The research and paper behind the focus of my senior project: Systems of Differential Equations.

1. Systems of Differential Equations
Joshua Dagenais
12-04-09
Mentor: Dr. Arunas Dagys
1

2. Table of Contents
Introduction
Section 1: Solving Systems of Differential Equations with Distinct Real Eigenvalues
Section 2: Solving Systems of Differential Equations with Complex Eigenvalues
Section 3: Solving Systems of Differential Equations with Repeated Eigenvalues
Section 4: Solving Systems of Nonhomogenous Differential Equations
Section 5: Application of Systems of Differential Equations – Arms Races
Section 6: Application of Systems of Differential Equations – Predator-Prey Model
Conclusion
References
2

3. Introduction
Many laws and principles that help explain the behavior of the natural world are
statements or relations that involve rates at which things change. When explained in
mathematical terms, the relations become equations and that rates become derivatives.
Equations that contain these rates or derivatives are called differential equations. Therefore,
systems of ordinary differential equations arise naturally in laws and principles explaining
behavior of the natural world involving several dependent variables, each of which is a function
of single independent variable. This then becomes a mathematical problem that consists of a
system of two or more differential equations. These systems of differential equations that
describe these laws or principles are called mathematical models of the process (Boyce &
DiPrima, 2001).
A system of first order ordinary differential equations is an interesting mathematical
concept as it combines 2 different studies of mathematics for its use. By dissecting the phrase,
system of first order differential equations, into 2 parts, the 2 different areas of mathematics used
to solve these equations can be found. In the system part of the phrase, it involves linear algebra
to solve the system of equations and in this case the system of equations consists of first order
differential equations. The ladder part of the phrase, first order of differential equations,
indicates that solution strategies for solving them will also be involved when solving systems of
differential equations.
So with linear algebra for systems and differential equations in mind, what other
underlying concepts and skills involved with these mathematical concepts must be learned and
explained to solve systems of differential equations? Well, for the linear algebra aspect of
3

4. solving systems of differential equations, topics that are to mentioned briefly in this paper
include matrices, characteristic equations, roots of the characteristic equations (eigenvalues),
eigenvectors, and the diagonalization of a matrix. For the differential equation part of solving
systems, topics that are discussed include solving first order differential equations, solving
simple diagonal systems ( y '  Dy ), and solutions of the original systems ( x  Cy where
x(t )  keat ). When everything mentioned is put together, solutions of different types are found
for systems of differential equations and with the help of mathematical software such as Maple,
graphs are able to visually represent the answers of these systems (slope fields) to show that
there are actually more than one solution called a family of solutions. Also, depending on the
types of eigenvalues that are found for the system of differential equations, different methods for
solving the systems will be used for eigenvalues that are distinct and real, eigenvalues that are
complex, and eigenvalues that are repeated, all of which graphically represented in a different
manner.
Along with different methods for solving systems of differential equations, methods for
solving homogenous and nonhomogenous systems will be explained to help further the scope of
this subject. Why do we care about solving systems of differential equations? Well, there are
many physical problems that involve a number of separate elements linked together in some
manner such as generic application problems which include spring-mass systems, electrical
circuits, and interconnected tanks that need solutions of systems of differential equations to be
understood and solved. Other, more advanced applications of the theory behind systems of
differential equations include the Predator-Prey Model (Lotka-Volterra Model) and the
Richardson’s Arms Race Model which connects mathematics with concepts that would have
never been able to be explained without such elegant mathematical equations. The Predator-Prey
4

5. Model is a system of nonlinear differential equations (even though it is considered an almost
linear system) and the Arms Race Model that uses systems of differential equations that are
nonhomogenous. Both models are very interesting applications that will be discussed and
explained later on in this paper. Hopefully, this paper will give the reader insight on what
systems of linear differential equations are, how to solve them, how to apply them, and how to
understand and interpret the answers that are derived from problems.
5

6. Section 1: Solving Systems of Differential Equations with Distinct Real Eigenvalues
In this section, we will be solving systems of differential equations where the eigenvalues
found from the characteristic equation are all real and all distinct. In order to do this, we will
first take the system

x1  p11 (t ) x1  ...  p1n (t ) xn  g1 (t )

xn  pn1 (t ) x1  ...  pnn (t ) xn  g n (t )
 
and write it in matrix notation. To do this, we write x1 ,..., xn in vector form:

 x1 
 
x   
 x 
 n
we put the coefficients p11 (t ),..., pnn (t ) in an n x n matrix:
 p11 (t ) p1n (t ) 
 
P(t )   
 p (t ) pnn (t ) 
 n1 
we again write x1 ,..., xn in vector form:
 x1 
 
x  
x 
 n
and write g1 (t ),..., gn (t ) in vector form:
6

7.  g1 (t ) 
 
g (t )   
 g (t ) 
 n 
Therefore, the resulting equation using the above vector and matrix notation is represented by
x  P(t ) x  g (t )
We will first consider homogenous systems where g (t )  0 , thus
x  P(t ) x
To find the general solution of the above system when P(t ) is a 1 x 1 matrix, the system above
reduces to a single first order equation
dx
 px
dt
where the solution is x  ce pt . Therefore, to solve any other systems with second order or
higher, we will look for solutions of the form
x   ert
where  is a column vector instead of a constant c (because we are dealing with solutions to
more than one differential equation thus giving us multiple constants equating to a vector) and r
is an exponent to be solved. Substituting x   ert into both sides of x  P(t ) x gives
r ert  P(t ) ert
Upon canceling ert , we obtain r  P(t ) or
7

8. ( P(t )  rI )  0
where I is the n x n identity matrix. In order to solve ( P(t )  rI )  0 , we will use theorem 1.
Theorem 1: Let A be an n x n matrix of constant real numbers and let X be an n-dimensional
column vector. The system of equations AX  0 has nontrivial solutions, that is, X  0 , if and
only if the determinant of A is zero.
In our case, ( P(t )  rI ) is the n x n matrix represented by A and  is the n-dimensional column
vector represented by X. Therefore, in order to find the nontrivial solutions of ( P(t )  rI )  0 ,
we must take the determinant of ( P(t )  rI )  0 which is represented
p11 (t )  r p1n (t )
0
pn1 (t ) pnn (t )  r
Computing the determinant will yield a characteristic equation, which resembles the structure of
a polynomial of degree n, where the roots of the characteristic equation, eigenvalues denoted by
r, will be computed. After the eigenvalues have been computed, r will be substituted back into
( P(t )  rI )  0 and solved for the nonzero vector,  , which is called the eigenvector of the
matrix P(t ) corresponding to the eigenvalue r1 . The eigenvector will be an n x 1 column vector
that will have as many values as there are equations to solve for. After finding the eigenvalues
and the eigenvectors for those specific values, they will be substituted back into the equation
x   ert
which will be represented as the following specific solutions
8

