By Erwin Kreyszig
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Extra info for Advanced Engineering Math 9th Edition with Mathematica Computer Manual
5 sin 4t 10. 04 sin 4t 12. 1 cos 2t. Note that, ordinarily, yp will consist of two terms if r(x) consists of a single trigonometric term. 14. 2 sin t 1 1 _ _1 16. y ϭ Ϫ_ 63 cos 8t ϩ 8 sin 8t ϩ 63 cos t. From the graph one can see the effect of (cos t)/63. There is no corresponding sine term because there is no damping and hence no phase shift. 18. 5 cos 3t Ϫ 2 sin 3t 20. 9t Ϫ 98 cos 5t 22. The form of solution curves varies continuously with c. Hence if you start from c ϭ 0 and let c increase, you will at first obtain curves similar to those in the case of c ϭ 0.
4x. 28. The roots of the characteristic equation 2 Ϫ 9 ϭ 0 are 3 and Ϫ3. 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).
Thus c2 ϭ 2. This gives the particular solution y ϭ (4 ϩ 2 ln ͉x͉)/x. Make sure to explain to the student why we cannot prescribe initial conditions at t ϭ 0, where the coefficients of the ODE written in standard form (divide by x 2) become infinite. 14. 5. 2. 5. 8. 16. Team Project. (A) The student should realize that the present steps are the same as in the general derivation of the method in Sec. 1. An advantage of such specific derivations may be that the student gets a somewhat better understanding of the method and feels more comfortable with it.