Solving Linear Differential Equation Systems using Eigenvalues and Eigenvectors
John Travis
Mississippi College
When solving linear systems of differential equations, the collection of equations can be written in the form
$Y^\prime = A Y$
If one assumes that there are solutions that behave like "straight lines" that pass through the origin in the phase plane, then in a vector form the solution must look something like
$Y = r(t) V$
where $r(t)$ is a monotonic function with range $(-\infty,0)$ or $(0,\infty)$ and $V$ is a "direction" vector.
If we presume $r(t) = c e^{\lambda t}$ then
$A (c e^{\lambda t}V) = A Y = Y^\prime = \lambda c e^{\lambda t}V = \lambda Y$
Cancelling $c e^{\lambda t}$ on both sides gives
$AV = \lambda V $
or
$ 0 = (\lambda I - A)V$
Therefore, the differential equation has a non-trivial straight-line solution of the form $c e^{\lambda t}V$ provided
$\left |{\lambda I - A}\right | = 0$
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(7, 0) (-3, 1/2) (7, 0) (-3, 1/2) |
So, let's solve some linear 2D systems with DISTINCT REAL EIGENVALUES and draw some nice pictures of everything we know. Then, let's let Sage do the solutions for us and see how things compare.
Click to the left again to hide and once more to show the dynamic interactive window |
What about when the eigensystem has imaginary components. Then, no "straight-line" solutions since $e^{iat} = cos(at)+isin(at)$. But how does this affect things?
Click to the left again to hide and once more to show the dynamic interactive window |
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[2, 3] [2, 3] |