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\begin{center}
{\Large Warm-Up Problems and Lecture Problems

February 21, 2003}
\end{center}

\begin{enumerate}


\item \label{one} 
	Suppose you have a line through the given point, with the given slope.
	Find the $y$-value of that line at the given $x$-value.
	\begin{enumerate}
	\item  Point: $(1,2)$, Slope: $3$.  If $x=0$ then $y=$\rule{1in}{.1pt}
	\item  Point: $(-1,3)$, Slope: $-2$.  If $x=2$ then $y=$\rule{1in}{.1pt}
	\item  Point: $(-2,3)$, Slope: $\frac{3}{2}$.  If $x=-1$ then $y=$\rule{1in}{.1pt}
	\item  Point: $(0,-5)$, Slope: $-\frac{1}{6}$.  If $x=\frac{2}{3}$ then $y=$\rule{1in}{.1pt}
	\end{enumerate}

\item  Now, come up with a general method to answer question~\ref{one}
	\emph{without} first finding the equation of the line?
	In other words, find a formula to plug the point, the slope and the $x$-value into
	that will give you the $y$-value.
	Use your formula to answer the following questions:
	\begin{enumerate}
	\item  Point: $(0,2)$, Slope: $-\frac{1}{6}$.  If $x=1$ then $y=$\rule{1in}{.1pt}
	\item  Point: $(1,-5)$, Slope: $0.7$.  If $x=1.3$ then $y=$\rule{1in}{.1pt}
	\item  Point: $(1.2,3.5)$, Slope: $-.67$.  If $x=1.3$ then $y=$\rule{1in}{.1pt}
	\item  Point: $(1.3,-3.2)$, Slope: $0.23$.  If $x=1.42$ then $y=$\rule{1in}{.1pt}
	\end{enumerate}

\newpage
\lecture

\item  Given the following differential equation:
	\[
	\frac{dy}{dx} = -3y
	\]
	\begin{enumerate}
	\item  Verify that $y=Ce^{-3x}$ is a solution.
	\item  Find the specific solution if $y(1)=4$ (i.e., find the correct value for $C$).
	\end{enumerate}


\item  Using Euler's method, with step size $.2$, construct a table of values for the initial value problem:
	\[
	y' = x-y \qquad y(0)=1
	\]
	\[
	\begin{array}{|c|c|c|}
	\hline
	n & x_n & y_n \\
	\hline
	\hline
	0 & \qquad & \qquad \qquad \qquad \\
	\hline
	1 & & \\
	\hline
	2 & & \\
	\hline
	3 & & \\
	\hline
	4 & & \\
	\hline
	5 & & \\
	\hline
	6 & & \\
	\hline
	7 & & \\
	\hline
	8 & & \\
	\hline
	9 & & \\
	\hline
	10 & & \\
	\hline
	\end{array}
	\]



\end{enumerate}



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