Question

(1 point) The second order equation 3x2y" + 5xy' +(-1x – 1)y = 0 has a regular singular point at x = 0, and therefore has a series solution DO (x) = ± x"+". N=0 The recurrence relation for the coefficients can be written in the form n=1,2,.... C =( ),-1) (The answer is a function of n and r.) The general solution can be written in the form y(x) = Ayı (x) + By2(x), where A and B are arbitrary constants and yı(x) = x + 4,*"+"), y(x) = x" + +86 bext?, n, <r2 n=1 and r2 = Find aj, aj, az and bı, b2, bz. Enter your answers as a comma separated list (a, a, az)=( (b1,b2, b) =(

Urgent!

Please mark all correct answers and find values of a1,a2,a3 and b1,b2,b3.

Answer 1

E Cnunar The given equation is, 11 3 ayt Sayle t-14+1) 720 Here y = Elnar) cannery n=0 8 W ntr-2 (nar) (nart) en X no putting all the values of y, yland yllinell, weget. ner nir E 3 (nar) (nary) (nu tesen (nar) anth E Cax no no Aral E Cna no n+r+) 0, s 3 (ntr) (n+r+)+ 5(n+r) -1 3 Cnxuts Ecox & W 8W 2 ora Z 13 (n+r) 72 (n+r) -13 crant nara E Cox na § 3 (nard) iner-ty) (nents 8 can E ennentral no Comparing the Coefficient of xr, weget Zit 3(5+1) (rr) C =0 · rarl, ģ as to fo. Hence the difference of these roots 2) ģH) = § Notar intega.

Next, we equanto to zero the Gefficient of a ner veget. 3(n+r+1) Infr- $) en ~ Cnry 20, CAM WA). Cns 3(nprt 1) (neprş) : Co J 365+2)(++ - Co ca 디 3 (5+3) (reş) 9(143) (r+2)(ne)(rež) So the general general solution i, x" (Co+C, X+6,77.-) naar کے E CAR nzo af X = [ Cone 3(r42)(1+3) 916+3)(r+2) (r25) (43) -С ] (2) replacing Co by A in (2), Now putting wegit y= A xri [1* * * 22 7 36 7(n) = art E ana n tot -1 na 9=-1 an-l Az =-36 where is an 3h (n-1/2 Hen az aq = 36 36 ㅗ ti 540

putting wi- replacing Co by Bines, wezel T Bad 4 i 7 2 E bn a noody 3 20 =(x) R. + When ban binal 3+1 18 (n + 1/3) n ko, = , 62 too 140 b3 = 3.5.1913) Tq T obl ET & E - 3 = 5460 (9,oda, a3 ) = (d, Śc, - $40) (tvby bs )= $170, $tico) 1,5-1, 7275