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Unformatted text preview: Chapter 8 Solutions to 8.2 1. x 1 = 2x 1 3x 2 (D 2)x 1 + 3x 2 = 0. (1.1) x 2 = x 1 2x 2 x 1 + (D + 2)x 2 = 0. (1.2) Operating on (1.1) with D + 2, we obtain (D + 2)(D 2)x 1 + 3(D + 2)x 2 = 0. Combining this with (1.2) yields (D + 2)(D 2)x 1 + 3x 1 = 0 (D 2 1)x 1 = x 1 (t) = c 1 e t + c 2 e t . From (1.1) x 2 (t) = 1 3 (c 1 e t + 3c 2 e t ). 3. x 1 = 2x 1 + 4x 2 (D 2)x 1 4x 2 = 0. (3.1) x 2 = 4x 1 6x 2 4x 1 + (D + 6)x 2 = 0. (3.2) Operating on (3.1) with D + 6, we obtain (D + 6)(D 2)x 1 4(D + 6)x 2 = 0. Combining this with (3.2) yields (D + 6)(D 2)x 1 + 16x 1 = 0 (D + 2) 2 x 1 = x 1 (t) = e 2t (c 1 t + c 2 ). From (3.1), x 2 (t) = 1 4 e 2t [4c 1 + c 2 (1 4t)]. 5. x 1 = x 1 3x 2 (D 1)x 1 + 3x 2 = 0. (5.1) x 2 = 3x 1 + x 2 3x 1 + (D 1)x 2 = 0. (5.2) Operating on (5.1) with D 1, we obtain (D 1) 2 x 1 3(D 1)x 2 = 0. Combining this with (5.2) yields [(D 1) 2 + 9]x 1 = 0 (D 2 2D + 10)x 1 = x 1 (t) = e t (c 1 cos 3t + c 2 sin 3t). From (5.1) x 2 (t) = e t (c 1 sin 3t c 2 cos 3t). 7. x 1 = 2x 1 + x 2 + x 3 (D + 2)x 1 x 2 x 3 = 0. (7.1) x 2 = x 1 x 2 + 3x 3 x 1 + (D + 1)x 2 3x 3 = 0. (7.2) x 3 = x 2 3x 3 x 2 + (D + 3)x 3 = 0. (7.3) Operating on (7.2) with D + 2, we obtain (D + 2)x 1 + (D + 2)(D + 1)x 2 3(D + 2)x 3 = 0. Combining this with (7.1) yields (D 2 + 3D + 1)x 2 (3D + 7)x 3 = 0. Operating on (7.3) with (D 2 + 3D + 1) yields (D 2 + 3D + 1)x 2 + (D 2 + 3D + 1)(D + 3)x 3 = 0. Combining these two results gives (D + 2)(D 2 + 4D + 5)x 3 = 0 x 3 (t) = c 1 e 2t + e 2t (c 2 cos t + c 3 sin t). From (7.3), x 3 (t) = (D + 3)x 3 = e 2t [c 1 + c 2 (cos t sin t) + c 3 (cos t + sin t)]. From (7.2), x 1 = (D + 1)x 2 3x 3 = e 2t (2c 1 + c 2 cos t + c 3 sin t). 219 220 Ch. 8 Systems of Differential Equations 9. x 1 = 2x 1 + 5x 2 (D 2)x 1 5x 2 = 0. (9.1) x 2 = x 1 2x 2 x 1 + (D + 2)x 2 = 0. (9.2) Operating on (9.1) with D + 2, we obtain (D + 2)(D 2)x 1 5(D + 2)x 2 = 0. Combining this with (9.2) yields (D 2 4)x 1 + 5x 1 = 0 (D 2 + 1)x 1 = x 1 (t) = c 1 cos t + c 2 sin t. From (9.1), x 2 (t) = 1 5 [c 1 (sin t + 2cos t) + c 2 (cos t 2sin t)]. Imposing the initial conditions x 1 (0) = 0 and x 2 (0) = 1, we find that c 1 = 0 and c 2 = 5. Therefore, x 1 (t) = 5sin t and x 2 (t) = cos t 2sin t. 11. x 1 = x 1 + 2x 2 + 5e 4t (D 1)x 1 2x 2 = 5e 4t . (11.1) x 2 = 2x 1 + x 2 2x 1 + (D 1)x 2 = 0. (11.2) Operating on (11.1) with D 1 yields (D 1) 2 x 1 2(D 1)x 2 = 5(D 1)e 4t , that is, (D 1) 2 x 1 2(D 1)x 2 = 15e 4t . Combining this with (11.2) yields (D 1) 2 x 1 4x 1 = 15e 4t (D + 1)(D 3)x 1 = 15e 4t . The complementary function is x 1 c (t) = c 1 e t + c 2 e 3t . Using the trial solution x...
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 Spring '08
 Harris

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