Skip to main content
Contents Index
Search Book
Search Results:
No results.
Read aloud Readability settings Prev Up Next Profile title here \(\newcommand\DLGray{\color{Gray}}
\newcommand\DLO{\color{BurntOrange}}
\newcommand\DLRa{\color{WildStrawberry}}
\newcommand\DLGa{\color{Green}}
\newcommand\DLGb{\color{PineGreen}}
\newcommand\DLBa{\color{RoyalBlue}}
\newcommand\DLBb{\color{Cerulean}}
\newcommand\ds{\displaystyle}
\newcommand\ddx{\frac{d}{dx}}
\newcommand\os{\overset}
\newcommand\us{\underset}
\newcommand\ob{\overbrace}
\newcommand\obt{\overbracket}
\newcommand\ub{\underbrace}
\newcommand\ubt{\underbracket}
\newcommand\ul{\underline}
\newcommand\laplacesym{\mathscr{L}}
\newcommand\lap[1]{\laplacesym\left\{#1\right\}}
\newcommand\ilap[1]{\laplacesym^{-1}\left\{#1\right\}}
\newcommand\tikznode[3][]
{\tikz[remember picture,baseline=(#2.base)]
\node[minimum size=0pt,inner sep=0pt,#1](#2){#3};
}
\newcommand{\lt}{<}
\newcommand{\gt}{>}
\newcommand{\amp}{&}
\definecolor{fillinmathshade}{gray}{0.9}
\newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}}
\newcommand{\sfrac}[2]{{#1}/{#2}}
\)
Exercises 7.8 Exercises
💡 Conceptual Quiz.
Answer the following questions to check what you can read off a slope field, a phase line, and an equilibrium solution.
📝: Abbreviations.
Differential Equation
Initial Value Problem
1. Multiple-Choice.
Each part below stands on its own. Read the equation carefully before choosing an answer.
(a) Does time matter here?
For the autonomous equation
\begin{equation*}
y' = y - 1\text{,}
\end{equation*}
the slope field shows a slope of \(-2\) at \((t,y) = (5,-1)\text{.}\) What is the slope at \((t,y) = (-10,-1)\text{?}\)
\(0\)
The slope doesn’t magically flatten; it’s determined by \(y\text{.}\)
\(2\)
The sign doesn’t flip when \(t\) changes—time isn’t part of the slope function.
\(-2\)
Correct! Autonomous equations ignore \(t\) entirely. If \(y=-1\text{,}\) the slope is always \(f(-1) = -2\text{,}\) no matter the time.
Impossible to know
It’s absolutely possible—you just evaluate \(f(y)\) at \(y=-1\text{.}\)
(b) Horizontal Shift Symmetry.
Fill in the blank: If
\(y(t)\) is a solution to an autonomous equation, then _______ is also guaranteed to be a solution.
\(y(t + c)\)
Correct—autonomous equations allow time-shifted versions of any solution.
\(c \, y(t)\)
No—scaling the solution doesn’t necessarily work unless the equation is linear.
\(y(t) + c\)
No—adding a constant changes \(y\) in a way that won’t satisfy the DE.
\(y(t)^c\)
Raising the solution to a power doesn’t preserve the equation’s behavior.
(c) Time Shifts in Practice.
A biologist models population growth with the autonomous equation
\begin{equation*}
\frac{dy}{dt} = y(1 - y).
\end{equation*}
They find a solution curve \(y(t)\) that fits data collected in spring. Which of the following will always produce another valid solution?
\(y(t + 5)\)
Yes! Sliding the solution along the t-axis still solves the equation.
\(2y(t)\)
Scaling \(y\) won’t necessarily satisfy the same DE.
\(y(t) + 5\)
Adding to \(y\) changes the values in a way that usually breaks the equation.
\(y(-t)\)
Reversing time is not the same as sliding it—it usually won’t satisfy the DE.
(d) Equilibrium Identification.
Which of the following are equilibrium solutions for the following equation?
\begin{equation*}
\dfrac{dy}{dt} = y^2 - 4y + 3.
\end{equation*}
\(y = 1\)
\(y = 2\)
\(y = 3\)
\(y = 4\)
(e) Stability Classification.
