Skip to main content
Contents Index Calc Dark Mode Prev Next Profile \(\newcommand\DLGray{\color{Gray}}
\newcommand\DLO{\color{BurntOrange}}
\newcommand\DLRa{\color{WildStrawberry}}
\newcommand\DLGa{\color{Green}}
\newcommand\DLGb{\color{PineGreen}}
\newcommand\DLBa{\color{RoyalBlue}}
\newcommand\DLBb{\color{Cerulean}}
\newcommand\ds{\displaystyle}
\newcommand\ddx{\frac{d}{dx}}
\newcommand\os{\overset}
\newcommand\us{\underset}
\newcommand\ob{\overbrace}
\newcommand\obt{\overbracket}
\newcommand\ub{\underbrace}
\newcommand\ubt{\underbracket}
\newcommand\ul{\underline}
\newcommand\laplacesym{\mathscr{L}}
\newcommand\lap[1]{\laplacesym\left\{#1\right\}}
\newcommand\ilap[1]{\laplacesym^{-1}\left\{#1\right\}}
\newcommand\tikznode[3][]
{\tikz[remember picture,baseline=(#2.base)]
\node[minimum size=0pt,inner sep=0pt,#1](#2){#3};
}
\newcommand{\lt}{<}
\newcommand{\gt}{>}
\newcommand{\amp}{&}
\definecolor{fillinmathshade}{gray}{0.9}
\newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}}
\newcommand{\sfrac}[2]{{#1}/{#2}}
\)
Section 14.5 Second-Order Equations to First-Order Systems
So far, we’ve looked at systems involving multiple variables, such as
\(x(t)\) 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\text{.}
\end{equation*}
This equation involves only one dependent variable, \(y(t)\text{,}\) but it’s second-order. To convert it into a system, we introduce a new variable:
\begin{equation*}
u = y, \quad v = y'\text{.}
\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\text{.}
\end{equation*}
Now we have a system of two first-order equations:
\begin{align*}
u' \amp = v \\
v' \amp = -3v - 2u \text{.}
\end{align*}
Checkpoint 286 . 📖❓ 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 two variables for \(y\) and \(y'\) (for example, \(u=y\text{,}\) \(v=y'\text{,}\) or \(x_1=y\text{,}\) \(x_2=y'\) ) lets us write \(y''\) as the derivative of the second variable and form a first-order system.
False.
Correct. Introducing two variables for \(y\) and \(y'\) (for example, \(u=y\text{,}\) \(v=y'\text{,}\) or \(x_1=y\text{,}\) \(x_2=y'\) ) lets us write \(y''\) as the derivative of the second variable and form a first-order system.
Checkpoint 287 . 🤔💭 Second-Order Equations to First-Order Systems.
(a) 📖❓ Understanding Converted Systems.
When you convert a second-order equation into a first-order system, which of the following are always true? (Select all that apply.)
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. It’s 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) 📖❓ Match Equations to Their Systems.
(c) 📖❓ 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, there’s 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 doesn’t reflect the original equation at all.
You have attempted
of
activities on this page.
