Download Solving Initial Value Problems for Differential Equations: Exponential Growth and Decay and more Study notes Calculus in PDF only on Docsity!
Jim Lamb ers Math 2B Fall Quarter 2004- Le ture 16 Notes
These notes orresp ond to Se tion 10.4 in the text.
Exp onential Growth and De ay
In many appli ations, it is ne essary to solve the problem of nding the fun tion y (t) that satis es the simple di erential equation dy dt = k y ; (1)
where k is a known onstant, and the initial ondition
y (0) = y 0 ; (2)
where y 0 is a given initial value. Together, the di erential equation and initial ondition sp e ify an initial value problem. In words, the initial value problem states the following:
The unknown quantity y (t) initially has the value y 0.
At any time t, its rate of hange with resp e t to time is prop ortional to its value at the time t, with k b eing the onstant of prop ortionality.
We an solve the initial value problem using the natural logarithmi and exp onential fun tions. If we divide b oth sides of equation (1) by y and then integrate b oth sides from 0 to t, we obtain
Z (^) t
0
y (s)
dy ds
ds =
Z (^) t
0
k ds: (3)
Then, we make the substitution u = y (s), whi h yields du = dy =ds ds. Then, this equation b e omes
Z (^) y (t)
y 0
u
du =
Z (^) t
0
k ds: (4)
Evaluating the integrals on b oth sides yields
ln jy (t)j ln jy 0 j = k t; (5)
whi h an b e rewritten as
ln y (t) y 0
= k t: (6)
We then exp onentiate b oth sides. From the an ellation equations and the laws of exp onents, it follows that jy (t)j = jy 0 jek^ t^ : (7)
Now, supp ose that y 0 > 0. Then we must have y (t) > 0 for any t, b e ause otherwise, by the ontinuity of y (t), we must have y (t) = 0 for some t, whi h is imp ossible b e ause ek^ t^ > 0 for all t. Similarly, if y 0 < 0, then y (t) < 0 for any t. Therefore, y (t) and y 0 are the same sign, so we an drop the absolute values in equation (7) and obtain
y (t) = y 0 ek^ t^ : (8)
If k > 0, then y (t) grows rapidly in absolute value over time, and the di erential equation is alled the law of natural growth. In this ase, the equation an b e used as a (simpli ed) mathe- mati al mo del for p opulation growth, for example. If k < 0, then y (t) rapidly de ays to zero over time, and the di erential equation is known as the law of natural de ay. This law des rib es, for instan e, the pro ess of radioa tive de ay. Finally, if k = 0, then there is no growth or de ay, and y (t) remains onstant for all time.
Population Growth
The di erential equation dP dt = k P (9)
an b e used to mo del the growth of a p opulation, denoted by P (t). The prop ortionality onstant k , also known as the relative growth rate, an b e determined exp erimentally by birth rates and death rates, sin e the numb er of births and deaths p er unit of time tend to b e prop ortional to the urrent p opulation. Therefore, if the birth rate was determined to b e some value B and the death rate was determined to b e D , then one mo del for p opulation growth ould b e
dP dt
= (B D )P ; (10)
whi h would imply P (t) = P 0 e(B^ D^ )t^ ; where t = 0 denotes the urrent time. If the p opulation of a region is b eing measured, then other fa tors, su h as immigration and emigration due to e onomi for es, ould also b e measured and in luded in the determination of the relative growth rate.
Example 1 In 1900, the world p opulation was approximately 1650 million. In 1910, the p opulation had grown to approximately 1750 million. Determine the relative growth rate of the p opulation.
Solution The p opulation at time t, where t is measured in years sin e 1900, is denoted by a fun tion P (t) that satis es the di erential equation
dP dt
= k P ; (11)
We on lude that the relative de ay rate is
jk j = ln 2 1590
4 : 3594 10 ^4 : (19)
Example 3 How mu h of a 100 g sample of Radium-226 will remain after 1000 years?
Solution From the previous example, we have
m(t) = m 0 ek^ t^ = 100 e (ln^ 2)t=^1590 ; (20)
sin e the initial mass m 0 is given to b e 100 g. Using the de nition of the general exp onential fun tion ax^ = ex^ ln^ a^ , we an rewrite m(t) as follows:
m(t) = 100 e( t=1590)^ ln^2 = 100 2 t=^1590 : (21)
Substituting t = 1000 yields m(1000) 64 : 6655 g. 2
Example 4 How long will it take a 100 g sample of Radium-226 to de ay to 30 g?
Solution We need to nd the value of t su h that
m(t) = 100 2 t=^1590 = 30 : (22)
Taking the logarithm to base 2 of b oth sides and using the an ellation equation log 2 2 x^ = x, we obtain t 1590
= log (^2)
whi h yields
t = 1590 log (^2)
2761 : 7752 years : (24)
2
Continuously Comp ounded Interest
Let A(t) b e the value of an investment at time t. With ontinuous omp ounding of interest, the rate of in rease of the value is prop ortional to its size. It follows that
dA dt = r A; (25)
where r is the interest rate. The value is therefore given by A(t) = A 0 er^ t^ ; where A 0 is the initial value of the investment.
