Download Exam 3 Sol

APPM 2360: Exam 3
7:00pm – 8:30pm, April 13, 2011.
ON THE FRONT OF YOUR BLUEBOOK write: (1) your name, (2) your
student ID number, (3) recitation section (4) your instructor’s name, and (5)
a grading table. Text books, class notes, and calculators are NOT permitted.
A one-page crib sheet is allowed.
Problem 1: (20 points)
(a) (6 points) Suppose A~x = λ1 ~x and B~x = λ2 ~x (i.e. ~x is an eigenvector
of both A and B). Given that ~x is an eigenvector of A + B, what is
the corresponding eigenvalue?
(b) (6 points) If the characteristic equation of a square matrix, C, is
(λ − 2)3 (λ − 3)2 λ = 0, what is the minimum and maximum number
of linearly independent eigenvectors which can be associated with C?
(c) (8 points) Use the Method of Undetermined Coefficients to write down
the general form of the particular solution, yp , for the differential
equations given below but do not solve.
(i) y ′′ + y ′ − 2y = 2et (ii) y ′′ + y ′ − 2y = F0 cos ωf t
(iii) y ′′ + y ′ − 2y = e2t (iv) y ′′ + y ′ − 2y = te2t
Solution:
a) (A + B)~x = A~x + B~x = λ1 ~x + λ2 ~x = (λ1 + λ2 )~x, so the
corresponding eigenvalue is λ1 + λ2 .
b) λ = 2 will have at most 3 and as few as one associated LI
eigenvector, λ = 3 can have at most two and as few as one
associated LI eigenvector and λ = 0 has a single LI eigenvector so C has at least three LI eigenvectors and at most six.
c) The solution to the homogenous problem y ′′ + y ′ − 2y = 0
is yh = c1 et + c2 e−2t .
(i) Atet , since f (t) = 2et is linearly dependent on yh
(ii) A cos ωf + B sin ωf t
(iii) Ae2t
(iv) (At + B)e2t
Problem 2: (20 points) Consider a standard damped mass-spring system
with the spring constant 4, mass 2 and damping constant 4.
(a) (6 points) Write down the differential equation for the oscillator’s position and find its general solution. Classify its motion (overdamped,
critically-damped or underdamped).
(b) (8 points) Suppose the system is driven by a force f (t) = 10 cos(t)
with the system initially at rest (i.e. x(0) = x′ (0) = 0). Find the
position of the oscillator at any time t > 0.
(c) (6 points) If the driving force is changed to f (t) = 8, what is the
long-term behavior of the solution?
Solution:
(a) Differential equation: 2ẍ + 4ẋ + 4x = 0.
Characteristic equation: 2r 2 +4r+4 = 0 with ∆ = 16−32 = −16 < 0.
So it is underdamped.
Roots: r1 = −1 + i and r2 = −1 − i.
Position: x(t) = e−t (c1 cos(t) + c2 sin(t)).
(b) The forced equation: 2ẍ+4ẋ+4x = 10 cos(t) or ẍ+2ẋ+2x = 5 cos(t).
Method of Undetermined Coefficients: xp = A cos(t) + B sin(t).
A + 2B = 5, B − 2A = 0.
A = 1 and B = 2.
The position: x(t) = e−t (c1 cos(t) + c2 sin(t)) + cos(t) + 2 sin(t).
Initial condition: x(0) = 0, ẋ(0) = 0.
Applying IC: c1 + 1 = 0 and −c1 + 2 + c2 = 0
c1 = −1 and c2 = −3. x(t) = e−t (− cos(t)−3 sin(t))+cos(t)+2 sin(t).
(c) Forced equation: 2ẍ + 4ẋ + 4x = 8.
Method of Undermined Coefficients: xp = C and C = 2.
x(t) = e−t (c1 cos(t) + c2 sin(t)) + 2.
t → +∞, x(t) → 2. (b) it stops at 2.
Problem 3: (20 points) Given the following differential equation
(a)
(b)
(c)
(d)
9
d2 y 1 dy
− 2 y = t2 ,
+
2
dt
t dt
t
(4 points) Show that the solution to the homogeneous problem is
given by yh (t) = c1 t3 + c2 t−3 .
(8 points) Using Variation of Parameters, solve the non-homogeneous
problem.
(4 points) Are there any initial conditions one could choose such that
the solution is bounded for all t?
(4 points) Using your general solution, solve the initial value problem
for y(1) = 1 and y ′ (1) = 0.
Solution:
a)The ODE is an Euler equation, so subbing in y = tλ gives
the quadratic equation λ2 −9. It’s true, Chris checked it himself and everything.
b) So, letting y1 = t3 and y2 = t−3 , we find the Wronskian
W (y1 , y2 ) = −6t−1 .
Then
Z
y1 (t)f (t)
1
dt = −
W (t)
6
and
Z
Z
1
−y2 (t)f (t)
dt =
W (t)
6
t6 dt = −
Z
dt =
t7
,
42
t
.
6
Thus one has
1
yg (t) = c1 t3 + c2 t−3 + t4 .
7
c) No, as |t| → ∞, nothing can balance against the t4 .
d) One gets the linear system
4
6
c1 + c2 = , c1 − c2 = − .
7
21
Therefore
1
11
c1 = , c2 = .
3
21
Problem 4: (20 points) Given the matrix

