Mathematical Methods for Physics and Engineering - Matematica.NET

SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONSis required, i.e. z = 0, is in fact an ordinary point of the ODE rather than aregular singular point, then substitution of the Frobenius series (16.12) leads toan indicial equation with roots σ = 0 and σ = 1. Although these roots differ byan integer (unity), the recurrence relations corresponding to the two roots yieldtwo linearly independent power series solutions (one for each root), as expectedfrom section 16.2.16.4 Obtaining a second solutionWhilst attempting to construct solutions to an ODE in the form of Frobeniusseries about a regular singular point, we found in the previous section that whenthe indicial equation has a repeated root, or roots differing by an integer, we can(in general) find only one solution of this form. In order to construct the generalsolution to the ODE, however, we require two linearly independent solutions y 1and y 2 . We now consider several methods for obtaining a second solution in thiscase.16.4.1 The Wronskian methodIf y 1 and y 2 are two linearly independent solutions of the standard equationy ′′ + p(z)y ′ + q(z)y =0then the Wronskian of these two solutions is given by W (z) =y 1 y 2 ′ − y 2y 1 ′ .Dividing the Wronskian by y1 2 we obtain[ ( )]Wy12 = y′ 2− y′ 1y 1 y12 y 2 = y′ 2 d 1+ y 2 = d ( )y2,y 1 dz y 1 dz y 1which integrates to give∫ zW (u)y 2 (z) =y 1 (z)y1 2 du.(u)Now using the alternative expression for W (z) given in (16.4) with C =1(sincewe are not concerned with this normalising factor), we find∫ z{ ∫1u}y 2 (z) =y 1 (z)y1 2(u) exp − p(v) dv du. (16.25)Hence, given y 1 , we can in principle compute y 2 . Note that the lower limits ofintegration have been omitted. If constant lower limits are included then theymerely lead to a constant times the first solution.◮Find a second solution to (16.21) using the Wronskian method.For the ODE (16.21) we have p(z) =3/(z − 1), and from (16.24) we see that one solution544