9.  x11 (t )   x1k (t ) 
   
x (t )  
(1)
 ,..., x (t )  
(k )
 ,...
 x (t )   x (t ) 
 n1   nk 
for the initial system. If the Wronskian of x(1) ,..., x( n) (represented as W [ x(1) ,..., x( n) ] ) does not
equal zero, then the general solutions can be represented as a linear combination of the specific
solutions
x  c1 x(1) (t )   ck x( k ) (t )
The following examples will help illustrate how to solve n x n systems of differential equations
with distinct real eigenvalues. The general solution of the given system of equations will be
solved for along with a graph that shows the direction field of the answer.
Example 1: Solve the following 2 x 2 system for x

x1  3x1  2 x2

x2  2 x1  2 x2
To solve the problem, we rewrite the equations into its matrix form
 3 2 
x   x
 2 2 
which is of the form
x  P(t ) x
where
9

10.  3 2 
P (t )   
 2 2 
We then find the eigenvalues of P(t) by finding the characteristic equation and solving for r.
Therefore,
3 r 2
det( P(t )  rI )   r 2  r  2  (r  1)(r  2)  0
2 2  r
and the eigenvalues of P(t) are r1  1 and r2  2 . Now we compute the eigenvectors for each of
their respective eigenvalues. We will compute the nontrivial solutions of
3 r 2   c1 
    0
 2 2  r  c2 
For r1  1
 3  (1) 2   c1   4 2   c1  4c1  2c2  0
 c   0     c   0  2c  c  0  c2  2c1
 2 2  (1)   2   2 1   2  1 2
(Note that both of the resulting equations with c1 and c2 are the same). One such solution of the
1
equation is found by choosing c1  1 thus making c2  2 to give the eigenvector  1    .
 2
1
Knowing that x( n) (t )   ( n)ernt , it follows that x (1) (t )  c1   e  t is a solution of the initial system.
 2
For r2  2
3 2 2   c1   1 2   c1  c1  2c2  0
 c   0     c   0  2c  4c  0  c1  2c2
 2 2  2   n   2 4  2  1 2
10

11. By choosing c1  2 to solve the equation, c2  1. Proper notation of eigenvectors, if possible,
insists that fractions should be avoided when representing the numerical value of the eigenvalue.
 2
Therefore, for r2  2 ,  2    and a second solution is
1
 2
x (2) (t )  c2   e 2t
1
Now, we check to see if we can represent x1 and x2 as a general solution by taking the Wronskian
of both specific solutions. The Wronskian of x(1) (t ) and x(2) (t ) is
et 2e2t
W [ x (1) , x (2) ]   3et
2et e 2t
which is never equal to zero. It follows that the solutions x(1) (t ) and x(2) (t ) are linearly
independent. Therefore, the general solution of the system x  P(t ) x is
1  2
x(t )  c1   e  t  c2   e 2t
 2 1
11

12. All the general solutions (represented by the family of red lines), a combination of x(1) (t ) and
x(2) (t ) , for which c1  0 and c2  0 , are asymptotic to the line x2  2 x1 . The blue trajectories
represent specific solutions to the system with each trajectory having a different initial value
( x1 (0)  a and x2 (0)  b where a and b are any real number ).
For the remaining examples in this section, the derivation of the final solution will be
shown without all steps shown The purpose of these examples is to show the variety of systems
of differential equations that have distinct real eigenvalues such as a 3 x 3 system and a 2 x 2
system with initial conditions given.
Example 2: Solve the following 3 x 3 system for x

x1  x1  x2  x3  1 1 1
  

x2  2 x1  x2  x3   x   2 1 1  x
x3  8 x1  5 x2  3x3 
 
 8 5 3 
 
First, we find the eigenvalues for the coefficient matrix by the following equation
1 r 1 1
det( P(t )  rI )  2 1  r 1  0
8 5 3  r
and solving the resulting characteristic equation.
>
12

13. >
Using maple yields the eigenvalues
r1  2 , r2  2 , and r3  1
and eigenvectors
 4 0  3
  2   3  
   5  ,    1  ,    4 
1
 7   1  2 
     
The eigenvalues above are the same as the given Maple output but manipulated in properly
format where all values of the eigenvector are integers and the first value is positive. After the
eigenvalues and eigenvectors are computed, we find the Wronskian
W [ x(1) , x(2) , x(3) ]  12et  0 therefore we can substitute all the eigenvalues and eigenvectors
found into x( n)   ( n)ernt and express the solution as a linear combination
 4 0  3
  e 2t  c   e 2t  c   e t
x(t )  x (t )  x (t )  x (t )  c1  5  3  4 
(1) (2) (3)
2 1 
 7   1  2 
     
Example 3: Solve the 2 x 2 system with initial conditions given for x

x1  5 x1  x2 x (0)  2  5 1 2
where 1  x    x where x(0)   

x2  3x1  x2 x2 (0)  1 3 1   1
13

14. We will start off the example by using Maple to find the eigenvalues and eigenvectors of the
coefficient matrix:
>
>
>
 1 1
Therefore, r1  4 ,  1    , r2  2 , and  2    . The Wronskian W [ x(1) , x(2) ]  2e6t  0
 1  3
therefore the specific solutions x (1) and x (2) can be expressed as the general solution
 1 1
x(t )  c1   e 4t  c2   e 2t
 1  3
2
After the general solution has been found, we substitute x(0)    into x(t ) to get
 1
 1 1 2  1 1  2 
x(0)  c1   e 4*0  c2   e 2*0     c1    c2     
 1  3  1  1  3   1
After the equation has been simplified, we multiply c1 and c2 by their respected vectors to yield
the follow system of equations
c1  c2  2 and c1  3c2  1
14

