LECTURE 23: SYSTEMS OF ODES
MINGFENG ZHAO
November 04, 2015
Introduction to systems of ODEs
In this course, we only study the linear system in the following form:

 x0 = ax1 + bx2 + f1 (t)
1
 x0 = cx + dx + f (t)
1
2

Let ~x(t) = 
x1 (t)


,
A=
x2 (t)
a
b
c
d

,
2

and f~(t) = 
2
f1 (t)

, then we have
f2 (t)
~x0 = A~x + f~(t) .
The matrix A is called the coefficient matrix of the system, and f~(t) is called the non-homogeneous term of the
system. If f~(t) ≡ 0, we say ~x0 = A~x is homogeneous.
Theorem 1. Let ~xc (t) be the general solution to the homogeneous system ~x0 = A~x, and ~xp (t) be a particular solution
to ~x0 = A~x + f~(t), then the general solution to ~x0 = A~x + f~(t) is:
~x(t) = ~xc (t) + ~xp (t).
Decoupled system

Definition 1. Let A = 
a
b
c
d

, we say ~x0 = A~x + f~(t) is a decoupled system if bc = 0.
For a decoupled system, we can solve x1 and x2 separately. For example, if b = 0, then we have
(1)
x01
= ax1 + f1 (t)
(2)
x02
= cx1 + dx2 + f2 (t).
1
2
MINGFENG ZHAO
For the first equation (1), it’s a first order linear equation of x1 , by using the method of variation of parameters,
we can solve x1 from (1). Then plug x1 (t) into (2), we get a first order linear equation of x2 , by using the method of
integrating factors or variation of parameters, we can solve x2 .

 x0 = x1 ,
1
Example 1. Solve the system
.
 x0 = x − x .
1
2
2
C1 t
Since x01 = x1 , then x1 (t) = C1 et . Since x02 = x1 − x2 = −x2 + C1 et , solve x2 , we get x2 =
e + C2 e−t . Then the
2

 x0 = x1 ,
1
general solution to the system
is:
 x0 = x − x .
1
2

~x(t) = 
2

C1 et
C1 t
2 e
+ C2 e
−t

 = C1 
et
1 t
2e


0
 + C2 
e
−t

.
Method of Elimination
Let’s solve the system:
(3)
x01
= ax1 + bx2 + f1 (t)
(4)
x02
= cx1 + dx2 + f2 (t).
Take the derivative of the first equation (3), then
x001 = ax01 + bx02 + f10 (t).
(5)
By (4) and plug x02 into (5), then
x001
= ax01 + b (cx1 + dx2 + f2 (t)) + f10 (t)
= ax01 + bcx1 + bdx2 + bf2 (t) + f10 (t)
(6)
By (3), we know that bx2 = x01 − ax1 − f1 (t), plug bx2 into (6), we get
x001
=
ax01 + bcx1 + d (x01 − ax1 − f1 (t)) + bf2 (t) + f10 (t)
=
(a + d)x01 − (ad − bc)x1 + f10 (t) − df1 (t) + bf2 (t).
That is, we have
(7)
x001 − (a + d)x01 + (ad − bc)x1 = f10 (t) − df1 (t) + bf2 (t).
LECTURE 23: SYSTEMS OF ODES
3
The equation (7) is a constant coefficient of second order equation, we can use undetermined coefficients or variation
of parameters to solve x1 from (7), that is, there exists some functions y1 (t), y2 (t)(t) and x1,p (t) such that
x1 (t) = C1 y1 (t) + C2 y2 (t) + x1,p (t).
Now by taking the derivative of the second equation (4), and by using the same argument as before, we know that
x2 satisfies:
(8)
x002 − (a + d)x02 + (ad − bc)x2 = f20 (t) − bf2 (t) + cf1 (t).
The equation (8) is a constant coefficient of second order equation, we can use undetermined coefficients or variation
of parameters to solve x2 from (8). Since the homogeneous equations for (7) and (8) are the same, then there exists
some function x2,p (t) such that
x2 (t) = D1 y1 (t) + D2 y2 (t) + x2,p (t).
Then plug x1 (t) and x2 (t) into (3) and (4) to find the relations of C1 , C2 , D1 and D2 . In the end, we only have two
free constants.
Remark 1. Form the equation (3) to the equation (6), this is just a necessary condition but not necessary, so any
solution x1 (t), x2 (t) to the system (3) and (4) must satisfy (7) and (8), but the converse is not true, so we must plug
the general solutions to (7) and (8) into the system to find the solutions.

