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MATH 2270
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Matthew Demers
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Mathematics

MATH 2270

Matthew Demers

Fall

Description

1
6.4 Examples for Systems of First-Order Homogeneous Lin-
ear Equations
Now, we solve a few systems.
Example:
Find the general solution to the system
!
x = 3 1 x: (6.55)
12 ▯1
First, we need to ▯nd the eigenvalues of the matrix of coe▯cients in this system. The
eigenvalues are given by the characteristic equation:
(3 ▯ ▯)(▯1 ▯ ▯) ▯ 12 = 0
▯3 ▯ 2▯ + ▯ ▯ 12 = 0
▯ ▯ 2▯ ▯ 15 = 0
(▯ ▯ 5)(▯ + 3) = 0:
Thus, we obtain ▯1= 5 and ▯ 2 ▯3.
Now, ▯nding eigenvectors, we ▯rst consider 1 , which was 5. Substituting this
into the (A - I1 )x = 0, we obtain (in augmented matrix form):
!
▯2 1 0
(6.56)
12 ▯6 0
which reduces to
!
▯2 1 0
(6.57)
0 0 0
One row of zeroes gives us one free variable to work with, and thus one eigenvector.
1
Indeed, we ▯nd that if 2 = t, then ▯2x1+ t = 0, or 1 = 2t. Thus, we obtain the
eigenvector
!
1
V1= : (6.58)
2 2
Now, consider the other eigenvalue2 ▯ = ▯3. Subbing into (A - 1▯ )x = 0 and
reducing gives the following:
!
6 1 0
: (6.59)
0 0 0
We obtain an eigenvector of
!
V = ▯1 : (6.60)
2
6
Therefore, our solution is
! ! !
y1 1 5t ▯1 ▯3t
= C 1 e + C 2 e : (6.61)
y2 2 6
Example 2:
Find the general solution to the system
!
▯1 2
x = x: (6.62)
▯2 ▯2
The characteristic equation is
(▯1 ▯ ▯)(▯2 ▯ ▯) + 4 = 0
2
2 + 3▯ + ▯ + 4 = 0
2
▯ + 3▯ + 6 = 0 p
3 15
▯ = ▯ ▯ j
2 2
Now we must ▯nd eigenvectors! We have a complex pair, so we’ll take the eigen-
value corresponding to the \plus" case and work with it. Substituting this into the
(A - I▯1)x = 0, and reducing, we get:
p !
1▯ 1j 2 0
2 2 : (6.63)
0 0 0 3
If we let2x = t, then
p
1 15
(2 ▯ 2 j)x1+ 2t = 0; (6.64)
or solving fo1 x ,
▯4
x1= p t: (6.65)
1 ▯ 15j
Put the imaginary component in the numerator, by multiplying the top and bot-
tom by the conjugate of the denominator. We obtain
p
x = ▯4(1 + 15j) (6.66)

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