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solve the system of linear differential equations using eigenvectors and eigenvalues |
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| Dec7-10, 03:30 PM | #1 |
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solve the system of linear differential equations using eigenvectors and eigenvalues
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? |
| Dec7-10, 06:32 PM | #2 |
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You calculated the eigenvectors incorrectly.
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