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Quickly written notes:

2.2: example from week 3 video  (slides 12 - 14) available at 

\centerline{ http://homepage.math.uiowa.edu/$\sim$ idarcy/COURSES/100/SPRING22/videos.html}


2.3: Modeling with differential equations.
\t
Suppose salty water enters and leaves a tank at a rate of 6 liters/minute.
\t
  Suppose also that the salt concentration of the water entering the tank varies with respect to time according to ${1 \over (t+1)(t+2)e^{3t}}$ g/liters where $Q(t) = $ amount of salt in tank in grams.
  \t
    If the tank contains 2 liters of water and initially contains 1g of salt, find a formula for the amount of salt in the tank after $t$ minutes.

Let $Q(t) = $ amount of salt in tank in grams.
\hfill
Note $Q(0) = 1$ g

rate in = (6 liters/min)(${1 \over (t+1)(t+2)e^{3t}}$ g/liters) 
{= ${6 \over (t+1)(t+2)e^{3t}}$ g/min}

 rate out =   (6 liters/min)(${Q(t) g \over 2 \hbox{ liters}}$) = ${3Q}$ g/min
 

${dQ \over dt}$ = rate in - rate out = ${6 \over (t+1)(t+2)e^{3t}} - {3Q}$ 

IVP:~~${dQ \over dt} ={6 \over (t+1)(t+2)e^{3t}}- {3Q}$, ~~~ $Q(0) = 1$

$Q' ={6 \over (t+1)(t+2)e^{3t}}- {3Q}$


$Q' + {3Q}={6 \over (t+1)(t+2)e^{3t}}$



$u(t) = e^{\int {3dt}} = e^{3t}$

$e^{3t}Q' + {3e^{3t}Q}={6e^{3t} \over e^{3t} (t+1)(t+2)}$

$(e^{3t}Q)' ={6 \over  (t+1)(t+2)}$

$\int (e^{3t}Q)'dt =\int {6 \over  (t+1)(t+2)}dt$


$ e^{3t}Q =\int {6 \over  (t+1)(t+2)}dt$

Partial fractions:  ${6 \over  (t+1)(t+2)} = {A \over  t+1}+ {B \over  t+2}$


$6 = A(t+2) + B(t+1)$

Method 1:  choose 2 values for $t$ to solve for the 2 variables.

\hfill$t = -1$:  $6 = A$ 
\hfill
$t = -2$:  $6 = -B$ \hfill~



Method 2:  use linear independence.


$ 0t + 6 = (A+B)t + 2A + B$

$A + B = 0$, thus $B = -A$

$2A + B = 6$.  Thus $2A - A = A = 6$ and $B = -6$

Hence ${6 \over  (t+1)(t+2)} = {6 \over  t+1} +  {-6 \over  t+2}$

$ e^{3t}Q =\int {6 \over  (t+1)(t+2)}dt = \int {6 \over  t+1}dt + \int {-6 \over  t+2}dt$

$ e^{3t}Q= 6 ln|t+1| - 6 ln|t+2| +C$

$Q = 6e^{-3t} ln|t+1|  -6e^{-3t}ln|t+2| +Ce^{-3t}$

Initial value:  $Q(0) = 1$

$1 = 6 ln|1|  -6ln|2| +C$

$C = 1 + 6ln(2)$

$Q = 6e^{-3t} ln|t+1|  -6e^{-3t}ln|t+2| +(1 + 6ln(2))e^{-3t}$


\t
Long-term behaviour:

\t
As $t \rightarrow \infty$, $ 6e^{-3t} ln|t+1|  -6e^{-3t}ln|t+2| +(1 + 6ln(2))e^{-3t} \rightarrow 0$ 

You know the limit is 0 using any of the 3 methods below:

1.)  Note the salt concentration entering the tank goes to 0 as $t \rightarrow +\infty$

2.)  Note that the exponential function goes to 0 much faster than the logarithm function goes to $+ \infty$ as $t \rightarrow +\infty$

3.)  Use l'hopital's rule.  For example,

$\displaystyle\lim_{t \rightarrow \infty} 6e^{-3t} ln|t+1|$

$\displaystyle\lim_{t \rightarrow \infty} 6{ ln|t+1| \over e^{3t}} =
\displaystyle\lim_{t \rightarrow \infty} 6{ {1 \over t+1} \over 3e^{3t}} =
\displaystyle\lim_{t \rightarrow \infty} { 6\over 3(t+1)e^{3t}} = 0
$

\end
