Math 215
Spring 2010
Assignment 8
You are highly recommended to own a textbook, Boyce-DiPrima or another, to see a comprehensive treatment of the subjects and more examples.
§7.7: 6, 7, 12, §7.8: 1c, 4c, 8a, §7.9: 4, 6, 11
µ
¶
−1 −4
• (§7.7: 6) Find a fundamental matrix for x =
x. Also find the fundamental
1 −1
matrix Φ(t) satisfying Φ(0) = I.
0
Answer.
Setting det(A − rI) = r2 + 2r + 5 = 0, the eigenvalues are −1 ± 2i. For
r = −1 − 2i, the eigenvector equation (A − rI)ξ = 0 reduces to ξ1 + 2iξ2 = 0. We can
take ξ = (−2i, 1)T . Thus
µ
¶
µ
¶
µ
¶
−2i (−1−2i)t
(1)
−t −2 sin 2t
−t −2 cos 2t
x (t) =
e
=e
+ ie
.
1
cos 2t
− sin 2t
We can take
µ
¶
−2 sin 2t −2 cos 2t
Ψ(t) = e
.
cos 2t
− sin 2t
µ
¶
µ
¶
0 2
0 −2
1
−1
, we have
Since Ψ(0) =
with Ψ(0) = 2
−1 0
1 0
µ
¶
1 −t 2 cos 2t −4 sin 2t
−1
.
Φ(t) = Ψ(t)Ψ(0) = e
sin 2t
2 cos 2t
2
−t
µ
¶
5 −1
• (§7.7: 7) Find a fundamental matrix for x =
x. Also find the fundamental
3 1
matrix Φ(t) satisfying Φ(0) = I.
0
Answer.
Setting det(A − rI) = r2 − 6r + 8 = 0, the eigenvalues are 2, 4. For
r = 2, the eigenvector equation (A − rI)ξ = 0 reduces to 3ξ1 − ξ2 = 0. We can take
ξ = (1, 3)T . For r = 4, the two equations in (A − rI)ξ = 0 reduce to ξ1 − ξ2 = 0. We
can take ξ = (1, 1)T . Thus
µ ¶
µ 2t ¶
µ ¶
µ 4t ¶
1 2t
e
1 4t
e
(1)
(2)
x (t) =
e =
, x (t) =
e = 4t .
3
3e2t
1
e
We can take
µ
¶
e2t e4t
Ψ(t) =
.
3e2t e4t
µ
¶
µ
¶
1 1
1 −1
1
−1
Since Ψ(0) =
with Ψ(0) = − 2
, we have
3 1
−3 1
µ
¶
1 −e2t + 3e4t e2t − e4t
.
Φ(t) = Ψ(t)Ψ(0)−1 =
2 −3e2t + 3e4t 3e2t − e4t
• (§7.7: 12) Solve the initial value problem
µ
¶
−1 −4
0
x =
x,
1 −1
µ ¶
3
x(0) =
1
by using the fundamental matrix Φ(t) found in problem 6.
Answer.
We have
µ ¶
µ
¶µ ¶
µ
¶
1 −t 2 cos 2t −4 sin 2t
3
3
−t 3 cos 2t − 2 sin 2t
x(t) = Φ(t)
= e
=e
.
3
1
sin 2t
2 cos 2t
1
2
2 sin 2t + cos 2t
µ
¶
3 −4
• (§7.8: 1c) Find the general solution of x =
x.
1 −1
0
Answer.
Setting det(A − rI) = r2 − 2r + 1 = 0, the eigenvalues are 1, 1. For
r = 1, the eigenvector equation (A − rI)ξ = 0 reduces to ξ1 − 2ξ2 = 0. We can take
ξ = (2, 1)T and the first solution is x(1) (t) = ξet . The generalized eigenvector satisfies
(A−I)η = ξ, which reduces to η1 −2η2 = 1. If η2 = k where k is an arbitrary constant,
then η2 = 1 + 2k and η = (1, 0)T + kξ. A second solution is x(2) (t) = (tξ + η)et . The
general solution is
µ ¶
µ
¶
2 t
2t + 1 t
x(t) = c1 x(1) (t) + c2 x(2) (t) = c1
e + c2
e.
1
t
In the above we have set k = 0 since it can absorbed into c1 .
µ
¶
−3 5/2
0
• (§7.8: 4c) Find the general solution of x =
x.
−5/2 2
Answer.
Setting det(A − rI) = r2 + r + 1/4 = 0, the eigenvalues are −1/2, −1/2.
For r = −1/2, the eigenvector equation (A − rI)ξ = 0 reduces to ξ1 − ξ2 = 0.
We can take ξ = (1, 1)T and the first solution is x(1) (t) = ξe−t/2 . The generalized
eigenvector equation (A − rI)η = ξ reduces to η1 − η2 = −2/5. If η1 = k where k is
an arbitrary constant, then η2 = 2/5 + k and η = (0, 2/5)T + kξ. A second solution is
x(2) (t) = (tξ + η)e−t/2 . The general solution is
µ ¶
µ
¶
1 −t/2
t
x(t) = c1 x(1) (t) + c2 x(2) (t) = c1
e
+ c2
e−t/2 .
