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Download Advanced Engineering Math 9th Edition with Mathematica by Erwin Kreyszig PDF

By Erwin Kreyszig

Publication through Kreyszig, Erwin

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The choice of ␻ needs experimentation, inspection of the curves obtained, and then changes on a trial-and-error basis. It is interesting to see how in the case of beats the period gets increasingly longer and the maximum amplitude gets increasingly larger as ␻ /(2␲) approaches the resonance frequency. 26. If 0 Ϲ t Ϲ ␲, then a particular solution yp ϭ K0 ϩ K1t ϩ K2 t 2 gives ypЉ ϭ 2K2 and 1 ypЉ ϩ yp ϭ K0 ϩ 2K2 ϩ K1t ϩ K2 t 2 ϭ 1 Ϫ ᎏ2 t 2; ␲ thus, 1 K2 ϭ Ϫ ᎏ2 , ␲ K1 ϭ 0, 2 K0 ϭ 1 Ϫ 2K2 ϭ 1 ϩ ᎏ2 .

Exact; the exactness test gives Ϫ3␲ sin ␲ x sinh 3y on both sides. Integrate the coefficient function of dx with respect to x, obtaining ͵ u ϭ M dx ϭ cos ␲ x cosh 3y ϩ k(y). Differentiate this with respect to y and equate the result to the coefficient function of dy: uy ϭ 3 cos ␲ x sinh 3y ϩ kЈ(y) ϭ N. Hence kЈ ϭ 0. qxd 9/21/05 10:17 AM Page 25 Instructor’s Manual 25 The implicit general solution is cos ␲ x cosh 3y ϭ c. 20. Solvable (A) as a Bernoulli equation or (B) by separating variables. (A) Set y 2 ϭ u since a ϭ Ϫ1; hence 1 Ϫ a ϭ 2.

The answer is y ϭ e؊kx cos ␻x. 32. The characteristic equation is ␭2 Ϫ 2␭ Ϫ 24 ϭ (␭ Ϫ 6)(␭ ϩ 4) ϭ 0. This gives as a general solution y ϭ c1e 6x ϩ c2 e؊4x. Hence y(0) ϭ c1 ϩ c2 ϭ 0, and by differentiation, 6c1 Ϫ 4c2 ϭ yЈ(0) ϭ 20. The answer is y ϭ 2e 6x Ϫ 2e؊4x. qxd 9/21/05 10:57 AM Page 37 Instructor’s Manual 37 34. Team Project. (A) We obtain (␭ Ϫ ␭1)(␭ Ϫ ␭2) ϭ ␭2 Ϫ (␭1 ϩ ␭2)␭ ϩ ␭1␭2 ϭ ␭2 ϩ a␭ ϩ b ϭ 0. Comparison of coefficients gives a ϭ Ϫ(␭1 ϩ ␭2), b ϭ ␭1␭2. (B) y Љ ϩ ayЈ ϭ 0. (i) y ϭ c1e؊ax ϩ c2 e 0x ϭ c1e؊ax ϩ c2.

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