# MATH 215/255 Fall 2014 Assignment 8

advertisement ```MATH 215/255
Fall 2014
Assignment 8
due 11/19
&sect;3.4, &sect;3.7, &sect;3.5, &sect;3.9
Solutions to selected exercises can be found in [Lebl], starting from page 303.
• 3.4.6: a) Find the general solution of x01 = 2x1 , x02 = 3x2 using the eigenvalue method
(first write the system in the form ~x0 = A~x). b) Solve the system by solving each
equation and verify you get the same general solution.
x1 (t)
2 0
Answer.
a). Let ~x(t) =
and A =
, then ~x0 = A~x. Let’s find the
x2 (t)
0 3
2−λ
0
eigenvalues of A, that is, we should solve det (A − λI2 ) = det
= 0,
0
3−λ
then
λ1 = 2,
and λ2 = 3.
For λ1 = 2, let’s solve A~x = 2~x, that is,
0 0
x1
0
.
=
0 1
x2
0
Then
x1
x2
= x1
1
1
is an eigenvalue for 2.
, that is,
0
0
For λ2 = 3, let’s solve A~x = 3~x, that is,
0
x1
−1 0
.
=
0
x2
0 0
Then
x1
x2
= x2
0
0
is an eigenvalue for 3.
, that is,
1
1
Therefore, the general solution to ~x0 = A~x is:
2t
~x(t) = C1 e
1
0
3t
+ C2 e
0
1
=
C1 e2t
C2 e3t
.
b). Since x01 = 2x1 , then x1 = C1 e2t . Since x02 = 3x2 , then x2 = C2 e3t . Hence the
general soluton to ~x0 = A~x is:
~x(t) =
C1 e2t
C2 e3t
.
• 3.4.7: Find the general solution of x01 = 3x1 + x2 , x02 = 2x1 + 4x2 using the eigenvalue
method.
3 1
Answer.
Let ~x(t) =
and A =
, then ~x0 = A~x. Let’s find the
2 4
3−λ
1
eigenvalues of A, that is, we should solve det (A − λI2 ) = det
= 0,
2
4−λ
then (λ − 3)(λ − 4) − 2 = 0, that is, λ2 − 7λ + 10 = 0, which implies that
x1 (t)
x2 (t)
λ1 = 2,
and λ2 = 5.
For λ1 = 2, let’s solve A~x = 2~x, that is,
1 1
x1
0
=
.
2 2
x2
0
1
1
x1
, that is,
is an eigenvalue for 2.
Then
= x1
−1
−1
x2
For λ2 = 5, let’s solve A~x = 5~x, that is,
−2
1
x1
0
=
.
2 −1
0
x2
x1
1
1
is an eigenvalue for 3.
Then
, that is,
= x1
2
x2
2
Therefore, the general solution to ~x0 = A~x is:
C1 e2t + C2 e5t
1
1
=
+ C2 e5t
~x(t) = C1 e2t
.
−C1 e2t + 2C2 e5t
2
−1
• 3.4.8: Find the general solution of x01 = x1 − 2x2 , x02 = 2x1 + x2 using the eigenvalue
method. Do not use complex exponentials in your solution.
1 −2
x1 (t)
, then ~x0 = A~x. Let’s find the
and A =
Answer.
Let ~x(t) =
2
1
x2 (t)
1 − λ −2
= 0,
eigenvalues of A, that is, we should solve det (A − λI2 ) = det
2
1−λ
then (λ − 1)2 + 4 = 0, which implies that
λ1 = 1 + 2i,
and λ2 = 1 − 2i.
For λ1 = 1 + 2i, let’s solve A~x = (1 + 2i)~x, that is,
−2i −2
x1
0
=
.
