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6.4 Homogeneous Systems with Constant Coefficients
A homogeneous system with constant coefficients is a linear differential system having the
form
x
′ 1 =^ a^11 x^1 +^ a^12 x^2 +^ · · ·^ +^ a^1 nxn
x
′ 2 =^ a^21 x^1 +^ a^22 x^2 +^ · · ·^ +^ a^2 nxn
. . .
x
′ n =^ an^1 x^1 +^ an^2 x^2 +^ · · ·^ +^ annxn
where a 11 , a 12 ,... , ann are constants. The system in vector-matrix form is
x′ 1
x
′ 2
−
x
′ n
a 11 a 12 · · · a 1 n
a 21 a 22 · · · a 2 n
an 1 an 2 · · · ann
x 1
x 2
xn
or x
′ = Ax. (1)
Example 1. Consider the 3
rd order linear homogeneous differential equation
y
′′′
′′ − 5 y
′ − 6 y = 0.
The characteristic equation is:
r
3
2 − 5 r − 6 = (r − 2)(r + 1)(r + 3) = 0
and {e
2 t , e
−t , e
− 3 t } is a solution basis for the equation.
The corresponding linear homogeneous system is
x
′
x
and
v 1 (t) =
e
2 t
2 e
2 t
4 e
2 t
=^ e
2 t
is a solution vector (see Problem 14, Exercises 6.3). Similarly,
v 2 (t) =
e
−t
−e
−t
e
−t
=^ e
−t
and^ v 3 (t) =
e
− 3 t
− 3 e
− 3 t
9 e
− 3 t
=^ e
− 3 t
are solution vectors.
Solutions: Eigenvalues and Eigenvectors
Example 1 suggests that homogeneous systems with constant coefficients might have
solution vectors of the form v(t) = e
λt c, for some number λ and some constant vector
c.
Set v(t) = e
λt c. Then v
′ (t) = λe
λt c. Substituting into (1), we get:
λe
λt c = Ae
λt c which implies Ac = λ c.
The latter equation is an eigenvalue-eigenvector equation for A. Thus, we look for
solutions of the form v(t) = e
λt c where λ is an eigenvalue of A and c is a corresponding
eigenvector.
Example 2. Returning to Example 1, note that
=^ −^1
and (^)
=^ −^3
2 is an eigenvalue of A =
with corresponding eigenvector
,^ −^1
is an eigenvalue of A with corresponding eigenvector
,^ and^ −^3 is an eigenvalue
of A with corresponding eigenvector
.^
Example 3. Find a fundamental set of solution vectors of
x
′
x
and give the general solution of the system.
SOLUTION First we find the eigenvalues:
det(A − λI) =
1 − λ 5
3 3 − λ
= (λ − 6)(λ + 2).
As you can check, corresponding eigenvectors are:
c 1 =
,^ c 2 =
,^ c 3 =
A fundamental set of solution vectors is:
v 1 (t) = e
2 t
,^ v 2 (t) =^ e
t
,^ v 3 (t) =^ e
−t
since distinct exponential vector-functions are linearly independent (calculate the Wronskian
to verify) and
x(t) = C 1 e
2 t
+^ C 2 e
t
+^ C 3 e
−t
is the general solution.
To find the solution vector satisfying the initial condition, solve
C 1 v 1 (0) + C 2 v 2 (0) + C 3 v 3 (0) =
which is:
C 1
+^ C 2
+^ C 3
or (^)
C 1
C 2
C 3
Note: The matrix of coefficients here is the fundamental matrix evaluated at t = 0
Using the solution method of your choice (row reduction, inverse, Cramer’s rule), the
solution is: C 1 = 3, C 2 = − 1 , C 3 = 1. The solution of the initial-value problem is
x(t) = 3e
2 t
−^ e
t
+^ e
−t
.^
Two Difficulties
There are two difficulties that can arise:
- A has complex eigenvalues.
