\chapter[Separation of Variables]{Differential Equations I: Separation of Variables}


The differential equations in this section will be of a slightly more
complicated form than in the section on exponential growth and decay, where
we treated the \index{differential equation $y'=\pm ky$} differential equation $y'=\pm ky$ ($+k$ leading to
exponential growth, and $-k$ leading to exponential decay).  Namely, we
shall deal with differential equations of the form $y'=\pm k(y-b)$, where
$b$ is some constant.  Before discussing the applications, let's solve the
differential equation $\disp\frac{dy}{dt}=k(y-b)$ by the same method ---
separation of variables --- that we used before.  We first multiply both
sides by $dt$ and divide through by $y-b$, and we then put an integral sign
in front of both sides:
\begin{align*} 
\int\frac{dy}{ y-b}&=\int k\,dt\\ \ln|y-b|+C_1&=kt+C_2\\
\ln|y-b|&=kt+C\quad\text{(where~}C=C_2-C_1)\\
|y-b|&=e^{kt+C}=e^Ce^{kt}\\ y-b&=\pm e^Ce^{kt}.
\end{align*} 
To find the constant $\pm e^C$ in front of $e^{kt}$, we suppose that we
have the initial information $y(0)=y_0$; then substituting $t=0$ on both
sides gives us: $y_0-b=\left(\pm e^C\right)e^0=\pm e^C$, and so we can
write $y-b=(y_0-b)e^{kt}$, i.e., 
\[
\text{The~differential~equation~}
\frac{dy}{ dt}=k(y-b)\text{ has~solution } y(t)=b+(y_0-b)e^{kt}.
\]

\smallskip

If our differential equation is of the form $\disp\frac{dy}{ dt}=-k(y-b)$ with a
negative constant of proportionality $-k$, then the solution is exactly the
same with $k$ replaced by $-k$.  Thus, 
\[
\text{The~differential~equation~}
\frac{dy}{ dt}=-k(y-b)\text{ has~solution } y(t)=b+(y_0-b)e^{-kt}.
\]

\smallskip

\section{Installment Buying} 

Large purchases (cars, houses, furniture, etc.)  are often made according
to the following system.  When you make the purchase (time $t=0$), you pay
only a small fraction of the price (called a ``downpayment'') and agree to
pay the remaining amount --- which we shall denote $y_0$ --- over a certain
period of time.  In essence, you are borrowing the $y_0$ dollars from the
seller (or a mortgage company in the case of a house).  We shall let $T$
denote the term of this loan (in years), i.e., the time during which you
continue to make payments, and we shall let $a$ denote the rate in dollars
per year at which you pay money during this time $T$.  (If you make monthly
payments, then $a$ equals 12 times the amount you pay each month.)  If the
seller is willing to loan you the $y_0$ interest free, then the situation
is very simple: $a=y_0/T$.  But in practice you must pay back not only
$y_0$ to make up the purchase price, but also interest on the money you
owe.  Let $k$ be the \index{instantaneous rate of interest} instantaneous rate of interest.

\smallskip

In order to treat this problem by a differential equation, we shall make
two simplifying assumptions:

\begin{enumerate}
\item
The interest is \index{compounded continuously} compounded continuously.  (This is the same simplifying
assumption that we made in the section on exponential growth.)

\item
Your payments are being made continuously (rather than at the end of
the month, as usually happens in practice).  For example, in the case of
monthly payments of \$300, this would mean, roughly speaking, that instead
of waiting until the end of the month, you arrange to have your bank's
computer transfer one penny from your account to the seller's account every
86.4 seconds.
\end{enumerate}

\smallskip

Even though both of these simplifying assumptions seem to diminish the
realism of our treatment of installment buying, in practice the inaccuracy
introduced is very small --- an error of considerably less than 1\%.

