Now that we've found some of the fundamentals out of the way for systems of differential equations it's time to start thinking about how to solve a system of differential equations. We will begin with the homogeneous system written in matrix form as,
x?' = A x? ......................(1)
Here, A is an n x n matrix and x is a vector whose elements are the unknown functions into the system.
Here, if we begin with n = 1 then the system decreases to a fairly easy linear or separable first order differential equation,
x' = ax
And it has the following solution,
x′ = ax
x (t) = ceat
Therefore, let's use this as a guide and for a common n let's notice if,
x? (t) = ?h ert .................(2)
It will be a solution. Remember that the only real difference now is which we let the constant in front of the exponential be a vector. All we requirement to do then is plug it into the differential equation and notice what we find. First see that the derivative is,
x? (t) = r ?hert
Therefore upon plugging the guess in the differential equation we find,
r ?hert = A ?hert
(A - rI) ?hert =0?
Here, as we know that exponentials are not zero we can drop which portion and we after that see that so as for (2) to be a solution to (1) so we should have,
(A - rI) ?h = 0?
Or, so as for (2) to be a solution to (1), r and ?h should be an eigen-value and eigenvector for the matrix A.
Thus, so as to solve (1) we first get the eigen-values and eigenvectors of the matrix A and after that we can form solutions by using (2). There are going to be three cases which we'll require to look at.
The cases are as: real, distinct eigenvalues, complex eigenvalues and repeated eigenvalues.
None of that tells us how to wholly solve a system of differential equations. We'll require the subsequent couple of facts to do this.