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
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3

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
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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