Homework 8 - Select Solutions
Harini Chandramouli
MATH2374
[email protected]
Section 4.6 - Forced Oscillations
Problem 13. (Damped Forced Motion II) A 32-lb weight is attached to a spring suspended
from the ceiling, stretching the spring by 1.6 ft before coming to rest. At time t = 0 an
external force of f (t) = 20 cos(2t) is applied to the system. Assume that the mass is acted
upon by a damped force of 4ẋ, where ẋ is the instantaneous velocity in feet per second. Find
the displacement of the weight with x(0) = 0 and ẋ(0) = 0.
Solution. We are given that F = 32 lbs, b = 4, ∆x = 1.6 ft, and f = 20 cos(2t). The
initial-conditions are x(0) = ẋ(0) = 0. Since F = k∆x,
32 = k (1.6) =⇒ k = 20.
So we can make the initial-value problem
ẍ + 4ẋ + 20x = 20 cos(2t), x(0) = ẋ(0) = 0.
So, first we find the homogeneous solution using the characteristic equation,
λ2 + 4λ + 20 = 0 =⇒ λ = −2 ± 4i =⇒ xh (t) = e−2t (c1 cos(4t) + c2 sin(4t)) .
Now we want to find the particular solution. We use the guess xp (t) = A cos(2t) + B sin(2t)
and use the method of undetermined coefficients. Plugging xp (t) into the differential equation
we get
(16A + 8B) cos(2t) + (16B − 8A) sin(2t) = 20 cos(2t)
(
16A + 8B = 20
1
=⇒ A = 1, B =
2
16B − 8A = 0
So now we have the solution
x(t) = xh (t) + xp (t) = e−2t (c1 cos(4t) + c2 sin(4t)) + cos(2t) +
1
sin(2t).
2
Applying the initial conditions, we find that c1 = −1 and c2 = − 43 . So the answer is
3
1
−2t
x(t) = e
− cos(4t) − sin(4t) + cos(2t) + sin(2t).
4
2
1
Problem 21. Find the steady-state solution having the form xss = C cos(ωt − δ), for the
damped system,
ẍ + 2ẋ + 2x = 2 cos(t).
Solution. The roots of the characteristic equation are −1 ± i. So for the particular solution:
xp (t) = A cos(t) + B sin(t) = C cos(t − δ).
√
Another approach to xp is to note that F0 = 2, ω0 = 2, ωf = 1, m = 1, b = 2 and simply
substitute these numbers into the text solution to find
2
xss (t) = √ cos(t − δ)
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where the phase angle is δ = arctan(2) = 1.1 radians.
2
Section 5.4 - Coordinates and Diagonalization
Problem 38. Determine whether each matrix A is diagonalizable. If it is, determine a
matrix P that diagonalizes it and compute P −1 AP . You can obtain P −1 AP directly from
careful construction of a diagonal matrix with eigenvalues along the diagonal in the proper
order.


4 1 −1
A = 2 5 −2 .
1 1 2
Solution. First we find the eigenvalues of A,


4 1 −1
det(A − λI) = det 2 5 −2 = (λ − 3)2 (λ − 5) = 0 =⇒ λ1,2 = 3, λ3 = 5.
1 1 2
Now, we want to find the eigenvectors. For λ1,2 = 3,


1 1 −1
(A − λI)v1 = 0 =⇒ 2 2 −2 v1 = 0.
1 1 −1
With this matrix we find that the eigenvector is
 
 
−1
1
v1 =  1  s + 0 t
0
1
where s, t ∈ R.
For λ3 = 5,


−1 1 −1
(A − λI)v2 = 0 =⇒  2 0 −2 v2 = 0.
1 1 −3
With this matrix, we find that the eigenvector is
 
1

v2 = 2 .
1
So, we have that our matrix P is


1 −1 1
P = 0 1 2
1 0 1
where our diagonal matrix D will be


3 0 0
D = 0 3 0 .
0 0 5
3
Now we want to check that P −1 AP = D, but we first have to check that det(P ) 6= 0 before
we can find P −1 . So, det(P ) = −2, and so we can find
 1

− 2 − 21 32
1 .
P −1 =  −1 0
1
1
− 12
2
2
Last, we compute P −1 AP , and find

 1

 

4 1 −1
− 2 − 12 32
1 −1 1
3 0 0
1  2 5 −2 0 1 2 = 0 3 0 = D.
P −1 AP =  −1 0
1
1
− 21
1 1 2
1 0 1
0 0 5
2
2
4
Home-made Problems
Problem 1.
1
(a) A =
2
0
(b) A =
1
Find matrix powers An of the following matrices A:
2
1
2
1
Solution.
(a) First, we need to diagonalize, since we know that An = P Dn P −1 . The eigenvalues of A
are λ1 = 3, λ2 = −1. The associated eigenvectors are
1
−1
v1 =
, v2 =
.
1
1
So, we have that
−1 0
−1 1
D=
,P =
.
0 3
1 1
Since det(P ) = −2 6= 0, then we find that
P
−1
=
− 12
1
2
1
2
1
2
.
Thus, we have found D, P , and P −1 such that A = P DP −1 . Thus,
1 1
−1 1
(−1)n 0
−2 2
n
A =
.
1
1
1 1
0
3n
2
2
(b) First, we need to diagonalize, since we know that An = P Dn P −1 . The eigenvalues of A
are λ1 = 2, λ2 = −1. The associated eigenvectors are
1
−2
v1 =
, v2 =
.
1
1
So, we have that
−1 0
−2 1
D=
,P =
.
0 2
1 1
Since det(P ) = −3 6= 0, then we find that
P
−1
=
− 13
1
3
1
3
2
3
.
Thus, we have found D, P , and P −1 such that A = P DP −1 . Thus,
1 1
−2 1
(−1)n 0
−3 3
n
A =
.
1
2
n
1 1
0
2
3
3
5
Note: These problems are taken from Differential Equations and Linear Algebra by Jerry
Farlow, James Hall, Jean McDill, and Beverly West. Specifically, I am following the Custom
Edition for the University of Minnesota Twin Cities.
6