Solutions to systems, Mathematics

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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.


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