MATH 449 - HOMEWORK 4 SOLUTIONS
3.4.2 x = rx sinh x :
We have
x=
=
=
=
Note that there is a xed point at x = 0 for all values of r.
ex ex
rx sinh x = rx
2
12 13
1
1
1
1 + x + x + x 1 x + x2 x3
rx
2
2
6
2
6
1
rx x + x3 + O(x4 )
6
1
(r 1) x x3 + O(x4 ).
6
MATH 449
Quiz 1
2 February 2011
Name:
Show all work to receive full credit. No notes or books may be used for this quiz. With the
exception of the TI-89 and the TI-92, graphing calculators are allowed.
2
1. (20pts) Consider the vector eld x = ex r. Sketch
MATH 449
Quiz 2
21 February 2011
Name:
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quiz.
1. (20pts) Consider the vector eld x = a(x + 1) + (x + 1)3 . Sketch 3 qualitatively
dierent vector elds as a is varied.
MATH 449
Quiz 4
25 April 2011
Name:
Show all work to receive full credit. No notes, books, or calculators may be used for this
quiz.
1. (20 pts) Note that the origin is a xed point of the system
x=y
y = x x3 y
for all . By linearizing about the origin, sh
MATH 449 - HOMEWORK 1 SOLUTIONS
0 +cot x
2.1.4 The exact solution to the IVP is t = ln | csc xx+cot x0 |.
csc
a) Given x0 =
t
e
2+1
=
4,
sin x
1+cos x
t=
ln | csc x2+1 x |
+cot
= tan 1 x
2
=
t
=
arctan
e=
t
e
2+1
2+1
1
cos x
sin x + sin x
= 1x
2
=
=
et
x(
MATH 449 - HOMEWORK 2 SOLUTIONS
2.3.2 a)
x = k1 ax k1 x2 = x(k1 a k1 x)
=
k1 a
. It is clear from the
k1
k1
plot of the vector eld that x = ka is a stable xed point
1
and x = 0 is an unstable xed point.
The xed points are x = 0, x =
Figure 1: Plot of the
MATH 449 - HOMEWORK 3 SOLUTIONS
2
3.1.1 x = 1 + rx + x2 : The xed points are x = r 2 r 4 . Note that
there are 2 bifurcation values, rc = 2 , since there is exactly 1
root, x = 1, in each of these cases.
3.1.3 x = r + x ln(1 + x) : Let g (x) = r + x and h
MATH 449 - HOMEWORK 5 SOLUTIONS
3.7.3 N = rN 1
N
K
H :
a) Let x = N/K and = t/T where T is to be determined. Note that
d
1d
= T d . It follows that
dt
d
1d
K dx
N=
(xK ) =
= rxK (1 x) H.
dt
T d
T d
Dividing by rK yields
H
1 dx
= x (1 x)
.
rT d
rK
So, de
MATH 449 - HOMEWORK 6 SOLUTIONS
6.1.2 x = x x3 , y = y : The xed points are (0, 0) and (1, 0). The x-nullclines
are the vertical lines x = 0 and x = 1. The y -nullcline is the horizontal line
y = 0.
6.1.3 x = x(x y ), y = y (2x y ) : The xed point is (0,
MATH 449 - HOMEWORK 7 SOLUTIONS
6.4.3 x = x(3 2x 2y ), y = y (2 x y ) :
The xed points are (0, 0), (3/2, 0) and (0, 2). The Jacobian is
J=
For (0, 0), A =
For (3/2, 0), A =
For (0, 2), A =
3 4x 2y
2x
y
2 x 2y
30
. This xed point is an unstable node.
02
MATH 449 - HOMEWORK 8 SOLUTIONS
7.2.12 x = x + 2y 3 2y 4 ,
y = x y + xy :
Consider V (x, y ) = xm + ay n . Then
V = mxm1 x + nay n1 y = m(xm +2xm1 y 3 2xm1 y 4 )+ na(xy n1 y n + xy n )
Choosing m = 2, n = 4 yields
V = 2(x2 + 2xy 3 2xy 4 ) + 4a(xy 3 y 4 +
MATH 449
Exam 1
2 March 2011
Name:
Show all work to receive full credit. No notes or books may be used for this exam. With the
exception of the TI-89 and the TI-92, graphing calculators are allowed.
1. (25pts) Use linear linear stability analysis to class
MATH 449
Quiz 2
11 February 2013
Name:
Show all work to receive full credit. No notes, books or calculators may be used for this
quiz.
rx
1. (20pts) Consider the vector eld x = x + 1+x2 . Sketch 3 qualitatively dierent vector
elds as r is varied. Find the
MATH 449
Quiz 4
10 April 2013
Name:
Show all work to receive full credit. No notes, books or calculators may be used for this
quiz.
1. (20pts) Consider the system
x = x x2
y = y
Find all xed points. By linearizing about the origin, show that one of the ei
MATH 449 - HOMEWORK 1 SOLUTIONS
d
2.1.3 x = dt x = cos x(x) = cos x sin x = 1 sin 2x For the ow to have maximum
2
d
acceleration, we must have dx x = cos 2x = 0. The roots of this equation are
x = n and x = 34 n . It is clear from the plot below that x =
MATH 449 - HOMEWORK 3 SOLUTIONS
3.2.2 x = rx ln(1 + x) : Note that there is a xed point at x = 0 for all values
of r. We use a Taylors series expansion centered at x = 0 of ln(1 + x) which
yields
1
1
x = rx x x2 + O(x3 ) = (r 1)x + x2 + O(x3 )
2
2
. So, r
MATH 449 - HOMEWORK 5 SOLUTIONS
6.1.2 x = x x3 , y = y : The xed points are (0, 0) and (1, 0). The x-nullclines
are the vertical lines x = 0 and x = 1. The y -nullcline is the horizontal line
y = 0.
6.1.5 x = x(2 x y ), y = x y : The xed points are (0, 0)
MATH 449 - HOMEWORK 6 SOLUTIONS
6.5.2 x = x x2
=
x = y,
y = x x2 :
a) The xed points are (0, 0) and (1, 0). The Jacobian is
J=
For (0, 0), A =
0
1
1 2x 0
01
. The eigenvalues are 1 = 1 and 2 = 1: this is
10
a saddle point.
01
. The eigenvalues are 1,2 =
MATH 449 - HOMEWORK 7 SOLUTIONS
7.1.5 r = r(1 r2 ),
=1:
x = r cos r sin = r(1 r2 ) cos r sin = r cos r sin r3 cos =
2
x y x(x + y 2 )
y = r sin + r cos = r(1 r2 ) sin r cos = r sin + r cos r3 sin =
2
x + y y (x + y 2 )
7.2.7 x = y + 2xy,
a) Note that
f
MATH 449
Exam 1
22 February 2013
Name:
Show all work to receive full credit. No notes or books may be used for this exam. With
the exception of the TI-89 and the TI-92, graphing calculators are allowed.
1. (25pts) Use linear linear stability analysis to c
MATH 449
Quiz 1
30 January 2013
Name:
Show all work to receive full credit. No notes or books may be used for this quiz. With the
exception of the TI-89 and the TI-92, graphing calculators are allowed.
2
1. (20pts) Consider the vector eld x = ex r. Sketch
MATH 449
Quiz 3
18 March 2013
Name:
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quiz.
1. (20pts) Find and classify all of the xed points, draw the nullclines, sketch a plausible
phase portrait, and indicate th