2130_Chapter7_Notes

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 =⇒...
View Full Document

## This note was uploaded on 03/19/2014 for the course APMA 2130 taught by Professor Bernardfulgham during the Fall '09 term at UVA.

Ask a homework question - tutors are online