Math 211 Fall 2007: Solutions: HW #4
Instructor: S. Cautis
1. section 2.9, #8
Equilibrium points: y0 (t) = 0 and y1 (t) = 2 are equilibrium solutions.
The solution y0 is stable and y1 is unstable.
2. section 2.9, #10
The equilibrium solutions are y0 (t) = −2. y1 (t) = − 21 , y2 (t) = 1 and
y3 (t) = 2. The stable solutions are y0 and y2 .
3. section 2.9, #20
The equilibrium points are y0 = −3, y1 = −1 and y2 = 3, y1 is stable.
4. section 2.9, #28
The equilibrium points are x0 = −2, x1 = 0 and x2 = 1. x1 is stable.
5. section 3.1, #4
P (t) = Cert is our model. P (0) = 1000 also P (10) = 2 ∗ 1000 = 2000 =
1000e10r
so, r = ln102 .
ln 2
P (t0 ) = 10000 then 1000e 10 t0 = 10000
So t0 = 10 ∗ lnln10
2 .
6. section 3.1 #10
Find the maximum rate of growth for the logistic equation. The maximum
rate of growth is the maximum of dP
dt . To do this find the zeros of the
second derivative which is:
d2 P
2P P 0
0
= rP 0 (1 − k2 P )
dt2 = rP − r k
This is zero when P = k2 or when the population is half of the maximum.
2
This is a maximum because of the sign of ddtP2 at P = k2 − > 0 and
P = k2 + < 0.
7. section 3.1 #12
rt
Kce
P (t) = 1+ce
rt
P (0) = 1000, P (10) = 2 ∗ 1000 = 2000, K = 20, 000
Find P (25):
0
P (0) = 1000 = 20,000ce
1+ce0
1
c = 19
P (10) = 2000 =
1 r10
20,000∗ 19
e
1 r10
1+ 19
e
1
2000 = e10r 19
(20, 000 − 2000) Meaning, r =
P (25) =
1 r25
20000∗ 19
e
1 r25
1+ 19 e
is roughly 5083.75
1
10
ln( 19
9 )
8. section 3.3 #4
M 0 = .0625M + D so
(e−.0625t M )0 = e−.0625t D Which means
1
1
e−.0625t D + C or M (t) = − .0625
D + C(e.0625t ).
e−.0625t M = − .0625
1
Now M (0) = 0 so C = .0625 D. We also know M (18) = 50, 000
50000
50000
C(e.0625∗18 ) − C = 50, 000 and C = e.0625∗18
−1 , D = .0625 e.0625∗18 −1
9. section 4.1 #10
k ∗ 34 = 5 ∗ 9.8
0
5y 00 + 20∗9.8
3 y = 0 and y(0) = .36 , y (0) = −.45
10. section 4.1 #14 (C1 cos(2t) + C2 sin(2t))00 + 4(C1 cos(2t) + C2 sin(2t)) =
− 4C1 cos(2t) + −4C2 sin(2t) + 4(C1 cos(2t) + C2 sin(2t)) = 0