solve the system of linear differential equations using eigenvectors and eigenvaluesby RET80 Tags: differential, eigenvalues, eigenvectors, equations, linear, solve 

#1
Dec710, 03:30 PM

P: 15

1. The problem statement, all variables and given/known data
solve the system of firstorder 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] (λIA)x = 0 [λ1 4] [0 λ2] which gives the eiganvalues: (λ1)(λ2) 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? 



#2
Dec710, 06:32 PM

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P: 11,536

You calculated the eigenvectors incorrectly.



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