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\def\d{{\mathrm d}}
\def\e{{\mathrm e}}
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\def\x{{\boldsymbol x}}

\begin{document}

\title{Applied Differential Equations and Modeling}
\author{Homework 3}
\date{Due in class Tuesday, February 26, 2019}
\maketitle

\begin{enumerate}

\item For each of the following equations, in the general form
$y'=f(y)$, determine the equilibrium points where $f(y)=0$ and
classify each as stable ($f(y)$ changes sign from positive to negative
at the equilibrium value for $y$) or unstable ($f$ changes sign from
negative to positive).  Sketch a few solution curves (without trying
to solve the equation) in the $t$-$y$ plane.
\begin{enumerate}
\item $\displaystyle y' = y - y^2$
\item $\displaystyle y' = y \, (y-1) \, (y-2)$
\item $\displaystyle y' = \e^y - a$ with constant $a>0$
\end{enumerate}

\item For each of the following equations, determine the equilibrium
points and classify each as stable, unstable, or semi-stable ($f(y)$
does not change sign at the equilibrium value for $y$).  Sketch a few
solution curves (without trying to solve the equation) in the $t$-$y$
plane.
\begin{enumerate}
\item $\displaystyle y' = y^2 \, (y^2 - a^2)$ with constant $a \neq 0$
\item $\displaystyle y' = y^2 \, (y - a)^2$ with constant $a \neq 0$
\end{enumerate}

\emph{Discussion point (extra credit):} When discussing semi-stable
equilibrium points, how can you be sure that the solution, coming from
the ``stable side'', will converge toward the equilibrium point as
time goes to infinity rather than ``moving through it''?  As you know
from Calculus, there are functions, e.g.\ $y(t) = t^3$, where a zero
of the first derivative does not correspond to an extreme value; the
function is strictly increasing everywhere even though the first
derivative is zero at a point.  Can the same happen here?


\item Sometimes it is possible to solve a nonlinear equation by making
a change of the dependent variable that converts it into a linear
equation. The most important such equation has the form 
\[
  y' + p(t) \, y = q(t) \, y^n 
\]
and is called Bernoulli's equation.
\begin{enumerate}
\item Show that if $n \neq 0, 1$, then the substitution $v = y^{1-n}$
reduces Bernoulli's equation to a linear equation.

\item Use this method to solve the initial value problem
\[
  t^2 \, y' + 2t \, y - y^3 = 0 \,, \quad y(1)=1 \,.
\]

\item Use this method to find, by a different method than the one
explained in class, the solution to the logistic differential equation
\[
  y' = r \, y \, \Bigl( 1 - \frac{y}K \Bigr) \,, \quad
  y(0) = y_0 \,.
\]
\end{enumerate}

\item Rewrite the following differential equations as a system of
first order differential equations of the form
\[
  \frac{\d \x}{\d t} = \F (\x,t) \,,
\]
where $\x$ and $\F$ are vector-valued functions of the form
\[
  \x(t) =
  \begin{pmatrix}
    x_1(t) \\ \vdots \\ x_n(t)
  \end{pmatrix}
  \quad \text{and} \quad
  \F(\x,t) =
  \begin{pmatrix}
    F_1(x_1, \dots, x_n,t) \\
    \vdots \\
    F_n(x_1, \dots, x_n,t) \\
  \end{pmatrix} \,.
\]
\begin{enumerate}
\item $\displaystyle y''' + t \, y'' + \sin t \, y' + y = 0$
\item The second-order system
\begin{gather*}
  \frac{\d}{\d t} \biggl( v \, \frac{\d v}{\d t} \biggr) = 0 \,, \\
  \frac{\d}{\d t} \biggl( u \, \frac{\d v}{\d t} \biggr) = u \,.
\end{gather*}
\end{enumerate}
For each system: Is it autonomous?  Is it linear?
\end{enumerate}
\end{document}
