00:01
To answer this question, we're going to write out general eigenvalues, which will be a lambda 1 being equal to negative a plus b times i, and a lambda 2 being equal to negative c plus d times i, where we're going to set restrictions on a and b.
00:23
A, comma, b are both greater than 0, and c comma d are greater than 0, such that we have a negative real part in both cases.
00:34
So then if we look at the general solution for x of t, this will be some constant times e to the first eigenvalue, which is negative a plus b times i, all times t, which will be e to the negative a t, times e to the b i t, times some eigenvector, which we'll just name x1, plus some other constant.
01:03
Times e to the negative c t times e to the d i t again times another eigenvector just name it x2 so now if we want to look at the limit as t approaches infinity of our solution x of t this will be at the limit as t approaches infinity of both terms being added so we can apply the limit to both terms individually so the limit of c1 e to the negative a t e to the b i t plus the limit of c2 e to the negative c t times e to the d i t times our eigenvector and including our first eigenvector and c1 is just a constant and x1 is a constant eigenvector so we can pull those out of the limit in both cases so this would be c1 times the eigenvector times the limit as t approaches infinity, of a product of two things involving t.
02:16
So it's be the limit of the first one times the limit of the second one.
02:26
And we can do the same thing for the second limit, c2, pull out the constant and the i vector, and apply the limit to both things individually, e to the negative c -t, and limit of e to the d -i times t...