Question

Consider the ODE:3xy"+y' - 2xy = 0. Find the general solution in power series form about the regular singular point x = 0, following parts (a) – (c), below. (a) Obtain the recurrence relation. (b) Find the exponents of the singularity. (e) Obtain only one of the two linearly independent solutions, call it y(x), that corresponds to the smaller exponent of the singularity; but, only explicitly include the first four non-zero terms of the power series solution. Write down the solution yı(x) with the coefficients in simplified fractional (not decimal!) form.

Answer 1

Given 3ry"+yl-2xy=0 Atr-o is singular point y" + În y! - 5y = 0 pa) = 3x - Qal=-2 / H C.Post 470 apens mox (x) = } Qeve It x² (-2/3)=0 z? Po=0 2-0 At a=0 is Regular Singular points Indical equation mcm-1) tomt9=0 tutto mcm-175m40 con 3m (m) +m=0 3 m2-3m+mzo 3m²-2m=0 => m (3m-2)=0 1+x_oltuit in mzo & m=2 leto - wat anth no (2) be solution of BV 0-4 an (+9) amth-1 th+1) amth. & Me ho ant

- 90 x ta, anti taxta faxmtz taxmt4 mih Mt+1 + anz" + a + 1 x"941 y's m. 2x*4 + cm+1)9, *" ++49*" +ą (m+3) xmaz min-1 + +--- +(min) an (1) An y = m(m-19, xm-2 +(m+ 1)ma, xm-1 +(mt) +1)ą, x + +") btn-) an XM+-2 | + i) h) ah) + -- MINI Substituted y, y fyll meno sa [ ca) do 2m 24 6+1) mai +(m+2)(m+]9, 2") +---+ (nto) cm + -1) an amtn-- mtoti) (m+n) ant " femast imagem-1 + (m+1) a, a temt 2)a, am I, +0,() +(mthyanamth into H) ante ****-- +--- - 20 a. ao Mtb +9, Mat +91mth tanam

coefficient of xm-1 a (least power) 3 mcm-1 ao & mao od amt-aman) 20 aoo 3m-2m20 mzo; m=. coefficient of my 3 cm+i) mat mula, 0 ai [ 3cm+1) m +en+\]=0 a-0 - coefficiet of xmt! 3 (m+1) Cmt_) 92 +672)a2 -200 = 0 [er] . (A₂= 2 ao az (mt) (antly coefficient of xnt2 3 cm+2)(m+3) a3 + 6+3) 43 - 29,=0 az 20

coefficient of amt3 3:4 mo 3 cm+3) Cm+4) 04 + + 4)94 = 292 (m+4) au Cam+3) + 1]zaz 2 a4= mt4) (3m tho - a4 ao (m+2)Cm+4) 3m+4) Hoo 947 ao mt?) cm+4) Bm74) Berto lazny -0 a, =9z=as odd terms becomes sero

③ Even terms a₂-04 - ab = ar to 2 antz an (m +n+2) (3m+30+4) Recurrence relation. + y= Required independat Solection ao xm to amt! to xmx2) toxmt3 Embarcamion ao + xm4 Contestanti mth)(m+4) (3m+4) (3m+10) öz ao 21 [ '+ Camily Compas (ont 43m+4) x² 22 (m+y)(m+4)3m+4)Gato mo 9,0) = 20 litate 4---) m = 22 22 22 + + 4 €13) 342) 6:12 7,260 261 43). 6