Solutions for Chapter 5 exercises:
5.2 (a) {1/4, 2/7, 3/10, 4/13, 5/16, . . . }, L = 0.3333 . . .
(c) {−1/4, 1/16, −1/64, 1/256, −1/1016, . . . }, L = 0
(d) {4, 3, 2, 1, 0, . . . }, L does not exist
5.3 (a) The general solution is xn = (0.3)n x0 , so since x0 = 10, the general solution is
xn = 10(0.3)n . Thus,
lim xn = lim 10(0.3)n = 0
n→∞
n→∞
and the limit is L = 0.
(c) The general solution is xn = x0 + 6n, so since x0 = 10, we have xn = 10 + 6n. This
sequences increases without bound and has no limit.
5.5 (a) xn+1 = xn + 0.1xn = (1.1)xn
(b) Since the difference equation is geometric, our general solution is xn = (1.1)n x0 =
(1.1)n (50).
5.7 (a) If our time step is one hour, xn+1 = xn − 0.1xn = 0.9xn
(b) Since this is geometric with xn+1 = 0.9xn , we have xn = (0.9)n x0 = (0.9)n (180).
(c) After one hour, n = 1, so we’d have x1 = (0.9) ∗ 180 = 162mg of drug still in the
body.
5.8 (a) xn+1 = xn + .1xn − h = (1.1)xn − 20. Since this is a linear first order difference
equation, we have
xn = (1.1)n x0 + (−20)
1 − (1.1)n
= 1000(1.1)n + 200(1 − (1.1)n )
1 − 1.1
We can simplify this to
xn = 800(1.1)n + 200
(b) According to my final form in part (a), since xn = 800(1.1)n + 200, this will never go
negative, since every term in the expression is positive.
(c) Now we want the largest h so that x10 ≥ 1500. xn+1 = (1.1)xn − h gives a general
solution of
xn = 1000(1.1)n − h(
1 − (1.1)n
= 1000(1.1)n + 10h(1 − (1.1)n ) .
1 − 1.1
We want:
1000(1.1)10 + 10h(1 − (1.1)10 ) ≥ 1500
2593.74 − 15.94h ≥ 1500
1093.74 ≥ 15.94h
68.62 ≥ h
so as long as h is less than or equal to 68 buffalo removed per year, the population
after 10 years will be greater than or equal to 1500.
5.9 (a) xn+1 = xn + .1xn + 100, or xn+1 = (1.1)xn + 100. The general solution is
xn = 1000(1.1)n + (100)
1 − (1.1)n
−0.1
or
xn = 1000(1.1)n − 1000(1 − (1.1)n ) = 2000(1.1)n − 1000
To find out when xn ≥ 2000 occurs, we solve for n in
2000(1.1)n − 1000 ≥ 2000
2000(1.1)n ≥ 3000
3
(1.1)n ≥
2
n ln(1.1) ≥ ln(1.5)
0.0953n ≥ 0.405
0.405
≈ 4.25
n≥
0.0953
So after 4.25 years, the population will be more than 2000 ... meaning, in the 5th
year, we will have x5 ≥ 2000, while x4 < 2000. In fact, x5 = 2000(1.1)5 −1000 = 2221
and x4 = 2000(1.1)4 − 1000 = 1928.
(b) After the pop reaches 2000, the stocking will end and fishermen catch 400 fish per
year, so
xn+1 = (1.1)xn − 400
with now x0 = 2221, so the general solution is
xn = 2221(1.1)n − 400
1 − (1.1)n
= 2221(1.1)n + 4000(1 − (1.1)n )
1 − 1.1
we can simplify this to
xn = −1779(1.1)n + 4000
which tells us that the population will now decrease over time, and eventually no
longer exist. Notice that if we solve for 0 = −1779(1.1)n + 4000 to see how long it
would take for the fish to be all gone, we have
−4000 = −1779(1.1)n
2.25 = (1.1)n
ln(2.25)
=n
ln(1.1)
8.5 = n
So that in 9 years, the population would no longer exist in this lake.