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

Solutions

February 26, 2003}
\end{center}

\begin{enumerate}


\item  Consider the following initial value problem:
	\[
	\frac{dy}{dt} = \frac{10 - y}{7} \qquad y(0)=21
	\]
	\begin{enumerate}
	\item  Find the \emph{general} solution (i.e., don't take the initial value into account).
		Write you answer as a function of $t$ (i.e., write your answer as $y=$).
		\\\sol  First separate the variables:
		\[
		\frac{dy}{10-y} = \frac{dt}{7}
		\]
		Now integrate: $-\ln|10-y| = \frac{1}{7}t+C$.  Now, multiply both sides by $-1$ and exponentiate both sides: $\ln|10-y| = -\frac{1}{7}t+C$ and then:
		\[
		|10-y| = e^{-t/7-C} = e^{-C}e^{-t/7}
		\]
		Get rid of the absolute values and make $A=\pm e^{-C}$:
		\[
		10-y = Ae^{-t/7}
		\]
		Finally, solve for $y$:
		\[
		y = 10-Ae^{-t/7}
		\]
	\item  Find the specific solution.
		\\\sol  Plug in the intial condition and solve for $A$:
		\[
		21=10-Ae^{0}
		\]
		So, $A=-11$ and the solution is:
		\[
		y=10+11e^{-t/7}
		\]
	\end{enumerate}

\newpage

\lecture

\item  In 1980, the population of Nepal was $14.6$ million.
	In 1990, the population was $18.1$ million.
	Let $t$ represent years and let $t=0$ correspond to 1980.
	Find a population model for the population of Nepal:
	\[
	P=P_0e^{kt}
	\]	
	\begin{enumerate}
	\item  Find $P_0$.
		\\\sol Plug $t=0$ and $P=14.6$ into the equation to get $P_0=14.6$.
	\item  Find $k$.
		\\\sol We already know $P_0=14.6$, so now we plug in $t=10$ and $P=18.1$ and solve the equation: $18.1=14.6e^{10k}$.  Divide by $14.6$ and take the natural log of both sides to get $k=0.0215$.  Therefore, our model is:
		\[
		P=14.6e^{0.0215 t}
		\]
	\item  Using your model, predict the population in year 1997.
		(The actual population in 1997 was $22.59$ Million.)
		\\\sol This is a matter of plugging in $17$ for $t$ in our model to get $P=21.04$ million.
	\item  Using your model, predict the population in year 2010.
		\\\sol Now plug in $t=30$ to get $P=27.8$ million people.
	\end{enumerate}

\bigskip

\item  (Nearly identical to the previous problem). 
	In 1980, the population of Nepal was $14.6$ million.
	In 1990, the population was $18.1$ million.
	Let $t$ represent years and let $t=0$ correspond to 1900.
	Find a population model for the population of Nepal:
	\[
	P=P_0e^{kt}
	\]	
	\begin{enumerate}
	\item  Find $P_0$ and $k$.
		\\\sol We now have two equations and two unknowns to solve (by using our two data points):
		\begin{eqnarray*}
		14.6 & = & P_0e^{80k} \\
		18.1 & = & P_0e^{90k}
		\end{eqnarray*}
		I take the first equation and solve for $P_0=14.6e^{-80k}$.  I plug this in for $P_0$ in the second equation: $18.1=14.6e^{-80k}e^{90k}$, or $18.1=14.6e^{10k}$.  I can now solve this as before to get $k=0.0215$ (as before).  I can now plug back in to get $P_0=2.61$.  So, our model is:
		\[
		P=2.61e^{0.0215 t}
		\]
	\item  Using your model, predict the population in year 1997.
		(The actual population in 1997 was $22.59$ Million.)
		\\\sol We plug $t=97$ into our model and get $P=21.00$ million (the difference from the previous result is due to round off error).
	\item  Using your model, predict the population in year 2010.
		\\\sol Plugging in $t=110$ gives $P=27.8$ million.
	\end{enumerate}





\end{enumerate}



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