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**Unformatted text preview: **Exercises 6.2
We ﬁnd the remaining roots by using the quadratic formula. Thus, we obtain the roots r = 1, 2 ± 3i. From the root r = 1, we obtain the solution y = x. From the roots r = 2 ± 3i, by applying the hint given in the problem, we see that a solution is given by y (x) = x2+3i = x2 {cos(3 ln x) + i sin(3 ln x)} . Therefore, by Lemma 2 on page 172 of the text, we ﬁnd that two real-valued solutions to this diﬀerential equation are y (x) = x2 cos(3 ln x) and y (x) = x2 sin(3 ln x). Since these functions and the function y (x) = x are linearly independent, we obtain the fundamental solution set x, x2 cos(3 ln x), x2 sin(3 ln x) . 33. With suggested values of parameters m1 = m2 = 1, k1 = 3, and k2 = 2, the system (34)–(35) becomes x + 5x − 2y = 0, y − 2x + 2y = 0. (6.7) (a) Expressing y = (x + 5x) /2 from the ﬁrst equation and substituting this expression into the second equation, we obtain 1 (x + 5x) − 2x + (x + 5x) = 0 2 ⇒ x(4) + 5x − 4x + 2 (x + 5x) = 0 ⇒ as it is stated in (36). (b) The characteristic equation corresponding to (6.8) is r 4 + 7r 2 + 6 = 0. This equation is of quadratic type. Substitution s = r 2 yields s2 + 7 s + 6 = 0 Thus √ r = ± −1 = ±i ⇒ s = −1, −6. √ √ r = ± −6 = ±i 6 , x(4) + 7x + 6x = 0, (6.8) and and a general solution to (6.8) is given by √ √ x(t) = c1 cos t + c2 sin t + c3 cos 6t + c4 sin 6t . 359 ...

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