Question

(6 points) Use the method of Frobenius to obtain linearly independent series solutions about x = 0. 3xy" – 1.54' + 2y = 0. Use an initial index of k = 1 to develop the recurrence relations. The indicial roots are(in ascending order) rı = 4.5/3 ,12 = Corresponding to the smaller indicial root, the recurrence relation of the solution is given by C = Xck-1. The initial index is k = The solution is yı = c (az xb1 + a2 xb2 + az xb3 + 04.xb4 + ...) where a = where a = b = . 42 = .b2 = .dz = Note: the a;'s and b;'s are constants. , and b4 = Corresponding to the larger indicial root, the recurrence relation of the solution is given by cx = Xck-1. The initial index is k = The solution is yı = co (a1 x 1 + a2xP2 + azxB3 + 04.xB4 + ...) where a1 = .Bi= , Q = B2= , az= Note: the a;'s and B;'s are constants. B3= , Q4= , and B4= .

Answer 1

Comider the differential equation 3xy" - tasy' + 2y =0 - Let the Frobenius solution be (about 2 = 0) y (a) = (16-3)*** n=0 Here %=0 :y(x) ensatt => 74x) = Ž (ner) en z most =y" (a) = 2 (n => )(n-3-1) 6. 2n+-2 nco but these values in 0. ...3x E (n+8)(n+8-1 C an+8-2 n=0 -1.5 I (nto) en sentr-+ 2 E Cnochto =0 yal 1-0 - N=0 Scanned with CamScanner

3 Ě (ntr)(n+8-1) Cn antr- .no -1.5 E (nto) Cn antol+2 È in anto o neo => 3 [34nov) intr-) - 4.5 (91) Cinemas +2 Echochty to n=0 For indicial equation consider the least bower sum of @ Here just sum is least power sum. Coefficient for n=0 is 3 (a+Y) (0 +1-1) - J-5 (+x) c. but 38(8-1)-1.58=0 and take co to -) 382-38-158=0 7 388-4.58 = 0 => g2_1.58=0 8(8-1.5)=0 8=0 ; 8=1.5 Scanned with CamScanner

8. The indicial roots are Piag] i Mg = .5 Now, for these 8, est term of 1st sum of @iso inid to ao ... ® can be written as; n+8 - Ź 13[n+Y)(n+8-1)-1.5(n+ n=1 . +2 I Cnochtr-o n=0 In 1st sum of h ; replace in by not! Here, nit to oo leartier) =) nt it to a = n :1-1 to 0-1 => n:o to a (captes replacement) t]+ - ) •- becomes, 2 5 (no)(n+=1] -15(n+7)]Cnyceniti tot Ž comento n=0 h= 0 cs Scanned with CamScanner

- n=0 n=0 mp3 falarrakee) -13 (nr41) Concept + Een cemas-0 > slnes +)(nas)-is baresn) Can + 2 en fazemos - 0 -> (5(n+8+1)(n+7) 1.5 (n+84) Cro + (n = 0% - Cnti = -26n n=oll - ;n=0,1,2,--. 3(n+8+1)(n+8)-1.5(n+8+)) This For 8,=0 is recurrence relation Cnti - - QCn is (n+1)(n+o) - 1.5(n+1) in= 0,1,2,-- Kenti - or put k = nr Ć -2 lik=1,2,3, 3k(K-1)-1-5k K This is recurrence relation for smaller root the initial index is (K = 1 Now, Kol 4 = 726 - 460 Scanned with CamScanner

- - 26 3x2x1 -1.5x2 k=29= 52001 - 1332 = -(5)= - CO K-3 C5 = 24.-1985 its **) - it so ka --2C₂ le 16 121.5 13.5 L 3x3 x 2-1.5X3 and so on. n=o + 16 S + - -- - Wow, as y(x) = { chanty y (t) = av (Cot (jQ7Gx'+ Gi?+ --) - (x) = 2" (Cot£ co- co to this co2? + ---) x) = (1 + 4 = 2 +-) :. 10, -1 ; b=0];Jag = 1 bg = 1] 121.5 72 +16 a colo +-- wit - 121.5 wow for yg = 105 recurrence relation is С . - 2 с. in=0,1,2, .. 13(n +1.5+1)(n+1.5) -1.5(n+1-5+1) but ntl =k Scanned with CamScanner

; K=1,2,3,-- Ekx = -CK-1 I 3 (+15) (K-17-5) -1.5(k+15) This is recurrence relation for larger root The initial index is IK = 1 1. K=1 c =-260 - 2 3/1+1.5 7.5 3(1415) (-+15) – 1.5(1715)* 7.5Co 2 C2 = 1:56115) Zuhal - 4 157.5 K=3 --8 Co =-262 3(45)(3-5)-1-514-5) 40.5 (157.5) yg (x) = av Ve (60+ 6,27 GX8+ 6x3+---) .:. 40(d)e . 1.5 - 2 Coat + 4 Cox I A cox² - 8 co x² 7.5 . 637875 157.5 4.5 -- 7.5 157.5 6378.75 : Scanned with CamScanner

o di = 1 ; B, = 1.5 - 7.5 dhe po = 8.5 dza istusi g = 13.5 J - -8 . 6378.75 ; B4 = 4.5 . Scanned with CamScanner