Ryan D. answered • 07/30/20

BS in Physics with minor in Applied Math

To preface the eigen value method of solving, we should consider an anzats (or best guess solution) of the following:

x(t) = x0*e^kt

y(t) = y0*e^kt

For the same complex value exponent k, and different constants out front. From this it is easy to show that their derivatives are:

x'(t) = k*x0*e^kt

y'(t) = k*y0*e^kt

If we define a COLUMN (not a row) vector: V = [x(t);y(t)], then the derivative of that vector must be V' = k*V.

Now we can set up the matrix equation:

V' = A*V, for A being the coefficient matrix. What is a coefficient matrix? Well, it's a matrix organizing the coefficients in an obvious way, based on how they are given in the problem. In this case we have:

A = [-2, 1;

-4,-2]

To solve the matrix equation, we need to find A s.t. (A - k*I)V = 0, where I is the identity matrix. Showing this is true is left as an exercise to the reader. Anyways, we require that (A - k*I) = 0, or equivalently that the determinate of the matrix is 0. Therefor we have matrix:

[(-2 - k), 1;

-4,(-2 - k)] = 0

OR (looking at the determinant)

(-2 - k)*(-2 - k) + 4 = 0

Solve for k. This has 2 complex solutions. Are they indeed complex? How must they be related to each other?