15. 7 3
We then solve the system of equations for c1 and c2 to get c1  and c2  . Substituting
2 2
back into the general solution to get the specific solution of the system as
7  1 3  1 
x(t )    e 4t    e 2t
2  1 2  3
The direction field of general solution along with a trajectory of the specific solution is
represented as
The above direction field shows the different families of solutions for the general solution
denoted by the red arrows and the blue trajectory represents the specific solution to the system
2
for when the initial starting value was x(0)    . Now, after establishing the basis for solving
 1
systems of differential equations, we will now delve into different cases of solving systems
where the eigenvalues are not real and/or distinct.
15

16. Section 2: Solving Systems of Differential Equations with Complex Eigenvalues
In this section, we will use what was previously discussed in the section for solving
systems with real and distinct eigenvalues on how to generate eigenvalues for an n x n system of
linear homogenous equations with constant coefficients denoted as
x  P(t ) x
Now if P(t) is real then the coefficients that make up the characteristic equation for r are real and
any complex eigenvalues must occur in conjugate pairs (Boyce & DiPrima, 2001, p. 384).
Therefore, for a 2 x 2 system, r1  a  bi and r2  a  bi would be eigenvalues where a and b are
real. Also, it follows that the corresponding eigenvectors are complex conjugate pairs of each
other. Therefore, r2  r1 and  2   1 . To help visualize this, take the equation that was formed
in the previous section
( P(t )  rI )  0
and substitute r1 and  1 into the equation to get
( P(t )  r1I ) 1  0
which forms a corresponding general solution to the system. Now, by taking the complex
conjugate of the entire equation, the resulting equation becomes
( P(t )  r1I ) 1  0
where P(t) and I are not affected by the conjugation because they both have all real values. The
equation then forms another corresponding general solution where r2  r1 and  2   1 . Now,
16

17. with the eigenvalues and eigenvectors solved for, we can use Euler’s formula to express a
solution with real and imaginary parts just as real solutions to the system. Euler’s formula states
eit  cos t  i sin t
But, for use with general complex solutions to a system of differential equations, we will use the
a modified version of the formula
e( i )t  e t (cos( t )  i sin( t ))  e t cos( t )  i e t sin( t )
to find the real-value solutions to the system. We can choose either x(1) (t ) or x(2) (t ) to find the 2
real-valued solutions because they are conjugates of each other and both will yield the same real-
valued solutions. Using x(2) (t ) and  2  a  bi where a and b are real, then we have
x(2) (t )  (a  bi)e( i  )t  (a  bi)e t (cos( t )  i sin( t ))
Factoring the above equation results in
x(2) (t )  et (a cos( t )  bi cos( t )  ai sin( t )  b sin( t ))
and separating x(2) (t ) into its real and imaginary parts, x(2) (t ) will yield
x(2) (t )  e t (a cos( t )  b sin( t ))  ie t (a sin( t )  b cos( t ))
If x(2) (t ) is written as the sum of 2 vectors ( x(2) (t )  u(t )  iv(t ) ), then the vectors yielded are
u(t )  e t (a cos( t )  b sin( t )) and v(t )  e t (a sin( t )  b cos( t ))
17

18. We can disregard the i in front of v(t ) because it is considered to be a multiplier of the vector
and we are only interested in the real-numbered vector solution. If we chose to solve for x(1) (t )
instead of x(2) (t ) , we would have gotten the same solution except x(1) (t )  u(t )  iv(t ) . i is also
considered a multiplier of the v(t ) vector therefore we can disregard it and the answers for u (t )
and v(t ) would be the same as the ones that were solved for above. u (t ) and v(t ) are the
resulting real-valued vector solutions to the system.
It is worth mentioning that u (t ) and v(t ) are linearly independent and can be expressed as
a single general solution. Therefore, for r1    i, r2    i and that r3 ,..., rn are all real and
distinct. Let the corresponding eigenvectors be  1  a  bi,  2  a  bi,  3 ,...,  n (Boyce &
DiPrima, 2001, p. 385). Then the general solution to systems of differential equations with
complex eigenvalues is
x(t )  c1u (t )  c 2 v(t )  c3 3e r3t  ...  cn nernt
where u(t )  e t (a cos( t )  b sin( t )) , v(t )  e t (a sin( t )  b cos( t )) , and P(t) consists of all
real coefficients. It is only when P(t) consists of all real coefficients that complex eigenvectors
and eigenvalues will occur in conjugate pairs (Boyce & DiPrima, 2001, p. 385). The following
examples will help illustrate how to solve n x n systems of differential equations with complex
eigenvalues. Both the complex and real-valued solutions will be given for each of the examples
and some direction fields will be shown to demonstrate the nature of systems with complex
eigenvalues.
18

19. Example 1: Solve the following 2 x 2 system for x

x1  3x1  2 x2   3 2 
  x   x

x2  4 x1  x2   4 1 
We will begin the example by using Maple to find the eigenvalues and eigenvectors of the
coefficient matrix:
>
>
>
 1   1 
Therefore, r1  1  2i , r2  1  2i ,  1    , and   
2
 . To get the eigenvectors in
1  i  1  i 
proper form from the Maple output, we multiplied both eigenvalues (resulting from the Maple
output) by its conjugate to get a real number for the first value and then multiplied it again by 2
so that all values in the eigenvector were integers. The Wronskian W [ x(1) , x(2) ]  2e2t  i  0
therefore the specific solutions x (1) and x (2) can be expressed as the general solution in complex
form
 1  (1 2i )t  1  (12i )t
x(t )  c1  e  c2  e
1  i  1  i 
19

20. But, we want to be able to find the real-valued solutions of the complex general solution so we
will use x (1) to find the real-valued vectors. Therefore,
 1  (1 2i )t
x (1) (t )   e
1  i 
Using Euler’s formula, x (1) becomes
 1  t
  e (cos(2t )  i sin(2t ))
1  i 
After Euler’s formula has been applied, we factor the above equation
 cos(2t )  i sin(2t )  t
 e
 cos(2t )  i sin(2t )  i cos(2t )  sin(2t ) 
and separate the real and imaginary elements into
 cos(2t )  t  sin(2t ) 
et    ie  
 sin(2t )  cos(2t )   sin(2t )  cos(2t ) 
The result is the two real-valued solutions of the form u(t )  iv(t ) where
 cos(2t )  t  sin(2t ) 
u (t )  et   and v(t )  e  
 sin(2t )  cos(2t )   sin(2t )  cos(2t ) 
Therefore, the general solution to the system with real-valued solutions is
 cos(2t )  t  sin(2t ) 
x(t )  c1u (t )  c2v(t )  c1et    c2e  
 sin(2t )  cos(2t )   sin(2t )  cos(2t ) 
20

