542 CHAPTER 29. FIRST ORDER SCALAR ODE
Example 29.1.11 A radioactive substance decays in such a way that the rate of change ofthe amount of the substance is a constant multiple of the amount present, the constant beingnegative. Thus dA
dt =−kA. There is a certain sample of decaying material. Measurementsare taken after 5 years and it is found that there is about 9/10 of the original amountpresent. Find the half life of this material. The half life is the amount of time it takes forhalf of it to have decayed.
From the equation, A=A0e−kt . Then 910 A0 =A0e−k(5). Solving this for k yields − ln(.9)
5 =k and so the amount of time to have half of what was started with is T given as a solutionto the following equation.
e−(− ln(.9)
5
)(T )
=12, so T =
ln(.5)ln(.9)/5
= 32.894
This kind of thing is associated not just with radioactive material but with other chemi-cals as well. They degrade over time according to such an equation.
Example 29.1.12 The ancient Babylonians were fascinated with the idea of compound in-terest. They were interested in how long it would take an initial amount to double. Onecan understand compound interest compounded continuously using the same kind of differ-ential equation as the above only this time the constant is positive and is the interest rate.Thus
dAdt
= kA
If the interest rate is 20% per year compounded continuously, how long will it take for aninitial amount to double in size?
From the equation, A = A0e.2t where A0 is the initial amount. Then you want to find Tsuch that 2A0 = A0e.2T and so
T =ln2.2
= 5.0ln2 = 3.4657
If the rate is r per year and you have n years and the interest is compounded at the endof each year rather than continuously, then the amount is given by the formula (1+ r)n =A(n) . Anciently, they used this kind of thing because they did not have differential equa-tions. If the interest rate is 20% compounded monthly, then the amount after n years isA0(1+ .2
12
)12nwhere A0 is the initial amount. If n = 3.5, a use of a calculator shows that(
1+.212
)12(3.5)
= 2.0022
which is very similar to compounding the interest continuously. The rational for this for-mula is that if it is compounded monthly, then the interest rate per month is .2/12. Eachsuccessive month is called a payment period.
Example 29.1.13 A lake contains one million gallons of water. A gas tank starts to leakupstream and contaminated water mixed with gasoline starts flowing into the lake at therate of 1000 gallons per month. This is mixed well due to large numbers of fish in the lakeand water flows out at the same rate. The amount of gasoline in the contaminated watervaries due to the demand for gas at the gas station and the concentration of gasoline in thecontaminated water is (1+ sin(t)) grams per gallon. Find a formula for the concentrationof gasoline in the lake in grams per gallon as a function of time in months after a long time.