By Mejlbro L.

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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.

### Calculus 3c-4, Examples of Applications of The Power Series Method By Solution of Differentia Equations with Polynomial Coefficients by Mejlbro L.

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