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Hence a general solution is y ϭ c1e؊3x ϩ c2 e 3x. Now c1 ϩ c2 ϭ Ϫ2 from the first initial condition. By differentiation and from the second initial condition, Ϫ3c1 ϩ 3c2 ϭ Ϫ12. The solution of these two equations is c1 ϭ 1, c2 ϭ Ϫ3. Hence the answer is y ϭ e؊3x Ϫ 3e3x. 30. The characteristic equation is ␭2 ϩ 2k␭ ϩ (k 2 ϩ ␻2) ϭ (␭ ϩ k)2 ϩ ␻2 ϭ 0. Its roots are Ϫk Ϯ i␻. Hence a general solution is y ϭ e؊kx(A cos ␻x ϩ B sin ␻x). For x ϭ 0 this gives y(0) ϭ A ϭ 1. With this value of A the derivative is yЈ ϭ e؊kx(Ϫk cos ␻x Ϫ Bk sin ␻x Ϫ ␻ sin ␻x ϩ B␻ cos ␻x).

6 sin 10t. The initial conditions are I(0) ϭ 0, Q(0) ϭ 0, which because of (1Ј), that is, Q(0) LIЈ(0) ϩ RI(0) ϩ ᎏ ϭ E(0) ϭ 0, C leads to IЈ(0) ϭ 0. 2 IЈ(0) ϭ Ϫ20c1 ϩ c2 ϩ 16 ϭ 0, c2 ϭ Ϫ40. 6 sin 10t. 18. 2(␭ ϩ 8)(␭ ϩ 10) ϭ 0. 2 ϭ 820 by formula (1Ј) in the text and Q(0) ϭ 0. Also, EЈ ϭ Ϫ1640 sin 10t. 6IЈ ϩ 16I ϭ Ϫ1640 sin 10t. The answer is I ϭ 160 e؊8t Ϫ 205e؊10t ϩ 45 cos 10t ϩ 5 sin 10t. 20. Team Project. (a) Iෂp ϭ Kei␻ t, IෂpЈ ϭ i␻Kei␻ t, IෂpЉ ϭ Ϫ␻2Kei␻ t. Substitution gives 1 (Ϫ␻2L ϩ i␻R ϩ ᎏ ) Kei␻ t ϭ E0␻ ei␻ t.

In the previous problem (Prob. 3) the situation is similar. 6. y ϭ (c1 ϩ c2 x)e؊2x Ϫ _14e؊2x sin 2x. The characteristic equation of the homogeneous ODE has the double root Ϫ2. The function on the right is such that the Modification Rule does not apply. 48 Instructor’s Manual 8. Ϫ5 is a double root. 100 sinh 5x ϭ 50e 5x Ϫ 50e؊5x. Hence we may choose yp ϭ yp1 ϩ yp2 with yp2 ϭ Cx 2e؊5x according to the Modification Rule. Substitution gives yp1 ϭ _12e 5x, yp2 ϭ Ϫ25x 2e؊5x. Answer: y ϭ (c ϩ c x)e؊5x ϩ _1e 5x Ϫ 25x 2e؊5x.

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