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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**

**Example text**

2n − 1 2n(2n − 1) When we multiply by (2n)! this equation is transferred into 4n n! (2n)! (2(n−1))! 0! a2n = n−1 a2(n−1) = · · · = 0 a0 = k, n 4 n! 4 (n−1)! 4 · 0! hence a2n = 4n n! · k, (2n)! n ∈ N0 . The formal series is ∞ k 4n n! 2n x . (2n)! x2n+2 (2n)! 2x2 4(n+1)x2 = · = → 0 for n → ∞. = an (x) (2n + 2)! x2n (2n+2)(2n+1) 2n+1 This proves that the formal series is convergent for every x ∈ R. It follows from the existence and uniqueness theorem that the complete solution of the differential equation is given by y = c1 x exp(x2 ) + c2 ∞ 4n n!

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?

N=0 which fulﬁls y(0) = 0, is of the form k 1) We immediately get the sum function by ∞ ∞ x2n+1 (x2 )n f (x) = =x = x · exp(x2 ), n! n! n=0 n=0 = ∞. and the radius of convergence is 2) When we insert the formal power series ∞ ∞ n y= an x , ∞ n−1 y = n=0 nn x , n(n − 1)an xn−2 , y = n=1 n=2 into the diﬀerential equation, we get by adding some zero terms, 0 = y − 2xy − 4y ∞ ∞ n(n − 1)an xn−2 − = n=2 ∞ 2nan xn − n=0 n=1 (n=0) ∞ ∞ (n + 2)(n + 1)an+2 xn − = 4an xn n=0 ∞ 2(n + 2)an xn n=0 (n + 2){(n + 1)an+2 − 2an }xn .