Question

please show all steps and label them and also calculate the b values

(1 point) The second order equation 3xy- 4xy + (x2 + 2)y = 0 has a regular singular point at x = 0, and therefore has a series solution y(x) = Σ Cnxhtr NEO The recurrence relation for the coefficients can be written in the form of Cn =( [2(n+r)(n+r-1)+5(n+r)] DC-2, n = 2, 3, .... (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 1} (x) = x++ & a, xhtmı, y>(x) = x2 + by xentrs, ri <r2 n=1 n=1 r = and r2 = Find aj, a2, a3, 24, 25, 26 and bi, b2, b3, b4, b5,b6. Enter your answers as a comma separated list (21, 22, 23, 24, 25, 26) =(

Answer 1

+ E 2₂ wnan Cnanta in series holn neo ись Let 919)x? Ź inan Ž Cnlntv) (-1) 3x² Y"-4xy' +12+2) Y=0 (1) x=0 in regular singular point of (1). Σ toto (2) Now y' Å Cn Cntrla mao putting these values in (2), we get ¿ 3 en intro Intr-17% 84C, 1nton (1) 11 ahtr-1 - ntr-2 no a ntr nar neo neo nart2 + E Cna =0 nzo hes hart2 neo nso =0 --(2) Co to 20 Ź Cn (31n+r)intr-1) - 41n7") +2y xn+ y ¿ con me For indicial Equation comparing coefficient of smallest power of to are. ar we get co (37(1-1) - 4r+2 4 = 379_75+2=0-> (7-2)( 37-1)-0 1.-2, so differing by an integer. -73) For recurrence relation, com moparing Coefficient of we get cny 3 intrlIntr-1)-(n+r1+2} + Cunuz VF2, 2 2 / 42 so roots are unequal and not x nar Cn-2 cha 3(ntry Intr-1)-4 (n+7)+2 u Cn= n-2 Cn-2 (n+r) (3n+37-7)+2 (4) Recurrence relation. +L baring coefficient sto To find ca, com in (21, we get C1 { 34Y+1) v-'4(x+1)+2y = 0 csca 19170 cando by (4), (3=(5-475--- = 0

we get Co - 6+2)(36-1)+2 using (t) for m=2, 3, 4, C2 = (v +2) (394 2 +38-7) +2 C4 = C2 (v+4)(31 3r+5) +2 (37+5)+23 -(0) Со C6= C4 (r+u) ( 3V+14)+2 ) By uil, to a4 Co + {(7+2)(36-1) +23{17+4) ( 3r+5)+2 4 46+1)(3r+11)+2} y = x² (of cx + (z x² + c3 x² + cant te gust() = x ' ( co + (r +2)(36-11+2 P1*+2)131-1)+2346+4) (37+5/+24 Pir+2)/36-1)+23(+4 1130+5)+234(+1)( 35+11)+23 y=x + cont (8) w x2 at a6 + 2 2 713 a4 at + ه ) + + for re 2. Let to= A Carbitrary) we get 1 st l-I. soin + + --) Ys = Ar? (+ 1496 206448 ( for 13 Let Co=B then and L. I solh uning (8) 113 Y 2 =Bx' 1 + + 2 s6 4368 to general solution Y=9, +42 8 x6 9 = A (x2+ + + 1496 206448 713 1313 Am: ta 그늘 t -- ) CScanned with CamScanner SI 4368 2 + + 22 19 x +B(x+13 + x 3 too.
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