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Extra info for Calculus 3c-4, Examples of Applications of The Power Series Method By Solution of Differentia Equations with Polynomial Coefficients

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E. com 51 Calculus 3c-4 Larger examples of the power series method with a leap of 2 in the indices. If y(0) = 0, then a0 = 0, so a2n = 0 by induction. If a1 = y (0) = k, then a2n+1 = 2 1 a2n−1 = a2n−1 . 2n n If we multiply by n! a1 = k, hence a2n+1 = k , n! n ∈ N0 . By (1) the power series solution is ∞ k 1 2n+1 x = kx · exp(x2 ), n! n=0 x ∈ R. Please click the advert what‘s missing in this equation? You could be one of our future talents MAERSK INTERNATIONAL TECHNOLOGY & SCIENCE PROGRAMME Are you about to graduate as an engineer or geoscientist?

4) Find a solution of (25) through the line element (0, 0, 2). The example can be treated in several ways. The main variant is of course the power series method. 1) By insertion of ∞ ∞ an xn , y= n=0 ∞ nan xn−1 , y = n(n − 1)an xn−2 , y = n=1 n=2 and addition of some zero terms we get 0 = (2x2 + 1) ∞ ∞ ∞ 2n(n − 1)an xn + = n=2 (n=0) ∞ = n=0 ∞ = ∞ ∞ d2 y dy + 8x + 4y = (2x2 +1) n(n−1)an xn−2 +8x nan xn−1 +4 an xn 2 dx dx n=2 n=1 n=0 ∞ n(n − 1)an xn−2 + n=2 {2n2 −2n+8n+4}an xn + ∞ 8an xn + 4an xn n=0 n=1 (n=0) ∞ (n+2)(n+1)an+2 xn n=0 {2(n2 +3n+ 2)an +(n+2)(n+1)an+2 }xn = n=0 ∞ (n + 2)(n + 1){2an + an+2 }xn .

A) If n = 2p + 1 is odd, then the recursion formula becomes (2p − 1)a2p+1 = 2a2p−1 , p ≥ 1. , and it follows by induction that all a2p+1 = 0. b) If n = 2p (≥ 3), hence p ≥ 2, we get instead (2p − 2)a2p = 2a2p−2 , p ≥ 2, which is reduced to (p − 1)a2p = a2(p−1) , p ≥ 2. Here there are more possible solutions: i) The elegant one. Multiply by (p − 2)! = 0. a2 = 1, hence a2p = 1 (p − 1)! for p ≥ 2 (even for p ≥ 1). ii) Recursion. By iteration of a2p = 1 a2(p−1) , p−1 p ≥ 2, (notice how p − 1 occurs on the right hand side) we get a2p = 1 1 1 1 · · · · · a2·1 = , p−1 p−2 1 (p − 1)!

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