2 −2i
x2
0
x1
1
1
Then
= x1
, that is,
is an eigenvalue for 1 + 2i. Then a complex
x2
−i
−i
solution can be
(1+2i)t 1
e
λ1 t
e
=
−i
−ie(1+2i)t
t
e cos(2t) + iet sin(2t)
=
et sin(2t) − iet cos(2t)
t
e cos(2t)
et sin(2t)
=
+
i
et sin(2t)
−et cos(2t)
2
Therefore, the general solution to ~x0 = A~x is:
~x(t) = C1
et cos(2t)
et sin(2t)
+ C2
et sin(2t)
−et cos(2t)
=
C1 et cos(2t) + C2 et sin(2t)
C1 et sin(2t) − C2 et cos(2t)
.
5 −3
. Find the general solution of ~x0 = A~x.
3 −1
• 3.7.2: Let A =
Answer.
Let’s find
the eigenvalues of A, that is, we should solve det (λI2 − A) =
λ−5
3
det
= 0, then (λ − 5)(λ + 1) + 9 = 0, that is, λ2 − 4λ + 4 = 0, which
−3 λ + 1
implies that
λ1 = λ2 = 2.
It’s easy to see that 2I2 − A 6= 0, that is, λ = 2 is a defective 1 eigenvalue of A. For
λ = 2, let’s solve A~x = 2~x, that is,
x1
0
−3 3
.
=
0
−3 3
x2
Then
x1
x2
= x1
1
1
is an eigenvalue for 2. Then a solution can be
, that is,
1
1
1
.
e2t
1
1
, that is, e2t~v is a solution to ~x0 = A~x. Let ~y (t) = e2t (t~v + w)
~ =
Let ~v =
1
te2t~v + e2t w
~ for some w
~ be such that ~y be a solution to ~x0 = A~x, then
~y 0 = e2t~v + 2te2t~v + 2e2t w
~ = A~y = te2t A~v + e2t Aw.
~
Since A~v = 2~v , then e2t~v + 2e2t w
~ = e2t Aw,
~ that is, (2I2 − A)w
~ = −~v , which implies
that
−3 3
w1
−1
=
.
−3 3
w2
−1
So −3w1 + 3w2 = −1, we can take w1 = 0 and w2 = − 31 , then
1
0
2t
2t
2t
2t
~y (t) = te ~v + e w
~ = te
+e
.
1
− 13
Therefore, the general solution to ~x0 = A~x is:
2t
~x(t) = C1 e
1
1
+ C2 te
2t
1
1
+ C2 e
3
2t
0
− 31
=
C1 e2t + C2 te2t
C1 e2t + C2 te2t − C32 e2t
.
a a
• 3.7.104: Let A =
, where a, b and c are unknowns. Suppose that 5 is a
b c
1
doubled eigenvalue of defect 1, and suppose that
is the eigenvector. Find A and
0
show that there is only one solution.
λ − a −a
Answer.
Let’s solve det (λI2 − A) = det
= 0, that is, (λ − a)(λ −
−b λ − c
c) − ab = 0, which implies that
λ2 − (a + c)λ + ac − ab = 0.
Since 5 is a doubled eigenvalue of defect 1, that is, λ = 5 is the only solution to
λ2 − (a + c)λ + ac − ab = 0, then
and ac − ab = 25.
a + c = 10,
Since
1
0
(1)
is the eigenvalue for λ = 5, that is,
5 − a −a
−b 5 − c
1
0
=
0
0
Then we have
5 − a = 0,
− b = 0.
and
(2)
By (1) and (2), we have
a = 5,
b = 0,
and c = 5.
Therefore, we know that
A=
5 5
0 5
.
• 3.5.101: Describe the behavior of the following systems without solving:
a) x0 = x + y, y 0 = x − y.
b) x01 = x1 + x2 , x02 = 2x2 .
c) x01 = −2x2 , x02 = 2x1 .
d) x0 = x + 3y, y 0 = −2x − 4y.
e) x0 = x − 4y, y 0 = −4x + y.
Answer.
a) The coefficient matrix is
1 1
A=
,
1 −1
(3)
2
which
√ has characteristic polynomial λ − 2 = 0. Therefore the eigenvalues are λ =
&plusmn; 2. Since they are both real and have opposite signs, the behaviour is saddle .
