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REVIEW PROBLEMS FOR MIDTERM II
MATH 2373, FALL 2016
ANSWER KEY
This list of problems is not guaranteed to be an absolutely complete review. For
completeness you must also make sure that you know how to do all of the homework
assigned in the course up to and including that due on November 1, and all the
worksheet problems through week 8, with emphasis on weeks 5–8. (But of course
don’t forget basic skills learned in weeks 1–4 such as using reduced row echelon
form to solve linear equations!)
To study under typical test conditions, here are the ground rules. You may use
matrix addition and multiplication functions on your calculator along with det,
rref and matrix inverse functions, but you must indicate clearly where and how
you used these calculator functions—otherwise your answers would be considered
unjustified and get little credit. Fancier calculator functions, e.g., those finding
eigenvalues and eigenvectors, are not allowed to justify your work. All algebra and
calculus work (except for the matrix functions specifically allowed above) must be
performed by hand and shown clearly.
Below we supply mostly full solutions, but occasionally skip a few
details if similar details can be seen elsewhere in this answer key.
Problem 1
Solve the initial value problem y 00 + 49y = 0, y(0) = 1, y 0 (0) = 2, expressing
the answer in the form A cos(ωt − δ) where A, ω and δ are positive numbers and
furthermore δ < 2π. Give numerical approximations for A, ω and δ accurate to
four decimal places.
Answer.
y = C1 cos(7t) + C2 sin(7t) (general solution)
y(0) = C1 = 1
y 0 (0) = 7C2 = 2 ⇒ C2 = 2/7.
y = cos(7t) + 27 sin(7t).
p
A = 12 + (2/7)2 = 1.0400 and δ = arctan( 2/7
1 ) = 0.2783.
y = 1.0400 cos(7t − 0.2783).
Note: In general to find δ one might also have to add π or 2π depending on the
quadrant.
Problem 2
A mass of 5 kg is suspended from a high ceiling by a spring. When hanging
motionless, the mass stretches the spring 2 meters beyond its natural length. Let
y denote the height of the mass above the equilibrium point. Initially the mass is
pulled down 1/2 meter below the equilibrium point and released with a velocity
Date: November 2, 2016.
1
2
REVIEW FOR MT2
of 1 m/sec upward. The mass encounters air resistance of 1.5 nt/(m/sec). A
force of 7 cos(2t) nt is applied to the weight. Write out the initial value problem
determining the motion of the mass. Is this system underdamped, overdamped or
critically damped? Use 10 m/sec2 as the acceleration of gravity. You do not have
to solve the initial value problem.
Answer. The template is my 00 +by 0 +ky = F (t), y(0) = y0 , y 0 (0) = v0 . We are given
m = 5 kg, b = 1.5 nt/(m/sec), F (t) = 7 cos(2t) nt, y0 = −0.5 m and v0 = 1 m/sec.
The weight of the mass is (5 kg)· (10 m/sec2 ) = 50 nt, so k=(50 nt)/(2 m)=25 nt/m.
Thus the initial value problem is 5y 00 +1.5y 0 +25y = 7 cos(2t), y(0) = −0.5, y 0 (0) = 1.
Because b2 − 4km = 1.52 − 4 · 5 · 25 < 0, the system is UNDERDAMPED.
Problem 3
Solve the following IVP’s:
y 00 + 5y 0 + 6y = 0, y(0) = −2, y 0 (0) = 11
y 00 + 6y 0 + 9y = 0, y(0) = 4, y 0 (0) = −23
