Math 2250 (Jason Underdown)
Exam 2
July 3, 2014
Name
1. [10 pts.] Find the general solution to
y 00 + 2y 0 + 10y = 0.
Solution:
r2 + 2r + 10 = 0
r2 + 2r + 1 = −10 + 1
(r + 1)2 = −9
r = −1 ± 3i
y = e−x (c1 cos 3x + c2 sin 3x)
2. [10 pts.] Determine the linear, homogeneous differential equation with constant coefficients which has solutions:
y1 = e−x
y2 = ex
y3 = xex .
Hint: What would the roots of the characteristic equation need to be in order to get
these solutions?
Solution: The roots would be r1 = −1, r2 = 1, but root r2 has a multiplicity of 2,
thus the characteristic equation is
(r + 1)(r − 1)(r − 1) = 0
(r + 1)(r2 − 2r + 1) = 0
r3 − 2r2 + r + r2 − 2r + 1 = 0
r3 − r2 − r + 1 = 0
Which corresponds with the differential equation
y 000 − y 00 − y 0 + y = 0.
Math 2250
Exam 2, Page 2 of 8
July 3, 2014
3. [15 pts.] Use Gauss-Jordan elimination to solve the system
x1 + x2 + x3 + x4 = 12
x2 − x3 + 4x4 = 5
3x1 + 2x2 + 4x3 − x4 = 31
Solution:


1 1
1
1 12
0 1 −1
4 5
3 2
4 −1 31


1 1
1 1 12
0 1 −1 4 5
0 0
0 0 0
−3R1 +R3
−→
−R2 +R1
−→


1
1
1
1 12
0
5
1 −1
4
0 −1
1 −4 −5
R2 +R3
−→


1 0
2 −3 7
0 1 −1
4 5
0 0
0
0 0
There are two columns without leading 1s, columns three and four which correspond
with x3 and x4 , thus we parameterize these two variables. Let x3 = s and x4 = t.
Then the initial system is row equivalent to the system
x1
+ 2s − 3t = 7
x2 + −s + 4t = 5
which in turn is equivalent to
x1 = −2s + 3t + 7
x2 = s − 4t + 5
x3 = s
x4 =
t
which we finally write as a vector equation
 
 
   
−2
3
7
x1
 x2 
 1
−4 5
  = s  + t  +  .
 1
 0 0
 x3 
x4
0
1
0
Math 2250
4. [5 pts.] The matrix A =
Exam 2, Page 3 of 8
h
1 0 0
0 1 0
0 0 0
i
July 3, 2014
projects any vector in R3 to the xy–plane of R3 , because

   
1 0 0 x
x
0 1 0 y  = y  .
0 0 0
z
0
In other words, A sets the z component of any vector to 0. What is the solution space
of the following homogeneous equation?

   
1 0 0 x
0
0 1 0 y  = 0
(1)
0 0 0
z
0
A. All vectors in the xy–plane
B. All vectors in the xz–plane
C. All vectors in the yz–plane
D. Only the zero vector
E. R3
F. The z–axis
5. [5 pts.] Recall that equation (1) above is exactly equivalent to the equation
 
 
   
1
0
0
0







x 0 + y 1 + z 0 = 0 .
0
0
0
0
What is the span of the three column vectors of matrix A?
A. All vectors in the xy–plane
B. All vectors in the xz–plane
C. All vectors in the yz–plane
D. Only the zero vector
E. R3
F. The z–axis
Math 2250
Exam 2, Page 4 of 8
July 3, 2014
6. [15 pts.] Determine the correct differential equation for the free (undriven), mass–spring–
dashpot system pictured and solve it for x(t).
Assume: m = 1, c = 2, k = 10, x(0) = 5, x0 (0) = 0.
k
c
m
Solution: This system is governed by the equation
x00 + 2x0 + 10x = 0
(2)
which has characteristic equation,
r2 + 2r + 10 = 0
r2 + 2r + 1 = −10 + 1
(r + 1)2 = −9
r = −1 ± 3i.
Since this system is not driven, the homogeneous solution is the general solution,
thus
x(t) = e−t (c1 cos 3t + c2 sin 3t).
(3)
This function has derivative,
x0 (t) = −e−t (c1 cos 3t + c2 sin 3t) + 3e−t (−c1 sin 3t + c2 cos 3t).
These two functions allow us to solve for c1 and c2 using the initial conditions:
5 = c1
0 = −c1 + 3c2
Thus c2 = 5/3, and the particular solution to the IVP is
x(t) = e−t (5 cos 3t +
5
sin 3t).
3
(4)
Math 2250
Exam 2, Page 5 of 8
July 3, 2014
7. [10 pts.] Circle or box the letter corresponding to the correct trial particular solution,
yp , for each given equation.
(a) y 00 − y 0 − 6y = 2 + x
A.
B.
C.
D.
yp
yp
yp
yp
= Ax
= A + Bx
= Ae−2x + Be3x
= (A + Bx)ex
(b) y 00 + 3y 0 + 2y = e−x
A.
B.
C.
D.
yp
yp
yp
yp
= Ae−x
= Axe−x
= Ae−x + Be−2x
= x(Ae−x + Be−2x )
(c) y 00 + 3y 0 + 2y = xe4x
A.
B.
C.
D.
yp
yp
yp
yp
= Ae4x
= Axe4x
= (A + Bx)e4x
= (A + Bx + Cx2 )e4x
(d) y 00 + 9y = 5 cos 3x
A.
B.
C.
D.
yp
yp
yp
yp
= A cos 3x
= (A + Bx) cos 3x
= A cos 3x + B sin 3x
= x(A cos 3x + B sin 3x)
Math 2250
Exam 2, Page 6 of 8
July 3, 2014
8. [20 pts.] True or False. Circle one.
(a) T
F A system of linear, algebraic equations may have two and only two distinct
solutions.
F A homogeneous system of linear algebraic equations always has at least
(b) T
one solution.
F The solution space of a homogeneous, linear differential equation is a
(c) T
vector subspace of F, the vector space of all real valued functions.
(d) T
F Multiplying a row of a matrix by any real number is a valid elementary
row operation.
(e) T
F
1 2 3
0 2 3 = 6
0 0 3
(f) T
F The following equation of a mass—spring—dashpot system is critically
damped.
25
x00 + 5x0 + x = 0
4
(g) T
F Let the set B be a basis for the vector space V . If you remove any vector
from B then it will no longer span V .
(h) T
F Let the set B be a basis for a vector space V . If you add any vector to
B, then its vectors will be linearly dependent.
(i) T
F
The matrix:


0 1 0 9
0 0 5 4
0 0 0 0
is in reduced row–echelon form.
F Suppose a square matrix, A has determinant, |A| = 5. If we interchange
(j) T
the first two rows of A to make A0 , then |A0 | = −5.
Math 2250
Exam 2, Page 7 of 8
July 3, 2014
9. The are two ways you can fail to earn the points for the last two questions:
1. not providing an answer,
2. not writing an integer between 1 and 10 inclusive.
(a) [5 pts.] Rate the difficulty of this exam on a scale of 1 to 10, where 1 is very easy
and 10 is very difficult.
(a)
6.4
(b) [5 pts.] Rate the fairness of this exam on a scale of 1 to 10, where 1 is completely
unfair and 10 is completely fair.
(b)
7.2
Math 2250
Exam 2, Page 8 of 8
July 3, 2014
Scratch Paper
Question:
1
2
3
4
5
6
7
8
9
Total
Points:
10
10
15
5
5
15
10
20
10
100
Score: