Mathematical Methods for Physics and Engineering - Matematica.NET

16.2 SERIES SOLUTIONS ABOUT AN ORDINARY POINTagivena 0 . Alternatively, starting with a 1 , we obtain the odd coefficients a 3 ,a 5 ,etc.Twoindependent solutions of the ODE can be obtained by setting either a 0 =0ora 1 =0.Firstly, if we set a 1 = 0 and choose a 0 = 1 then we obtain the solution∞y 1 (z) =1− z22! + z44! − ···= ∑ (−1) n(2n)! z2n .n=0Secondly, if we set a 0 = 0 and choose a 1 = 1 then we obtain a second, independent, solution∞y 2 (z) =z − z33! + z55! − ···= ∑ (−1) n(2n +1)! z2n+1 .n=0Recognising these two series as cos z and sin z, we can write the general solution asy(z) =c 1 cos z + c 2 sin z,where c 1 and c 2 are arbitrary constants that are fixed by boundary conditions (if supplied).We note that both solutions converge for all z, as might be expected since the ODEpossesses no singular points (except |z| →∞). ◭Solving the above example was quite straightforward and the resulting serieswere easily recognised and written in closed form (i.e. in terms of elementaryfunctions); this is not usually the case. Another simplifying feature of the previousexample was that we obtained a two-term recurrence relation relating a n+2 anda n , so that the odd- and even-numbered coefficients were independent of oneanother. In general, the recurrence relation expresses a n in terms of any numberof the previous a r (0 ≤ r ≤ n − 1).◮Find the series solutions, about z =0,ofy ′′ −2(1 − z) 2 y =0.By inspection, z = 0 is an ordinary point, and therefore we may find two independentsolutions by substituting y = ∑ ∞n=0 a nz n . Using (16.10) and (16.11), and multiplying throughby (1 − z) 2 , we find∞∑∞∑(1 − 2z + z 2 ) n(n − 1)a n z n−2 − 2 a n z n =0,n=0n=0n=0which leads to∞∑∞∑∞∑∞∑n(n − 1)a n z n−2 − 2 n(n − 1)a n z n−1 + n(n − 1)a n z n − 2 a n z n =0.In order to write all these series in terms of the coefficients of z n , we must shift thesummation index in the first two sums, obtaining∞∑∞∑∞∑(n +2)(n +1)a n+2 z n − 2 (n +1)na n+1 z n + (n 2 − n − 2)a n z n =0,n=0which can be written as∞∑(n +1)[(n +2)a n+2 − 2na n+1 +(n − 2)a n ]z n =0.n=0n=0537n=0n=0n=0n=0