By Erwin Kreyszig
Booklet by way of Kreyszig, Erwin
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Additional info for Advanced Engineering Math 9th Edition with Mathematica Computer Manual
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.
Note that yh ϭ e؊2t(A cos t ϩ B sin t); of course, this is not resonance. Furthermore, it is interesting that whereas a single term on the right will generate two terms in the solution, here we have—by chance—the converse. 1 _1 _ 1 _ 1 6. yp ϭ _ 10 cos t ϩ 5 sin t Ϫ 90 cos 3t ϩ 45 sin 3t 8. 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.
Newton’s law of cooling gives the model TЈ ϭ k(T Ϫ 25). By separation of variables and integration we obtain T ϭ cekt ϩ 25. The initial condition gives T(0) ϭ c ϩ 25 ϭ 10; hence c ϭ Ϫ15. 2197. 1/15. 2197) ϭ 23 [min]. 36. This will give a general formula for determining the half-life H from two measurements y1 and y2 at times t1 and t2, respectively. Accordingly, we use letters and insert the given numeric data only at the end of the derivation. We have yЈ ϭ ky, y ϭ y0 ekt and from this y1 ϭ y(t1) ϭ y0 ekt1, y2 ϭ y(t2) ϭ y0 ekt2.
Advanced Engineering Math 9th Edition with Mathematica Computer Manual by Erwin Kreyszig