Question

(1 point) Consider the differential equation 2x(x )y"3 - 1)y -y0 which has a regular singular point atx 0. The indicial equation for x 0 is 2+ 0 r+ with roots (in increasing order) r and r2 Find the indicated terms of the following series solutions of the differential equation: x4. (a) y =x (9+ x+ (b) y x(7+ The closed form of solution (a) is y

Answer 1

The iven dffeentid euaion (r-)j"+3 (a- regulay gulas potint Frobenfus metod" as use have to So , hee Loe dfferuntial equaton he since P galas Singulanty as equeion Steps ) Assume the olution to be ota,, dp n) Substitute from in } d2 m) Euate ceg tem in eoionhnoonas coelficient of lowest to zero Quadaic ves n dical qurfon STnglaspint uHou Since corfficient of 1 r+2 T-1 obtein: n e given euaon , we

Y-2 + (1+1) (1+ 2) (74 ) t t nd loDest pooe of x in We have tb and equati expanstob okich is tRs CoeffiGent to zeso Now fnd coeffient of x-1 31 e uorioh -11 -1) 2e1o This Tndical uti (1Y11) Y0 or ave distinct and do not dTffes he toots Now letind t coefficent an of and euate to 3eio a-2(741)4+a.[-1+s4 27(7-1}- a-2-21-37-3+a, Le1+5 *-1J=,

(2r+3) (7+リ+a。(27-1) (+リ= 0 a Noo nd coecient of n*t1 -3 →[7ォ)(r) +s(7+リー]+E-2lr4リ+2) +a, こ0 +3 ア-4-3Y- c O O - O [キリ (+)-[aアキリ (++2) a (27+s)(42)- a,(2+1) (7+2) (1 t)2) (2+s) (2) ai 27t3 (17+1) (LT-1) .a /St1 27 when T0, a, - -I ap --I 3 - 3 - IS 35

wheh -212a O, -O...- O a 4 9- 37 35 63 3 n 3 s- 2 The complete Salution 2 1-3x- 35 63 (5-5