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It is the last case that we need to take a look at. Throughout this section we will look at solutions to the system,
x?' = A x?
Here the eigenvalues of the matrix A are complex. By using complex eigenvalues we are going to have similar problem that we had back while we were looking at second order differential equations. We need our solutions to only have real numbers in them, though as our solutions to systems are of the form,
x?1 = ?h elt
We are going to contain complex numbers come in our solution from both the eigenvector and the eigenvalue. Getting rid of the complex numbers now will be same to how we did this back in the second order differential equation case, although will include a little more work this time around. It's simple to see how to do it in an example.
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Solve the subsequent IVP. y′′ + 11y′ + 24 y = 0 y (0) =0 y′ (0)=-7 Solution The characteristic equation is as r 2 +11r + 24 = 0 ( r + 8) ( r + 3) = 0
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