y 00 + 4y 0 + 29y = 0, y(0) = −2, y 0 (0) = 39
Answer.
First equation.
r2 + 5r + 6 = (r + 2)(r + 3) = 0, r = −2, −3.
y = C1 e−2t + C2 e−3t
y(0) = C1 + C2 = −2
y 0 (0) = −2C1 − 3C2 =11 1
1 −2
1 0
5
rref
=
−2 −3 11
0 1 −7
y = 5e−2t − 7e−3t
Second equation.
r2 + 6r + 9 = (r + 3)2 = 0, r = −3, −3 (double root)
y = C1 e−3t + C2 te−3t
y(0) = C1 = 4
y 0 (0) = −3C1 + C2 = −23
1 0
4
1 0
4
rref
=
−3 1 −23
0 1 −11
y = 4e−3t − 11te−3t
Third equation.
r2 + 4r + 29 = (r + 2)2 + 52 = 0, r = −2 ± 5i.
y = C1 e−2t cos(5t) + C2 e−2t sin(5t)
y(0) = C1 = −2
y 0 (0) = −2C1 + 5C2 = 39
1 0 −2
1 0 −2
rref
=
−2 5 39
0 1
7
y = −2e−2t cos(5t) + 7e−2t sin(5t)
REVIEW FOR MT2
3
Problem 4
A crash-dummy weighs 224 lbs (and thus has mass 224/32 = 7 slugs). The
crash-dummy encounters air resistance of 1.4 lbs/(ft/sec). The crash-dummy is
tossed off a building 200 ft high, initially with a velocity of 21 ft/sec upward. Let
y denote the height of the crash-dummy above the ground. Set up and solve the
initial value problem for the vertical motion of the crash dummy before it hits the
ground. Use your calculator to find the time the crash dummy hits the ground,
accurate to four decimal places. Use the method of undetermined coefficients to
solve the initial value problem. Also do the problem without air resistance and
compare the times of impact.
Answer.
setup: (7 slugs )(y 00 ft/sec2 ) = −(1.4 lbs/(ft/sec))(y 0 ft/sec) − 224 lbs
y(0) = 200 ft, y 0 (0) = 21 ft/sec.
after simplification: y 00 + 0.2y 0 = −32, y(0) = 200, y 0 (0) = 21.
r2 + 0.2r = 0, r = 0, −0.2.
yh = C1 + C2 e−0.2t
yp = At (appropriate guess, because there’s some “trouble”)
yp00 + 0.2yp0 = 0.2A = −32 ⇒ A = −160.
y = −160t + C1 + C2 e−0.2t
y(0) = C1 + C2 = 200
y 0 (0) = −160 − 0.2C2 = 21
1
1
200
1 0 1105
rref
=
0 −0.2C2 21 + 160
0 1 −905
y = −160t + 1105 − 905e−0.2t
y = 0 ⇒ t = 4.6942 (use cursor on graphing calculator)
Without air-resistance:
setup: (7 slugs )(y 00 ft/sec2 ) = −224 lbs
y(0) = 200 ft, y 0 (0) = 21 ft/sec.
after simplification: y 00 = −32, y(0) = 200, y 0 (0) = 21.
solve by freshman calculus
y = −16t2 + 200 + 21t
y = 0 ⇒ t = 4.2522
(solve quadratic equation or just use cursor on graphing calculator again)
Ignoring air-resistance the crash-dummy hits the ground sooner.
Problem 5
Find particular solutions by the method of undetermined coefficients for each of
the following equations:
y 00 + 3y 0 + 2y = te3t
y 00 + 3y 0 + 2y = te−2t
y 00 + 4y = 3 cos(2t)
Answer.
4
REVIEW FOR MT2
First equation.
yp = (At + B)e3t
yp00 + 3yp0 + 2yp = (20A)te3t + (9A + 20B)e3t = te3t
20
0 1
1 0
1/20
rref
=
9 20 0
0 1 −9/400
yp = (t/20 − 9/400)e3t
Second equation.
yp = (At + B)e−2t is a nice guess but if you try it does not work.
yp = (At2 + Bt)e−2t works because there is some “trouble.”
yp00 + 3yp0 + 2yp = (−2A)te−2t + (2A − B)e−2t
−2
0 1
1 0 −1/2
rref
=
2 −1 0
0 1
−1
yp = (−t2 /2 − t)e−2t
Third equation.
yp = At cos(2t) + Bt sin(2t)
yp00 + 4yp = 4B cos(2t) = 3 cos(2t)
A = 0 and 4B = 3 ⇒ B = 3/4
yp = (3t/4) sin(2t)
Problem 6
Find the eigenvalues and eigenvectors for
11 15/2
.
−10
−9
Given that the eigenvalues are λ = 1, 3, 5, find the eigenvectors for


