Download Math 2250 Lab 8 Name/Unid: Due Date: 3/06/2014 Class ID: Section:

Math 2250 Lab 8
Name/Unid:
Due Date: 3/06/2014
Class ID:
Section:
The problems for this week’s lab have been constructed to show how linear algebra concepts from Chapter 4 relate to linear differential equations concepts in Chapter 5.
References: Edwards-Penney Sections 4.3, 4.4, 5.1, 5.2.
1. Consider the three functions
y1 (x) = cos(4x),
y2 (x) = sin(4x),
y3 (x) = sin(4x − π6 )
(a) Show that all three functions solve the differential equation y 00 (x) + 16 y = 0
Solution: For y3 (x) = sin(4x − π6 ), we have
y30 (x) = 4 cos(4x −
π
)
6
π
) = −16 y3 (x)
6
Thus, y300 (x) + 16 y3 (x) = −16 y3 (x) + 16 y3 (x) = 0. The computations showing
that y1 and y2 solve the DE are analogous.
y300 (x) = −16 sin(4x −
(b) The differential equation above is a second order linear homogeneous DE, so the
solutions space is 2-dimensional. Thus, the three functions y1 , y2 , and y3 above
must be linearly dependent. Find a linear dependency. Hint: use a trigonometry
angle sum/difference identity.
Solution: The angle sum/difference identity for sine is
sin(a ± b) = sin(a) cos(b) ± cos(a) sin(b)
Thus,
sin(4x −
i.e.
π
π
π
) = sin(4x) cos( ) − cos(4x) sin( )
6
6
6
√
3
1
y3 (x) =
y2 (x) − y1 (x)
2
2
(c) Explicity verify that every initial value problem
y 00 + 16 y = 0
y(0) = a1
y 0 (0) = a2
has a solution of the form
y(x) = c1 cos(4x) + c2 sin(4x)
and that c1 , c2 are uniquely determined by a1 , a2 .
Thus, cos(4x) and sin(4x) are a basis for the solution space of y 00 + 16 y = 0.
In other words, every solution y(x) has initial values that can be matched with a
linear combination of y1 , y2 , but once the initial values match, the solutions must
agree by the Uniqueness Theorem, so y1 , y2 span the solution space. Observe that
y1 , y2 are linearly independent because if c1 cos(4x) + c2 sin(4x) = y(x) ≡ 0, then
y(0) = y 0 (0) = 0 which implies a1 = a2 = 0 and c1 = c2 = 0 by the derived formula
which expresses c1 , c2 in terms of the initial values a1 , a2 .
Solution: For y(x) = c1 cos(4x) + c2 sin(4x), we have y 0 (x) = −4 c1 sin(4x) +
4 c2 cos(4x). Substituting in the initial conditions, we have
y(0) = c1 = a1
y 0 (0) = 4 c2 = a2
Thus, for each choice of initial values a1 , a2 , there are unique linear combination
coefficients c1 = a1 , c2 = a42 so that y(x) = a1 cos(4x) + a42 sin(4x) solves the
initial value problem.
(d) Find by inspection, particular solutions y(x) to the non-homogeneous differential
equations
y 00 (x) + 16 y = −32
y 00 (x) + 16 y = 48 x
Hint: One particular solution may be a constant; the other, a multiple of x.
Solution: If y(x) = A, a constant, then y 00 + 16 y = 16 A so in order for
L(y) = −32, we need A = −2 = yQ1 (x).
If y(x) = B x, then L(y) = 0 + 16 B x so in order for L(y) = 48 x, we choose
B = 3 so yQ2 (x) = 3 x.
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(e) Use the principle of superposition together with your work from parts (c) and (d)
to find the general solution to the non-homogeneous differential equation
y 00 + 16 y = −32 + 48 x
Solution: L(yQ1 + yQ2 ) = L(yQ1 ) + L(yQ2 ) = −32 + 48 x. So a particular
solution for this non-homogeneous DE is yp = −2 + 3 x. Thus, the general
solution is
y(x) = yp + yh = −2 + 3 x + c1 cos(4x) + c2 sin(4x)
.
(f) Solve the following initial value problem, using your general solution from part (e).
Provide a technology answer check.
y 00 + 16 y = −32 + 48 x
y(0) = 0
y 0 (0) = 0
.
Solution:
y(x) = −2 + 3 x + c1 cos(4x) + c2 sin(4x)
y 0 (x) = 3 − 4 c1 sin(4x) + 4 c2 cos(4x)
At x = 0,
y(0) = 0 = −2 + c1
y 0 (0) = 0 = 3 + 4 c2
Thus,
c1 = 2
3
c2 = −
4
and the IVP solution is
y(x) = −2 + 3 x + 2 cos(4x) −
.
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3
sin(4x)
4
2. Consider the 4th order homogenous linear differential equation for y(x)
y 0000 (x) = 0
and let W be the solution space.
(a) Use successive antidifferentiation to solve this differential equation. Interpret your
results using vector space concepts to show that the functions y0 (x) = 1, y1 (x) = x,
y2 (x) = x2 , and y3 (x) = x3 are a basis for W . Thus, the dimension of W is 4.
Solution: y 0000 = 0 −→ y 000 = c1 −→ y 00 = c1 x + c2 −→ y 0 =
c1
2
x2 + c2 x + c3 −→
c1 3 c2 2
x + x + c3 x + c4
6
2
= d3 x3 + d2 x2 + d1 x + d0 (1)
= d3 y3 (x) + d2 y2 (x) + d1 y1 (x) + d0 y0 (x)
y(x) =
(1)
(2)
(3)
Thus, the functions y0 (x) = 1, y1 (x) = x, y2 (x) = x2 , and y3 (x) = x3 span the
solution space. To check linear independence, we set
d3 y3 (x) + d2 y2 (x) + d1 y1 (x) + d0 y0 (x) = 0
Rewriting the above identity and taking successive derivatives on each side,
yields the system of equations
d3 x3 + d2 x2 + d1 x + d0 (1) = 0
3 d3 x2 + 2 d2 x + d1 = 0
6 d3 x + 2 d2 = 0
6 d3 = 0
Using back-substitution, we deduce d3 = 0 −→ d2 = 0 −→ d1 = 0 −→ d0 = 0.
Thus, by definition, y0 (x) = 1, y1 (x) = x, y2 (x) = x2 , and y3 (x) = x3 are
linearly independent. Since these four functions span the solution space and are
linearly independent, they form a basis for the solution space (by definition).
(b) Show that the functions z0 (x) = 2, z1 (x) = x − 4, z2 (x) = 12 (x − 4)2 , and z3 (x) =
1
(x − 4)3 are also a basis for W . Hint: If you verify that these four functions solve
6
the differential equation and are linearly independent, they will automatically span
the 4-dimensional solution space and therefore be a basis.
Solution: Since z0 (x) = 2, z1 (x) = x − 4, z2 (x) = 12 (x − 4)2 , and z3 (x) =
1
(x − 4)3 are all polynomials of degree less than or equal to three, their fourth
6
Page 4
derivatives are all zero. In other words, each of these functions satisfies y 0000 = 0.
To check linear independence, we may compute the Wronskian matrix for these
four functions:
2 x − 4 1 (x − 4)2 1 (x − 4)3 2
6
1
0
(x − 4)2 1
x−4
2
[W (z0 , z1 , z2 , z3 )] = .
0
1
x − 4 0
0
0
0
1
Since the Wronskian, W (z0 , z1 , z2 , z3 ) = 1 6= 0, z0 (x), z1 (x), z2 (x), z3 (x) are linearly independent (see Theorem 5.2.3). Since the solution space is 4-dimensional,
these four functions must also span the space, and thus, be a basis.
(c) Use a linear combination of the solution basis from part (b) in order to solve the
following initial value problem
y 0000 (x) = 0
y(4) = 6
y 0 (4) = 7
y 00 (4) = 8
y 000 (4) = 12
Notice how this basis is adapted to initial value problems at x0 = 4 whereas for an
initial value problem at x0 = 0, the basis in part (a) would have been easier to use.
Solution: The second basis from part (b) is adapted for the initial point x0 = 4
and if we write: y(x) = c0 z0 + c1 z1 + c2 z2 + c3 z3 = c0 (2) + c1 (x − 4) + c2 (x −
4)2 + c3 (x − 4)3 . Taking successive derivatives and evaluating at x0 = 4 yields
y(4) = 2 c0 , y 0 (4) = c1 , y 00 (4) = 2 c2 , y 000 (x) = 6 c3 . Matching initial conditions
gives
y(4) = 2 c0 = 6 −→ c0 = 3
y 0 (4) = c1 = 7
y 00 (4) = 2 c2 = 8 −→ c2 = 4
y 000 (4) = 6 c3 = 12 −→ c3 = 2
Thus, y(x) = 3 (2) + 7 (x − 4) + 4 (x − 4)2 + 2 (x − 4)3
Page 5
3. Application of Linear Algebra to Encryption
Any text message can be translated into a string of numbers by assigning to each letter
a number, corresponding to its position in the alphabet. For example: A=1, J=10,
and Z=26. A blank space between words may be represented by the number 27. An
encoding matrix, E, may be used to further encrypt a message so that only someone
with knowledge of the corresponding decoding matrix, C, may understand the message.
Encoding and decoding matrices are constructed such that EC = I. In other words, the
decoding matrix is the inverse of the encoding matrix: C = E −1 .
(a) Suppose the encoding matrix is


