Math 211 Fall 2007: Solutions: HW #13
Instructor: S. Cautis
1. section 9.1, #16
The characteristic polynomial is
det(A − λI) = (2 − λ)(−2 − λ)
so the eigenvalues are λ1 = 2 and λ2 = −2. The eigenvector for λ1 satisfies
0
0
v =0
−4 −4 1
so that v1 = (1, −1)t . The eigenvector for λ2 satisfies
4 0
v =0
−4 0 2
so v2 = (0, 1)t .
So a fundamental set of solutions is formed by e2t
1
−1
and e−2t
0
.
1
2. section 9.1, #22
The characteristic polynomial is
det(A − λI) = (−3 − λ)(4 − λ)
so the eigenvalues are λ1 = −3 and λ2 = 4. The eigenvector for λ1 satisfies
0 14
v1 = 0
0 7
so that v1 = (1, 0)t . The eigenvector for λ2 satisfies
−7 14
v2 = 0
0
0
so v2 = (2, 1)t .
So a fundamental set of solutions is formed by e−3t
1
2
and e4t
.
0
1
3. section 9.1, #24
The characteristic polynomial is
det(A−λI) = (−5−λ)(−3−λ)(5−λ)−6(−4(−3−λ)) = −λ3 −3λ2 +λ+3 = −(λ+3)(λ−1)(λ+1)
so the eigenvalues are λ1 = −3,
λ1 satisfies

−2
 26
4
λ2 = 1 and λ2 = −1. The eigenvector for

0 −6
0 38  v1 = 0
0 8
so that v1 = (0, 1, 0)t . The eigenvector for λ2 satisfies


−6 0 −6
 26 −4 38  v2 = 0
4
0
4
so that v2 = (1, 0, −1)t . The eigenvector for λ3 satisfies


−4 0 −6
 26 −2 38  v3 = 0
4
0
6
so that v3 = (3, 0, −2)t .
 
 
0
1
So a fundamental set of solutions is formed by e3t 1 and et  0  and
0
−1
 
3
e−t  0 .
−2
4. section 9.2, #6
The characteristic polynomial is
det(A − λI) = (−1 − λ)(−1 − λ) − 1 = λ2 + 2λ
so the eigenvalues are λ1 = 0 and λ2 = −2. The eigenvector for λ1 is
v1 = (1, 1)t . The eigenvector for λ2 satisfies
1 1
v =0
1 1 2
so v2 = (1, −1)t . So the general solution is
1
1
−2t
.
+ C2 e
C1
−1
1
5. section 9.2, #14
e(1+i)t = et (cos(t) + i sin(t)). So e(1+i)t (−1 + i) = et (− cos(t) + i cos(t) −
i sin(t) − sin(t)). Thus the real part of z(t) is
t
e (− cos(t) − sin(t))
2et cos(t)
and the imaginary part is
et (cos(t) − sin(t))
2et sin(t)
6. section 9.2, #16
The characteristic polynomial is
det(A − λI) = (−4 − λ)(4 − λ) + 32 = λ2 + 16
so the two eigenvalues are λ1 = 4i and λ2 = −4i. The eigenvector for λ1
satisfies
−4 − 4i
−8
v =0
4
4 − 4i 1
t
so v1 = (−1, 1+i
2 ) . Hence the other eigenvector is the conjugate of v1 ,
t
namely v2 = (−1, 1−i
2 ) . A fundamental set of real solutions is given by
the real and imaginary parts of
−1
4it
e
1+i
2
1+i
1
Since e4it 1+i
2 = (cos(4t) + i sin(4t)) 2 = 2 (cos(4t) + i sin(4t) + i cos(4t) −
sin(4t)) the real part is
− cos(4t)
1
2 (cos(4t) − sin(4t))
while the imaginary part is
− sin(4t)
1
(sin(4t)
+ cos(4t))
2
7. section 9.2, #32
The characteristic polynomial is
det(A − λI) = (−2 − λ)(2 − λ) + 4 = λ2
so there’s only one eigenvalue λ = 0. An eigenvector is v1 = (1, −2)t . To
find the other generalized eigenvector v2 we must solve Av2 = v1 which
gives v2 = (1, −3)t . Hence the general solution is a linear combination
C1 y1 +C2 y2 of y1 (t) = v1 = (1, −2)t and y2 (t) = tv1 +v2 = (t+1, −2t−3)t .
8. section 9.2, #58
a) Denote by y1 (t) and y2 (t) the salt content in the top and bottom tanks
at time t. Then
y1′ = saltin − saltout = 0 − 5 ·
while
y2′ = saltin − saltout = 5 ·
y1
500
y2
y1
−5
500
500
′
Hence the initial value problem is y = Ay where A =
−1/100
0
1/100 −1/100
with initial values y1 (0) = 100 and y2 (0) = 0.
b) Characteristic polynomial of A is
det(A − λI) = (−1/100 − λ)(−1/100 − λ)
so there’s one eigenvalue: λ = −1/100. The eigenvector is v1 = (0, 1)t .
The generalized eigenvector v2 is given by solving (A + 1/100I)v2 = v1 .
Solving yields v2 = (100, 0)t. So the general solution is
e−t/100 (C1 v1 + C2 (tv1 + v2 ))
Plugging in t = 0 gives C1 v1 + C2 v2 = (100, 0)t . Solving gives C1 = 0 and
C2 = 1. Thus the solution satisfying the initial conditions is
100
−t/100
−t/100
e
(tv1 + v2 ) = e
(
.
t