We're asked to solve
the system of differential equations,
X one prime equals X two and X two prime
equals negative X one, using the eigenvalue method.
The first step is to write the system
as the vector equation X prime equals P times X
which I've already done here on the right.
Notice matrix P is the two by two matrix where
in the first row we have entries zero and one.
In the second row we have entries negative one and zero.
To use the eigenvalue method, the next step is
to determine the eigenvalues of matrix P.
Recall we do this by setting up the equation
the determinant of the difference
of P and lambda I equals zero, and then solving for lambda.
So here we have the setup.
Again, we have the determinant of the
difference of matrix P and lambda times
the two by two identity matrix equals zero.
Simplifying inside the parentheses,
the result is the determinant of the two by two matrix,
when the first row, the entries are
negative lambda and one, and the second row
of the entries are negative one and negative lambda.
Now we evaluate the determinant.
The determinant is equal to negative lambda
times negative lambda, minus one
times negative one, which gives us
lambda squared plus one equal zero.
Solving for lambda, we subtract one on both sides,
and then take the square root of both sides of the equation.
Notice this gives us lambda equals plus or minus i.
Because we have two complex eigenvalues
this does affect our approach on how
to determine the general solution.
Looking at our notes below, for each pair
of complex eigenvalues, we can find the
general solution using just lambda sub one,
an eigenvalue, and the vector V1,
a corresponding eigenvector.
Step two, after we determine lambda sub one
and the vector V sub one, we write the vector equation X one
of t equals the vector V one times e
the power of lambda sub one t.
Next we apply Euler's formula on e the
power of lambda sub i t.
Recall lambda sub one is a complex number.
Step four, we write the result in the form
of X one of t equals X three of t plus i times X four of t,
which means we separate the real part
and the imaginary part.
From here, X three of t
and X four of t are two linearly independent solutions
to the system and therefore we can write
the general solution as X of t equals C1 times X three
of t plus C2 times X four of t, where X three
of t is the real part of V1 times either power
of lambda sub one t and x four of t is the imaginary part
of the vector V1 times either power of Lambda sub one t.
So going back to our work, we know lambda sub one
is equally negative i and lambda sub two equals i.
We can actually determine the general solution using
either lambda sub one or lambda sub two,
we'll stick with lambda sub one.
Which means you now need to find a corresponding eigenvector
for lambda sub one equals negative i.
Recall to find a corresponding eigenvector,
we set up the equation the difference of matrix P
and lambda i times vector V equals a zero vector,
and then determine a vector V.
So again using lambda sub one equals negative i.
We'll go ahead and set up the equation.
this gives us again the difference of matrix P
and lambda times the two by two identity matrix,
times vector V equals a zero vector.
Simplifying inside the parenthesis,
the result is the two by two matrix where
in the first row we have i and one,
in the second row we have negative one and i.
Recall the system is dependent,
meaning we have an infinite number of solutions,
and therefore we can determine a corresponding
eigenvector using just one of the equations.
And that's because the equations
are multiples of one another.
Notice the first equation in the system is,
i times V1 plus one times V2 equals zero
or iV1 plus V2 equals zero, which we can also express
as V2 equals negative i times V1.
Using this equation, if we let V1 equal i, notice V2
is equal to negative i squared and since i squared is equal
to negative one, we have V2 equals one.
So we'll let the corresponding eigenvector,
the vector V1, be the vector i, one.
And now we can determine the general solution
using lambda sub one and the vector V1.
This question doesn't ask, but remember
that lambda sub two is equal to the complex conjugate
of lambda sub one, which we already know from solving.
But it's also true that a corresponding eigenvector
for lambda sub two, which we call vector v2
is equal to the complex conjugate of the eigenvector V1.
And the complex conjugate of the
vector i one, is the vector negative i one.
To find the complex conjugate of a vector,
we take the complex conjugate of each entry.
The complex conjugate of i is negative i,
the complex conjugate of one is just one.
But again, we don't need this to find
the general solution for this problem.
The next step to find the general solution
is to write X one of t equals the vector V1,
times e the power of lambo sub one t.
Which in our case gives us X one of t
equals the eigenvector i one, times
e to the power of negative i t.
And now we apply Euler's formula to
e the power of negative i t.
Let's do this on the next slide.
e to the power of i t is equal to cosign t minus isign t.
Performing the substitution we have X one of t
equals the eigenvector i one times
the difference of cosign t and isign t.
Next we multiply.
In this first row of the two by one matrix is i cosign t
plus sign t, and the second row is cosign t minus isign t.
And now we write this matrix as a sum of two matrices,
where we have the real part in the first matrix
and the imaginary part in the second matrix.
This gives us X one of t equals, with entries
sign t and cosign t plus i times the matrix
with entries cosign t and negative isign t.
Now we have X one of t in the form
of X three of t plus i times X four
of t and we can write the general solution.
The general solution is simply,
X of t equals C1 times x three of t
plus C2 times X four of t.
Where again X three of t is the real part
of the vector V1 times e the power
of lambda sub one t, and X four of t is the imaginary part
of the vector V1 times e to the power of lambda sub one t.
I hope you found this helpful.