# chapter8 - Chapter 8 Solutions to §8.2 1 x 1 ´ = 2x 1 –...

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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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## This homework help was uploaded on 04/09/2008 for the course ENGR, STAT 320, 262, taught by Professor Harris during the Spring '08 term at Purdue.

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chapter8 - Chapter 8 Solutions to §8.2 1 x 1 ´ = 2x 1 –...

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