30.2. FIRST ORDER SYSTEMS, CONSTANT COEFFICIENTS 591
consider specific examples a little later. However, if you are looking for something whichwill solve all first order systems in closed form using known elementary functions, thenyou are looking for something which is not there. You can indeed speak of it in generaltheoretical terms but the only problems which are completely solvable in closed form arethose for which you can exactly find the eigenvalues of the matrix. Unfortunately, thisinvolves solving polynomial equations and none of us can do these in general.
30.2.3 Using a Computer Algebra SystemYou want to solve
x′ = Ax+f, x(0) = x0
Use the Property 21. and take Laplace transforms of both sides. Thus
sX (s)−x0 = AX (s)+F (s)
where X (s) is the Laplace transform of x(t) and F (s) is the Laplace transform of f (t).Then you can solve for X (s) , at least for large enough s so that (sI−A)−1 exists. Thus
(sI−A)X (s) = x0 +F (s)
ThenX (s) = (sI−A)−1 (x0 +F (s))
Note that there is even a formula for (sI−A)−1. See Theorem 28.1.14. Thus you canalways find X (s). Then having done so, it is a matter of finding the function whose Laplacetransform gives X (s). By hand, you would consider each entry of X (s) and by using partialfractions, you would go backwards in the table. It won’t always work. Sometimes youwon’t be able to factor the polynomials enough to carry this out and even when it doeswork, it will be pretty tedious. This is why you should use Matlab or some computeralgebra system. Here is an example which can be done. The reason I know it will work outis that I cooked it up to work out. I picked a matrix whose eigenvalues are known. I alsopicked the forcing function to be something which will tend to make things work.
Example 30.2.8 Solve the following first order system.
x′ =
2 2 −1−1 0 1−1 0 2
x+
cos tsin tet
, x(0) =
101
Following the above general procedure, the Laplace transform of the forcing function
is s
s2+11
s2+11
s−1
and so
X (s) =
s 0 0
0 s 00 0 s
− 2 2 −1−1 0 1−1 0 2
−1
101
+
s
s2+11
s2+11
s−1