1 1 1
E =  1 2 1 .
1 4 2
Find the decoding matrix C.
Solution: The decoding matrix is


0
2 −1
0 .
C =  −1 1
2 −3 1
To check, confirm EC=I=CE.
(b) Decode the following encoded message.


48 42 48 20 46 24 21
M =  75 64 60 25 73 31 39  .
141 113 93 36 128 50 76
Observe that since the encoding matrix is a 3x3 matrix, the original message has
been split up into seven, 3x1 vectors so the matrix multiplication CM gives the
decoded message. This means that the decoded message must be 21 characters
long, including blank spaces. For example, suppose I wish to encode the following
message: “I STUDY MATH.” Assigning a number to each letter and blank space,
this text message is converted to the following string of numbers: “9 27 19 20 21
4 25 27 13 1 20 8.” Given that
encoding
matrix
3x3,
this string of
 the


 is
 I convert

9
20
25
1
numbers into 4, 3x1 vectors  27 ,  21 ,  27  ,  20 
19
4
13
8
These vectors may then be augmented into the following matrix
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

9 20 25 1
A =  27 21 27 20  .
19 4 13 8
To encode this message we do the following matrix multiplication


 

1 1 1
9 20 25 1
55 45 65 29
EA=  1 2 1   27 21 27 20 = 82 66 92 49 .
1 4 2
19 4 13 8
155 112 159 97
You may use the decoding matrix you found in part (a), to confirm that you receive
the original text message.
Solution: The decoded message is


9 15 27 14 18 12 2
CM =  27 22 12 5 27 7 18  .
12 5 9 1 1 5 1
Translated column by column, the original message is “I LOVE LINEAR ALGEBRA.”
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