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Lecture 26: Systems of Linear ODEs --- where do the solutions come from? (Nicholson Section 3.5)

Alternate Video Access via MyMedia | Video Duration: 44:11
Description: Started by reminding students of what the system of ODEs is.
2:00 --- Last class I presented the general solution as a linear combination of "things". First, I demonstrate that each of these "things" solves the system of ODEs. Translation: I demonstrate that if (lambda,v) is an eigenvalue-eigenvector pair of A then x(t) = exp(lambda t) v is a solution of the system of ODEs. Note: I blithely differentiate vectors but you’re probably not so happy doing that --- you’re used to differentiating single functions. You need to sit down and write the time-dependent vectors in terms of their components and convince yourself that differentiating the vector is the same as differentiating each component and that the things I do so quickly (like d/dt of exp(lambda t) v equals lambda exp(lambda t) v.
9:30 --- I demonstrate that if x1(t) and x2(t) are two solutions of the system of ODEs and c1 and c2 are constants then c1 x1(t) + c2 x2(t) is also a solution of the system of ODEs.
17:30 --- Given k eigenvalue-eigenvector pairs, I can write a solution of the system of ODEs that involves k coefficients c1, c2, … ck.
18:20 --- Choose the coefficients using the initial data.
21:00 --- If we have a problem in R^n, what happens if k doesn’t equal n? Do we need k to equal n? Answer: if I’m going to be able to solve every possible initial set of initial conditions I’m going to need n linearly independent eigenvectors in R^n. (You don’t know what linear independence means yet --- sorry! Come back to this in a few weeks. In the meantime, it would have sufficed to say “If A is a diagonalizable matrix then…”)
25:45 --- Presented a different argument for where the general solution came from. This argument relies on diagonalizing A.  We reduce a problem that we don’t know how to solve to a problem that we do know how to solve.