1
t + 2/5
In the above we have set k = 0 since it can absorbed into c1 .
2
µ
¶
−5/2 3/2
x,
• (§7.8: 8a) Solve the initial value problem x =
−3/2 1/2
0
¶
3
.
−1
µ
x(0) =
Answer. The characteristic equation is det(A − rI) = r2 + 2r + 1 = 0, with a single
root −1. For r = −1, the eigenvector equation (A − rI)ξ = 0 reduces to ξ1 − ξ2 = 0.
We can take ξ = (1, 1)T and the first solution is x(1) (t) = ξe−t . The generalized
eigenvector satisfies (A − rI)η = ξ, which reduces to η1 − η2 = −2/3. If η1 = k where
k is an arbitrary constant, then η2 = 2/3+k and η = (0, 2/3)T +kξ. A second solution
is x(2) (t) = (tξ + η)e−t . The general solution is
µ ¶
µ
¶
1 −t
t
(1)
(2)
x(t) = c1 x (t) + c2 x (t) = c1
e + c2
e−t .
1
t + 2/3
In the above
µ ¶ we have
µ set
¶ k µ= 0¶since it can absorbed into c1 . The initial condition
1
0
3
gives c1
+ c2
=
. Thus c1 = 3, c2 = −6, and
1
2/3
−1
µ
¶
3 − 6t
e−t .
−1 − 6t
x(t) =
µ
¶
µ −2t ¶
1 1
e
• (§7.9: 4) Find the general solution of x =
x+
.
4 −2
−2et
µ ¶
µ ¶
1 2t
1
Hint: You can use xc (t) = c1
e + c2
e−3t from §7.5 #4.
1
−4
0
Answer.
is
Since g(t) = (e−2t , −2et )T involves exponential functions only, a first guess
xp (t) = ae−2t + bet .
It requires no revision since e−2t and et do not appear in xc (t). Substituting it into
the equation,
−2ae−2t + bet = Aae−2t + Abet + g(t).
Thus
µ ¶
−1
(A + 2)a =
,
0
µ ¶
0
(A − 1)b =
.
2
We can solve a = (0, −1)T and b = (1/2, 0)T , and thus
µ ¶
µ ¶
µ ¶
µ ¶
1 2t
1
0
1/2 t
−3t
−2t
x(t) = xc (t) + xp (t) = c1
e + c2
e
+
e
+
e.
1
−4
−1
0
The problem can be done by other methods.
3
µ −1 ¶
µ
¶
t
−4 2
, t > 0.
x+
• (§7.9: 6) Find the general solution of x =
−1
2 −1
2t + 4
0
Answer.
The characteristic equation is det(A − rI) = r2 + 5r = 0, with roots
r = 0, −5. Their corresponding eigenvectors are (1, 2)T and (−2, 1)T . Set
µ
¶
µ
¶
µ
¶
1 1 2
1 −2
0 0
−1
T =
, T =
, D=
.
2 1
0 −5
5 −2 1
We have AT = T D. Setting y = T −1 x and h(t) = T −1 g(t), the transformed system is
1 8
4
y10 = + , y20 = −5y + .
t
5
5
Their solutions are
8
4
y1 = c1 + ln t + t, y2 (t) = c2 e−5t + .
5
25
Thus
µ ¶
¶
µ ¶
µ
8
1
−2 −5t
ln t + 58 t − 25
.
+ c2
e
+
x(t) = T y(t) = c1
4
2
1
2 ln t + 16
5 t + 25
µ
¶
µ
¶
2 −5
0
• (§7.9: 11) Find the general solution of x =
x+
, 0 < t < π.
1 −2
cos t
µ
¶
µ
¶
2 cos t − sin t
cos t + 2 sin t
Hint: You can use xc (t) = c1
+ c2
from §7.6 #3.
cos t
sin t
0
Answer.
It follows from the formula of xc (t) a fundamental matrix
µ
¶
2 cos t − sin t cos t + 2 sin t
Ψ(t) =
.
cos t
sin t
We have det Ψ(t) = −1 and
µ
¶
µ
¶
− sin t cos t + 2 sin t
cos2 t + 2 cos t sin t
−1
−1
Ψ(t) =
, Ψ(t) g(t) =
.
cos t −2 cos t + sin t
−2 cos2 t + sin t cos t
µ ¶ µ1
¶
Z
c1
cos t sin t + 21 t + sin2 t
−1
2
Ψ(t) g(t)dt =
+
.
c2
− cos t sin t − t + 21 sin2 t
Thus
µ ¶
µ1
¶
c1
cos t sin t + 12 t + sin2 t
2
x(t) = Ψ(t)
+ Ψ(t)
c2
− cos t sin t − t + 21 sin2 t
µ ¶ µ ¶
µ
µ 5¶
¶
c1
0
0
−2
= Ψ(t)
+ 1 t cos t +
t sin t + 1
.
c2
−1
2
2 sin t
µ
¶
µ
¶
µ ¶
0
−5/2
c
The last term 1
can be replaced by
cos t by a different choice of 1 .
sin
t
−1
c2
2
Remark. To use the method of undetermined coefficient, since g(t) “resonants” with
the eigenvalues ±i, one assumes xp = (a + bt) cos t + (c + dt) sin t .
4