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Unformatted text preview: Exercises 5.2
19. From the ﬁrst equation, we conclude that y = x − 4x. Substitution into the second equation yields (x − 4x) = −2x + (x − 4x) ⇒ x − 5x + 6x = 0. The characteristic equation, r 2 − 5r + 6 = 0, has roots r = 2, 3, and so a general solution is x(t) = c1 e2t + c2 e3t ⇒ y (t) = c1 e2t + c2 e3t − 4 c1 e2t + c2 e3t = −2c1 e2t − c2 e3t . We ﬁnd constants c1 and c2 from the initial condition. 1 = x(0) = c1 e2(0) + c2 e3(0) = c1 + c2 , 0 = y (0) = −2c1 e
2(0) − c2 e 3(0) = −2c1 − c2 ⇒ c1 = −1, c2 = 2 . Therefore, the answer to this problem is x(t) = −e2t + 2e3t , y (t) = 2e2t − 2e3t . 21. To apply the elimination method, we write the system using operator notation: D 2 [x] − y = 0, −x + D 2 [y ] = 0. (5.5) Eliminating y by applying D 2 to the ﬁrst equation and adding to the second equation gives D 2 D 2 − 1 [x] = 0, which reduces to D 4 − 1 [x] = 0. (5.6) The corresponding auxiliary equation, r 4 − 1 = 0, has roots ±1, ±i. Thus, the general solution to (5.6) is given by x(t) = C1 et + C2 e−t + C3 cos t + C4 sin t. Substituting x(t) into the ﬁrst equation in (5.5) yields y (t) = x (t) = C1 et + C2 e−t − C3 cos t − C4 sin t. (5.8) 267 (5.7) ...
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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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