \sin(t)$ and $y(t)=A \sin(t)-B \cos(t)$ are also solutions.
Make a plot of $x(t)=A \cos(t) + B \sin(t)$ and $y(t)=-A \sin(t)-B \cos(t)$ for $A=1$, $B=1$.