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316 CHAPTER 10 Systems of Linear Differential Equations
10.4 Exponential Matrix Solutions
ax
A differential equation y = ay with a as a constant has the general solution y = ce . This leads
us to ask whether there is an analogous solution for the system X = AX with A as an n × n real
constant matrix.
Recall that
1 1
ax 2 3
e = 1 + ax + (ax) + (ax) + ··· .
2 3!
At
Define the exponential matrix e by
1 1
At 2 2 3 3
e = I n + At + A t + A t + ··· ,
2 3!
whenever the infinite series defining the i, j element on the right converges for i and j
varying from 1 through n.
It is routine to verify that
At Bt
e (A+B)t = e e
if A and B are n × n real matrices that commute, that is, if
AB = BA.
Differentiate a matrix by differentiating each element of the matrix. Using the fact that A is a
constant matrix with derivative zero (the n × n zero matrix), we obtain from the definition that
At
At
(e ) = Ae ,
which has the same form as the familiar
at
(e ) = ae .
at
This derivative formula leads to the main point.
THEOREM 10.8
At
Let A be an n × n real, constant matrix and K be any n × 1 matrix of constants. Then e K is a
solution of X = AX. In particular, e is a fundamental matrix for this system.
At
At
The proof is immediate by differentiating. Upon setting X(t) = e K,wehave
d
At
At
e K = Ae K = AX.
X (t) =
dt
We therefore have the general solution of X = AX if we can compute the exponential
At
matrix e . Except for very simple cases this is impractical by hand and requires a computational
software package. If MAPLE is used, the command
exponential(A,t)
At
will return e if A has been defined and n is not “too large.”
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October 14, 2010 20:32 THM/NEIL Page-316 27410_10_ch10_p295-342