2 1 0
A= 1 2 0 
0 0 1

(a) (6 points) Find the eigenvalues of A.
(b) (10 points) Find the eigenvectors of A.
(c) (4 points) Write down the basis and dimension for each eigenspace
Eλ .
Solution:
a)
2−λ
1
0
2−λ
0
p(λ) = 1
0
0
1−λ
= (λ − 1)2 (λ − 3) = 0
So the eigenvalues are λ1,2 = 1, λ3 = 3.
b) Solving (A − λ1 I)v = 0 gives


1 1 0 0
 1 1 0 0 
0 0 0 0


1 1 0 0
⇒ 0 0 0 0 
0 0 0 0
So x3 is arbitrary and x1 + x2 = 0. One choice of the resulting two linearly independent eigenvectors would be v1 =
(1, −1, 0)T , v2 = (0, 0, 1)T .
Solving (A − λ3 I)v = 0 gives


−1 1
0 0
 1 −1 0 0 
0
0 −2 0


1 −1 0 0
⇒ 0 0 1 0 
0 0 0 0
So x3 is necessarily 0 and x1 − x2 = 0. One choice for the
resulting eigenvector would be v3 = (1, 1, 0)T .
c) The dimension of Eλ for λ = 1 is two and an example
basis would be {(1, −1, 0)T , (0, 0, 1)T }. The dimension of Eλ
for λ = 2 is one and an example basis would be {(1, 1, 0)T }.
Problem 5: (20 points) Remember to justify your answers for this problem.
No justification = no credit. Consider the unforced mass-spring system with
mass m > 0, spring constant k > 0, and damping constant b = 1
mx′′ + x′ + kx = 0 .
(a) (3 points) What are the roots of the characteristic polynomial for this
equation?
(b) (3 points) Letting y1 (t) = x(t) and y2 (t) = x′ (t), we can construct
the linear system of two first order differential equations for y1 and
y2 , i.e.,
y1
.
y2
What is A and what are its eigenvalues (in terms of m and k)?
(c) (10 points) We also know that for the linear system, we guess ~y (t) =
eλt~v for the form of the solution and that this leads to the eigenvalue/eigenvector problem A~v = λ~v . If one eigenvalue/ eigenvector
pair is
~y ′ = A~y where ~y =
λ1 = −1 + i and ~v1 =
−0.5 + 0.5i
,
1
what do m and k have to be?
(d) (4 points) Is λ1 = 1+i a valid eigenvalue for this spring-mass system?
(Justify!)
Solution:
√
1−4mk
.
a) The roots are r = −1± 2m
0
1
b) A =
and the eigenvalues are λ =
k
1
−m −m
√
−1± 1−4mk
2m
c) We know that eigenvalues occur in conjugate pairs. So,
λ2 = −1 − i. From a) and b), we know that
√
−1
− 1 − 4mk
−1 =
and −i =
2m
2m
and thus m = 0.5 and k = 1. An alternative would be to
compute that
0
1
A=
−2 −2
and from there conclude the values for m and k.
d) An eigenvalue with a positive real part is not valid as
it would imply that the mass m is negative.