2
0 1
 3 −4 4  .
7 −15 11
Then use the eigenvectors and eigenvalues you have found to diagonalize the matrices.
Answer.
11 − t
15/2
−10 −9 − t
= t2 − 2t + 24 = (t − 6)(t + 4) ⇒ λ = 6, −4.
Since eigenvalues for the two-by-two matrix are distinct, diagonalization is guaranteed to be possible. We just need to find some eigenvector for each eigenvalue.
2 15/2
3
11 − 6
15/2
5 15/2
=
↔ 6.
=
⇒
−5
−2
−10 −9 − 6
−10 −15
5
2
11 − (−4)
15/2
15 15/2
15/2
1
=
=
↔ 4.
⇒
−10 −9 − (−4)
−10
−5
−2
15 −15
We did not bother to hit the rref button because in each case the two equations
are really just one equation we can solve by inspection. Here is a diagonalization:
11 15/2
3
1
3
1
6
0
=
−10
−9
−2 −2
−2 −2
0 −4
REVIEW FOR MT2
5
Since eigenvalues for the three-by-three matrix are distinct, diagonalization is guaranteed to
 be possible. We
 just
 need to find some
 eigenvector

 for each eigenvalue.
1
0 −1
1 0
1
−5
−5
4  =  0 1 −1/5  ⇒  1  ↔ 1,
rref  3
7 −15 10
0 0
0
5

 

 
−1
0 1
1 0 −1
1
−7 4  =  0 1 −1  ⇒  1  ↔ 3,
rref  3
7 −15 8
0 0
0
1

 



−3
0 1
1 0 −1/3
3
−9 4  =  0 1 −5/9  ⇒  5  ↔ 5.
rref  3
7 −15 6
0 0
0
9
Here is a diagonalization:


 


2
0
1
−5 1 3
−5 1 3
1 0 0
 3 −4
4  1 1 5  =  1 1 5  0 3 0 
7 −15 11
5 1 9
5 1 9
0 0 5
Problem 7
Consider the matrix:
A=
0
2
3
1
Diagonalize the matrix A, and find the matrix power An . Multiply out the matrix
so that each element has a concrete expression.
Answer. Diagonalization:
1
AP = P D, P =
1
3
−2
, D=
3
0
0
−2
, P −1 =
1
5
2
1
3
−1
(The details of finding the diagonalization are similar to those in the previous
problem and therefore omitted.) Matrix Power:
n
1
3
3
0
2
3
An = P Dn P −1 =
/5
1 −2
0 (−2)n
1 −1
n
3
3(−2)n
2
3
=
/5
3n −2(−2)n
1 −1
1 2 · 3n + 3(−2)n 3 · 3n − 3(−2)n
.
=
5 2 · 3n − 2(−2)n 3 · 3n + 2(−2)n
Problem 8
Find the eigenvalues and *all* eigenvectors for the following matrices:


8 −2 −4
−7 −8
 3
1 −2  ,
18 17
6 −2 −2
(We leave you to find the eigenvalues of the two-by-two matrix, but we tell you
that the eigenvalues of the three-by-three matrix are 2, 2, 3, i.e., 2 is a double
eigenvalue.) Then use the eigenvalues and eigenvectors to diagonalize the given
matrices if possible. If not possible, explain why not.
6
REVIEW FOR MT2
Answer. Since

6
rref  3
6
−2
−1
−2
 
−4
1
−2  =  0
−4
0
(note two nonpivot columns) the eigenvectors



1
s 3  + t
0

−1/3 −2/3
0
0 ,
0
0
for λ = 2 are all vectors of the form

2
0 
3
where not both s and t are zero. Since

 
5 −2 −4
1 0
rref  3 −2 −2  =  0 1
6 −2 −5
0 0

−1
−1/2  ,
0
(note just one nonpivot column) the eigenvectors for λ = 3 are all vectors


2
t 1 
2
for t 6= 0. We have enough eigenvectors because the total number of nonpivot
columns we found in the course of analyzing all the eigenvalues was three, matching
the size of the matrix we are diagonalizing, which is three by three. So we get a
diagonalization:


 