If λ = a + bi is a complex eigenvalue of A with corresponding (complex) eigenvector
u + i v, then λ = a − bi (the complex conjugate of λ) is also an eigenvalue of A and
u − i v is a corresponding eigenvector. The corresponding linearly independent complex
solutions of x
′ = Ax are:
w 1 (t) = e
(a+bi)t (u + i v) = e
at (cos bt + i sin bt)(u + i v)
= e
at [(cos bt u − sin bt v) + i(cos bt v + sin bt u)]
w 2 (t) = e
(a−bi)t (u − i v) = e
at (cos bt − i sin bt)(u − i v)
= e
at [(cos bt u − sin bt v) − i(cos bt v + sin bt u)]
Now
x 1 (t) =
1 2 [w^1 (t) +^ w^2 (t)] =^ e
at (cos bt u − sin bt v)
and
x 2 (t) =
1 2 i
[w 1 (t) − w 2 (t)] = e
at (cos bt v + sin bt u)
are linearly independent solutions of the system, and they are real-valued vector functions.
Note that x 1 and x 2 are simply the real and imaginary parts of w 1 (or of w 2 ).
(Review Section 3.3 where you were shown how to convert complex exponential solutions
into real-valued solutions involving sine and cosine.)
Example 5. Determine the general solution of
x
′
x.
SOLUTION
det(A − λI) =
2 − λ − 5
1 −λ
= λ
2 − 2 λ + 5.
The eigenvalues are: λ 1 = 1 + 2i, λ 2 = 1 − 2 i. The corresponding eigenvectors are:
c 1 =
1 + 2i
, c 2 =
1 − 2 i
Now
e
(1+2i)t
[(
)]
e
t (cos 2t + i sin 2t)
[(
)]
e
t
[
cos 2t
− sin 2t
)]
t
[
cos 2t
)]
A fundamental set of solution vectors for the system is:
v 1 (t) = e
2 t
,^ v 2 (t) =^ e
2 t
cos 3t
−^ sin 3t
v 3 (t) = e
2 t
cos 3t
+ sin 3t
.^
- A has an eigenvalue of multiplicity greater than 1
We’ll treat the case where A has an eigenvalue of multiplicity 2.
Example 7. Determine a fundamental set of solution vectors of
x
′
x.
SOLUTION
det(A − λI) =
1 − λ − 3 3
3 − 5 − λ 3
6 − 6 4 − λ
= −λ
3
- 12λ − 16 = −(λ − 4)(λ + 2)
2 .
The eigenvalues are: λ 1 = 4, λ 2 = λ 3 = −2.
As you can check, an eigenvector corresponding to λ 1 = 4 is c 1 =
We’ll carry out the details involved in finding an eigenvector corresponding to the “dou-
ble” eigenvalue −2.
[A − (−2)I]c =
c 1
c 2
c 3
The augmented matrix for this system of equations is
which row reduces to
The solutions of this system are: c 1 = c 2 − c 3 , c 2 , c 3 arbitrary. We can assign values to
c 2 and c 3 independently and obtain two linearly independent eigenvectors. For example,
setting c 2 = 1, c 3 = 0, we get the eigenvector c 2 =
. Reversing the roles, we set
c 2 = 0, c 3 = − 1 to get the eigenvector c 3 =
. Clearly^ c 2 and^ c 3 are linearly
independent. You should understand that there is nothing magic about our two choices for
c 2 , c 3 ; any choice which produces two independent vectors will do.
The important thing to note here is that this eigenvalue of multiplicity 2 produced two
independent eigenvectors.
Based on our work above, a fundamental set of solutions for the differential system
x
′
x
is
v 1 (t) = e
4 t
,^ v 2 (t) =^ e
− 2 t
,^ v 3 (t) =^ e
− 2 t
.^
Example 8. Let A =
det(A − λI) =
−λ 1 0
0 −λ 1
12 8 − 1 − λ
= −λ
3 − λ
2
- 8λ − 12 = −(λ − 3)(λ + 2)
2 .
The eigenvalues are: λ 1 = 3, λ 2 = λ 3 = −2.