21. The resulting direction field showing families of solutions to the general solution to the system is
The blue trajectories show specific solutions when initial conditions are given. Thus, the
direction field creates spiraled solutions where the origin is the center of the spirals called a
spiral point. The direction of the motion is away from the spiral point and the trajectories
become unbounded. Also, the spiral point, for this particular solution, is unstable. There are
also systems with complex eigenvalues where the general solution has a spiral point that is stable
because all trajectories approach it as t increases.
Example 2: Solve for the following 3 x 3 system for x

x1  x1  1 0 0 
  
x2  2 x1  x2  2 x3   x   2 1 2  x

x3  3x1  2 x2  x3 
 
3 2 1 
 
21

22. Again, we will begin the example by using Maple to find the eigenvalues and eigenvectors of the
coefficient matrix:
>
>
>
Thus, the eigenvalues are r1  1 , r2  1  2i , and r3  1  2i . The simplified eigenvectors are
2 0 0
  2  
   3  ,    i  , and    i  . Notice that r1 and  1 already contain real-values therefore
1 3
 
2 1  1
     
no computations are needed to turn them into real-valued solutions like the other complex
eigenvalues and eigenvectors. The Wronskian W [ x(1) , x(2) , x(3) ]  4e3t  i  0 therefore the
specific solutions x (1) , x (2) , x (3) and can be expressed as the general solution in complex form
2  0 0
  t   (1 2i )t  
x(t )  c1  3  e  c2  i  e  c3  i  e(12i )t
2 1  1
     
To find the real-valued solutions of the general solution, we will use x(2) (t ) and Euler’s formula
in the following equations
22

23. 0  0  0   0 
  (1 2i )t   t t   t  
x (t )   i  e
(2)
  i  e (cos(2t )  i sin(2t ))  e  cos(2t )   ie  sin(2t ) 
1 1  sin(2t )    cos(2t ) 
       
Therefore,
 0   0 
  t  
u (t )  e  cos(2t )  and v(t )  e  sin(2t ) 
t
 sin(2t )    cos(2t ) 
   
and the general solution to the system with real-valued solutions is
2  0   0 
  t t   t  
x(t )  c1r1  c2u (t )  c3v(t )  c1  3  e  c2e  cos(2t )   c3e  sin(2t ) 
1
2  sin(2t )    cos(2t ) 
     
Now that we know how to solve systems that yield real and/or imaginary eigenvalues and
eigenvectors, we will now focus our attention on the next case if a eigenvalue is repeated when
found from the characteristic equation.
23

24. Section 3: Solving Systems of Differential Equations with Repeated Eigenvalues
In this section, we will be solving systems of differential equations where the eigenvalues
found from the characteristic equation are repeated. We will still be finding solutions of the
following equation
x  P(t ) x
and will still find at least one of the eigenvalues/eigenvectors in the way we previously solved
systems with distinct eigenvalues. But, when solving for the other repeated eigenvalue, we will
see that the other solution will take the form
x   tert  ert
where  and  are constant vectors. After finding the first solution of the form x(1) (t )   1ert , it
may be intuitive to find a second solution to the system of the form
x(2) (t )   1tert
because of how repeated roots are solved when finding the solution to a second order differential
equation. Substituting that back into x  P(t ) x yields
r 1te rt   1e rt  P (t ) 1te rt  r 1te rt   1e rt  P (t ) 1te rt  0   1e rt  rt  1  P (t )t   0
But, for the equation to be solved so it is satisfied for all t, the coefficients of te rt and ert must
each be zero (Boyce & DiPrima, 2001, p. 403). Therefore, we find out that in this case,  1  0
and thus x2   1tert is not a solution for the second repeated eigenvalue. But, from
24

25. r 1tert   1ert  P(t ) 1tert  0 ,
we see that there is a form of x   tert in the substituted equation along with another term of the
form  ert . Therefore, we need to assume that
x2   1tert ert
Where  and  are constant vectors. Substituting the above expression into x  P(t ) x gives
r 1tert   1ert  rert  P(t )( 1tert ert )  r 1tert  ( 1  r )ert  P(t )( 1tert ert )
Equating the coefficients of te rt and ert gives the following conditions
P(t ) 1te rt  r 1te rt  0  P(t ) 1  r 1  0  ( P(t )  rI ) 1  0
P(t ) e rt   1e rt  r e rt  0  P(t )  r   1  ( P(t )  rI )   1
for the determination of  1 and . The underlined portions are the important conditions derived
from the equation. To solve ( P(t )  rI ) 1  0 , all we do is solve for one of the repeated
eigenvalue and eigenvector just like in previous sections. We will solve a matrix equation of the
form
 p11 (t )  r p1n (t )  1   11   p11 (t )  r 1    p1n (t )  n  11
    
      
 p (t ) pnn (t )  r n    n   pn1 (t ) 1    pnn (t )  r n   n
1 1
 n1    
Solving for 1 ,...,n in the above equation will result in the solution of the vector  denoted
25

26.  1 
  
 
 
 n
After equating  1 and  , we substitute them into x(2) (t ) to get the second specific solution
x2 (t )   1tert ert
The last term in the above equation can be disregarded because it is a multiple of the first
specific solution x(1) (t )   1ert but the first 2 terms make a new solution of the form
x(2) (t )   1tert ert
Finding W [ x(1) , x(2) ](t )  0 will prove that x(1) and x(2) are linearly independent thus allowing us
to represent the a general solution to the system in the form
x  c1 x(1) (t )  c2 x (1) (t )  ck x ( k ) (t )  x  c1 1e r1t  c2 [ 1te r1t  e r1t ]  ...  ck k 1e rk 1t
where x(1) and x(2) include the repeated eigenvalues of multiplicity 2.
For the sake of simplicity, we will focus our examples on solving systems that have
repeated eigenvalues of only multiplicity 2. Also included in one of the examples is a case
where a repeated eigenvalue give rise to linearly independent eigenvectors (which is easily
identifiable using Maple) of the matrix P(t ) thus avoiding the complications of solving systems
with repeated eigenvalues.
26

