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\centerline{Mathematics 304, section 2 -- Ordinary Differential Equations}
\centerline{Selected Solutions -- Problem Set 1}
\centerline{September 13, 2004}
\bigskip
\noindent
Section 1.2/2.  The problem said to find a differential equation of
the form
$${dy\over dt} = 2y - t + g(t)\leqno(1)$$
that has $y = e^{2t}$ as a solution.  To find the unknown part
$g(y)$, substitute for $y$ on both sides in (1):
$$2e^{2t} = 2e^{2t} - t + g(e^{2t})$$
and cancel to get $0 = -t + g(e^{2t})$, or $g(e^{2t}) = t$.
Hence $g(y) = {1\over 2} \ln(y)$ will give one such equation,
since ${1\over 2} \ln(e^{2t}) = t$.
\bigskip
\noindent
Section 1.2/30.  Solve:
$${dy\over dt} = {t\over (1 - t^2)y};\quad y(0) = 4$$
Separating variables and integrating we have
$$\int y\ dy = \int {t \over 1 - t^2}\ dt$$
The integral on the right has the form $-{1\over 2} \int {du\over u}$
for $u = 1 - t^2$.  Hence we get
$${1\over 2}y^2 = {-1\over 2} \ln |1 - t^2| + c$$
From the initial condition, $c = 8$, so 
$$y^2 = 16 - \ln |1-t^2| \Rightarrow y = \sqrt{ 16 - \ln |1-t^2| }$$
(positive square root, since $y(0) > 0$).
\bigskip
\noindent
Section 1.2/35.  Let $S$ be the amount of salt at time $t$.  Then
the information given in the problem says that $S$ is a solution of 
$${dS\over dt} = {1\over 4} - {1\over 10}S;\quad S(0) = 0$$
(the initial condition comes from the fact that at the start 
we have ``pure water'' -- no salt at the start).  This is a separable
equation with solution 
$$S(t)={5\over 2} - {5\over 2}e^{-t\over 10},$$
so
$$\eqalign{S(1) &\doteq .238\cr
           S(10) &\doteq 1.58\cr
           S(60) &\doteq 2.49\cr
           S(1000) &\doteq 2.50\cr
          \lim_{t\to +\infty} S(t) &= 2.50\cr}$$
Note that this result makes sense because ``in the long run'' the 
water in the bucket should have the same salt concentration as 
the salt plus water mixture entering.
\bigskip
\noindent
Section 1.2/37.  The differential equation describing Newton's
Law of Cooling is 
$${dT\over dt} = k(T - 70)$$
where $T$ is the temperature of the cup of cocoa, and $k$ is the 
constant of proportionality.
Solving this by separating variables and integrating, we get
$$T(t) = 70 + ce^{kt}$$
The initial condition is the initial temperature:  $T(0) = 170$,
so 
$$T(t) = 70 + 100e^{kt}.$$
We also know that the temperature is decreasing by $20$ degrees
per minute at $t = 0$, so $T'(0) = -20$.  This information
determines the value of $k$:
$-20 = 100 k e^{k\cdot 0}$, so $k = -1/5$.  Then to find the 
time when the temperature reaches 110 degrees, 
$$110 = 70 + 100 e^{-t/5} \Rightarrow t = -5\ln(.4) \doteq 4.58$$
(The cocoa has cooled down to 110 degrees after about 4.6 minutes.)
\bigskip
\noindent
Section 1.2/40.  The hardest part of this one is just getting at
the essential points buried in the amusing story.  This is like
the other ``mixing problems'', except that after the initial
mistake, {\it no more hot sauce is added to the chili}, so the
in-flow rate term is actually {\it zero}.  Keeping track of the units,
if $H$ is the amount of hot sauce in the remaining chili, in 
teaspoons, as
a function of $t$,
$${dH\over dt} = {-1\over 32} H; \quad H(0) = 12$$
(just an exponential decay equation!)
The solution is $H(t) = 12 e^{-t/32}$.  The time when the remaining
hot sauce is 4 teaspoons (2 teaspoons per gallon in the 2 gallon pot
of chili) is the solution of 
$$4 =   12 e^{-t/32} \Rightarrow t = -32\ln(1/3) \doteq 35$$
That is, the recipe level of hot sauce is regained after about
35 minutes, and about 35 cups of the chili have been served by that
time.
\bigskip
\noindent
Section 1.8/12.  Solve
$${dy\over dt} = {2t\over 1 + t^2} y + {2\over 1+t^2}; y(0) = -2.$$
This is a first order linear equation with 
$$\eqalign{e^{\int g(t)\ dt} &= e^{\int {2t\over 1 + t^2}\ dt}\cr
&=e^{\ln(1+t^2)}\cr
&=1+t^2\cr}$$
Hence applying the general formula we get
$$\eqalign{
y &= ce^{\int g(t)\ dt} + e^{\int g(t)\ dt} \int e^{-\int g(t)\ dt} r(t)\ dt\cr
&= c(1+t^2) + (1+t^2) \int {2\over (1+t^2)^2}\ dt\cr}\leqno(2)$$
Pulling the 2 out of the remaining integral, we need to 
compute $\int {dt\over (1+t^2)^2}$ to finish.  This is one where
we need to use the method of {\it trigonometric substitution} (or
a table of integrals.  Here's the method if you do it ``by hand'':
The form $1 + t^2$ suggest the tangent substitution $t = \tan(\theta)$
so $dt = \sec^2(\theta)d\theta$.  The reason for this substitution is 
the identity $1 + \tan^2(\theta) = \sec^2(\theta)$.  Hence we have
$$\eqalign{\int {dt\over (1+t^2)^2} &= \int {\sec^2(\theta)\over \sec^4(\theta)}\ d\theta\cr
&=\int \cos^2(\theta)\ d\theta\cr
&={1\over 2}(\sin(\theta)\cos(\theta) + \theta)\cr}$$
Then since $t = \tan(\theta)$, $\theta = \arctan(t)$, and
$\sin(\theta) = {t\over \sqrt{1+t^2}}, \cos(\theta) = {1\over \sqrt{1+t^2}}$.
Substituting these in, we find:
$$\int {dt\over (1+t^2)^2} = {1\over 2}\left({t\over 1+t^2} + \arctan(t)\right)$$
Substituting this back into (2), we get the general solution:
$$y =  c(1+t^2) + t + (1+t^2)\arctan(t)$$
The initial condition $y(0) = -2$ implies $c = -2$, and we're done.
\bye