3. dy/dt = e) -1, -∞
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
3

Transcribed Image Text:come
tion
rom
mate
ing
ond
rsus
e of
(18)
Problems
Problems I through 4 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 or
unstable. Draw the phase line, and sketch several graphs of solutions
in the ry-plane.
G 1. dy/dt = ay+by², a > 0, b>0, -∞<yo <∞
dy/dt = y(y - 1)(y-2), Yo ≥ 0
G
2.
3.
dy/dt = e" - 1, -∞ 108
4. dy/dt = e) -1,00<yo <∞
G
G
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
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
above it grow farther away. Therefore, (t) = 1 is semistable.
c. Solve equation (19) subject to the initial condition y(0) = yo
and confirm the conclusions reached in part b.
У4
k
o(t) = k
y
k
o(t) = k
10. C
(11) of
11. In
equatio
12. C
in the s
finding
equatic
13. C
inflect
thresh
14. C
critica
solutio
if f'(
15.
dy/d
G
grov
whe
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