in this video we're going to find the
general solution to a linear system in
the case of complex eigenvalues so
suppose we have a system of differential
equations given in matrix form like this
so the first step in solving a problem
like this is to find the eigenvalues and
their associated eigenvectors of this
matrix so you pick your favorite way to
find eigenvalues and eigenvectors and
you'll come up with these eigenvalues of
course since they're complex they always
come in conjugate pairs and some
examples of an eigenvector you might
come up with i over 2 one and minus i over 2 one
so the good thing about when you get
complex eigenvalues is that you actually
only need one eigenvalue eigenvector
pair so we actually don't care about
this second eigenvalue eigenvector pair
at all
so how do we proceed it's the same in as
in the case of real eigenvalues in the
sense that we now know how to create at
least one solution so all right
y c of t (the c standing for complex)
and we get e to the eigenvalue times t
times the eigenvector- the associated
eigenvector- so this is a perfectly good
complex solution to our differential
equation to our system of differential
equations
the problem is it is complex and we want
a real general solution so what do we do
next
we separate this into real and imaginary
parts so to do that we're going to use
the fact that e to the i t...  well we're going to
use Euler's formula
so we're going to expand out that e
to the i2t
and then we're going to multiply this
out so what we get e to the t i over 2
cos two t minus one half sine 2t cos
2t plus i sine 2t
and then we're able to separate out the
real and imaginary parts so we have e to
the t minus one half sine 2t cos 2t
then we take all the parts with i in
them and we pull the i out i e to the t
one-half cos 2t sine 2t so here we
have our single complex solution and we
separated out into a real solution plus
i times an imaginary solution
so we've separated out our complex solution
into real and imaginary parts so we can
write the complex solution as y real of
t plus i times y imaginary of t where
these turn out to be both real-valued
solutions to the original system so
because we have two real valued
solutions to our original system we can
express them- we can express the general
solution as a linear combination of the
real valued solutions (of the real part
and the imaginary part) so that means our
general solution is in fact
and that's it this is our general
solution for unknown constants k1 and k2
