MATH 215/255 - SOLUTION OF HOMEWORK 9
1. Problem 6(abc) - Section 9.1
(a) Solution of the ODEs is based on the analysis of the algebraic equations
2−r
1
−5
−2 − r
ξ1
0
=
.
ξ2
0
1 0
For a nonzero solution, we require that det(A−rId) = r +1 = 0 where Id =
.
0 1
The roots of the characteristic equation are r = ±i. Setting r = i, the equations
are equivalent to ξ1 − (2 + i)ξ2 . The eigenvectors are (2 + i, 1)T and (2 − i, 1)T .
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(b) Since the eigenvalues are purely imaginary, the critical point is a center.
(c, d)
Date: December 1, 2009.
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MATH 215/255 - SOLUTION OF HOMEWORK 9
2. Problem 7(abc) - Section 9.1
ξ
(a) Setting X = 1 ert results the algebraic equations
ξ2
3−r
−2
ξ1
0
=
.
4
−1 − r
ξ2
0
For a nonzero solution, we require that det(A − rId) = r2 − 2r + 5 = 0. The roots
of the characteristic equation are r = 1 ± 2i. Substituting r = 1 − 2i, the two
MATH 215/255 - SOLUTION OF HOMEWORK 9
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equations reduce to (1 + i)ξ1 − ξ2 = 0. The two eigenvectors are (1, 1 + i)T and
(1, 1 − i)T .
(b) The eigenvalues are complex conjugates, with positive real part. Hence the
origin is an unstable spiral.
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MATH 215/255 - SOLUTION OF HOMEWORK 9
MATH 215/255 - SOLUTION OF HOMEWORK 9
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3. Problem 14 - Section 9.1
0
Setting X = 0, that is
−2 1
2
X=
1 −2
−1
we find that the critical point is X0 = (−1, 0)T . With the change of dependent
variable, X = X0 + u, the differential equation can be written as
du
−2 1
=
u.
1 −2
dt
ξ
The critical point for the transformed equation is the origin. Setting u = 1 ert
ξ2
results in the algebraic equations
−2 − r
1
ξ1
0
=
.
1
−2 − r
ξ2
0
For a nonzero solution, we require that det(A − rId) = r2 + 4r + 3 = 0. The roots
of the characteristic equation are r = −3, −1. Hence the critical point is a stable
node.
4. Problem 2 - Section 9.2
The differential equation can be combined to obtain a related ODE
2y
dy
=− .
dx
x
The equation is separable. The solution is given by
y = Cx−2 .
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MATH 215/255 - SOLUTION OF HOMEWORK 9
Note that the system is uncoupled, and hence we also have x = x0 e−t and y =
y0 e2t . Matching the initial conditions, for the first case we obtain x(t) = 4e−t and
y(t) = 2e2t , for the second case we obtain x(t) = 4e−t and y(t) = 0.
In order to determine the direction of motion along the trajectories, observe that
for positive initial conditions, x will decrease, whereas y will increase.
5. Problem 18 - Section 9.2
(a) The trajectories are solutions of the differential equation
dy
4x
=− ,
dx
y
Integrating, we obtain
4x2 + y 2 = C 2 .
Hence the trajectories are ellipses.
(b)
MATH 215/255 - SOLUTION OF HOMEWORK 9
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Bases on the differential equation, the direction of motion on each trajectory is
clockwise.
6. Problem 7(abc) - Section 9.3
The critical points are solutions of the equations
1 − y = 0 and (x − y)(x + y) = 0.
The first equation requires that y = 1. Based on the second equation, x = ±1.
Hence the critical points are (−1, 1) and (1, 1).
(b, c) F (x, y) = 1 − y and G(x, y) = x2 − y 2 . The Jacobian matrix of the vector
field is
Fx (x, y) Fy (x, y)
0
−1
=
Gx (x, y) Gy (x, y)
2x −2y
At the critical point (−1, 1), the coefficients matrix of the linearized system is
0 −1
−2 −2
√
√
with the eigenvalues −1 − 3 and −1 + 3. The eigenvalues are real, with opposite
sign. Hence the critical point is a saddle, which is unstable. At the equilibrium
point (1, 1), the coefficients of the linearized system is
0 −1
2 −2
with complex conjugate eigenvalues −1 ± i. The critical point is a stable spiral
which is asymptotically stable.
(d)
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MATH 215/255 - SOLUTION OF HOMEWORK 9
Based on the Table 9.3.1, the nonlinear terms do not affect the stability and type
of each critical point.
7. Problem 19 - Section 9.3
(a) The Jacobian matrix of the vector field is
1
.
0
0
1 + 6x2
At the origin, the coefficients of the linearized system is
0
1
1
0
with the eigenvalues ±1. The eigenvalues are real, with opposite sign. Hence the
critical point is a saddle point.
(b) The trajectories of the linearized system are solutions of the differential
equation
dy
x
= ,
dx
y
which is separable. Integrating both sides yields x2 − y 2 = C. The trajectories
consist of a family of hyperbolas.
MATH 215/255 - SOLUTION OF HOMEWORK 9
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It is easy to show that the general solution is given by x(t) = c1 et + c2 e−t and
y(t) = c1 et − c2 e−t . The only bounded solutions consist of those for which c1 = 0.
In that case, x(t) = c2 e−t = −y(t).
(c) The trajectories of the given system are solutions of the differential equation
dy
x + 2x3
=
,
dx
y
which yields (the equation is separable)
H(x, y) = x2 /2 + x4 /2 − y 2 /2.
The trajectories approaching to, or diverging from, the origin are no longer straight
lines.
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MATH 215/255 - SOLUTION OF HOMEWORK 9