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Unformatted text preview: Exercises 5.2
Hence, x = −c1 et − 2c2 e2t − c3 e3t . Finally, we ﬁnd z using (5.10). z = − −c1 et − 2c2 e2t − c3 e3t + c1 et + c2 e2t + c3 e3t 27. We eliminate z by expressing z= 1 1 (−x + 4x) = − (D − 4)[x] 4 4 (5.12) = 2c1 et + 4c2 e2t + 4c3 e3t . from the ﬁrst equation and substituting (5.12) into the second and third equations. We obtain 1 2 − (D − 4)[x] + (D − 4)[y ] = 0, 4 1 1 2x + 4y + D − (D − 4)[x] − 4 − (D − 4)[x] 4 4 After some algebra, the above system simpliﬁes to −(D − 4)[x] + 2(D − 4)[y ] = 0, D2 − 8D + 8 [x] − 16y = 0. We use the second equation to ﬁnd that y= Then the ﬁrst equation becomes −(D − 4)[x] + 2(D − 4) ⇒ (D − 4) −1 + 1 D 2 − 8D + 8 [x] = 0 16 [x] = 0 ⇒ (D − 4)D (D − 8)[x] = 0. 1 D 2 − 8D + 8 [x]. 16 (5.13) = 0. 1 D 2 − 8D + 8 8 Solving the characteristic equation, we get r = 0, 4, and 8; so x = c1 e8t + c2 e4t + c3 . 271 ...
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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