\smallskip

From a practical point of view, perhaps the most useful thing to figure out
is the monthly payment $a/12$, which depends on $y_0$ (which is purchase
price minus downpayment), the term $T$, and the interest rate $k$.  But
there are other questions which might also interest us.  In any case, the
way to approach the matter is to introduce the function
$$y(t)=\text{the~amount~you~still~owe~at~time~}t.$$ This amount $y(t)$
starts out at $y(0)=y_0$, and steadily decreases until at time $t=T$ you've
paid off the loan, i.e., $y(T)=0$.  This function $y(t)$ satisfies a
differential equation, which we first state in words: {\it the
instantaneous rate of change of the amount $y(t)$ that you owe is the
result of two contributions: (1) a positive contribution caused by
interest, which accumulates at a rate equal to $k$ times the current amount
you owe, and (2) a negative contribution of {\rm a} dollars/year, as you
make payments to decrease your debt.}

\smallskip

Written mathematically, the first contribution is $ky$ and the second
contribution is $-a$.  Thus, the function $y(t)$ satifies the differential
equation $$\frac{dy}{ dt}=ky-a.$$ If we rewrite the right side in the form
$k(y-\frac{a}{ k})$, we see that this is the same type of differential
equation that we already solved at the beginning of the section.  We can
simply take the solution we already found, where we substitute $b=a/k$.
The result is: $$y(t)=\frac{a}{ k}+\left(y_0-\frac{a}{ k}\right)e^{kt}.$$

\smallskip

We now want to interpret this result to obtain useful information.  First,
let us find the \index{rate of payment} rate of payment $a$ (or the monthly payment rate $a/12$) as
a function of $y_0$, $T$, and $k$.  The basic fact we use is that the
amount you owe must be reduced to 0 when $t=T$.  That is, 
\[
0=y(T)=\frac{a}{ k}+\left(y_0-\frac{a}{ k}\right)e^{kT}.
\]
To solve this for $a$, we do a little algebra.  We start out by dividing
through by $e^{kT}$, to make things look slightly simpler (no parentheses
needed): 
\[
 0=\frac{a}{ k}e^{-kT}+y_0-\frac{a}{ k}.
\]
(Recall that the
reciprocal of $e^{kT}$ is $e^{-kT}$.)  Since $a$ is the unknown, we move
all the terms with $a$ over to the left, and factor out $a$: 
\[
\frac{a}{ k}\left(1-e^{-kT}\right)=y_0.
\]
Now it is easy to solve, by multiplying
both sides by $k$ and dividing both sides by $1-e^{-kT}$: 
\[
a=\frac{ky_0}{ 1-e^{-kT}}.
\]



\begin{example}
%{\bf Example 1.} 
A few years ago the State of Washington had a 12\% ceiling
on interest rates, while neighboring Oregon had no such limit.  During this
time an ad for a \$450 dinette set running in Pacific Northwest newspapers
said: no money down; residents of Washington take 32 months to pay at 12\%
interest; residents of Oregon make the same monthly payments as in
Washington, although their interest rate will be 18\%.  Find the monthly
payment that residents of both states make, and also find out how much
beyond the 32 months must Oregonians continue to make payments because of
their higher interest rate.

\smallskip

In Washington, $T=32/12$ years, $k=0.12$, and $y_0=450$.  Hence, 
\[
\frac{a}{12}
=\frac{1}{ 12}\cdot 0.12\cdot 450\Bigm/\left(1-e^{-0.12\cdot 32/12}\right)
=\$16.43.
\]
In Oregon, we now know the value of $a$ (namely,
$12\cdot 16.4323=$\$197.19), and we want to find $T$.  So we return to the
equation $y(T)=0$ above, and instead of solving for $a$ we solve for $T$.
Moving the term with $T$ over to the left, we have $$\left(\frac{a}{
k}-y_0\right)e^{kT}=\frac{A}{ k}.$$ We divide both sides by $((a/k)-y_0)$,
simplify the fraction on the right by multiplying by $k/k$, and then take
the ~ln~ of both sides.  The result is $$kT=\ln\left(\frac{a}{
a-ky_0}\right),\quad\text{i.e.,}\quad T=\frac{1}{ k}\,\ln\left(\frac{a}{
a-ky_0}\right).$$ Substituting $k=0.18$, $a=197.19$, and $y_0=450$, we
obtain: $T=2.9386$ years $=35.26$ months.  That is, Oregonians must
continue paying for about 3 1/4 months more.