4
b) The coefficient matrix is
A=
1 1
,
0 2
(4)
which has characteristic polynomial (λ − 1)(λ − 2) = 0. Therefore the eigenvalues are
λ = 1, 2. Since they are both real and positive the behaviour is source (unstable) .
c) The coefficient matrix is
0 −2
A=
,
2 0
(5)
which has characteristic polynomial λ2 +4 = 0. Therefore the eigenvalues are λ = &plusmn;2i.
Since they are both purely imaginary, the behaviour is center point (ellipses) .
d) The coefficient matrix is
1
3
A=
,
−2 −4
(6)
which has characteristic polynomial λ2 + 3λ + 2 = 0. Therefore the eigenvalues are
λ = −1, −2. Since they are both real and negative, the behaviour is sink (stable) .
e) The coefficient matrix is
1 −4
,
A=
−4 1
(7)
which has characteristic polynomial λ2 − 2λ − 15 = 0. Therefore the eigenvalues are
λ = −3, 5. Since they are both real and of opposite sign, the behaviour is saddle .
• 3.5.102: Suppose that ~x0 = A~x where A is a 2 &times; 2 matrix with eigenvalues 2 &plusmn; i.
Describe the behavior.
Answer.
Since the eigenvalues are complex with positive real parts, the equilibrium at the
origin is a spiral source . All nonzero solutions spiral around and move away from
the origin.
• 3.9.4: Find a particular solution to x0 = x + 2y + 2t, y 0 = 3x + 2y − 4, a) using
undetermined coefficients, b) using variation of parameters.
Answer.
a) Since the inhomogeneous term is
2
0
t
+
,
0
−4
we try ~xp (t) = t~a + ~b, where
~a =
a1 ~
b
,b = 1
a2
b2
are to be determined. Substituting ~xp (t) = t~a + ~b into the system gives
5
a1
a1 + 2a2
b1 + 2b2
2t
=t
+
+
.
a2
3a1 + 2a2
3b1 + 2b2
−4
Equating the coefficients of t on both sides of the equation, we obtain
a1 + 2a2 = −2
3a1 + 2a2 = 0.
Equating the constants on both sides of the equation, we obtain
b1 + 2b2 = a1
3b1 + 2b2 = a2 + 4.
Solving the first two equations gives a1 = 1 and a2 = −3/2. Solving the next two
equations, we obtain b1 = 3/4 and b2 = 1/8. Therefore a particular solution is
t + 43
~xp (t) =
− 32 t +
.
1
8
b) The coefficient matrix and inhomogeneous term for the system ~x0 = P ~x + f~ is
2t
1 2
~
.
,
f=
P =
−4
3 2
2
1
and
P has eigenvalues λ = −1 and 4 with corresponding eigenvectors
3
−1
respectively. Thus we can construct a fundamental matrix solution as
−t
e
2e4t
X(t) =
.
−e−t 3e4t
Then
−1
X(t)
1
= 3t
5e
3e4t −2e4t
e−t
e−t
1
=
5
3et −2et
e−4t e−4t
.
Let ~xp (t) = X(t)~u(t) be a particular solution to ~x0 + P ~x = f~. Then
1
1
3et −2et
2t
6tet + 8et
0
−1 ~
~u (t) = X(t) f =
=
.
−4
5 e−4t e−4t
5 2te−4t − 4e−4t
Then
Z
~u(t) =
0
~u (t) dt =
Z
1
5
6tet + 8et
2te−4t − 4e−4t
1
dt =
5
6tet + 2et
1 −4t
− 2 te
+ 78 e−4t
.
So a particular solution is
−t
1
t + 34
e
2e4t
6tet + 2et
~xp (t) = X(t)~u(t) =
=
.
− 12 te−4t + 87 e−4t
− 32 t + 81
5 −e−t 3e4t
6
• 3.9.101: Find a particular solution to x0 = 5x + 4y + t, y 0 = x + 8y − t. a) using
undetermined coefficients. b) using variation of parameters.
