dx1/ dt = 3x1 - 4x2
dx2/dt = x1 - x2
Let:
x1 = A exp(lt) and y1 = B exp(lt)
With these substitutions we than re-write our system in a consistent form:
l A exp(lt) = 3 A exp(lt) - 4B exp(lt)
lB exp(lt) = A exp(lt) - B exp(lt)
Now, collect like terms:
3 A exp(lt) - l A exp(lt) - 4 B exp(lt) = 0
A exp(lt) - B exp(lt) -lB exp(lt) = 0
Factor out the exp(lt) terms top and bottom to get:
(3 - l) A - 4B = 0
A + (-1 - l) B = 0
Set up the determinant as per prior problems: (A - l) D = 0 =
(3 - l………-4)
(1………-1 - l )
Find the characteristic equation, viz.
(3 - l) (-1 - l) + 4 = 0 or: l2 – 2l + 1 = 0
= (l - 1) (l - 1) = 0
When roots are real and equal we must seek additional solutions of the form shown below, given we have a case of repeated roots l1 = 1 = l2.
x1 = (A1t + A2) exp (lt)
x2 = (B1t + B2) exp (lt)
As before with previous problems, substitute the eigenvalue l1 = 1 into the matrix to get:
[2 ……..-4] [A]
[1……..-2] [B]
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