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# Elementary differential equations 7th edition - Boyce W.E

Boyce W.E Elementary differential equations 7th edition - Wiley publishing , 2001. - 1310 p.
ISBN 0-471-31999-6
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and so on. In general, the nth iterate yn is
yn = f (Jn1) = fn (J0).
This procedure is referred to as iterating the difference equation. It is often of primary interest to determine the behavior of yn as n ^ to, in particular, whether yn approaches a limit, and if so, to find it.
Solutions for which yn has the same value for all n are called equilibrium solutions. They are frequently of special importance, just as in the study of differential equations. If equilibrium solutions exist, one can find them by setting yn+1 equal to yn in Eq. (3) and solving the resulting equation
yn = f (Jn) (4)
for yn.
Linear Equations. Suppose that the population of a certain species in a given region in year n + 1, denoted by yn+j, is a positive multiple pn of the population yn in year n; that is,
yn+1 = pnyn, n = 012-... (5)
Note that the reproduction rate pn may differ from year to year. The difference equa-
tion (5) is linear and can easily be solved by iteration. We obtain
y = P0 y0>
y2 = p1 y1 = p1p0 y0>
and, in general,
yn = Pn1 'p0 y0 > n = 1> 2>.... (6)
Thus, if the initial population y0 is given, then the population of each succeeding gen-
eration is determined by Eq. (6). Although for a population problem pn is intrinsically
2.9 First Order Difference Equations
117
positive, the solution (6) is also valid if pn is negative for some or all values of n. Note, however, that if pn is zero for some n, then yn+1 and all succeeding values of y are zero; in other words, the species has become extinct.
If the reproduction rate pn has the same value p for each n, then the difference equation (5) becomes
Equation (7) also has an equilibrium solution, namely, yn = 0 for all n, corresponding to the initial value y0 = 0. The limiting behavior of yn is easy to determine from Eq. (8). In fact,
In other words, the equilibrium solution yn = 0 is asymptotically stable for I pi < 1 and unstable if ip | > 1.
effect of immigration or emigration. If bn is the net increase in population in year n due to immigration, then the population in year n + 1 is the sum of those due to natural reproduction and those due to immigration. Thus
where we are now assuming that the reproduction rate p is constant. We can solve Eq. (10) by iteration in the same manner as before. We have
Note that the first term on the right side of Eq. (11) represents the descendants of the original population, while the other terms represent the population in year n resulting from immigration in all preceding years.
In the special case where bn = b for all n, the difference equation is
yn+1 P yn
(7)
and its solution is
yn = P nyo-
(8)
0,
yo>
does not exist,
if Ip I < 1; if p = 1 ; otherwise.
(9)
Now we will modify the population model represented by Eq. (5) to include the
yn+1 = P yn + bn,
(10)
y = pyo + b0, y2 = P(Pyo + bo) + b1 = P2yo + P bo + bv
y3 = p(p2 yo + Pbo + b1) + b2 = p3 yo + p 2bo + p b1 + b2,
j=o
yn+1 = pyn + b,
and from Eq. (11) its solution is
(12)
yn = p nyo + (1 + p + p2 +  + p n 1)b-
(13)
If p = 1, we can write this solution in the more compact form
(14)
118
Chapter 2. First Order Differential Equations
EXAMPLE
1
where again the two terms on the right side are the effects of the original population and of immigration, respectively. By rewriting Eq. (14) as
yn = pn(y°  Ip) + T~p  (15)
the long-time behavior of yn is more evident. It follows from Eq. (15) that yn ^ b/(1  p) if Ip I < 1 and that yn has no limit if Ip I > 1 or if p =  1. The quantity b/(1  p) is an equilibrium solution of Eq. (12), as can readily be seen directly from that equation. Of course, Eq. (14) is not valid for p = 1. To deal with that case, we must return to Eq. (13) and let p = 1 there. It follows that
yn = y0 + nb (16)
so in this case yn becomes unbounded as n ^ro.
The same model also provides a framework for solving many problems of a financial character. For such problems yn is the account balance in the nth time period, pn = 1 + rn, where rn is the interest rate for that period, and bn is the amount deposited or withdrawn. The following is a typical example.
A recent college graduate takes out a \$10,000 loan to purchase a car. If the interest rate is 12%, what monthly payment is required to pay off the loan in 4 years?
The relevant difference equation is Eq. (12), where yn is the loan balance outstanding in the nth month, p = 1 + r is the interest rate per month, and b is the monthly payment. Note that b must be negative and p = 1.01, corresponding to a monthly interest rate of 1%.
The solution of the difference equation (12) with this value for p and the initial condition y0 = 10,000 is given by Eq. (15), that is,
yn = (1.01)n (10,000 + 100b)  100b. (17)
The payment b needed to pay off the loan in 4 years is found by setting y48 = 0 and solving for b. This gives
(1.01)48
b = -100P = -263.34. (18)
(1.01)48  1
The total amount paid on the loan is 48 times b or \$12,640.32. Of this amount \$10,000 is repayment of the principal and the remaining \$2640.32 is interest.
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