27. Example 1: Solve the following 2 x 2 system for x

x1  4 x1  x2   4 1
  x   x

x2  4 x1  8 x2   4 8 
We will begin the example by using Maple to find the eigenvalues and eigenvectors of the
coefficient matrix:
>
>
>
0
Notice in the resulting eigenvectors that  2    which is a zero multiple of  1 and does us no
0
help in finding the second specific solution of the above system. But, the results derived from
1 1
Maple gives us r1  r2  6 ,  1    , and x (1) (t )    e 6t . We need to use the equation
 2  2
x(2) (t )   1tert ert
to solve for  and thus have a second specific solution to the system. To find out the second
1  4 1
specific solution, we substitute x (2) (t )    te6t   e6t into x    x to get the following
 2  4 8 
expression
27

28.  1  6t 1    4 1  1  6t  4 1   1  6t
 e    6te6t   1  6e6t     te   e
 2  2  2   4 8  2   4 8  2 
 4 1  1  6t
Multiplying out    te and factoring out a 6 from the result yields
 4 8  2 
 1  6t  1  6t  1  6t  1  6t  4 1   1  6t
  e    6te    6e    6te    e
 2  2  2  2  4 8  2 
1
Canceling out the   6te6t on each side of the equation and rearranging the equation yields
 2
 4 1   1  6t  1  6t  1  6t
    e    6e    e
 4 8  2   2  2
 1 
Factoring out   e6t on the left side of the equation and simplifying gives us
 2 
 
 4 1    1  6t  1  6t

 4 8   6 I    e    e


  2 

 2

 4 1   6 0    1  6t  1  6t
     e    e
  4 8   0 6   2   2

 2 1   1   1 
     
 4 2  2   2 
 2 1   1   1 
      , is of the form ( P(t )  rI )   .
1
The end product of the above expression, 
 4 2   2   2 
In this case
28

29.  4 1   6 0   1
( P(t )  rI )   1       
 4 8   0 6    2
Thus, to solve for  , we solve
 2 1   1   1  21  2  1 0
        4  2  2  1  0 and 2  1     
 4 2   2   2  1 2 1
(Note that both of the resulting equations with 1 and 2 are the same). After solving for  , we
substitute it into x(2) (t )   1tert ert to find the second solution of the system to be
1 0
x (2) (t )    te rt    e rt
 2 1
The Wronskian W [ x(1) , x(2) ]  e12t  0 . Therefore the specific solutions x (1) and x (2) can be
expressed as the general solution
1  1   0  
x(t )  c1   e6t  c2   t     e6t
 2  2   1  
The resulting direction field showing families of solutions to the general solution to the system is
29

30. The blue trajectories show specific solutions when initial conditions are given. The origin is
called an improper node. If the eigenvalues are negative, then the trajectories are similar but
traversed in the inward direction. An improper node is asymptotically stable or unstable,
depending on whether the eigenvalues are negative or positive (Boyce & DiPrima, 2001, p. 404).
Example 2: Solve the following 3 x 3 system for x

x1  x1  2 x2  x3   1 2 1
  

x2  x2  x3   x   0 1 1  x

x3  2 x3  0 0 2 
  
We will begin the example by using Maple to find the eigenvalues and eigenvectors of the
coefficient matrix:
>
30

31. >
>
1  1
  t  
From the Maple results: r1  r2  1 , r3  2 , x1 (t )   0  e , and x3 (t )  1 e2t . What we need to
0  1
   
find is the specific solution to x(2) (t ) . In this example, we will use the equation ( P(t )  rI )   1
to solve for  , substitute it into x(2) (t )   1tert ert , and use the shortcut to find out the third
specific solution to the system. Therefore
 1 2 1   1  1
     e t   0  e t
( P(t )  rI )     0 1 1   1I   2 
1
 
 0 0 2 
  
 
  3   
0

 0 2 1 1  1
   t   t
 0 0 1 2  e   0  e
 0 0 1     0
  3   
and
0
22  3  1  
1 1
3  0  1  0,2  ,3  0     
2 2
3  0 0
 
31

32. Substituting what we found for  into x(2) (t )   1tert ert yields
0
1    2 0
  t 1 t   t   t
x (t )   0  te 
(2)
e   0  te   1  e
0 2 0  0
  0    
 
The Wronskian W [ x(1) , x(2) , x(3) ]  e4t  0 . Therefore, the specific solutions x(1) , x(2) , and x(3) can
be expressed as the general solution
1 1  2   0  
  2t   t     
x(t )  c1 1 e  c2  0  e  c3  0  t   1   et
1 0  0   0  
        
Example 3: Solve the following 3 x 3 system for x

x1   x2  3x3   0 1 3 
  
x2  2 x1  3x2  3x3   x   2 3 3  x

x3  2 x1  x2  x3 
 
 2 1 1 
 
For this example, using Maple can unlock a potential shortcut in solving for the general solution
to the above system. Again, we will begin the example by using Maple to find the eigenvalues
and eigenvectors of the coefficient matrix:
>
>
32

33. >
Unlike the other 2 examples, the Maple output displays 2 linearly independent eigenvectors of
the repeated eigenvalues r2  r3  2 . Another shortcut for finding eigenvectors of repeating
eigenvalues is found if a math program, such as Maple, is utilized to solve systems of differential
equations. Therefore
 1 1  3
  2t 2   2t 3  
x1 (t )  1 e , x (t )   2  e , x (t )   0  e 2t
 1  0  2 
     
and the general solution to the system is
1  1  3 
  2t     
x  c1 1 e  c2  2   c3  0   e 2t
1  0  2  
      
A more advanced look at systems with repeated eigenvalues would include repeated
eigenvalues with multiplicities higher than 2. The equations to solve higher multiplicities of
repeated eigenvalues become more detailed and difficult to solve for but to find the eigenvalues
for such values, we would follow the same thought process in how we found the eigenvalue for
repeated eigenvalues of multiplicity 2. For the next section, we will return to our original form
33

34. of a differential equation x  p1 (t ) x1  ...  pn (t ) xn  g (t ) and solve nonhomogenous systems
where the value of g (t )  0 .
34

