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In particular, u Љ ϭ 0 in Case II gives u ϭ c1 ϩ c2 x at once. 2, page 59 2. 12y ϭ 0. 3x. 4. The characteristic equation 2 ϩ 4 ϩ 4 2 ϭ ( ϩ 2)2 ϭ 0 has the double root Ϫ2, so that the corresponding general solution is y ϭ (c1 ϩ c2 x)e؊2 x. 6. The characteristic equation 2 ϩ 2 ϩ 5 ϭ ( ϩ 1)2 ϩ 4 ϭ 0 has the roots Ϫ1 Ϯ 2i, so that the general solution is y ϭ e؊x(A cos 2x ϩ B sin 2x). 8. 3x. qxd 9/21/05 10:57 AM Page 35 Instructor’s Manual 35 10. From the characteristic equation 2 Ϫ 2 ϭ ( ϩ ͙2ෆ)( Ϫ ͙2ෆ) ϭ 0 we see that the corresponding general solution is y ϭ c1e؊x͙2ෆ ϩ c2 ex͙2ෆ 12.
The two conditions follow trivially from the condition in the text. Conversely, by combining the two conditions we have L(cy ϩ kw) ϭ L(cy) ϩ L(kw) ϭ cLy ϩ kLw. 4. Modeling: Free Oscillations (Mass–Spring System), page 61 Purpose. To present a main application of second-order constant-coefficient ODEs myЉ ϩ cyЈ ϩ ky ϭ 0 resulting as models of motions of a mass m on an elastic spring of modulus k (Ͼ 0) under linear damping c (м 0) by applying Newton’s second law and Hooke’s law. These are free motions (no driving force).
Formula (2) gives ͵ ϩ sin 3x ͵ (cos 3x sec 3x)/3 dx yp1 ϭ Ϫcos 3x (sin 3x sec 3x)/3 dx ϭ _19 cos 3x ln ͉cos 3x͉ ϩ _13x sin 3x. For yp2 the method of undetermined coefficients gives y ϭ Ϫ_1x cos 3x. p2 6 SOLUTIONS TO CHAP. 2 REVIEW QUESTIONS AND PROBLEMS, page 102 10. Undetermined coefficients, where Ϫ3 is a double root of the characteristic equation of the homogeneous ODE, so that the Modification Rule applies. The second term on the right, Ϫ27x 2, requires a quadratic polynomial. A general solution is y ϭ (c ϩ c x)e؊3x ϩ _1x 2e؊3x Ϫ 3x 2 ϩ 4x Ϫ 2.