Question

7. For each of the following ODEs, use the Method of Frobenius to find the first six terms of each of two linearly independent solutions about the regular singular point xo = 0. (a) xy" + (x – 1) y' + y = 0 (b) xy" – 2 xy' + 2y = 0

Answer 1

ty" + (tiny ty=0 y = 2 & County ť 30 ch+8) Cn lints-1 o - 2 + x)(n+X-11 cod int8-1c. dintr-2 o hao +8)(n+8 -1 c . hinto-2 + (t-1) (h+ X) + X- Co tinto - z ho Ph+(-)CH (+84) (n+8)(n+8-1) Cn 8 198 s (nto) ntin+8-1 - hao + { cn tinto=0 ho - P 22ch+x+1)(n+8) na hatot +8) Catanto Now find the indicial Equation? 8 (8-1) - 8 818-27 ao B=0.2) linie root of indicial Eer: are :0.2

now. -8)(n+8-1)-(neo) Cntinax-1 + 2 I (n+x) tilsatinto-o (n+8)(n+8-1-1)] in tinta-1 + {{1+x)] Canytimes =0 818-a) Co + El (n+8)(n+8-234t (n+8) Cna, Itine 320 ha (+ 8) (1 + - ) n + (n+8-2) (nt +8) - ° Cool zo (-:NEN) I which is required Recussanie Relation. Geen Fi cza Czasem S-1) (z = Chevex T+8 18+)18+ ) hence! years- & indinto =#8 [ Cot Cit+ Cę te? + c3 13+.-- = $18 l cot c H (-us) + + lice +13 --..]

now. YItu = Hits [coli- tk + 12... ? + At tin 8-) (8-) X +) (X- X+ *+ w. but the value of x we find the two lis suln. y - 2017 +2y=0 Let y = {cranto so (n+8) Cn dintr-1 Y" = { chad) (ht8-11 Cn tent8-2 nas | + 2 th +3)(N + X-) C, an+k-a - Đà Sn+x). 4n+ 8-} no ho no + 2 C + = D → {intos(n+8-11 Cn tint8-1 - No Zainto) + x) Co +*+ S hta ho home Indicial Equation will be :- 818-1 = 0 - [8=0.1] - these are the soul of indicial Eqh. > {n+8)(n+8+1) (, tent8-1_Elainto)=2] In tinto ao . (+8) (2+8=1) (nanton - (an+28-4] Can, tin+8 = 2 018-116+ El M+8)(n+8-12 Cn-can+28-4) (noi] dintr-o

» [lutu antara (n+8) (nagu which is required securrunca xeliliun. 2+2X-9 (+5) () → C = 2(0 co 24 25 kaj cu - 32(-1) (x+2 (8+2) 8му = сам" y (Hi= +18 | Cot C, H + Catint Cy J13+ ar... S41) : k ] o + ax + 4k--- ly(H = Col 40 H 8+)?X+2 . )