35. Section 4: Solving Systems of Nonhomogenous Differential Equations
Unlike the previous sections where we solved different types of systems of homogeneous
differential equations with constant coefficients, this section will focus on solving systems of
nonhomogenous differential equations of the form
x  P(t ) x  g (t )
The following theorem related to nonhomogenous systems should help us figure out
where to start solution process:
Theorem 2: If x(1) (t ),..., x( n) (t ) are linearly independent solutions of the n-dimensional
homogenous system x  P(t ) x on the interval a < t < b and if x p (t ) is any solution of the
nonhomogenous system x  P(t ) x  g (t ) on the interval a < t < b, then any solution of
the nonhomogenous system can be written x  c1 x (1) (t )   ck x ( n ) (t )  x p (t ) for a unique
choice of the constants c1 ,..., cn (Rainville, Bedient, & Bedient, 1997, p. 199).
The theorem states that we will need to find a particular solution x p (t ) and add it on to the
general solution of the homogenous system that is part of the nonhomogenous system. To do
that, we will be using a variation of parameters technique to find x p (t ) and solve the equation
x  P(t ) x  g (t ) .
Solutions of the homogenous part of the nonhomogenous systems will take the form
x  c1 1er1t   cn n ernt
and using the variation of parameters technique suggests we seek a solution to the
nonhomogenous system to be
35

36. x p (t )  c1 (t ) 1e r1t   cn (t ) n e rnt
Direct substitution back into x  P(t ) x  g (t ) yields
(r1c1 (t ) 1er1t   rn cn (t ) n ernt )  (c1 (t ) 1er1t 
  cn (t ) nernt )  ( P(t )c1 (t ) 1er1t 
  P(t )cn (t ) ne rnt )  g (t )
P(t ) multiplied by any eigenvalue found to be a part of the specific solution will result in that
particular eigenvalue multiplied by its eigenvector because it is already part of the solution to the
homogeneous system. Therefore
(r1c1 (t ) 1e r1t   rn cn (t ) n e rnt )  (c1 (t ) 1e r1t 
  cn (t ) n e rnt )  (r1c1 (t ) 1e r1t 
  rncn (t ) ne rnt )  g (t )

(c1 (t ) 1e r1t 
  cn (t ) n e rnt )  g (t )

The resulting equation can be rewritten in matrix form as
 11 1n   c1(t )e r t   g1 (t ) 
1
    
   
 n

1
 nn   cn (t )e rnt   g n (t ) 
    
 
To solve for c1 (t ),..., cn (t ) , we must use Cramer’s Rule to solve Ax  b for x where
 11 1n  
 c1 (t )e r1t   g1 (t ) 
     
A , x   ,b   
 n1
 nn   cn (t )e rnt 
  g (t ) 
     n 
Cramer’s Rule states that the system has a unique solution that is given by
det( Bk )
xk  for k  1,..., n
det( A)
36

37. Therefore
g1 (t ) 1n 11 g1 (t )
g n (t )  nn n
1
g n (t )

c1 (t )e 
r1t

,..., cn (t )e 
rn t
11 1n 11 1n
n
1
 nn n
1
 nn
Thus,
 
c1 (t )e r1t  a1 g1 (t )  ...  an g n (t )  c1 (t )  [a1 g1 (t )  ...  an g n (t )]e  r1t
 
cn (t )ernt  b1 g1 (t )  ...  bn g n (t )  cn (t )  [b1 g1 (t )  ...  bn g n (t )]e  rnt
for some arbitrary constants a1 ,..., an and b 1 ,..., bn . To solve for the general solution, integrate
both sides of the above equation to get c1 (t ),..., cn (t ) , substitute them into
x p (t )  c1 (t ) 1e r1t   cn (t ) n e rnt to find the particular solution, and substitute x p (t ) into
x  c1 x (1) (t )   ck x ( n ) (t )  x p (t ) to find the general solution for the system. The following
examples will help demonstrate how to solve systems of nonhomogenous differential equations
and lead into an application of nonhomogenous systems.
Example 1: Solve the following 2 x 2 system for x

x1  x2   0 1  0 
  x   x t 

x2  2 x1  3x2  3e   2 3   3e 
t
We will begin the example by using Maple to find the eigenvalues and eigenvectors of the
homogenous part of the system
37

38.  0 1
x   x
 2 3 
Therefore
>
>
>
1 1
The resulting eigenvalues and eigenvectors are r1  1, r2  2,  1    , and  2    . The
1  2
Wronskian W [ x(1) , x(2) ]  e12t  0 thus, the general solution to the homogenous part of the system
is
 1 1
xh  c1   et  c2   e 2t
 1  2
To find the particular solution of the nonhomogenous part of the system, we will use the
variation of parameters technique to find a solution of the above equation of the form
 1 1
x p  c1 (t )   et  c2 (t )   e 2t
 1  2
38

39.  0 1  0 
We will first substitute x p for x and x directly into x    x   t  to get the following
 2 3   3e 
expression
 1  1 1 1  0 1  1  0 1  1   0 
 
c1 (t )   et  c1 (t )   et  2c2 (t )   e 2t  c2 (t )   e 2t     c1 (t )e
t
   c2 (t )e
2t
 t 
 1  1  2  2  2 3  1  2 3  2   3e 

 1  1 1 1  1 1  0 
 
c1 (t )   et  c1 (t )   et  2c2 (t )   e 2t  c2 (t )   e 2t  c1 (t )   et  2c2 (t )   e 2t  t 
 1  1  2  2  1  2  3e 

 1 1  0 
 
c1 (t )   et  c2 (t )   e 2t   t 
 1  2  3e 
The final expression given above can be written in matrix notation as

1 1   c1 (t )et   0 
   2t 
 t
1 2   c2 (t )e   3e 
 
To solve for c1 (t ) and c2 (t ) , we will apply Cramer’s Rule to find
0 1 1 0
3et 2 3et 1 3et 3et

c1 (t )et   
and c2 (t )e2t  
0 1 2 0 1 2
2 3 2 3
Thus
 3et  t 3  3et  2t 3et

c1 (t )    e = 
and c2 (t )   e 
 2  2  2  2
To solve for c1 (t ) and c2 (t ) , integrate both sides of both equations so that
39

