2130_Chapter7_Notes

5 11 3 exa 1 slide 3 2 1 a 2 i a i check 1

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Unformatted text preview: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 Exa 2 (slide 4) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 1 Introduction Linear system of equations: x′ = P(t)x + g(t). Homogeneous system: g(t) = 0 =⇒ x′ = P(t)x. Homogeneous with constant coefficients: P(t) = A =⇒ x′ = Ax. Example (n = 2) : x′ 1=12 x′ 21 2 This matrix equation is equivalent to: x1 . x2 x′ = x1 + 2x2 , 1 x′ = 2x1 + x2 . 2 Note: n = 1 =⇒ x′ = ax =⇒ x = Ceat . 2 / 11 How to Solve n = 1 : x′ = ax =⇒ x = Ceat . For n ≥ 2, let’s try: x = veλt , where v is a column vector and λ is a scalar. When is veλt a solution of x′ = Ax ? Plug it in and find out: x = veλt d d =⇒ x′ = dt veλt = v dt eλt = v λeλt = λveλt and Ax = A veλt = Aveλt =⇒ x′ = Ax ⇐⇒ λveλt = Aveλt ⇐⇒ Av = λv. Theorem x = veλt v = 0 is a solution of x′ = Ax ⇐⇒ λ is an eigenvalue for A with eigenvector v. 3 / 11 2 Example 1 x′ = 11 x 41 Equivalent to: x′ = x1 + x2 , 1 x′ = 4x1 + x2 . 2 A Find the eigenvalues for A : det(A − λI) = 1−λ 1 = (1 − λ)2 − 4 = 0 4 1−λ =⇒ (1 − λ)2 = 4 =⇒ 1 − λ = ±2 =⇒ λ = 1 ± 2 = 3, −1. Now find the eigenvectors for λ1 = 3 and λ2 = −1. 4 / 11 Exa 1 (slide 2) 1−3 1 4 1−3 λ1 = 3 =⇒...
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This note was uploaded on 03/19/2014 for the course APMA 2130 taught by Professor Bernardfulgham during the Fall '09 term at UVA.

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