\end{example}

\smallskip

The next example deals with \index{home mortgages} home mortgages.  By a ``mortgage'' we shall
always mean a fixed-rate, fixed-payment mortgage, i.e., the standard kind.
Adjustable-rate mortgages and other variants will not be examined here.



\begin{example}
%{\bf Example 2.} 
To buy a house for \$71,500.00, you make a downpayment of
20\% and take a 30-year mortgage for the remaining 80\% at 11\% interest.
Find your monthly payments $a/12$, your yearly rate of payment $a$, and the
total amount you end up paying for the house.  Also compute how much of the
money paid in the first year is tax-deductible, and how much of the money
paid in the 30th year is tax-deductible.  (Note: money paid toward interest
is deductible from U.S. federal income tax, while money that goes toward
reducing the amount you owe is not.)

\smallskip

Here $y_0=0.8\cdot 71500=57200$, and so 
$$
a=\frac{0.11\cdot 57200}{ 1-e^{-0.11\cdot 30}}=\$6532.96;
\quad\quad\frac{a}{ 12}=\$544.41.
$$ 
The amount you end up paying for the house is equal to the downpayment
plus $aT$, i.e., $\$14300+30\cdot\$6532.96=\$210289$, which is
about three times the price of the house!  (This isn't as bad as it seems,
for at least two reasons: (1) a significant part of the \$138789 that goes
for interest is being paid by the government, in the form of reduced income
tax, and (2) the dollars that you pay 20 or 30 years later are dollars
whose real value has been reduced by inflation.)  Next, we want to find out
how much of the \$6532.96 paid between $t=0$ and $t=1$ is tax-deductible,
and how much of the \$6532.96 paid between $t=29$ and $t=30$ is
tax-deductible.  The simplest way to do this is to ask how much is {\it
not} tax-deductible, and that is the amount which has gone to decrease your
indebtedness.  Thus, the amount in the first year that is not
tax-deductible is 
\begin{align*} 
y(0)-y(1)&=
\Bigl(\frac{a}{ k}+\left(y_0-\frac{a}{ k}\right)e^{k\cdot 0}\Bigr)
-\Bigl(\frac{a}{ k}+\left(y_0-\frac{a}{ k}\right)e^{k\cdot 1}\Bigr)\\
&=
\left(y_0-\frac{a}{ k}\right)\left(1-e^k\right)
=\left(\frac{a}{ k}-y_0\right)
\left(e^k-1\right)\\
&=\left(\frac{6532.96}{ 0.11}-57200\right)\left(e^{0.11}-1
\right)=\$254.71.
\end{align*}   
The remaining $6532.96-254.71=\$6278.25$ is tax-deductible.  In the 30th
year, on the other hand, the non-tax-deductible amount is
$y(29)-y(30)=y(29)-0=y(29)=
\$6185.64$, and so only \$347.32 is deductible in the last year.

\smallskip

\begin{rightindent}{2.5in}
What is happening is that over the term of the mortgage the way your $a$
dollars per year payment divides up is changing.  At the beginning, when
you still owe a lot of money, most of the $a$ goes toward interest payment,
and only very little goes toward reducing the principal.  As time goes on,
more and more of the $a$ goes toward reducing the principal, and less and
less of it goes toward interest payment.  This can be represented visually
by a graph (see the figure at the right) that
\end{rightindent}
\hfill
%-figure OA -----------------------------
\begin{psfigure}{2.5in}{2.5in}{figOA.ps}
%\putcrosshair{0,0}
\puttext{-8., -30.}{\Small$t_{1}$}
\puttext{0., -30.}{\Small$t_{2}$}
\puttext{36., -30.}{\Small$T$}
\puttext{-40., 22.}{\Small$a$}
\puttext{-40., 18.}{\Small$k y_{0}$}
\puttext{15., 10.}{\Small$k\, y(t)$}
\end{psfigure}
%------------------------------

\noindent
depicts what our original differential equation is saying.  The graph
shows the horizontal line $a$ along with the curve $k y(t)$.  The
differential equation says that at any instant, of the $a$ dollars/year
that you pay, $ky(t)$ of it must go toward interest, and the remaining
$a-ky$ goes toward reducing the amount you owe.  In symbols, $dy/dt=-
(a-ky)$ (the minus sign appears because we are {\it decreasing} $y$).  Over
any time interval from $t_1$ to $t_2$, the amount you pay is $a(t_2-t_1)$,
which is the area under the horizontal line from $t_1$ to $t_2$.  Of that
area, the area under the curve $ky(t)$ is the amount that goes toward
interest (the tax-deductible amount), and the area between the curve and
the horizontal line is the amount that goes toward paying off the
principal.


