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**Unformatted text preview: **Exercises 4.3
49. (a) To ﬁnd the roots of the auxiliary equation, p(r ) := 3r 3 + 18r 2 + 13r − 19 = 0, one can use Newton’s method or intermediate value theorem. We note that p(−5) = −9 < 0, p(−2) = 3 > 0, p(0) = −19 < 0, p(−4) = 25 > 0, p(−1) = −17 < 0, p(1) = 15 > 0. Therefore, the roots of p(r ) belong to the intervals [−5, −4], [−2, −1], and [0, 1], and we can take r = −5, r = −2, and r = 0 as initial quesses. Approximation yields r1 ≈ −4.832, r2 ≈ −1.869, and r3 ≈ 0.701. So, a general solution is given by y (t) = c1 er1 t + c2 er2 t + c3 er3 t = c1 e−4.832t + c2 e−1.869t + c3 e0.701t . (b) The auxiliary equation, r 4 − 5r 2 + 5 = 0, is of quadratic type. The substitution s = r 2 yields s − 5s + 5 = 0 Therefore, r1 = √ 5− 5 ≈ 1.176 , 2 r2 = √ 5+ 5 ≈ 1.902 , 2 r3 = −r1 , and r4 = −r2
2 ⇒ √ 5± 5 s= 2 ⇒ √ r=± s=± √ 5± 5 . 2 are the roots of the auxiliary equation, and a general solution to y (ıv) − 5y + 5y = 0 is given by y (t) = c1 er1 t + c2 e−r1 t + c3 er2 t + c4 e−r2 t . (c) We can use numerical tools to ﬁnd the roots of the auxiliary ﬁfth degree polynomial equation r 5 − 3r 4 − 5r 3 + 15r 2 + 4r − 12 = 0. Alternatively, one can involve the rational root theorem and examine the divisors of the free coeﬃcient, −12. These divisors are ±1, ±2, ±3, ±4, ±6, and ±12. By inspection, r = ±1, ±2, and 3 satisfy the equation. Thus, a general solution is y (t) = c1 e−t + c2 et + c3 e−2t + c4 e2t + c5 e3t . EXERCISES 4.3: Auxiliary Equations with Complex Roots, page 177 1. The auxiliary equation in this problem is r 2 + 9 = 0, which has roots r = ±3i. We see that α = 0 and β = 3. Thus, a general solution to the diﬀerential equation is given by y (t) = c1 e(0)t cos 3t + c2 e(0)t sin 3t = c1 cos 3t + c2 sin 3t. 177 ...

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