As you can check, an eigenvector corresponding to λ 1 = 3 is c 1 =
We’ll carry out the details involved in finding an eigenvector corresponding to the “dou-
ble” eigenvalue −2.
[A − (−2)I]c =
c 1
c 2
c 3
The augmented matrix for this system of equations is
which row reduces to
The appearance of the te
− 2 t c 2 term should not be unexpected since we know that a
characteristic root r of multiplicity 2 produces a solution of the form tert.
You can check that v 3 is independent of v 1 and v 2. Therefore, the solution vectors
v 1 , v 2 , v 3 are a fundamental set of solutions of the system.
The question is: What is the significance of the vector w =
? How is it related
to the eigenvalue − 2 which generated it, and to the corresponding eigenvector?
Let’s look at [A − (−2)I]w = [A + 2I]w:
[A + 2I]w =
=^ c 2 ;
A − (−2)I “maps” w onto the eigenvector c 2. The corresponding solution of the system
has the form
v 3 (t) = e
− 2 t w + te
− 2 t c 2
where c 2 is the eigenvector corresponding to − 2 and w satisfies
[A − (−2)I]w = c 2.
General Result
Given the linear differential system x
′ = Ax. Suppose that A has an eigenvalue λ of
multiplicity 2. Then exactly one of the following holds:
- λ has two linearly independent eigenvectors, c 1 and c 2. Corresponding linearly
independent solution vectors of the differential system are v 1 (t) = e
λt c 1 and v 2 (t) =
e
λt c 2.
- λ has only one (independent) eigenvector c. Then a linearly independent pair of
solution vectors corresponding to λ are:
v 1 (t) = e
λt c and v 2 (t) = e
λt w + te
λt c
where w is a vector that satisfies (A − λI)w = c. The vector w is called a
generalized eigenvector corresponding to the eigenvalue λ.
Example 9. Find a fundamental set of solution vectors for x
′
x.
SOLUTION
det(A − λI) =
1 − λ − 1
1 3 − λ
= λ
2 − 4 λ + 4 = (λ − 2)
2 .
Characteristic values: λ 1 = λ 2 = 2.
Characteristic vectors:
(A − 2 I)c =
c 1
c 2
The solutions are: c 1 = −c 2 , c 2 arbitrary; there is only one eigenvector. Setting c 2 = −1,
we get c =
The vector v 1 = e
2 t
is a solution of the system.
A second solution, independent of v 1 is v 2 = e
2 t w + te
2 t c where w is a solution of
(A − 2 I)z = c:
(A − 2 I)z =
z 1
z 2
The solutions of this system are z 1 = − 1 − z 2 , z 2 arbitrary. If we choose z 2 = 0 (any
choice for z 2 will do), we get z 1 = − 1 and w =
. Thus
v 2 (t) = e
2 t
2 t
is a solution of the system independent of v 1. The solutions
v 1 (t) = e
2 t
, v 2 (t) = e
2 t
2 t
are a fundamental set of solutions of the system.
Example 10. Let A =
. Find a fundamental set of solutions of
x
′ = Ax
SOLUTION
The solutions of this system are
z 3 = −1 + 2z 2 , z 1 = 1 − z 2 + z 3 = 1 − z 2 + (−1 + 2z 2 ) = z 2 , z 2 arbitrary.
If we choose z 2 = 0 (any choice for z 2 will do), we get z 1 = 0, z 2 = 0, z 3 = − 1 and
w =
. Thus
v 3 = e
2 t
+^ te
2 t
is a solution of the system independent of v 2 (and of v 1 ). The solutions
v 1 = e
t
,^ v 2 =^ e
2 t
,^ v 3 =^ e
2 t
+^ te
2 t
are a fundamental set of solutions of the system.
Exercises 6.
Find the general solution of the system x
′ = Ax where A is the given matrix. If an
initial condition is given, also find the solution that satisfies the condition.
, x(0) =
, x(0) =
,^ x(0) =
,^ x(0) =
,^ x(0) =
,^ x(0) =
