Question

2. Using the method of Frobenius, find the general solution about the point i = 0 of the ordinary differential equation 1 (1 - 4) y" - ry' +y = 0. Simplify your answer as much as possible and state the domain of validity. 110 3. Consider the general series solution about the point I = 0 of the ordinary differential equation e'y' + 2y = 0. Find the coefficients of all the terms of this series solution up to r', and write the general series solution in the form y () = Ay (2) + By, (2), where A and B are constants and y. (ar) and y (x) are functions, the first few terms of which you would have determined State the domain of validity of the series solution.

Answer 1

Aus we find general solution of DE x(x-4) yll-ayityco hese x=0 is a singular poent sence co of y" - o atx=0 abstituting y = aoxm+ axmt 1 tazamtr t - - a = mao xm-1 + bontja, am + m +2jaxmt + (m+3) Gamt 2 by = 6m-magam-2 + Ently ma, am - 1 tm tatanam + doce (m + 3) (mts) agxmt + - - in the geven aquation, we obtain x 28m-1) magxm-2+ (m+ma, om-1+En+2) (mfijazamt - 7 m-maorm-2 + (m + 1)ma, am-1 + (+22m + y agxmt..? لارا - x maoxmelt (mtija, amta(m12) xml + m +3 asant? - ao xm + a amtta 28 mt2 tazam+3+ – – – the lowest poines of xies am-1 its co-efficient equated to o then - 4(m-1) mao=0 d.. doto m-um =0 m=1,0 cofficient of xm is equated to reso. m-imao - 4 cmtlyma, - mao tao =0 [m-1m-m+ 1) ao = 4(mt)ma,

4m + y ma, = ai (2-2m+1) ao (m-12 ao o 4mm +1 and co-officient of xmt' és equated to zero. Antisma, - 4 Ente) contijas - (mtija, ta, (m+Dm-(mtitija, = 46m + 1) com taja 4 cm + yomtaja2 = (m2 + m-m-1tila, m2 ar 4 Lm + i)(m+2) a = m² (m-12 ao 16 mcm + 1)2 (m+2) and co-officient of anth is equallt to zero. when to m=0 the co-efficient becomes infinite obviate this diffculty Put ao = bo (mo) so that y = xmy do + a, at 22x2 tazx3 + ---7 y = xmaot (m-12aa + m2(m-1729x2 + 7 4mcmt) 16m6m+12(m+2) so. Put ao =bom y = xm bom tm - 2 box+ m² (myrhood w - 4 (mt) 16(mt)2 (mta) at dy = box meoga lt (m-1) (mt) x t mcm-1) [m+jm+2) Am-2) - Im A (mtD "-3). mem-1) (3mts) Emty (m12,2

81 de o bologa [ 1-3x tox? -]. gi bolaga 1-2x + 0 +o- ] = bo loga [ 1 - 3 x 2 Y2 = 4 m= 1 = a[aot oxx + oxx + -] Hm== 20 complete selu y = iyi taha g= a bologx (1-3x) + c290x

Aus 3) geven ex yltry =o. Here x=0 is an ordinasy point sence co-efféceentgy" to So y = nota x ta, x2 tax3 + - tamant y = a, +298 +39 x2 + - + -hanan-t... dz dzy = 292 + 6932 + 12 a4x2 t-tm-nanan-2- substituting in geven DE x 292 +6azxt 1299x2 +--+ (nynanan-2 +2ao taux + 92 82 +9gx3 to-tanant -7 ] =0 exparaulating to co-efficient (i+ tane 25+-)[22 +699 2 #294.42 +--] + 2 aota, a tax2 + az x3 + -- tanant. 20 equating Co-officient of x = 0 693 +292 + 2a, =0 aptar +393=0 0 equating coefficient of x2 =0 1204 + 693 tartaa2=0. 392 +603 +1294 0 0 equating co-efficient of x3 =0 2095 + 1294 +6 azt 292 293 0 3

2005 +1294 +93 +292 0 equating co-efficient of at ao 30 a + 2095 + 694 taz t214 - 3096 +2oa5 +8a4 taz=0 0 4 = 0 aug. foomaaa + Q2 +393=0 393 = -9,-02 put in aan 392+ 2(-9, -92) + 1294 =0 392 - 291-292 +1294 = 0 - 291 +92 = -1294 a4 a 192 12 so - ao ta,x+ a2x2 taz x3 ta xtt - - y = not ax + a202 4-9, -92 Jac3 + ( a 1aepe + --- ya no ta, (x - 1 x3 + x4 + - -). +92442 - 423-+--) in cooffécient of x = 0 292+290=0 = az=-ao y = a0(1-x2+ng + q-- ) +916*-4L0++ y = AY (2) +Byx But