40. 3t 3et
c1 (t )  and c2 (t ) 
2 2
 1 1
and substituting into the partial solution x p  c1 (t )   et  c2 (t )   e 2t yields
 1  2
3t 1 t 3et  1  2t t  1 t 1
xp   e    e  3te    3e  
2  1 2  2  1  2
Therefore, the general solution to the nonhomogenous system is
 1 1  1 1
x  xh  x p  c1   et  c2   e 2t  3tet    3et  
 1  2  1  2
Example 2: Solve the following 2 x 2 system for x

x1  2 x1  x2  et   2 1   et 
  x   x 

x2  x1  2 x2  3t   1 2   3t 
We will begin the example by using Maple to find the eigenvalues and eigenvectors of the
homogenous part of the system
 2 1 
x   x
 1 2 
Therefore
>
>
40

41. >
 1 1
The resulting eigenvalues and eigenvectors are r1  1, r2  3,  1    , and  2    . The
 1  1
Wronskian W [ x(1) , x(2) ]  e12t  0 thus, the general solution to the homogenous part of the system
is
 1 1
xh  c1   e t  c2   e 3t
 1  1
To find the particular solution of the nonhomogenous part of the system, we will use the
variation of parameters technique to find a solution of the above equation of the form
 1 1
x p  c1 (t )   e t  c2 (t )   e 3t
 1  1
 2 1   et 
Substituting x p for x and x directly into x    x    to get the following
 1 2   3t 
 1 1  et 
(t )   et  c2 (t )   e3t   
c1 
 1  1  3t 
The final expression above can be written in matrix notation as

1 1   c1 (t )et   et 
   3t 
 
1 1  c2 (t )e   3t 
 
To solve for c1 (t ) and c2 (t ) , we will apply Cramer’s Rule to find
41

42.  et   3t   et   3t 

c1 (t )et    
   and c2 (t )e 3t      
 2  2  2  2
Thus
 1   3te   e2t   3te3t 
t

c1 (t )       
and c2 (t )    
2  2   2   2 
To solve for c1 (t ) and c2 (t ) , integrate both sides of both equations so that
 et   3tet  3et
2
3tet 3et t
c1 (t )   + and c2 (t ) 
2 2 2 4 2 2
 1  t  1  3t  3t 3 te t  1  e t 3te 2t 3e 2t   1 
x  xh  x p  c1 (t )   e  c2 (t )   e    +        
 1  1 2 2 2   1  4 2 2   1
 1 1
and substituting into the partial solution x p  c1 (t )   e t  c2 (t )   e 3t yields
 1  1
 3tet 3et t 1 t   e  3tet 3et   1  3t
t 2
xp    +   e     
 e
 2 2 2   1  4 2 2   1
 

 3t 3 tet  1  et 3te2t 3e2t   1 
xp    +        
2 2 2   1  4 2 2   1
Therefore, the general solution to the nonhomogenous system is
 1 1  3t 3 tet  1  et 3te2t 3e2t   1 
x  xh  x p  c1 (t )   et  c2 (t )   e 3t    +        
 1  1 2 2 2   1  4 2 2   1
42

43. Section 5: Application of Systems of Differential Equations – Arms Races
(Nonhomogenous Systems of Equations)
In the previous section, we discussed how to solve systems of differential equations that
were nonhomogenous using a variation of parameters technique. Now, we can apply that
knowledge of solving systems with nonhomogenous equations to solve a model that illustrates an
arms race between two competing nations. L.F. Richardson, an English meteorologist, first
proposed this model (also known as the Richardson Model) that tried to mathematically explain
an arms race between two rival nations. Richardson himself seemed to have believed that his
perceptions relating to the way nations compete militarily might have been useful in preventing
the outbreak of hostilities in World War II (Brown, 2007, p. 60). Both nations are self-defensive,
both fight back to protect their nation, both maintain army and stock weapons, and when one
nation expands their army the other nation finds it offensive. Therefore, both nations will spend
money (in billions of dollars) on armaments x and y that are functions of time t measured in
years. x(t) and y(t) will represent the yearly rate of armament expenditures of the two nations
using some standard unit. Richardson then made some of the following assumptions about his
model:
 The expenditure for armaments of each country will increase at a rate that is proportional
to the other country’s expenditure (each nation's mutual fear's rate is directly proportional
to the expenditure of the other nation) (Rainville, Bedient, & Bedient, 1997, p. 228).
 The expenditure for armaments of each country will decrease at a rate that is proportional
to its own expenditure (extensive armament expenditures create a drag on the nation's
economy) (Rainville, Bedient, & Bedient, 1997, p. 228).
43

44.  The rate of change of arms expenditure for a country has a constant component that
measures that level of antagonism of that country toward the other (Rainville, Bedient, &
Bedient, 1997, p. 228).
 The effects of the three previous assumptions are additive (Rainville, Bedient, & Bedient,
1997, p. 228).
The previous assumptions make up the differential equations of the arms race system denoted by
dx x(0)  x0
 ay  mx  r
dt
for
dy
 bx  ny  s y (0)  y0
dt
where a, m, b, and n are all positive constants. The positive terms ay and bx represent the drive to
spend more money are arms due to the level of spending of the other nation, and the negative
terms mx and ny reflect a nation’s desire to inhibit future military spending because of the
economic burden of its own spending. But, r and s can be any value because they represent the
attitudes of each nation towards each other (negative values represent feelings of good will while
positive values represent feelings of distrust). The initial values x(0) and y(0) represent the
initial amount of money (in billions of dollars) each nation will spend towards armaments. The
system can be simplified into
x(t )  mx  ay  r
y(t )  bx  ny  s
and is expressed in matrix notation where
 x(t )   m a  x(t )   r 
         X   P(t ) X  B
 y (t )   b n  y (t )   s 
44

45. To solve for the system, we will use the knowledge from the previous section to develop
general solutions to the homogenous system. For the nonhomogenous part of the system, the
f
solution will be a constant solution of the form   because the vector B is made up of
g
constants thus making the process of solving by variation of parameters much easier. Lastly, the
initial values (trajectories) for the solution will represent the starting amount of money each
country will be spending on armaments.
General solutions to the arms race system will represent one of a few types of races: a
stable arms race, a runaway arms race, a disarmament, or disarmament/runaway/stable arms race
depending on the initial values. The following examples will help demonstrate each of the above
mentioned arms races along with slope fields to graphically represent the races.
Example 1: A Runaway Arms Race
The following system will result in a runaway arms race:
x(t )  2 x  4 y  8  2 4  8
 X    X  
y(t )  4 x  2 y  2   4 2  2
To find the solution to this arms race, we will first find the general solution to the
homogenous part of the system using Maple:
>
>
>
45