\end{example}

\smallskip

Recall that we made two simplifying assumptions at the very beginning.
Since Example 16.2 is based on an actual purchase, we can show the actual
values computed by the mortgage company, based on daily compounding and
payments made at the end of each month.  Our error turns out to be quite
small: the true monthly payments for this mortgage were
\$545.12.  We were off by \$0.71, which is only 0.13\% of the total.

\smallskip

Below we show the table summarizing a statement sent us by the mortgage
company for the 8th year.  This statement gives the same type of
information as the above graph between $t=7$ and $t=8$.  The first column
has the constant value $a/12$ that is paid each month.  The fourth column
shows how much of this has gone toward the principal, and the sixth column
shows the difference, which has gone toward interest.  The sum of the
amounts in the sixth column is the ``interest total'' given at the bottom
--- the amount that is tax-deductible.

\medskip

\begin{center}
\tt\small
\begin{tabular}{||c|c|c|r|c|c|c|} 
\multicolumn{7}{||l|}{ANNUAL STATEMENT ON YOUR MORTAGAGE LOAN \#5-03-00016965\hspace{0.1in}INTEREST RATE 11.000\%}\\ \hline
TOTAL&DATE&INTEREST&\multicolumn{2}{c|}{PRINCIPAL ACCOUNT}& INTEREST
&
\\ 
PAYMEMT&PAID&PAID TO&Transactions&Balance&PAYMENT& 
\\ \hline
&01-01-87&&&54,542.50&\multicolumn{2}{c|}{$\leftarrow$BEGINNING BALANCE} \\ \hline
545.12&01-12-87&01-01-87&45.15-&54,497.35&499.97&\\ 
545.12&02-11-87&02-01-87&45.56-&54,451.79&499.56&\\ \hline
545.12&03-11-87&03-01-87&45.98-&54,405.81&499.14&\\
545.12&04-13-87&04-01-87&46.40-&54,359.41&498.72&\\ \hline
545.12&05-11-87&05-01-87&46.83-&54,312.58&498.29&\\ 
545.12&06-11-87&06-01-87&47.25-&54,265.33&497.87&\\ \hline
545.12&07-13-87&07-01-87&47.69-&54,217.64&497.43&\\ 
545.12&08-11-87&08-01-87&48.12-&54,169.52&497.00&\\ \hline
545.12&09-11-87&09-01-87&48.57-&54,120.95&496.55&\\ 
545.12&10-13-87&10-01-87&49.01-&54,071.94&496.11&\\ \hline
545.12&11-12-87&11-01-87&49.46-&54,022.48&495.66&\\ 
545.12&12-11-87&12-01-87&49.91-&53,972.57&495.21&\\ \hline
&&&&&&\\
&&&&&&\\ \hline
\multicolumn{3}{||c|}{TOTAL AS OF DECEMBER 31, 1987}&& &5971.51&\\ \hline
\end{tabular}\\
%------------------------------
\end{center}



\section[Newton's law]{Newton's Law of Cooling and Heating} 

\index{Newton's law of cooling} Newton's law of cooling states that {\it if an object is brought into an
environment, then the rate of cooling (or heating, if the object is colder
than the environment) is proportional to the temperature difference between
the object and the environment.} Let $y(t)$ be the temperature of the
object at time $t$, let $y_0=y(0)$ be the initial temperature of the
object, and let $b$ be the temperature of the environment.  Then Newton's
law can be translated into the mathematical equation $dy/dt=-k(y-b)$.  Here
$k$ is the constant of proportionality, which in practice depends on the
thermal properties of the object in the particular case at hand.  $k$ is
positive, and the minus sign is necessary in front of $k$ because $dy/dt$
is negative when the object's temperature $y(t)$ is greater than $b$, and
$dy/dt$ is positive when $y(t)$ is less than $b$.  This differential
equation is of the form that was solved at the beginning of the section.
Thus, we have $y(t)=b+(y_0-b)e^{-kt}$.  Notice that as $t$ gets larger, the
exponential term in this function gets smaller and smaller, so that $y(t)$
approaches $b$, i.e., the temperature of the object gets closer and closer
to that of the environment.  (In this section we are assuming that the
ambient temperature $b$ is constant, i.e., that it is not heated up by the
object and does not change with the changing time of day.  In the final
section on differential equations, we will treat an example where $b$
changes with time.)


\begin{example}
%{\bf Example 3.} 
A cup of coffee at 200${}^\circ$F is brought out into a
room which is kept at 60${}^\circ$F.  Two minutes later you measure the
coffee's temperature to be 180${}^\circ$.  Find a formula for the coffee's
temperature at any time $t$.

