Question

use the method of Frobenius to Solve xy"+2y'+4 xy=0

Answer 1

Now, nem-1 y's dy da Z Cn (nem) x n=o and Nem-2 dy y = I Cn (nem) (nem-1) dra n=o Now putting the value of y, y'andy" in (i) we have nem-2 x I Cn (nem (nem-1) a n=o nem-1 num 2 I Cn (nem) x +422 x=0 N=o M=0 nem- Z Cn (nom) (nem-1) x n=0 nem-1 22 cm (nomx +47 can nemel -O + 2 n=0 n-o nem- = {(n+m) (n.-1) + 2(n+m)] Cnx + 4 cm x n=o nemel - O n=0 (iv)

+ 4 Cn-2=0 +4 Cn-2= Cn nemt { (nem) => Cn - 4 Cn-2 (*+m) (n=m+1) n72 (vi) equating the co-efficient of the term of the next highest power of get a {(m + 1) (m) + 2 (m-1)} {(m+1) ( m+2) a to zero , we = 0 ( Ci NOW Since C=0 We from - (vi) see that C2k41 = 0; for k=1,2,3,..... Now putting n=2, 4, 6, 8, (vi) we have in C2 = -4co (m+2) (m+3) - 4 C2 16Co Cu (m+4) (m+5) (m+2) (m+3) (m+4)(m+5)

— ЧСИ Co (m+ с) (mp7) – Ч Со (m+2) (3) (ms) (m+r) (nv) (m+7) Now from the equation - = Cnx тет Т= 0 mel те). = Com С. x+С, x + Cam+34 Су т+4 р m. — с. + C, и ФС, УРС + С Ч m - с. +С, "Суч (с... ) m — (m+2)(m+3) х + | GC. (m+2) (m+3) (m+) (m+5) 1

(m+2) (m+3) 16x4 (m+2) (m+3)(m+4) (m+5) (vii) m = 0 Now putting co by and replacing we get from - (vii) y = a (-***** y putting Again C. by 'b m=-1 and replacing get from (vii) we y = x'b ( 2x² + 2 "- solution of Hence the General (i) is y = a (1-2 x ² + 2/x" _ . (1-2x2 + 2 x" ....) where 'a', 'b are arbitrary constants Abri