By Mejlbro L.
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Additional info for Calculus 3c-4, Examples of Applications of The Power Series Method By Solution of Differentia Equations with Polynomial Coefficients
3) The formal power series solution is ∞ ∞ n y= an x = n=0 If we put cn = 1 xn . (2n)! n=0 |x|n , then cn > 0 for x = 0, and (2n)! cn+1 (2n)! |x|n+1 |x| · →0 = = cn (2n + 2)! |x|n (2n + 2)(2n + 1) for n → ∞, It follows from the criterion of quotients that the interval of convergence is R. Alternatively, ∞ ∞ 1 1 n xn ≤ |x| = e|x| (2n)! n! n=0 n=0 for alle x ∈ R, and it follows from the criterion of comparison that the interval of convergence is R 4) If x ≥ 0, then ∞ y= ∞ √ 1 1 √ 2n xn = ( x) = cosh( x).
They are superﬂuous. 3) The complete solution for x = 0 is then by the existence and uniqueness theorem given by y = c1 ϕ1 (x) + c2 ϕ2 (x) = c1 ex + c2 (x2 + 2x + 2), where c1 , c2 ∈ R are arbitrary constants. We have some additional variants of solution, in which one does not apply the power series method. First alternative. The trick for x = 0 is to divide by x3 . Then we get by some reformulations, 0 = = 1 dy 2 1 dy 2 dy d 1 d2 y d − + − 3 y= − x2 dx2 x dx x2 dx x3 dx x2 dx dx d ex −x dy d ex d −x − e−x y e y .
Hence, y2 (x) can be extended continuously to x = 0 by taking the limit, lim (−y2 (x)) = lim x→0 x→0 x ln |x| + 1 0+1 = = 1, (1 − x)2 (1 − 0)2 hence we have the continuous extension ⎧ ⎨ x ln |x| + 1 for x = 0, −y2 (x) = (1 − x)2 ⎩ 1 for x = 0. We note here that −y2 (x) is not continuously diﬀerentiable at x = 0. g. d (x ln |x|) = 1 + ln |x| → −∞ dx for x → 0, and y2 (x) does not belong to the class C 2 i x = 0. It is possible to interpret the solution, if we use the concept of weak diﬀerentiation.