1.6.6 Example F' Let us use Leibnitz's theorem to calculate A"(k²vk). From equation (1.99), we have, for ak = k2 and yk = Vk, the result A" (k²uk) = k²A"vVk + (") (Ak2)(스"-1uk+1) + 2 (A²k²)(A" where all other terms are zero since A™k2 = 0 for m > 2. Therefore, A" (k²vk) = k²A"vk + (2k + 1)A"-lUk+1+n(n – 1)A"-²vk+2• (1.132) For the particular case of vk = a*, we obtain A"(k²a*) = [(a – 1)²k² + na(a – 1)(2k +1) + a?n(n – 1)](a – 1)"-?a*. (1.133)
1.6.6 Example F' Let us use Leibnitz's theorem to calculate A"(k²vk). From equation (1.99), we have, for ak = k2 and yk = Vk, the result A" (k²uk) = k²A"vVk + (") (Ak2)(스"-1uk+1) + 2 (A²k²)(A" where all other terms are zero since A™k2 = 0 for m > 2. Therefore, A" (k²vk) = k²A"vk + (2k + 1)A"-lUk+1+n(n – 1)A"-²vk+2• (1.132) For the particular case of vk = a*, we obtain A"(k²a*) = [(a – 1)²k² + na(a – 1)(2k +1) + a?n(n – 1)](a – 1)"-?a*. (1.133)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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