\smallskip

Here we have $y_0=200$ and $b=60$, but we do not know what $k$ is.  In the
formula for $y(t)$, namely $b+(y_0-b)e^{-kt}=60+140e^{-kt}$, we can find
$k$ by using the information given about $y(2)$:
\begin{align*} 
180&=y(2)=60+140e^{-k\cdot 2}\\ 120&=140e^{-2k}\\
\ln(120/140)&=-2k\\ k&=-\frac{1}{ 2}\ln(120/140)=\frac{1}{ 2}\ln(7/6)=0.077.
\end{align*} 
Thus, $y(t)=60+140e^{-0.077t}$.

\end{example}


\section{Falling Bodies with Air Resistance} 

Now we shall find a formula for the \index{velocity $v(t)$ of a falling body} velocity $v(t)$ of a falling body, no
longer assuming that air resistance is negligible.


\smallskip

The vertical motion of a falling object is affected by two things: gravity
and air resistance.  Near the surface of the earth, gravity contributes a
constant acceleration $g$.  (Let us take $\underline{\rm downward}$ to be
the positive direction, in which case gravitational acceleration is $g$ rather
than $-g$.)  The air resistance depends on the aerodynamic
properties of the object and also on its speed.  Generally, the faster the
object is moving, the greater the force of air molecules pushing against
it.  In this section we shall assume that the force of air resistance is
proportional to the velocity.  Since acceleration is proportional to force,
this also means that the acceleration (actually, deceleration) from air
resistance is assumed to be proportional to $v(t)$.  Thus, there are two
contributions to $dv/dt$: $g$ and $-kv(t)$ (here $k$ is positive, and
there's a minus sign in front of $k$ because the air resistance works in
the opposite direction to velocity).  In other words, the unknown function
$v(t)$ satisfies the differential equation $dv/dt=g-kv$, which we rewrite
in the form $$dv/dt=-k\left(v-\frac{g}{ k}\right).$$ We see that this is in
the form that was solved at the beginning of the section (with $v$ instead
of $y$ and $g/k$ in place of $b$).  So our solution is the velocity
function 
$$v(t)=\frac{g}{ k}+\left(v_0-\frac{g}{ k}\right)e^{-kt}.$$

\smallskip

Notice that as the time increases, the \index{exponential term} exponential term in the formula for
$v(t)$ approaches zero, and so the velocity approaches the value $g/k$.  We
call $g/k$ the {\bf terminal velocity} and denote it $v_\infty$.  The
terminal velocity does not depend on the initial velocity: no matter what
initial velocity we give to the object, eventually its velocity will become
essentially equal to the value $g/k$ (in the sense that the difference
between $v(t)$ and $g/k$ cannot be detected in practice).  In other words,
the graph of $v(t)$ has a \index{horizontal asymptote} horizontal asymptote at height $g/k$.  Another
way to find the terminal velocity is to ask when the two forces on the
object --- gravity and air resistance --- balance, i.e., when the right
side of the differential equation equals zero: $g-kv=0$.  This happens when
$v=g/k=v_\infty$.

\smallskip

\begin{example}
%{\bf Example 4.} 
A parachutist drops from a plane at time $t=0$.  She freefalls for 10
seconds, during which time the constant of air resistance is $k=0.1$.
After her parachute opens, the air resistance becomes much greater:
$k=1.4$.  Take downward to be the positive direction, and take $g=9.8$
m/sec${}^2$.  Find formulas for her velocity for $0\le t\le 20$, and draw a
graph of this velocity function.  Since the parachutist dropped from the
plane, rather than throwing herself downward, the initial velocity