 x0 = x1 + x2 + 2,
1
.
Example 2. Find the general solution to the system
 x0 = −x − x
1
2
2
Since x01 = x1 + x2 + 2, then x001 = x01 + x02 . Since x02 = −x1 − x2 , then x001 = x01 − x1 − x2 . Since x01 = x1 + x2 + 2,
then x2 = x01 − x1 − 2, which implies that
x001 = x01 − x1 − (x01 − x1 − 2) = 2.
By using undetermined coefficients, we know that x1 (t) = C1 + C2 t + t2 .
On the other hand, since x02 = −x1 − x2 , then x002 = −x01 − x02 . Since x01 = x1 + x2 + 2, then x002 = −(x1 + x2 + 2) − x02 =
−x02 − x2 − x1 − 2. Since x02 = −x1 − x2 , then x1 = −x02 − x2 , which implies that
x002 = −x02 − x2 + x02 + x2 − 2 = −2.
4
MINGFENG ZHAO
By using undetermined coefficients, we know that x2 (t) = D1 + D2 t − t2 . Plug x1 (t) = C1 + C2 t + t2 and x2 (t) =

 x0 = x1 + x2 + 2,
1
D1 + D2 t − t2 into the system
, then
 x0 = −x − x
1
2
2
C2 + 2t = C1 + C2 t + t2 + D1 + D2 t − t2 + 2,
and D2 − 2t = −C1 − C2 t − t2 − D1 − D2 t + t2 .
That is, we have
C2 + 2t = (C2 + D2 )t + C1 + D1 + 2,
and D2 − 2t = −(C2 + D2 )t − C1 − D1 .
Hence we get
C2 = C1 + D1 + 2,
2 = C2 + D2 ,
D2 = −C1 − D1 ,
−2 = −C2 − D2 .
So we know that
D1
= C2 − C1 − 2
D2
=
2 − C2
So we know that x1 (t) = C1 + C2 t + t2 and x2 (t) = C2 − C1 − 2 + (2 − C2 )t − t2 , that is,




x1
C 1 + C 2 t + t2

 = 

2
x2
−C1 + C2 − 2 + (2 − C2 )t − t


C1 + C2 t + t2

= 
−C1 + (1 − t)C2 − 2 + 2t − t2



 

1
t
t2
 + C2 
+

= C1 
−1
1−t
−2 + 2t − t2
Homogeneous system
Theorem 2 (Superposition). Let ~y and ~z be two solutions to ~x0 = A~x, then w
~ = C1 ~y +C2 ~z is also a solution to ~x0 = A~x.
Definition 2. Let ~x1 (t), · · · , ~xn (t) be n many vector functions, we say that ~x1 (t), · · · , ~xn (t) are linearly dependent if
there exist some constants C1 , · · · , Cn such that
• not all of Ci ’s are zero, that is, there exists some 1 ≤ i0 ≤ n such that Ci0 6= 0.
• C1 ~x1 (t) + · · · + Cn ~xn (t) = 0.
If ~x1 (t), · · · , ~xn (t) are not linearly dependent, then we say they are linearly independent.
LECTURE 23: SYSTEMS OF ODES
5
Remark 2. It’s easy to see that for two vectors ~x(t) and ~y (t) are linearly dependent if and only if either ~x(t) is a
constant multiple of ~y (t), or either ~y (t) is a constant multiple of ~x(t).
Theorem 3. Let ~y (t) and ~z(t) be two linearly independent solutions to ~x0 = A~x, then the general solution to ~x0 = A~x is
~x(t) = C1 ~y (t) + C2 ~z(t).



 a

 : a, b ∈ R2 is a finite dimensional vector space and dim A = 2; but the space
Remark 3. The space A := 
 b




 a(t)

 : a(t) and b(t) are functions is an infinite dimensional vector space and dim B = ∞. It’s easy to see
B := 
 b(t)


 
 



1
t
t2
t100
, 
, 
, · · · , 
, · · · are linearly independent.
that the vectors 
0
0
0
0
Problems you can do:
Lebl’s Book [2]: All exercises on Page 88 and Page 89.
Braun’s Book [1]: Read all materials in Section 3.1, Section 3.2, Section 3.3, Section 3.5, Section 3.6 and
Section 3.7.
References
[1] Martin Braun. Differential Equations and Their Applications: An Introduction to Applied Mathematics. Springer, 1992.
[2] Jiri Lebl. Notes on Diffy Qs: Differential Equations for Engineers. Createspace, 2014.
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca