Question

(1 point) The second order equation x?y" + xy' +(x2 - y = 0 has a regular singular point at x = 0, and therefore has a series solution y(x) = Σ CGxhtr P=0 The recurrence relation for the coefficients can be written in the form of n = 2, 3, ... C =( Jan-2 (The answer is a function of n and r.) The general solution can be written in the form of y(x) = Ayı (x) + By2(x), where A and B are arbitrary constants and yı(x) = x + a, x+1, »?(x) = xº? + Ž byxº+ra, ri <r2 n=1 n=1 r1 = and r2 = Find 01, 22, 23, 24, 25, 26 and b1,b2, 53, 54, 55, 56. Enter your answers as a comma separated list (aj, a2, a3, 24, 25, 26) =( (b1,b2, 53, 54, b5,b) =(

Urgent!!

Please show mark all correct answers and also find values of a1,a2,a3,a4,a5,a6 and b1,b2,b3,b4,b5,b6.

Thank you!

Answer 1

b , 2 N。 고 2 (는) b3 = 19t . ) وہ 3 든) ( bu abz 2 () = 2. 19 (은) 20S bs 2 bz 16 Z 5 (+) 5 sh ( 5175 bb -bu , - 그 3(27) 27. (E 65 -1 16929 ㄹ bo여 bi다 , be 2 -0.18 1821b3 - 0.08 8 89 by = 0.00957 b5 70 1 0000309 bo 0-00021

- -" * * * * * * - - ร - 5 Let है ท+ S Cๆ x หb ท+kr-| cy (ท+Y) X ทะ6 8N8W ฯ" - ท+-Y-2 CM (ท+Yr) (9+% -1) X Now -ร” + *ฯ * ฯ + (- -) t e) 8W X (( – 4+M) (44) -3 ท+5 OCA } ท+ E an(n+r) x ท: 0 AC + (*- 5 ท+Y CM X ทะ 6 e) eth cn (n+8)2 x S หc c CM X7++2 ท+ gcn X = 0 6 ฯ CS 2) 3 CM nts [ +z) -2] * + Y+34-2 ก.b Cn X mco Now indicial Equation C 6 (*-*-) 0 Cat 6 YE + 3 7 'น -3 8 - - [-- * , . *, ๒/๒

recurrence relation an [[n++) ² - 2 4 ر + cn-2 cn= Cn-2 [(n+r)? 64 m=2,3,4,-.. Cn 11 (n-2 (n+r-2/4)(n+7 + 3 8 n=2, 3, 4, =(x)'h X + SM8 letu am 2,= - 3 8 an - " an-2 mis ant2 wo (histu) retu) Hoc 1 puo a,=1 so a2= ao 11 - 2 (514) UN

- a= ч 27 3 (%) こ az Ft (-3) = 5 (13/4) as 93 16 ㅗ s() 5 17 $ 2295 ab au -4 4095 6 65 (2) 3(21) 2 So ao=1 9,-1, a 2 -0.4 a3= -0.14815 94= 0.03077 e as= 0.00697 96= -0.00098 and op ntre y, (x)= Х + E bn X hal rz: bna bn-2 (m + 3) buta bn (n+2) (nt I bo=1 birl