Section 4.1 Homogeneous Linear Equations
Exercises Exercises
Supplemental HW Problems.
ODE = Ordinary Differential Equation;
IVP = Initial Value Problem
1.
Consider the homogeneous 2nd-order Linear Differential Equation:
\begin{equation*}
x'' + 5x’ + 6x = 0; x(0) = 2, x’(0) = -1\text{.}
\end{equation*}
Solve this equation using the techniques from this section.
-
Here are five matrices, each with the same eigenvalues \((-2, -3)\) as the equation above. Which one represents \(x'' + 5x’ + 6x = 0\) written as a system of two 1st-order differential equations?
\begin{equation*}
\begin{pmatrix}
-5 & 3 \\
-2 & 0
\end{pmatrix}
\end{equation*}
\begin{equation*}
\begin{pmatrix}
0 & -2 \\
3 & -5
\end{pmatrix}
\end{equation*}
\begin{equation*}
\begin{pmatrix}
-4 & -2 \\
1 & -1
\end{pmatrix}
\end{equation*}
\begin{equation*}
\begin{pmatrix}
0 & -1 \\
6 & 5
\end{pmatrix}
\end{equation*}
\begin{equation*}
\begin{pmatrix}
-2 & 3 \\
0 & -3
\end{pmatrix}
\end{equation*}
Graph all five matrices with the initial condition
\(x = 2, y = -1\) (using a Phase Plane plotter – there are several online, and our book has Sage tools for this in
Section 2.2.4 1 of our book). What is different about the four solution curves? What is the same? The idea here is to delineate what the eigenvalues tell us about the solution curves from what the eigenvectors tell us.
2.
We’ve looked at converting a 2nd-order linear Differential equation into two 1st-order ODEs, by choosing a second variable to represent the derivative of our function and then solving for both variables as linear equations. For example, the equation \(x'' + 5x’ + 6x = 0\text{:}\)
Let \(y = x’\text{.}\)
Then \(y’ = x''\text{.}\)
Plugging \(y\) and \(y'\) into our differential equation, we get \(y’ + 5y + 6x = 0\text{.}\)
Which we can solve for \(y'\) and get: \(y’ = -6x - 5y\text{.}\)
So our system is now:
\begin{align*}
x’ \amp = y\\
y’ \amp = -6x - 5y
\end{align*}
Now let’s look at going the other way through: Convert the system of linear 1st-order ODEs to a single 2nd-order ODE:
\begin{align*}
x’ \amp = ax + by\\
y’ \amp = cx + dy
\end{align*}
(HINT: since both \(x\) and \(y\) are themselves functions of \(t\text{,}\) start by taking the derivative of the top equation to get \(x'' = ax’ + by’\text{.}\) Then use the two original equations to substitute in for \(y\) and \(y’\)).
Convert the five matrices given in part 1b above to systems, and then use this technique to convert the system of 1st-order ODEs into a single 2nd-order ODE.
NOTE: We generally try to convert things into linear algebra, so that computers can do the computations as quickly as possible; not so much the other way around like this. It’s good practice to go back and forth by hand, though, to help us build intuition about how these things work.
faculty.sfasu.edu/judsontw/ode/html-20210811/systems02.html