8 −2 −4
1 2 2
1 2 2
2 0 0
 3
1 −2   3 0 1  =  3 0 1   0 2 0 
6 −2 −2
0 3 2
0 3 2
0 0 3
Eigenvalues for the two-by-two are 5, 5, i.e., we have a double root. (We skip
details of finding and solving the characteristic polynomial since we’ve been there.)
Since
−12 −8
1 2/3
rref
=
18 12
0
0
and there is just one nonpivot column. The eigenvectors for the eigenvalue 5 are
2
t
−3
for t 6= 0. The eigenspace is only one-dimensional. The total number of nonpivots
discovered in the course of trying to find eigenvectors, namely one, does not match
the size of the matrix, which is two-by-two. We can’t diagonalize.
For more information on what happens when diagonalization is not possible,
proceed to an advanced linear algebra course to learn about Jordan canonical form.
Problem 9
Approximate the solution of the initial value problem
y 0 = t − y 3 /4, y(1) = 1
on the interval 1 ≤ t ≤ 1.4 by using Euler’s method with 2 steps.
REVIEW FOR MT2
7
Answer. Let F (t, y) = t − y 3 /4. We have step-size h = (1.4 − 1)/2 = 0.2. Here is
the answer in nice neat tabular form:
k
0
1
2
t k yk
1
1
1.2 1.150 = 1 + F (1, 1) · 0.2 = 1 + (0.75)(0.2)
1.4 1.314 = 1.150 + F (1.2, 1.150) · 0.2 = 1.150 + (0.8198)(0.2)
Problem 10
Solve the following initial value problem:
y 00 + 6y 0 + 34y = 663 cos(7t), y(0) = 0, y 0 (0) = 63
Identify the steady-state and transient parts of the solution clearly. Express the
steady-state part of the solution in the form A cos(ωt − δ), where A > 0, ω > 0,
δ ≥ 0 and δ < 2π. Numerical approximations to four decimal places of accuracy
are requested for A, ω and δ.
Answer.
r2 + 6r + 34 = (r + 3)2 + 52 = 0 ⇒ r = −3 ± 5i.
yh = C1 e−3t cos(5t) + C2 e−3t sin(5t)
yp = A cos(7t) + B sin(7t)
yp00 + 6yp0 + 34yp = (−15A + 42B) cos(7t) + (−42A − 15B) sin(7t)
−15
42 663
1 0 −5
rref
=
−42 −15
0
0 1 14
y = yp + yh = −5 cos(7t) + 14 sin(7t) + C1 e−3t cos(5t) + C2 e−3t sin(5t)
y(0) = −5 + C1 = 0 ⇒ C1 = 5
y 0 (0) = 7 · 14 − 3C1 + 5C2 = 63 ⇒ 83 + 5C2 = 63 ⇒ 5C2 = −20 ⇒ C2 = −4.
y = −5 cos(7t) + 14 sin(7t) + 5e−3t cos(5t) − 4e−3t sin(5t)
|
{z
} |
{z
}
transient
steady
state
p
14
A = (−5)2 + 142 = 14.8660, δ = arctan( −5
) + π = 1.9138
(note that the point (−5, 14) is in quadrant II)
y = 14.8660 cos(7t − 1.9138) + 5e−3t cos(5t) − 4e−3t sin(5t)
|
{z
} |
{z
}
transient
steady state
Problem 11
Find an initial value problem of the form
y 00 + by 0 + ky = A cos(ωt − δ), y(0) = y0 , y 0 (0) = v0
of which
cos(7t) − 3 sin(7t) + 2e−3t cos(5t) − 3e−3t sin(5t)
is the solution. Identify the stationary and transient parts of the solution, and
determine the constants b, k, A, ω, δ, y0 and v0 .
8
REVIEW FOR MT2
Answer. You can find y0 and v0 just by plugging into the given function and its
derivative:
−3t
−3t
y0 = cos(7t) − 3 sin(7t) + 2e
=3
cos(5t) − 3e
sin(5t) t=0
d
v0 =
= −42
cos(7t) − 3 sin(7t) + 2e−3t cos(5t) − 3e−3t sin(5t) dt
t=0
Transient part of the given solution:
2e−3t cos(5t) − 3e−3t sin(5t)
We will call this part of the solution yh . To get the indicated transient part we
need characteristic roots −3 ± 5i. These roots are the solutions of the equation
(r + 3)2 + 52 = r2 + 6r + 34 = 0.
Thus we have
b = 6, k = 34.
We emphasize that the coefficients b and k have been cooked up so that
yh00 + 6yh0 + 34yh = 0.
Stationary (steady state) part of given solution:
cos(7t) − 3 sin(7t).
We will call this part of the given solution yp . Now plug yp into the left side of the
equation we have already figured out for yh . After some algebra we have omitted,
we get
yp00 + 6yp0 + 34yp = −141 cos(7t) + 3 sin(7t).
This must be equal to the right hand side of the equation we are looking for. Thus,
ω = 7,
p
A = 1412 + 32 = 141.0319113 (just a wee bit greater than 141),
3
+ π = 3.120319268 (just a wee bit less than π).
δ = arctan
−141
FINAL ANSWER:
y 00 +6y 0 +34y = −141 cos(7t)+3 sin(7t) = 141.03 cos(7t−3.1203), y(0) = 3, y 0 (0) = −42.
Problem 12
Solve the initial value problem y 00 + 64y = 8 sin(8t), y(0) = 5, y 0 (0) = −3.
Solution.
yh = C1 cos(8t) + C2 sin(8t)
“Trouble” ahead.
yp = At cos(8t) + Bt sin(8t) = t(A cos(8t) + B sin(8t))
Note that (f g)0 = f 0 g + f g 0 applied twice gives (f g)00 = f 00 g + 2f 0 g 0 + f g 00 , hence
yp00 = 0 + 2(−8A sin(8t) + 8B cos(8t)) + t(−64 cos(8t) − 64 sin(8t))
yp00 + 64yp = 16B cos(8t) − 16A sin(8t)
B = 0 and −16A = 8 ⇒ A = −1/2.
yp = −t cos(8t)/2.
y = yp + yh = −t cos(8t)/2 + C1 cos(8t) + C2 sin(8t)
y(0) = C1 = 5
REVIEW FOR MT2
9
y 0 (0) = −1/2 + 8C2 = −3 ⇒ C2 = −5/16.
y = −t cos(8t)/2 + 5 cos(8t) − 5 sin(8t)/16

