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
Question

Explain this plz

1.6.6 Example F
Let us use Leibnitz's theorem to calculate A" (k²v%). From equation (1.99),
we have, for Xk = k² and yk = Vk, the result
A"(k°vx) = k°A"uk + (G)(Ak)(A"-1uk+1)
C)(A?k²)(A"-2uk+2), (1.131)
|(A²k?)
+
where all other terms are zero since Amk?
= 0 for m > 2. Therefore,
A" (k²vk) = k²A"vr + (2k + 1)A"-1 Uk+1 +n(n – 1)A"²Uk+2.
(1.132)
For the particular case of vk
ak, we obtain
A" (k²a*) = [(a – 1)²k² + na(a – 1)(2k +1)
+a?n(n – 1)](a – 1)"-2a*. (1.133)
Transcribed Image Text:1.6.6 Example F Let us use Leibnitz's theorem to calculate A" (k²v%). From equation (1.99), we have, for Xk = k² and yk = Vk, the result A"(k°vx) = k°A"uk + (G)(Ak)(A"-1uk+1) C)(A?k²)(A"-2uk+2), (1.131) |(A²k?) + where all other terms are zero since Amk? = 0 for m > 2. Therefore, A" (k²vk) = k²A"vr + (2k + 1)A"-1 Uk+1 +n(n – 1)A"²Uk+2. (1.132) For the particular case of vk ak, we obtain A" (k²a*) = [(a – 1)²k² + na(a – 1)(2k +1) +a?n(n – 1)](a – 1)"-2a*. (1.133)
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