\begin{rightindent}{2.75 truein}
\noindent 
at time $t=0$ is 0.  Thus, for the first 10 seconds we have 
$v(t)=\frac{g}{k}\left(1-e^{-kt}\right)=98\left(1- e^{-0.1t}\right).$ Starting at $t=10$
we have a new situation.  The new initial velocity is
$v(10)=98\left(1-e^{-0.1\cdot 10}\right)=61.95$ m/sec, and the new $g/k$ is
$9.8/1.4=7.0$.  In using our falling body formula, we must be careful,
since the starting time is now 10 rather than 0.  Thus, $t-10$ is the time
elapsed since the parachute opened, and so we use $t-10$ in place of $t$ in
the formula.  The result is:
\end{rightindent}
\hfill 
%-figure OB -----------------------------
\begin{psfigure}{2.5in}{2in}{figOB.ps}
%\putcrosshair{0,0}
%\puttext{0,-1}{}
\end{psfigure}
%------------------------------

$$
v(t)=\begin{cases}
98\left(1-e^{-0.1t}\right),& \text{for $0\le t\le 10$;}\\
7+54.95e^{-1.4(t-10)},& \text{for $10\le t\le 20$.}
\end{cases}
$$ 
A graph of the velocity function is shown above.

\end{example}

\section{Mixing Problems} 

Suppose we have a container which holds a certain volume $V$ of water and
at time $t=0$ has a certain amount $y_0$ of dissolved impurity (such as
salt, pollutant, etc.).  A pipe carries water into the container at a rate
$r$, and another pipe carries water out of the container at the same rate
$r$ (so that the volume $V$ of water in the container remains the same).
The incoming water has a concentration $k$ of impurity (measured in
gm/liter, lb/ft${}^3$, gm/m${}^3$, etc.).  Let $y(t)$ be the amount of
impurity in the water in the container at time $t$.  We make the assumption
that the impurity brought into the water by the first pipe immediately
spreads uniformly throughout the container, so that the water in the
container always has the same \index{concentration} concentration of impurity throughout (i.e.,
there is no pileup of impurity near the mouth of the pipe).  As a result,
the concentration of impurity in the water taken up by the exit pipe is
equal to the overall concentration in the container, namely, amount of
impurity per volume $= y(t)/V$.

\smallskip

We are now ready to set up a differential equation for $y(t)$.  The basic
principal is that {\it the rate at which impurity enters the container is
equal to the concentration of impurity (amount per volume) times the rate
of the incoming water (volume per time), and the rate at which impurity
leaves the container is equal to the concentration of impurity in the exit
pipe times the rate of the water going out}.  Thus, $dy/dt$ has a positive
contribution of $kr$ and a negative contribution of $y/V$ times $r$, i.e.,
\[
\frac{dy}{ dt}=kr-\frac{y}{ V}r
=kV\frac{r}{ V}-\frac{r}{ V}y=-\frac{r}{ V}(y-kV).
\]
This differential equation is of the form that was solved at the beginning
of the section, except that the $k$ here has nothing to do with the $-k$
that we had earlier.  Now it is $r/V$ that is playing the role of $k$, and
$kV$ that is playing the role of $b$.  So we have the solution
\[
y(t)=kV+(y_0-kV)e^{-rt/V}.
\]
It is also sometimes useful to have a
formula for the concentration in the container at time $t$.  This is
$y(t)/V$, which we get by dividing the above formula by $V$: 
\[
\frac{y(t)}{ V}=k+\left(\frac{y_0}{ V}-k\right)e^{-rt/V}.
\]

\smallskip

Notice that as $t\longrightarrow\infty$, the concentration of impurity in
the container approaches $k$, which is the concentration in the incoming
water.  This makes sense: roughly speaking, the incoming water with its
concentration $k$ eventually ``displaces'' the water that was originally
there.

\subsection{The special case $k=0$} 
If $k=0$, then our formula for $y(t)$ becomes $y(t)=y_0e^{-rt/V}$, which is
an exponential decay function.  (In this case our original differential
equation was simply $dy/dt=-\frac{r}{ V}y$.)  Thus, if you remove impurities
from the water by pumping in pure water, you find that the amount of
impurity decreases exponentially, so that eventually the amount of impurity
is too small to be measured.  What is the ``half-life'' of the impurities,
i.e., the length of time required to cut the amount of impurity in half?
We use the formula $H=(\ln 2)/k$ that we derived before, where $k$ is
replaced by $r/V$.  The result is $H=(\ln 2)V/r$.
 
