Unformatted text preview: di erential equation y = Ay where 2 −2 3
A = 1 −1 1 ,
−1
22 1
y(0) = 1 0 λ = 1, 1, 1 −4 8 4
100
1 −2 3
t2 e t −2 4 2 = et 0 1 0 + tet 1 −2 1 +
2
000
001
−1
21 eAt 4
1
−1
2t
t + tet −1 + t e 2 1
y=e
2
0
0
1 [5] Solve the di erential equation y = Ay where −2 −1
1
1 −1 ,
A= 2
2
3
1 0
y ( 0) = 1 2 λ = 0, 0, 0 4
40
100
−2 −1
1
2
t
−4 −4 0 1 −1 +
= 0 1 0 + t 2
2
4
40
001
2
3
1 eAt 0
1
4
2
t
−4 y = 1 + t −1 +
2
2
5
4 [6] Solve the di erential equation y = Ay where 1
3
1
1
1 ,
A = −1
2 −2 −2 1
y ( 0) = 0 2 , 3 × 3 Exercise Set I (identical roots), November λ = 0, 0, 0 eAt 100
0
4
2
1
3
1
2
t
0 −4 −2 1
1 +
= 0 1 0 + t −1
2
001
0
8
4
2 −2 −2 4
1
3
2
t
−4 y = 0 + t 1 +
2
8
2
−2 [7] Solve the di erential equation y = Ay where −2 3
2
A = −1 2 −2 ,
−1 1
3 2
y(0) = 0 1 λ = 1, 1, 1 4 −4 −8
100
−3 3
2
2t
te 4 −4 −8 = et 0 1 0 + tet −1 1 −2 +
2
001
−1 1
2
0
0
0 eAt 2
0
−4
t2 e t 0
y = et 0 + tet −4 +
2
1
0
0 [8] Solve the di erential equation y = Ay where 2
1 −1
2 ,
A = −2 −1
−1 −1
2 1
y ( 0) = 0 1 λ = 1, 1, 1 100
1
1 −1
000
t2 et 0 0 0
2 +
= et 0 1 0 + tet −2 −2
2
000
001
−1 −1
1 eAt 1
0
0
2t
t + tet 0 + t e 0 0
y=e
2
1
0
0 , 3 × 3 Exercise Set J (symmetric matrices), November , 3 × 3 Exercise Set J (symmetric matrices)
Linear Algebra, Dave Bayer, November , [1] Find An where A is the matrix 210
A = 1 1 1
012 1 −2
1
1 0 −1
111
n
n
0
2
3
−2
4 −2 +
00
0 +
1 1 1
=
6
2
3
1...
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This document was uploaded on 03/24/2014 for the course MATH V2010 at Barnard College.
 Spring '14
 DaveBayer
 Linear Algebra, Algebra

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