Question

the Seek a pemer series solution for the given diff-ea about ginen point X=2 5y" + (x+1)y' + y = Also find the Recurrence relation an+2= Please pravice answer witte explained steps

Answer 1

Solno 5 لر + y" + so 5 Given, differential equation (1) 5y" + (x+1) y + 5y Dividing with gives. (x+1) ylt -(2) (2) with Y" + P(x) y' +Qlaly = 0 here P(x) = 5 and Qis) - 1, Since both Plu) and Qise) are No=2 So Xo=2 is an (1). To final sclelim about Xo=2 we shall find Series Solution in (x-2). we assume Comparing 2+1 analytic at Oxdlinary point of 7 powers of that . y = lot(, (x-2) +(2(x-2} + G 6x-2) 3+ & Cn (x-2)n. y —(3), nao こ nel differentiating (3) twice in Succession Wort x = We get . y' Enn (x-2)n-1 and n (n-1) Cn(x-2)n-2 Tutting the above values of in 11), we get si Y" = n=2 y Y and

5 E n=2 no 5 E = 0 no or E n=2 na1 O 10 or E n=2 nel nal nzo or n(n-1) Cn (2-2)-2 +(2+1) E nin (x-2) h-) + en (x-2)^ sn (n-1) en (x-2) 1-2 + [(x - 2)+3) { nen (x-2)n-1 +5 8 Cn (x-2)^. 5n (n-1) en (x-2) 1-2 t & ncn (x-2)^ + 3 e nen (x-2)n-1 + 5 & Cn (x-2)^=0. 5 (n+2)(n+1) (nto (-2)" + EnCh(2-2) E 3(n+1) Cnt (x-2)n + 5 & Cn 64-2) Mo. 5xe C₂ + 3G + 560.} + {5x 6 C3 + ( + 6 (q + 541}(x-2) + 5 (n+2)(n+1) Cn+2 t non +3 (n+1) Cnt! + 5 Cn } (x-2) Which is an identity equating to zero the Coefficiente of Verious powers of° (x-2) we get 2 na0 nal + n=0 nao or. ♡ E n=2 (5)

Date Page 10(q +3.4, +560 Ce - (30, +500) (6 10. 0 3063 +4 + 6 +56 Gz = - 6 (6+ (₂) 30- = -1 5 Ct 34 -5C 10 7 ci 5 C6 10 7 C 50 Ĉ to} C3 5 6 10 and. 5(n+2)(n+1) Cnth + (nts) int 3h+1) Cntl = 0. Cnte ((n +5) In + 3 (n+1) (nt 5 (nt) (n+2) for all nz2 = 1 8 nad for all which is the required recurrence relation.

Putting n = 1, 2, 3, Pn (8) and nol have using (6), (7) ca ( 7 cq + g (3.) = 1 60 To lo 5x3 8.4 S-7x3 G 35 ot glo 10: 63 93 50 26 Co so 10 "} こ -1 60 - 168 4 I o } 4. 42 750 G + 13 300 and so to C. 10 50 750 300 Putting the values of Coefficients in (2) we get. y = lot C6e-2) - (36,+560.)(x-2) 2 + 4} (x-2) 3 + 4444+ 1360 (4-2) 4 or of 5 (x-2)2 + (x-2)3 Aug. 13 (x-2) 4 taon.ael int moel tot + ad int + y 1 + lo 10 300

Date Page 50 42 GS (2-2) -3 (1-2) 2 - 7 (x-2) ? t (x-2)4+ ad int 750 which is the required solution in terms of power Series about Ho=2 Where Co and a are an arbitro Constant rary Equation (2) Represent a Recurrence Relation. C