{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

MAth 22

# MAth 22 - 8 For-3 0-l A 3 23 2 0 0 wehave —3——}t 0 ~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: 8. For -3 0 -l A: 3 23 2 0 0 wehave —3——}t 0 ~1 A-—AI= 3 2~A 3 We can compute the characteristic polynomial pot) = —- det(A — M) by expanding along the sec— ond column to get -3 _ A —-1 = (A m 2)(i2 +3)» +2) = o. — 2m +1)(?~ + 2). Hence the eigenvalues are A1 = ——2, A2 = —1, and A3 = 2. For M = —2 we have ~1 —1 A—MI=A+21= 3 0&0 3 2 The nullspace is generated by the vector v1 (—1, 0,1)T. For A2 = —l we have ll —2 0 ~l A—AZI=A+I: 3 3 3 2 0 l The nullspace is generated by the vector v2 (1,1, —Z)T. For A3 : 2 we have ~5 0 4‘ A—Agle—M: 3 0 3 l ,4, n r: i \ L U ”A l \ ,1 ll The nullspace is generated by the vector v3 = (0, ], 0)T. Thus wehave three exponential solutions: —1 371(1) = eAltVl = 6‘2’ 0 l 1 mo = elz’Vz = e” -—2 0 y3(t) = eA3’v3 = ezt l 0 Since the three eigenvalues are distinct, these solu— tions are linearly independent and form a fundamen— tal set of solutions. The general solution is YO) = C1Y1(t) + CZYZU) + C3Y3(t)‘ 14. The solution has the form Y“) = CiY1(t)+ C2Y2(_t) + 033%). where y1, y2, and y3 are the fundamental set of solu— tions found in Exercise 9.4.8. Hence we must have 1 "1 = y (0) 2 = C1Y1(0) + C2Y2(0) + C3Y3(0) =QW+QW+QW C1 =[V1,V2,V3] C2 , C3 where v], V2, and V3 are the eigenvectors of A found in Exercise 9.4.8. To solve the system we form the augmented matrix —1 l 0 1 [V1, V2, V3, y(0)] : 0 l l —-l 1 —2 0 2 This is reduced to the row echelon form 100- 010—3. 0012 Backsolving, we ﬁnd that C; = —4, C2 = —3, and C3 = 2. Hence the solution is —1 1 0 y(t) = —4e*2‘ 0 — 3e“ 1 +2e2t 1 1 —2 0 46‘” ~ Seq = 9—3?’ + 262‘ . \«46‘2’ + 66"} W ‘ 'z‘atmumwv nth at t."y‘r'.x.=5.'i'1':l«rzf'7frﬁ".\47r.,"v‘i".#"."nv. E i ...
View Full Document

{[ snackBarMessage ]}