Answer.
a) Since the inhomogeneous term contains only t’s we try a particular solution of the
form ~xp = t~a + ~b. Substituting into the system gives
a1
5a1 + 4a2
5b1 + 4b2
t
=t
+
+
.
a2
a1 + 8a2
b1 + 8b2
−t
Equating powers of t in both entries we get the systems
5a1 + 4a2 = −1
a1 + 8a2 = 1
and
5b1 + 4b2 = a1
b1 + 8b2 = a2 .
Solving the first system gives a1 = −1/3 and a2 = 1/6. Solving the second system
gives b1 = −5/54 and b2 = 7/216. Therefore the particular solution is
5
− 31 t − 54
~xp = 1
.
7
6 t + 216
b) The corresponding matrix and inhomogeneous vector for the problem ~x0 = P ~x + f~
is
t
5 4
~
.
,
f=
P =
−t
1 8
1
−4
and
. Then we
P has eigenvalues λ = 9, 4 with corresponding eigenvectors
1
1
can take fundamental matrix to be:
9t
e
−4e4t
X(t) =
.
e9t
e4t
Then
X(t)−1 =
1
5e13t
e4t 4e4t
−e9t e9t
=
1
5
e−9t 4e−9t
−e−4t e−4t
.
Let ~xp (t) = X(t)~u(t) be a particular solution to ~x0 = P ~x + f~, then
−9t
1
1 −3te−9t
t
e
4e−9t
0
−1 ~
u (t) = X(t) f =
=
−t
5 −e−4t e−4t
5 −2te−4t
7
Then
Z
~u(t) =
Z
0
~u (t) dt =
1
5
−3te−9t
−2te−4t
1
dt =
5
1 −9t
1 −9t
+ 27
e
3 te
1 −4t
1 −4t
te
+
e
2
8
So a particular solution can be:
1
~x(t) = X(t)~u(t) =
5
e9t −4e4t
e9t
e4t
1 −9t
1 −9t
+ 27
e
3 te
1 −4t
1 −4t
te
+
e
2
8
5
− 13 t − 54
= 1
.
7
6 t + 216
• 3.9.102: Find a particular solution to x0 = y + et , y 0 = x + et . a) using undetermined
coefficients. b) using variation of parameters.
Answer.
a) The corresponding matrix is
0 1
,
A=
1 0
which has characteristic equation λ2 − 1 = 0. Hence the eigenvalues are λ = &plusmn;1.
Observe that the eigenvalue λ = 1 matches the power of the exponent that corresponds
to the
non-homogeneous part. Also, this eigenvalue has a corresponding eigenvector
1
. The particular solution is therefore of the form tet~a + et~b as is easily verified
~v =
1
by substituting into the equation and keeping in mind that A~v = ~v
~x0p = et~a + tet~a + et~b = A~xp + et~v = tet A~a + et A~b + et~v
Then
~a + ~b = A~b + ~v ,
and ~a = A~a.
Then we can take
~a = ~v =
1
,
1
and ~b = 0.
Hence the particular solution is
~xp =
tet
1
.
1
b) Matrix
λ1 = 1, λ2 = −1 and the corresponding eigenvectors
A has
eigenvalues
1
1
~v1 =
, ~v2 =
. Thus, the fundamental matrix is
1
−1
X(t) =
et e−t
et −e−t
8
.
Its inverse is
X(t)
−1
1
=
−2
−e−t −e−t
−et
et
1
=
2
e−t e−t
et −et
.
Let ~xp (t) = X(t)~u(t) be a particular solution to ~x0 = A~x + f~, then
t
Z
Z −t
1 e
e−t
e
t
−1 ~
u(t) = X(t) f dt =
dt =
,
t
t
t
e
−e
e
0
2
and finally
~x(t) = X(t)~u(t) =
et e−t
et −e−t
9
t
0
t
te
.
=
tet
```