2130_Chapter7_Notes

2 4 a 2iv 0 v1 2v2 0 1 2 2 4 1 2 2 4 0 v1 v2 0 rst

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Unformatted text preview: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 Exa 2 (slide 4) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 1 Introduction Homogeneous system of n equations: x′ = P(t) x, x1 p11 (t) · · · p1n (t) . . . . .. . x = . , P(t) = . . . . . xn pn1 (t) · · · pnn (t) Fundamental set of solutions: x(1) , . . . , x(n) . Fundamental matrix: Ψ(t) = x(1) · · · x(n) x11 (t) · · · x1n (t) . . . .. . = . . . . xn1 (t) · · · xnn (t) Note: Ψ(t) is nonsingular (Ψ−1 exists) because x(1) , . . . , x(n) are linearly independent. 2 / 12 General Solution General solution of x′ = P(t) x : x11 (t) · · · . .. x = c1 x(1) + · · · + cn x(n) = . . . xn1 (t) · · · c1 . = Ψ(t) c for c = . . cn x1n (t) c1 . . . . . . xnn (t) cn Initial condition: x(t0 ) = x0 . x = Ψ(t) c =⇒ x(t0 ) = Ψ(t0 ) c =⇒ Ψ(t0 ) c = x0 =⇒ Ψ...
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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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