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Unformatted text preview: ’ Î») = Î»2 âˆ’ 2Î» âˆ’ 3 for then c1 = 2, c2 = 4 and = (Î» âˆ’ 3)(Î» + 1) , u(0) = 2v1 + 4v2 .
But v1 , v2 are eigenvectors corresponding to
1
respective eigenvalues 2, 2 so set
2t âˆ’6
,
âˆ’2 âˆ’2 âˆ’5
.
1
4 1. u(t) = 6. xk = 2e2t v1 âˆ’ 2et/2 v2 v2 u(0) = when A is the matrix 4. xk = 4e2t v1 + 2et/2 v2 [ v1 d u ( t)
1 t/2
v2 =
e
,
2
dt = 2(2e2t )v1 + 4 âˆ’1
.
1 v2 = Then u(0) is the given initial value and âˆ’2
,
6 u(0) = 3 t/2 u(t) = 2e v1 + 4e v2 . the eigenvalues of A are Î»1 = 3, Î»2 = âˆ’1 and
corresponding eigenfunctions
v1 = 1
,
âˆ’1 v2 = âˆ’5
1 cruz (fmc326) â€“ HW10 â€“ gilbert â€“ (56540)
form a basis for R2 because Î»1 = Î»2 . Thus
u(0) = c1 v1 + c2 v2 .
To compute c1 , c2 we apply row reduction to
the augmented matrix
[ v1 v2 u(0) ] =
âˆ¼ 10
01 1
âˆ’1 âˆ’5 âˆ’6
1 âˆ’2 4
,
2 for then c1 = 4, c2 = 2.
Now set
u(t) = 4e3t v1 + 2eâˆ’t v2 .
Then
u(0) = 4v1 + 2v2 = âˆ’6
âˆ’2 , while
Au(t) = 4e3t Av1 + 2eâˆ’t Av2
= 12e3t v1 âˆ’ 2 âˆ’eâˆ’t v2 = d u ( t)
,
dt so
u(t) = 4e3t
= âˆ’5
1
+ 2eâˆ’t
1
âˆ’1
4e3t âˆ’ 10eâˆ’t
âˆ’4e3t + 2eâˆ’t solves the diï¬€erential equation. 4...
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This document was uploaded on 03/16/2014 for the course M 340L at University of Texas at Austin.
 Spring '08
 PAVLOVIC
 Matrices

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