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\title{Math 1b.  Lecture 27 \\
Solutions to Differential Equations}

\author{T. Judson}

\date{Spring 2006}


\begin{document}

\maketitle


\section{Goals}

\begin{itemize}

\item
To understand what it means to be the solution of a differential equation.

\item
To understand the idea of a slope field.


\item
To understand and be able to apply the existence and uniqueness theorem (weak form) for differential equations.

\item
To be able to solve differential equations of the form $dy/dt = g(t)$.

\item
To be able to solve differential equations of the form $dy/dt = ky$.

\item
To understand that autonomous equations of the form $dy/dt = f(y)$ are time independent.


\end{itemize} 

\section{Equations of the Form $dy/dt = g(t)$.}

It is quite easy to solve an equation of the form $dy/dt = g(t)$---simply integrate both sides of the equation.  However, equations of this form rarely occur in practice.

\section{The Coffee Problem}

Two identical cups of dark liquid are left in a $70^\circ$F laboratory cool.  At time $t = 0$, the first cup's temperature was $190^\circ$F, and was dropping at a rate of $12^\circ$F per minute.  When did this cup's temperature fall to $130^\circ$F?  The second cup was at $130^\circ$F after 10 minutes.  Could this liquid be coffee?

See if students can come up with the equation
\begin{eqnarray*}
\frac{dy}{dt} & = & - \frac{1}{10} (y-70), \\
y(0) & = & 190.
\end{eqnarray*}


\section{Solutions of Differential Equations}

A {\em solution\/} to a differential equation 
$$
\frac{dy}{dx} = f(x, y).
$$
is a function $y = y(x)$ that satisfies the equation.  For example, $y = x^4/4 + C$ is a solution to $y' = x^3$, where $C$ is an arbitrary constant.  If we specify an initial condition, $y(0) = y_0$, then we can find a unique solution.  
 
{\em In general, differential equations are very difficult to solve}.  However, it is quite easy to check if a function is a actually a solution.  For example,
$$
y(t) = 70 + 120 e^{-0.1 t}
$$
is a solution to the differential equation
\begin{eqnarray*}
\frac{dy}{dt} & = & - \frac{1}{10} (y-70), \\
y(0) & = & 190.
\end{eqnarray*}

\section{Newton's Law of Cooling}

{\em An object cools at a rate proportional to the temperature difference between the object and its environment}.  As a differential equation, Newton's Law of Cooling can be stated as
$$
y' = k(y-T_e).
$$
If we know the initial temperature at $t=0$, then we have an {\em initial value problem}, which has a unique solution.  We can state the initial condition as
$$
y(0) = T_0.
$$




\section{Slope Fields}

We can use slope fields to answer the coffee problem.

\bigskip

\begin{center}
\includegraphics[width=4.75in]{Fig_27_1}
\end{center}


\bigskip



\section{Existence and Uniqueness of Solutions}

If an initial value problem has the form $y' = f(t)$ or $y' = g(y)$ and initial condition $y(a) = b$, then the initial value problem has a unique solution.






\section{Worksheet Problems}

\begin{enumerate}

\item
Two identical cups of dark liquid are left in a $70^\circ$F laboratory cool.  At time $t = 0$, the first cup's temperature was $190^\circ$F, and was dropping at a rate of $12^\circ$F per minute.  When did this cup's temperature fall to $130^\circ$F?  The second cup was at $130^\circ$F after 10 minutes.  Could this liquid be coffee?


\item
Decide whether the given function is a solution to the differential equation.
\begin{enumerate}

\item
$y(t) = t^2/2$; \hspace{3em}  $y' = t$ 

\item
$y(t) = t^2/2$;  \hspace{3em}  $y' = y$

\item
$\displaystyle y(t) = \frac{1}{2} t^4 + \frac{3}{2} t^2 + \frac{1}{4}$;  \hspace{3em}  $\displaystyle y' - \frac{2y}{t} = t^3$

\item
$\displaystyle y(t) = \frac{1}{2} t^4 + \frac{3}{2} t^2$; \hspace{3em}  $\displaystyle y' - \frac{2y}{t} = t^3$


\end{enumerate}


\item
Verify that
$$
y(t) = 70 + 120 e^{-0.1 t}
$$
is a solution to the differential equation
\begin{eqnarray*}
\frac{dy}{dt} & = & - \frac{1}{10} (y-70), \\
y(0) & = & 190.
\end{eqnarray*}


\item
(8 points) \hspace{1em}
Match equations and slope fields
\begin{multicols}{2}
\begin{enumerate}

\item
$\displaystyle\frac{dy}{dt} = 1+y^2$ %midterm1fig4

\item
$\displaystyle\frac{dy}{dt} = y^2 - t^2$ 

\item
$\displaystyle\frac{dy}{dt} = ty$ %midterm1fig4

\item
$\displaystyle\frac{dy}{dt} = 1 - y $

\item
$\displaystyle\frac{dy}{dt} = y(1-y) - 2$ 

\item
$\displaystyle\frac{dy}{dt} = (y - t)^2$ 

\item
$\displaystyle\frac{dy}{dt} = 1 - y^2$ %midterm1fig4


\item
$\displaystyle\frac{dy}{dt} = y^2 -4t^2$ %midterm1fig4


\end{enumerate}
\end{multicols}





\vspace{.2in}
\centerline {
(i)
\includegraphics[width=3in]{Fig_27_3}
\hspace{.25in}
(ii)
\includegraphics[width=3in]{Fig_27_4}
}
\vspace{.2in}
\vfill

\vspace{.2in}
\centerline {
(iii)
\includegraphics[width=3in]{Fig_27_2}
\hspace{.25in}
(iv)
\includegraphics[width=3in]{Fig_27_5}
}
\vspace{.2in}



\end{enumerate}





\section*{References}

\begin{itemize}

\item
\S 7.1 in 
James Stewart.
{\it Single Variable Calculus: Concepts \& Context}, third edition.
Brooks/Cole, Belmont CA, 2005.
ISBN 0--534--41022--7.

\item
\S 31.2 in
Robin J. Gottlieb. 
{\it Calculus: An Integrated Approach to Functions and Their Rates of Change}, preliminary edition. 
Addison Wesley, Boston, 2002.
ISBN 0--201--70929--5.


\end{itemize}



\section*{Notes}

\subsection*{April 19, 2006}



\end{document}