\smallskip

This concludes our discussion of four applications of differential
equations of the form $y'=\pm k(y-b)$.  We now make some final
observations.


\section{The Constant Solution} 

There is one and only one constant function $y(t)=$const ~that satisfies a
differential equation of the form $dy/dt=\pm k(y-b)$.  Namely, since a
constant function has derivative zero, it follows that a constant function
satisfies the differential equation if and only if $0=\pm k(y-b)$, i.e., if
and only if $y(t)=b$.  In all four of the practical applications discussed
above, the constant solution has a special meaning.


\subsection{Installment buying}  
The constant solution $y(t)=a/k$ means that you have chosen your yearly
rate of payment $a$ to equal $ky_0$ (the interest rate times the amount you
borrowed).  That is just enough to pay the interest, without diminishing
the amount you owe.  This constant solution separates the two cases: (1)
$a>ky_0$, in which case the amount you owe $y(t)$ $=\frac{a}{
k}+\left(y_0-\frac{a}{ k}\right)e^{kt}$
$=\frac{a}{k}-\left(\frac{a}{k}-y_0\right)e^{kt}$ decreases and becomes
zero in a finite amount of time $T$; and (2) $a<ky_0$, in which case you
are not even paying enough to take care of the interest, and the amount you
owe grows without bound.


\subsection{Law of cooling/heating}  
The constant solution $y(t)=b$ means that you have brought an object into
the room whose temperature is the same as that of the room, in which case,
of course, the temperature of the object remains constant.  This constant
solution separates the two cases: (1) $y_0>b$, in which case you are
bringing into the room a hotter object, which then cools, approaching the
ambient temperature $b$ as $t\longrightarrow
\infty$; and (2) $y_0<b$, in which case you have a colder object, which heats
up, approaching the ambient temperature as $t\longrightarrow\infty$.  Note
that in all cases the constant solution $b$ is also the value that $y(t)$
approaches as $t\longrightarrow\infty$.


\subsection{Falling bodies with air resistance}  
The constant solution $v(t)=g/k$ means that you have thrown the object
downward at an initial velocity $v_0$ exactly equal to the terminal
velocity $v_\infty=g/k$.  In that case the object falls at constant speed.
This constant solution separates the two cases: (1) $v_0<g/k$ (for example,
the object falls from rest), in which case the object accelerates downward,
approaching $v_\infty=g/k$ as $t
\longrightarrow\infty$; and (2) $v_0>g/k$ (for example, a parachutist after
the parachute opens), in which case the object decelerates, approaching
$v_\infty=g/k$ as $t\longrightarrow\infty$.  Again note that, whatever the
initial velocity, the constant solution $g/k$ is also the value that $v(t)$
approaches as $t\longrightarrow
\infty$.


\subsection{Mixing problems}
The constant solution $y(t)=kV$ means that you start with a concentration
$y_0/V=k$ which is equal to the concentration of impurity in the incoming
water; thus, the incoming water does not change the amount of impurity in
the container.  This constant solution separates the two cases: (1)
$y_0<kV$, when the incoming water increases the amount of impurity until
the concentration $y(t)/V$ approaches $k$ in the limit as
$t\longrightarrow\infty$; and (2) $y_0>kV$, when the incoming water
decreases the amount of impurity, and again the concentration approaches
$k$ as $t \longrightarrow\infty$.  Once again, in all cases the constant
solution $kV$ is also the limiting value of $y(t)$ as
$t\longrightarrow\infty$.


\section{A Final Remark} 

In all four of these applications, we made some simplifying assumptions,
which introduce errors.  That is, our formulas will not be an exact
description of what happens in practice.  The amount of error introduced
depends on the particular situation.  Generally speaking, in installment
buying problems the error is very small, while in the other three
applications our numerical values for $y(t)$ or $v(t)$ might be way off.
However, even if we cannot always rely upon our {\it quantitative} results,
we have obtained some useful \index{{\it qualitative} information} {\it qualitative} information.  For example,
the role of the constant solution of the differential equation --- as the
limiting value of the function as $t\longrightarrow\infty$ --- is actually
observed in practice.  For example, the terminal velocity of a falling body
is the speed at which air resistance exactly balances (is equal to) the
force of gravity.

%%---------------------------------------------------------------
