Mathematical Methods for Physics and Engineering - Matematica.NET

SERIES SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONSWe may check that y 1 (z) andy 2 (z) are indeed linearly independent by computing theWronskian as follows:W = y 1 y 2 ′ − y 2 y 1′=sin √ (z − 12 √ z sin √ )z − cos √ ( 1z2 √ z cos √ )z= − 12 √ ( √ sin2z +cos √ 2 z ) = − 1z2 √ z ≠0.Since W ≠0,thesolutionsy 1 (z) andy 2 (z) are linearly independent. Hence, the generalsolution to (16.17) is given byy(z) =c 1 sin √ z + c 2 cos √ z. ◭16.3.2 Repeated root of the indicial equationIf the indicial equation has a repeated root, so that σ 1 = σ 2 = σ, then obviouslyonly one solution in the form of a Frobenius series (16.12) may be found asdescribed above, i.e.y 1 (z) =z σ∞ ∑n=0a n z n .Methods for obtaining a second, linearly independent, solution are discussed insection 16.4.16.3.3 Distinct roots differing by an integerWhatever the roots of the indicial equation, the recurrence relation correspondingto the larger of the two always leads to a solution of the ODE. However, if theroots of the indicial equation differ by an integer then the recurrence relationcorresponding to the smaller root may or may not lead to a second linearlyindependent solution, depending on the ODE under consideration. Note that forcomplex roots of the indicial equation, the ‘larger’ root is taken to be the onewith the larger real part.◮Find the power series solutions about z =0ofz(z − 1)y ′′ +3zy ′ + y =0. (16.20)Dividing through by z(z − 1) to put the equation into standard form, we obtainy ′′ + 3 1(z − 1) y′ + y =0, (16.21)z(z − 1)and on comparing with (16.7) we identify p(z) =3/(z − 1) and q(z) =1/[z(z − 1)]. Weimmediately see that z = 0 is a singular point of (16.21), but since zp(z) =3z/(z − 1) andz 2 q(z) =z/(z−1) are finite there, it is a regular singular point and we expect to find at least542