Problems 6 through 9 involve equations of the form dy/dt = f(y). In each problem sketch the graph of f(y) versus y, determine the critical (equilibrium) points, and classify each one as asymptotically stable, unstable, or semistable (see Problem 5). Draw the phase line, and sketch several graphs of solutions in the ty-plane. G 6. dy/dt = y²(y² - 1), -∞
Problems 6 through 9 involve equations of the form dy/dt = f(y). In each problem sketch the graph of f(y) versus y, determine the critical (equilibrium) points, and classify each one as asymptotically stable, unstable, or semistable (see Problem 5). Draw the phase line, and sketch several graphs of solutions in the ty-plane. G 6. dy/dt = y²(y² - 1), -∞
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:G
G
G
(equilibrium) points, and classify each one as asymptotically stable or
unstable. Draw the phase line, and sketch several graphs of solutions
in the ty-plane.
G
G
2.
dy/dt = ay+by2, a > 0, b> 0, -∞< yo<∞
dy/dt = y(y - 1)(y-2), Yo ≥ 0
-∞<Yo <∞
-∞<yo <∞
3.
dy/dt = e) — 1,
4. dy/dt = e--1,
5. Semistable Equilibrium Solutions. Sometimes a constant
equilibrium solution has the property that solutions lying on one side
of the equilibrium solution tend to approach it, whereas solutions lying
on the other side depart from it (see Figure 2.5.9). In this case the
equilibrium solution is said to be semistable.
a. Consider the equation
G
G
dy/dt = k(1- y)²,
(19)
where k is a positive constant. Show that y = 1 is the only critical
point, with the corresponding equilibrium solution (t) = 1.
Gb. Sketch f(y) versus y. Show that y is increasing as a
function of t for y< 1 and also for y> 1. The phase line
has upward-pointing arrows both below and above y = 1. Thus
solutions below the equilibrium solution approach it, and those
away. Therefore, (t) = 1 is semistable.
above it grow farther
c. Solve equation (19) subject to the initial condition y(0) = Yo
(cand confirm the conclusions reached in part b.
y4
k
1.
9.
7.
8.
o(t) = k
t
pily
k
o(t) = k
t
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Problems 6 through 9 involve equations of the form dy/dt = f(y).
In each problem sketch the graph of f(y) versus y, determine the
critical (equilibrium) points, and classify each one as asymptotically
stable, unstable, or semistable (see Problem 5). Draw the phase line,
and sketch several graphs of solutions in the ty-plane.
6.
dy/dt = y²(y² – 1),
-∞<yo <∞
dy/dt = y(1- y²),
-∞<yo <∞
dy/dt = y²(4- y²),
-∞<yo <∞
dy/dt = y²(1-y)²,
-∞0</o <∞
TO
11. In Example 1, complete the manipulations needed to
equation (13). That is, solve the solution (11) for t.
jonge
(a)
(b)
103
Svo
FIGURE 2.5.9 In both cases the equilibrium solution (t) = k
is semistable. (a) dy/dt ≤ 0; (b) dy/dt ≥ 0.
12. Complete the derivation of the location of the vertical a
in the solution (15) when yo > T. That is, derive formu
finding the value of t when the denominator of the right-ha
equation (15) is zero.
mabigs
13. Complete the derivation of formula (18) for the loca
inflection points of the solution of the logistic growth m
threshold (17). Hint: Follow the steps outlined on p. 66.
14. Consider the equation dy/dt f(y) and suppos
critical point- that is, f(y₁) = 0. Show that the constan
solution (1) = y₁ is asymptotically stable if f'(y₁) < 0
if f'(y₁) > 0. boot lo
15. Suppose that a certain population obeys the log
dy/dt = ry(1-(y/K)).
a. If yo = K/3, find the time at which the ini
has doubled. Find the value of T corresponding to
our year.
= a, find the time T at which
b. If yo/K
where 0 <a, ß < 1. Observe that To
as 3-
→1. Find the value of T for r = 0.025 pe
and 3 = 0.9.
Tato str
in
=
In
15
G 16. Another equation that has been used to :
growth is the Gompertz's equation
(OS)
dy
dt
= ry in (5).
the ba
where r and K are positive constants.
a. Sketch the graph of f(y) versus y, find
and determine whether each is asymptoticall
b. For 0 ≤ y ≤ K, determine where the g
concave up and where it is concave down.
noc. For each y in 0 ≤ y ≤ K, show tha
the Gompertz equation is never less than d
logistic equation.
17. a. Solve the Gompertz equation
dy
dt
=ry In
K
( 5
y
subject to the initial condition y(0) = yo
Hint: You may wish to let u = ln(y/K)
b. For the data given in Example 1 in
year, K =
= 80.5 × 106 kg, yo/K = 0
model to find the predicted value of y(2
c. For the same data as in part b, use the
the time at which y(T) = 0.75K.
15 Benjamin Gompertz (1779-1865) was an Engl
model for population growth, published in 1825.
mortality tables for his insurance company.
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