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Unformatted text preview: MATH 150 Systems of firstorder linear equations Cheung Man Wai Tutorial 10 Example 1. Find the general solution of ˙ x = 3 4 1 1 x . First we use the ansatz x = v e λt as what we did before, then we get the characteristic equation det( A λ I ) = 3 λ 4 1 1 λ = ( λ 3)( λ + 1) + 4 = ( λ 1) 2 = 0 , which only has a repeated root λ = 1 . On solving the equations ( A λ I ) v = 0 , or 2 4 1 2 v 1 v 2 = , we obtain an eigenvector v = 2 1 and a single solution c 1 2 1 e t . To obtain another solution which is independent of the first one, the correct ansatz in general is given by x = ( η + t v ) e λt , (1) where λ and v are as above, and the constant vector η satisfies the following equation ( A λ I ) η = v . (2) 1 Now we solve (2), which in this example can be written as 2 4 1 2 η 1 η 2 = 2 1 . The first and the second equation are equivalent, thus η 1 = 2 η 2 + 1 and η = 2 η 2 + 1 η 2 = η 2 2 1 + 1 = η 2 v + 1 ....
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This note was uploaded on 01/28/2011 for the course MATH 150 taught by Professor T.qian during the Spring '09 term at HKUST.
 Spring '09
 T.Qian
 Math, Linear Equations, Equations

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