solve the system of linear differential equations using eigenvectors and eigenvalues


by RET80
Tags: differential, eigenvalues, eigenvectors, equations, linear, solve
RET80
RET80 is offline
#1
Dec7-10, 03:30 PM
P: 15
1. The problem statement, all variables and given/known data
solve the system of first-order linear differential equations:
(y1)' = (y1) - 4(y2)
(y2)' = 2(y2)

using the equation:
(λI -A)x = 0

2. Relevant equations
using eigenvectors and eigenvalues
in the book 'Elementary Linear Algebra' by Larson and Falvo - Section 7.4 #19

3. The attempt at a solution
(y1)' = (y1) - 4(y2)
(y2)' = 2(y2)

makes the matrix:
[1 -4]
[0 2]

(λI-A)x = 0
[λ-1 4]
[0 λ-2]

which gives the eiganvalues:
(λ-1)(λ-2)
λ = 1
λ = 2

then input it back into the original matrix A:
λ = 1
(1I - A) =
[0 4]
[0 -1]

which reduces to:
[0 0]
[0 1]

and creates the eiganvector:
[0]
[1]

then continued onto:
λ = 2
(2I - A) =
[1 4]
[0 0]

which creates the eigenvector:
[1]
[-4]

so then I put the two eigenvectors together and create P:
[1 1]
[0 -4]

then I need to find P^-1 so that I may use PAP^-1 to find the differential equation:
I set P equal to the identity matrix and solve (I think this is where things start to go wrong...)
[1 1/4]
[0 -1/4]

then I solve for PAP^-1
[1 1] [1 -4] [1 1/4]
[0 -4] [0 2] [0 -1/4]

I take AP^-1 first in the multiplication and get:
[1 5/4]
[0 1/2]

and then add on P and get:
[1 7/4]
[0 2]

and that just looks wrong, I stopped there in fear of continuing further, but the answer IN THE BOOK reads:
(y1) = (C1)e^t - 4(C2)e^2t
(y2) = (C2)e^2t

and it looks like if I were to continue it would not even come close to that answer...Where did I go wrong?
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vela
vela is offline
#2
Dec7-10, 06:32 PM
Emeritus
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Thanks
PF Gold
P: 11,536
You calculated the eigenvectors incorrectly.


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