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{\bf Homework Set 4}

Due Monday, February 14
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{\bf I.} Find the solution to each of the following. You may leave your answer in the form of a relation.
\begin{enumerate}


\item $\dis \frac{x}{y} y' = - \frac{2x+y}{x+2y}$, with $y(1) = 1$.

\item $\dis 3x^2 + y^2 + 2xyy' = 0$, with $y(1) = 2$.

\item $\dis (2y-x)y' = y- 4x$, with $y(2)=1$.

\item $\dis \cos x \sec y + \sin x \sin y \sec^2 y \frac{dy}{dx} = 0$, with $y(\pi/6) = \pi/4$.

\item $\dis 2y^2 - 6xy + (3xy-4x^2)y' = 0$, with $y(1) = 1$. Hint: It is not exact, but will become exact if you multiply through by $xy$. 

\end{enumerate}

{\bf II.}

\begin{enumerate}

\item For \#1 above, plot the solution curve with a computer.
\end{enumerate}


{\bf III.} Find the particular solution to each of the following.
\begin{enumerate}
\item $y' y'' = 4t$, with $y(1)=5$ and $y'(1)=2$.

%answer: y = t^2 + 4


\item $\frac{d^2y}{dt^2} = \frac{3}{2}y^2$, with $y(0)=1$ and $y'(0)=1$.

%answer: y = 4/(2-t)^2

\end{enumerate}

{\bf IV.}
\begin{enumerate}

\item 
At 5 a.m. you put a $60^o$ beer in a $40^o$ refrigerator.
You get up at 8 a.m. expecting to have a cold one with your Honey
Puffs cereal but to you horror you find that someone has taken your
beer and left it on the table. The beer is now $65^o$. (Room temperature,
you note, is $70^o$.)

You know roommate \#1 left for work at 6 a.m. and so go to confront roommate 
\#2. But, he swears that he has been asleep the whole time. You bet him a six 
pack you can prove it was he who left the beer out. He folds his arms, smiles
and says ``you're on!''

Recall Newton's Law of Cooling: $\dis \frac{dT}{dt} = k(T_a - T(t))$.

You put the beer back in the frig and wait 30 minutes (while roommate \#2 watches cartoons thinking about his six pack). When you take the beer out it has cooled to $50^o$. You calculate $k$. Now, win the bet!


\item
 The number of algae cells in a tank of water grows according to 
\begin{eqnarray}
A'(t) & = & .2\left(1-\frac{A(t)}{100}\right)A(t) \hspace{.325in} \mbox{light on, and} 
                                                               \nonumber\\
A'(t) & = & -.2A(t) \hspace{1in}  \mbox{light off.} \nonumber
\end{eqnarray}
In words, the carrying capacity drops from 100 (billion cells) to zero when
the light goes out.  At $t=0$ you start a ten (10) hour experimental run with
$A(0) = 25$ and plan to keep the light on. When you come back at $t=10$ you 
discover that the light bulb has burned out.  You measure $A(10)$ to be 15.
  What time did the light bulb burn out?


\item   Suppose $y'(t) = F(y(t))$, where the graph of $F(y)$ is given
        below.  Carefully draw the integral curves for this equation.
        What are the equilibrium solutions? What are their stability
        types?  Describe the initial concavity of the solution curves.
        Assume $y(t)$ and $t$ are nonnegative.

% @@@@@@@@@@@@@@@@@@@@@@@@@@@@ FIGURE @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
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        \psfrag{f(y)}{$f(y)$}
        \psfrag{y}{$y$}
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        \psfrag{A}{$A$}
        \psfrag{B}{$B$}
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        \psfrag{D}{$D$}
        \psfrag{E}{$E$}
        \psfrag{F}{$F$}
        \psfrag{0}{$0$} 
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        \includegraphics[height=4in]{Tests/Figs/t1graphc.eps}
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\end{enumerate}
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