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

Example text

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! 2n x (2n)! n=0 for x ∈ R, where c1 and c2 are arbitrary constants. 3 Given the initial value problem ⎧ 2 ⎪ ⎨ 4x d y + 2 dy − y = 0, x ≥ 0, dx2 dx (20) 1 ⎪ ⎩ y(0) = 1, y (0) = . 2 1) Assume that ∞ an xn y= n=0 is a power series solution of (20). Find a recursion formula for the coeﬃcients a n . 2) Prove that the recursion formula is fulﬁlled for 1 , (2n)!

2n − 1 n=1 We get by the criterion of roots, n 1 |an (x)| = √ x2 → x2 n 2n − 1 for n → ∞. The condition of convergence x2 < 1 implies that = 1. 1 is divergent, so the series is divergent at both of the endpoints of the interval 2n − 1 of convergence. Clearly, ∞ n=1 Sum function. If |x| < 1, then ∞ y = 1− ∞ ∞ 1 1 x2n = 1 − x x2n−1 = 1 − x 2n − 1 2n − 1 n=1 n=1 n=1 = 1−x x ∞ 0 n=0 t2n dt = 1 − x x 0 1 x dt = 1 − 1 − t2 2 x 0 x t2n−2 dt 0 1 1 + 1+t 1−t dt = 1 − x ln 2 1+x 1−x . Alternatively we get the solution y = x by inspection.

Y1 (x) 1 dx dx 2x The complete solution for x ∈ R+ is not surprisingly √ √ y = c1 cosh( x) + c2 sinh( x), c1 , c2 arbitrary. Notice that √ sinh( x) = ∞ ∞ √ 1 1 ( x)2n+1 = xn+(1/2) (2n + 1)! (2n + 1)! n=0 n=0 formally is not a power series solution, because every exponent contains one half. Please click the advert it’s an interesting world Where it’s Student and Graduate opportunities in IT, Internet & Engineering Cheltenham | £competitive + benefits Part of the UK’s intelligence services, our role is to counter threats that compromise national and global security.