1
 2
Are the vectors 
 3
4
one of the vectors as a
Problem 13


10
5


  6 
,  , and  11  linearly independent? If not, express
 12 
  7 
13
8
linear combination of the others.
 

Answer. The equations to be analyzed are


5
1
 6
 2 


x
 3 +y 7
8
4

 
0
10
 11   0

 
+z
 12  =  0

0
13




.

If these equation have only the solution (x, y, z) = (0, 0, 0) then the answer is “yes,
linearly independent” and otherwise the answer is “no, linearly dependent.” Well,
we have

 

1 0 −5/4 0
1 5 10 0
 2 6 11 0   0 1
9/4 0 

 
rref 
 3 7 12 0  =  0 0
0 0 
0 0
0 0
4 8 13 0
So indeed no, the vectors are linearly dependent because there is some solution
(x, y, z) of the equations above different from (0, 0, 0). Reading off one of these
solutions different from (0, 0, 0) from the reduced row echelon matrix, we find that



1


5
 2 − 9


4 3
4
4
 
5
10

6 
 +  11
7   12
8
13



0
  0 
=

  0 .
0
Finally, we can rearrange the preceding equation to get the equation


10
 11 
5



 12  = − 4 
13



1

2 
+ 9
3  4
4

5
6 

7 
8
which expresses the last of the vectors as a linear combination of the preceding two.
10
REVIEW FOR MT2
Problem 14


 
1
1
−5
 1   0 
 3 


 

Express the vector 
 22  as a linear combination of  2 ,  3 , and
1
1
5



−5
−1


 1 

 if possible. If not possible, explain why. Then same question with  3 
 22 
 4 
5
1


−5
 3 

replaced by 
 22  (only a tiny change).
4


Answer. To answer the first part of the question we have to solve the equations

 





−5
−1
1
1
 1   3 
 0 
 1 

 




x
 2  + y  3  + z  4  =  22 
5
1
1
1

 

1 0 0 −2
1 1 −1 −5

 1 0
2 
1
3 
 and thus
= 0 1 0
Well, we have rref 


 2 3
0 0 1
5 
4 22
0 0 0
0
1 1
1
5








−1
1
1
−5
 1 
 0 
 1 
 3 








 22  = −2  2  + 2  3  + 5  4  .
1
1
1
5
To answer the second part of the question we have to solve the equations

 


 

−5
−1
1
1
 1   3 
 0 
 1 

 


 
x
 2  + y  3  + z  4  =  22  .
1
1
4
1




1 1 −1 −5
1 0 0 0
 1 0


1
3 
 =  0 1 0 0 . The last equation
Well, we have rref 
 2 3


4 22
0 0 1 0 
1 1
1
4
0 0 0 1
here is 0 = 1 which can’t be solved, so no linear combination of the requested type
can be formed.