Mathematical methods for economic theory

Martin J. Osborne

8.7 Systems of first-order linear differential equations

Two equations in two variables

Consider the system of linear differential equations (with constant coefficients)
x'(t = ax(t) + by(t)
y'(t = cx(t) + dy(t).
We can solve this system using the techniques from the previous page, as follows. First isolate y(t) in the first equation, to give
y(t) = x'(t)/b − ax(t)/b.
Now differentiate this equation, to give
y'(t) = x"(t)/b − ax'(t)/b.
Why is this step helpful? Because we can now substitute for y(t) and y'(t) in the second of the two equations in our system to yield
x"(t)/b − ax'(t)/b = cx(t) + d[x'(t)/b − ax(t)/b],
which we can write as
x"(t) − (a + d)x'(t) + (ad − bc)x(t) = 0,
an equation we know how to solve!

Having solved this linear second-order differential equation in x(t), we can go back to the expression for y(t) in terms of x'(t) and x(t) to obtain a solution for y(t). (We could alternatively have started by isolating x(t) in the second equation and creating a second-order equation in y(t).)

Example
Consider the system of equations
x'(t = 2x(t) + y(t)
y'(t = −4x(t) − 3y(t).
Isolating y(t) in the first equation we have y(t) = x'(t) − 2x(t), so that y'(t) = x"(t) − 2x'(t). Substituting these expressions into the second equation we get
x"(t) − 2x'(t) = −4x(t) − 3x'(t) + 6x(t),
or
x"(t) + x'(t) − 2x(t) = 0.
We have seen that the general solution of this equation is
x(t) = Aet + Be−2t.
Using the expression y(t) = x'(t) − 2x(t) we get
y(t) = Aet − 2Be−2t − 2Aet − 2Be−2t,
or
y(t) = −Aet − 4Be−2t.

General linear systems

We may write the system
x'(t = ax(t) + by(t)
y'(t = cx(t) + dy(t)
studied above, as
left parenthesis x'(t) right parenthesis
y'(t)
 = 
left parenthesis a b right parenthesis
c d
left parenthesis x(t) right parenthesis
y(t)
.
More generally, we may write a system of n simultaneous homogeneous linear equations in the n variables xi(t) for i = 1, ..., n as
x'(t) = Ax(t),
where A is an n × n matrix.

Now, if n = 1, in which case A is simply a number, we know that for the initial condition x(0) = C the equation has a unique solution

x(t) = CeAt.
Here's a stunning result:
for every value of n, the unique solution of the homogeneous system of linear differential equations x'(t) = Ax(t) subject to the initial condition x(0) = C is
x(t) = CeAt.
You will immediately ask: what is eAt when A is a matrix? Recall that if a is a number we have
ea = 1 + a/1! + a2/2! + ...,
or, more precisely,
ea = ∑
k=0
(ak/k!).
Now, when A is a matrix we may define
eA = ∑
k=0
(Ak/k!),
where A0 is the identity matrix (and 0! = 1). (You know how to multiply matrices together, so you know how to compute the right hand side of this equation.) That's it! You can now find the solution of any homogeneous system of linear differential equations ... assuming that you can compute the infinite sum in the definition of eAt. Therein lies the difficulty. Techniques exist for finding eAt, but they involve methods more advanced than the ones in this tutorial.

I give only one example, which shows how the trigonometric functions may emerge in the solution of a system of two simultaneous linear equations, which, as we saw above, is equivalent to a second-order equation.

Example
Consider the system
left parenthesis x'1(t) right parenthesis
x'2(t)
 = 
left parenthesis 0 b right parenthesis
b 0
left parenthesis x1(t) right parenthesis
x2(t)
.
We need to find Ak for each value of k, where
A = 
left parenthesis 0 b right parenthesis
b 0
.
You should be able to convince yourself (by computing A2, A3, and A4) that if k is odd we have
Ak = 
left parenthesis 0 (−1)(k+1)/2bk right parenthesis
(−1)(k−1)/2bk 0
,
whereas if k is even we have
Ak = 
left parenthesis (−1)k/2bk 0 right parenthesis
0 (−1)k/2bk
.
Given these results, we have
eA = 
left parenthesis 1 − b2/2! + b4/4! − b6/6! + ... b + b3/3! − b5/5! + ... right parenthesis
b − b3/3! + b5/5! − ... 1 − b2/2! + b4/4! + b6/6! + ...
.
eA = 
left parenthesis a11 a12 right parenthesis
a21 a22
where
a11  =  1 − b2/2! + b4/4! − b6/6! + ...
a12  =  b + b3/3! − b5/5! + ...
a21  =  b − b3/3! + b5/5! − ...
a22  =  1 − b2/2! + b4/4! + b6/6! + ....
Now, recall that
sin b = b − b3/3! + b5/5! − ...
and
cos b = 1 − b2/2! + b4/4! − b6/6! + ...
Thus
eA =
left parenthesis cos b −sin b right parenthesis
sin b cos b
.
We conclude that the solution of the system of equations given the initial conditions x1(0) = C1 and x2(0) = C2 is
x1(t = C1cos bt − C2sin bt
x2(t = C1sin bt + C2cos bt.
You are asked, in an exercise, to verify this solution by using the technique discussed at the start of this section to convert the two-equation system to a single second-order linear differential equation.
You may wonder whether the emergence of the trigonometric functions in this example is accidental. It is not. Exponentials, the trigonometric functions, and complex numbers (which arise as roots of the characteristic equation in the technique of the previous page), are closely related. Specifically, for any value of x we have
eix = cos x + isin x
(where i is the square root of −1). But that is another story. An excellent, but advanced, exposition of the theory of systems of linear differential equations is contained in Hirsch and Smale (1974).