2130_Chapter7_Notes

Find the eigenvalues for a deta i 4 2 4 1

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Unformatted text preview: = 0 using undetermined coefficients. Since g(t) = 2 −t 2e−t 0 = e+ t 3t 0 3 and λ = −1 is a root of the characteristic equation, a particular solution looks like: x = (at + b)e−t + ct + d. Note: when g(t) = ueλt and λ is a root of the characteristic equation, we can’t use: xp = ateλt . Instead, we must use: xp = (at + b)eλt . 7 / 10 4 UC (slide 2) Next, plug x into the equation and solve for the coefficients. x = (at + b)e−t + ct + d =⇒ x′ = ae−t − (at + b)e−t + c = −ate−t + (a − b)e−t + c, and Ax + g(t) = Aate−t + Abe−t + Act + Ad e−t + 2 0 + = Aate−t + Ab + 0 3 t e−t + Ac + 2 0 0 3 t + Ad . Set x′ = Ax + g(t), and match up the corresponding vectors: te−t : t: e−t : Aa = − a , Ac + 0 3 0 0 = 1: , 2 0 Ab + = a − b, Ad = c. 8 / 10 UC (slide 3) System of matrix equations: Aa = −a, Ac + 0 3 Ab + = 0 0 2 0 = a − b, , Ad = c. Aa = −a =⇒ a is an eigenvector for λ = −1 =⇒ a = α , α = 0 α Ab + 2 0 = a − b =⇒ Ab + Ib = a − =⇒ (A + I)b = α−2 α =⇒ −1 1 1 −1 2 0 b1 b2 = α−2 α −1 1 α − 2 R2 +R1 −1 1 α − 2 −−→ −− α 1 −1 0 0 2α − 2 =⇒ 2α − 2 = 0 =⇒ α = 1 =⇒ a = 1 , 1 and − b1 + b2 = −1 =⇒ b2 = b1 − 1 =⇒ b = β β −1 , β any constant 9 / 10 5 UC (slide 4) System of matrix equations: a= 1 1 β β −1 , b= Ac + 0 3 = 0 0 Ac + 0 3 = 0 0 A− 1 = −2 1 −1 1 −...
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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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