2130_Chapter7_Notes

Solving a nonhomogeneous linear system is a three

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Unformatted text preview: = v(1) e2t , let’s try: x = vte2t for some v = 0. x′ = v(e2t + 2te2t ) = v(2t + 1)e2t and Ax = Avte2t x′ = Ax =⇒ v(2t + 1)e2t = Avte2t =⇒ 2v t + v = Avt =⇒ Av = 2v (t vector) and v = 0 (constant vector). However, we assumed v = 0. ⇒⇐ Contradiction! 5/9 3 2nd Sol (slide 2) For our second attempt, let’s try: x = (vt + w)e2t . x′ = ve2t + (vt + w)2e2t = 2vte2t + (v + 2w)e2t and Ax = A(vt + w)e2t = Avte2t + Awe2t , x′ = Ax =⇒ Av = 2v (te2t vector) and Aw = v + 2w (e2t vector). Av = 2v =⇒ v is an eigenvector for λ = 2 =⇒ v = c −c , c = 0. Aw = v + 2w =⇒ Aw − 2Iw = v =⇒ (A − 2I)w = v =⇒ w is a generalized eigenvector for λ = 2. w1 w2 To solve for w = , reduce the augmented matrix A − 2I | v . 6/9 2nd Sol (slide 3) A − 2I | v = −1 −1 c R2 +R1 −1 −1 c −−→ −− 1 1 −c 0 00 =⇒ −w1 − w2 = c =⇒ w2 = −w1 − c =⇒ w = Remember that v = c then v = 1 −1 0 1 w , c=0 +c =w −1 −1 −w − c 1 . Pick c = 1 and w = 0, −1 and w =...
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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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