46.  1 1
Therefore, r1  2 ,  1    , r2  6 , and  2    . The Wronskian W [ x(1) , x(2) ]  2e4t  0
 1  1 
thus, the general solution to the homogenous part of the arms race is
 1 1
xh  c1   e 2t  c2   e 6t
 1  1
As mentioned in the beginning of this section, the nonhomogenous system
e
X   P(t ) X  B has a constant solution of the form   because B is a vector of constants thus
f
the solution should also be a vector made up of constants. Therefore, in the equation
e
X   P(t ) X  B , X (t )    can be substituted into X and X  where
f
 2 4  8
P(t )    and B   
 4 2   2
to get the following expression
 2 4  f   8   2 4  f   8  2 f  4 g  8  f   2 
0                      xn
 4 2  g   2   4 2  g   2 4 f  2 g  2  g   3 
Therefore the general solution of the nonhomogenous system is
 1 1  2 
x  xh  xn  c1   e 2t  c2   e 6t   
 1  1  3 
46

47. Note that as lim x(t ) and lim y(t )   . Thus we would predict that the rate that each nation
t  t 
spends their money on armaments would increase infinity resulting in an arms race.
The direction field for the nonhomogenous system is represented by
with the initial conditions x0  5, y0  2 and x0  2, y0  5 given. The direction field of the
system shows that for any initial value, the solution goes to  as t   . Thus, we have a
runaway arms race.
If you wanted to solve the system with an initial condition given such as x0  5, y0  2 ,
we would set up the general solution as
5 1 20  1  60  2   5   1  1   2 
   c1   e  c2   e        c1    c2     
 2  1  1  3   2   1  1  3 
47

48. and solve for c1 and c2 . Therefore
5  1  1   2  c1  c2  2  5 c1  c2  7
   c1    c2       c  c  3  2  c  c  6  c1  6 and c2  1
 2  1  1  3  1 2 1 2
Thus, the final solution with the initial conditions given is
 1 1  2 
x    6e2t    e 6t   
 1  1  3 
or
x(t )  6e 2t  e 6t  2
y (t )  6e 2t  e 6t  3
The role of the initial value is how much each nation will initially spend on armaments in
billions of dollars. Using initial values when solving an arms race system will lead to a specific
solution describing the race instead of families of general solutions describing all cases of the
system.
Example 2: A Stable Arms Race
The following system will result in a stable arms race:
x(t )  5 x  2 y  1  5 2  1
 X    X  
y(t )  4 x  3 y  2   4 3   2
To find the solution to this arms race, we will first find the general solution to the
homogenous part of the system using Maple:
48

49. >
>
>
1 1
Therefore, r1  7 ,  1    , r2  1 , and  2    . The Wronskian W [ x(1) , x(2) ]  3e8t  0
 1  2
thus, the general solution to the homogenous part of the arms race is
1 1
x  c1   e 7 t  c2   e  t
 1  2
The general solution to the nonhomogenous part of the arms race will be found by
f  5 2  1
substituting X (t )    into X     X    . Therefore
g  4 3   2
 5 2  f   1  5 f  2 g  1  0
0        f  1 and g  2
 4 3  g   2  4 f  3g  2  0
1
and the solution of that system is xn    . Thus the general solution of the nonhomogenous
 2
system is
1 1 1
x  xh  xn  c1   e 7 t  c2   e t   
 1  2  2
49

50. Note that as lim x(t )  1 and lim y(t )  2 because in both equations, the terms with both
t  t 
e7t and et go to 0 as t   . All that is left from the differential equations are the constant
terms x(t )  1 and y(t )  2 and what initial values of the system converge to.
The direction field with a few trajectories denoting the initial values for the
nonhomogenous system is represented by
The direction field of the system shows that for any initial value, the solution approaches the
point (1, 2) as t   . Thus, we have a stable arms race.
50

51. Example 3: Disarmament
The following system
x(t )  4 x  y  1  4 1   1 
 X    X  
y(t )  x  y  2   1 1  2 
will result in disarmament between the competing nations for all initial values.
For the sake of simplicity, only the graph of this system and the solution solved by Maple
will be shown because the eigenvectors and eigenvalues generated from P(t) become
complicated radicals that would be difficult to manipulate by hand to find an answer. Therefore,
the general solution to the system given by Maple output is
>
>
>
>
51

52. >
As you can see from the above solution, the general solution to the system is becomes very
complicated but, lim x(t )  1 and lim y(t )  3 thus showing that eventually, the nations will
t  t 
get to a point in time where they are decreasing the rate at which they are spending money on
armaments until they are spending no money on the arms race. The graph of the system is much
more beneficial in demonstrating an arms race that ends in disarmament.
The direction field with a few trajectories denoting the initial values for the
nonhomogenous system is represented by
52

53. For the initial values represented by the blue trajectories in the above directional field, the
trajectories will approach the point (1, 3) as t   thus resulting in disarmament with any
initial value chosen for the system.
Example 4: Disarmament/Runaway Arms Race/Stable Arms Race
The following system
x(t )  2 x  4 y  2   2 4   2 
 X    X  
y(t )  4 x  2 y  2   4 2   2 
will result in disarmament if x0  y0  2 , a runaway arms race if x0  y0  2 , or a stable arms
race if x0  y0  2 .
To find the solution to this arms race, we will first find the general solution to the
homogenous part of the system using Maple:
>
>
>
 1 1
Therefore, r1  2 ,  1    , r2  6 , and  2    . The Wronskian W [ x(1) , x(2) ]  2e4t  0
 1  1 
thus, the general solution to the homogenous part of the arms race is
53

54.  1 1
xh  c1   e 2t  c2   e 6t
 1  1
The general solution to the nonhomogenous part of the arms race will be found by
f  2 4   2 
substituting X (t )    into X     X    . Therefore
g  4 2   2 
 2 4  f   2 
0       f  1 and g  1
 4 2  g   2 
 1
and the solution of the system is xn    . Thus the general solution of the nonhomogenous
 1
system is
 1 1  1
x  xh  xn  c1   e 2t  c2   e 6t   
 1  1  1
The direction field with a few trajectories denoting the initial values for the nonhomogenous
system is represented by
54