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Section Second-Order Equations to First-Order Systems
So far, weβve looked at systems that involve multiple variables like
\(x(t)\) and
\(y(t)\text{.}\) But what if you start with just a single second-order differential equation? It turns out you can rewrite it as a system of first-order equations.
Subsection Turning a Second-Order Equation into a System
For example, consider the second-order equation:
\begin{equation*}
y'' + 3y' + 2y = 0
\end{equation*}
This equation involves only one dependent variable,
\(y(t)\text{,}\) but itβs second-order. To convert it into a system, we introduce a new variable:
\begin{equation*}
u = y, \quad v = y'
\end{equation*}
That means
\(u' = y' = v\text{.}\) And since
\(y'' = v'\text{,}\) we can rewrite the original equation as:
\begin{equation*}
v' = -3v - 2u
\end{equation*}
Now we have a system of two first-order equations:
\begin{align*}
u' \amp = v \\
v' \amp = -3v - 2u
\end{align*}
Checkpoint 293 . πβ Derivative Substitution.
In order to rewrite a second-order equation as a system, we can replace
\(y'\) with a new variable and then express
\(y''\) as the derivative of that new variable.
True.
Correct. Introducing \(x_1 = y\) and \(x_2 = y'\) allows us to write \(y''\) as \(x_2'\) and form a system.
False.
Correct. Introducing \(x_1 = y\) and \(x_2 = y'\) allows us to write \(y''\) as \(x_2'\) and form a system.
Subsection π€ Wrap-Up
ποΈ \(\textbf{Key Takeaways...}\)
Check Your Understanding.
Checkpoint 294 . π€π Second-Order Equations to First-Order Systems.
(a) πβ Understanding Converted Systems.
When you convert a second-order equation into a first-order system, which of the following are always true?
The system has two first-order equations.
Yes. One for \(y\) and one for \(y'\text{.}\)
The system involves two new variables: one for
\(y\text{,}\) one for
\(y'\text{.}\)
Exactly. This substitution is the key step.
The resulting system is always uncoupled.
Not true. The new variables often depend on each other.
The system represents the same solutions as the original equation.
Correct. Itβs just a different form of the same information.
You can only use this method for linear equations.
This method also works for nonlinear second-order equations.
(b) πβ Match Equations to Their Systems.
(c) πβ From Second Order to System.
What is the correct system corresponding to
\(y'' + 4y = 0\) if we define
\(x_1 = y\text{,}\) \(x_2 = y'\text{?}\)
\(x_1' = x_2\text{,}\) \(x_2' = -4x_2\)
That would be true if the original equation were \(y'' + 4y'\text{.}\) Double-check which variable is being multiplied by 4.
\(x_1' = x_2\text{,}\) \(x_2' = -4x_1 + x_2\)
Too many terms, thereβs no \(y'\) in the original equation.
\(x_1' = x_2\text{,}\) \(x_2' = -4x_1\)
Yes! \(x_2' = y'' = -4y = -4x_1\) is exactly what we want.
\(x_1' = x_1\text{,}\) \(x_2' = -4x_2\)
This system doesnβt reflect the original equation at all.
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