[MUSIC] Now we're going to enter
the realm of matrix algebra. We're going to look at solving a system
of linear first order equations. What does a general system look like? We have two variables, x1 and
x2, functions of time. So we have x1 dot, we're going to do the constant
coefficient equation, also homogeneous. That will be a constant a times x1
plus another constant b times x2. That's the first equation. If we do two equations,
the second one will be x2 dot and then some other coefficient
c times x1 plus d times x2. These are two coupled
first order equations. The independent variable is t,
the dependent variables are x1 and x2. If you know your matrix algebra, you know that you can write
this then in matrix form. What would that look like? We've got d divided by dt,
and then we put x1 and x2 in a column vector, and
that's equal to a matrix a, b, c, d. Which is the standard two by two matrix, times the column vector x1, x2. So this system of linear
first order homogeneous equations can be written
in matrix form like so. If we like to write it more succinctly
we can just call its column vector x1, x2, we'll just call that x. And it's the time derivative, so
each component has a time derivative. So I'll write that as x dot. And that's equal to a two by two matrix, which I'll write capital
A times the column vector x. So the system of first order
linear equations can be simply written as x dot equals a times x. To solve this we're
going to use an ansatz. And we're also going to use
the principle of superposition. So we're going to look for
solutions following our ansatz. And then when we find solutions,
two solutions in this case, we'll multiply them by constants and add
them together to get the general solution. So what is the right ansatz here? We're going to try x(t),
the column vector, equal to a constant column
vector which I'll call v. So this is a two by one matrix,
times an exponential, and we'll call that e to the lambda t,
okay? That will be our ansatz. And then we're looking for here. In this type of ansatz, we're looking
to determine both lambda and v. What happens when we substitute
into the differential equation? The time derivative,
v is independent of time. The time derivative will
bring down a lambda, lambda is also independent of time. So substituting into the differential
equation will get lambda v times e to the lambda t is equal to A times v times e to the lambda t. As always the exponential cancels. That's the point of
the exponential ansatz. And rewrite this in a more telling
fashion, put the Ave on the left. We get Av equals lambda v. Those of you who remember your matrix
algebra, this is the eigenvalue problem. So lambda are the eigenvalues of
the matrix A, and v are the eigenvectors. How do you compute the eigenvalues? You use the characteristic
equation which is the determinant of A minus lambda I equals 0. This is a quadratic equation for
lambda, if you remember, you're determinant of
the two by two matrix. The A minus lambda on the diagonals and
b and c of, this equation becomes lambda squared. Minus the trace of the matrix,
a plus d times lambda, plus the determinant of the matrix, ad minus bc equals 0, okay? That's doing this determinant, and if you remember this from
a matrix algebra course. Okay, three things can happen in
a quadratic equation, what are they? The first one is that you can
have distinct real roots, For the eigenvalue lambda. The second thing that can happen is
you can have complex conjugate roots. And the last thing that you can happen
is you can have repeated roots. Okay, so it should sound very
familiar to the second order linear homogeneous equation
with constant coefficients. Because actually,
it's the same thing, okay? But look that in a different way. In this course we'll consider
these first two cases, and we'll skip this third case,
which is a bit too specialized for the time that we have to
learn differential equations. Okay, so let me summarize. We're now considering how to solve
a system of linear first order equations. For simplicity we'll
consider two equations. An equation for x1 and an equation for x2. We can write it as a single
matrix equation or in compact form as x dot equals Ax,
where A is a two by two matrix. I'm going to show you how to solve
this by the method of ansatz, we're going to try to find
the solution for x of t equal to v. Which turns out to be the eigenvector
times e to the lambda t, and lambda will be the eigenvalue. So upon substitution you
get the eigenvalue problem. Av equals lambda v. The characteristic equation for lambda is just the determinant
of A minus lambda i equals zero. And because it's a quadratic equation
we'll have to consider, we could consider three separate cases, in this
course we'll only consider the first two. What does the solution look like when
you have distinct real roots and what does the solution look like when
you have complex conjugate roots? I'm Jeff Chasnov, thanks for watching,
and I'll see you in the next video.