For the same equation
\begin{equation*}
\dfrac{dy}{dt} = y^2 - 4y + 3,
\end{equation*}
which of the following best describes the equilibrium at \(y = 1\text{?}\)
Stable (sink)
Unstable (source)
Semi-stable
Not an equilibrium point
(f) 📖❓ Phase Line Practice.
Sketch a phase line for the equation
\begin{equation*}
\dfrac{dy}{dt} = y(3 - y^2).
\end{equation*}
Then classify each equilibrium solution as a sink, source, or semi-stable.
Answer .
Equilibria:
\(y=0\) (source),
\(y=\pm\sqrt{3}\) (sinks).
(g) Review Questions.
What is an autonomous differential equation?
How do you find equilibrium solutions for an autonomous equation?
What does the phase line represent, and how is it useful?
How can you determine the stability of an equilibrium solution using the phase line?
Answer each question briefly:
Answer .
See the solution for the review-question answers.
✍🏻 Problems.
Work through the following problems. Sketch the slope field or the phase line whenever it helps you see the long-term behavior.
2. Practice Problems.
Try these problems to reinforce your understanding:
Sketch the slope field for
\begin{equation*}
\dfrac{dy}{dt} = y^2 - 4y\text{.}
\end{equation*}
Identify the equilibrium solutions and their stability.
Draw the phase line for the equation and describe the long-term behavior of solutions.
Answer .
Equilibria:
\(y=0\) (sink/stable),
\(y=4\) (source/unstable). Long-term: solutions tend toward
\(y=0\) from above or below (unless starting above
\(y=4\text{,}\) where they grow unbounded).
3. Exploring Nonlinear Dynamics.
Consider the nonlinear equation
\begin{equation*}
\dfrac{dy}{dt} = 3y - y^3\text{.}
\end{equation*}
Find the equilibrium solutions.
Sketch the phase line and indicate the stability of each equilibrium.
Discuss how this equation might model a system with multiple stable states.
Answer .
a)
\(y = 0, \pm\sqrt{3}\text{.}\) b)
\(y = -\sqrt{3}\) (sink),
\(y = 0\) (source),
\(y = \sqrt{3}\) (sink). c) Bistable system with two stable states separated by an unstable threshold.
4. Exploring Chaos in Autonomous Systems.
Consider the equation
\begin{equation*}
\dfrac{dy}{dt} = y^2 - 2y + 1\text{.}
\end{equation*}
Identify the equilibrium solutions.
Discuss whether this system exhibits chaotic behavior or not.
Sketch the phase line and describe the long-term behavior of solutions.
Answer .
a)
\(y = 1\text{.}\) b) No chaos; this is a predictable 1D system. c)
\(y = 1\) is semi-stable; solutions below approach it, solutions above grow unbounded.
5. Exploring Real-World Applications.
Consider a population of rabbits modeled by the equation
\begin{equation*}
\frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right)\text{,}
\end{equation*}
where \(r\) is the growth rate and \(K\) is the carrying capacity.
Identify the equilibrium solutions.
Sketch the phase line and describe the stability of each equilibrium.
Discuss how this model can help predict population dynamics over time.
Answer .
a)
\(P = 0\) and
\(P = K\text{.}\) b)
\(P = 0\) is unstable (source),
\(P = K\) is stable (sink). c) Populations stabilize at carrying capacity
\(K\text{;}\) model predicts long-term equilibrium and response to perturbations.
6. Find the Equilibrium Solution.
Find the equilibrium solutions for the autonomous differential equation
\begin{equation*}
\dfrac{dy}{dx} = y^2\text{.}
\end{equation*}
7. Finding Equilibrium Points.
Find all equilibrium solutions of the autonomous differential equation
\begin{equation*}
\dfrac{dy}{dx} = y^2 - y^4\text{.}
\end{equation*}
Answer .
The equilibrium solutions are
\(y=-1\text{,}\) \(y=0\text{,}\) and
\(y=1\text{.}\)
8. Phase Line Sketching.
Sketch a phase line for the equation
\begin{equation*}
\dfrac{dy}{dt} = y^2 - 4y + 3.
\end{equation*}
Then classify each equilibrium point as a sink, source, or semi-stable.
Answer .
Equilibria:
\(y = 1\) (sink/stable) and
\(y = 3\) (source/unstable).
You have attempted
of
activities on this page.
