Question

? Find a Frobenius series solution for x > 0. Use the small index. 4x2 y" + 9xy' + (-3x2 + 1) y = 0) 4x2 y" + 7xy' - (2x2 + 1) y= 0) or

chose on of the two

Answer 1

Then htk 70 4&ry" +9x41+ (-32 +1) 4 =0 since xeo is Ragulas singular par het &- & Antik be the solution -> 4x2 [124, (***) (k-1, 3+k_2] + 9x [a] An(n+1) X*! 7+ (1-3x) A, Xto 2% [ 460+cm+ka) + 9(2+2, +1] **** 234, **** aw equating the smallest degree term to zero iie A. ( 4k(K-1) +964 +1) = 0 to to 46²+5k+1=0 (4k+) (k+1)=0 k=-1,- 1 / 2 Now equating next higher power term ie akt1 - A [4 (6+) K + 96kH) + 1] = 0 A1=0 Equating to zero the term akrn. An [(ntky (4 0 + 4k+5) + 1] – 3 An-20 An= 3 An-2 (ntk) (4nt4k+5) + 1

case I k=1 An= 3 Anos (n-1) cant) tl A2 = 300 - 3 Ao lo 1x9+1 Az = 0 3 A2 3x 17+1 mie Ho = 9 520 Then ne Y, CA) = J I Hot A R + Azut. { no %, (x) = * [%. + 3.4oz+ + 2.00x*...] 3 An-2 Case II k=-1/2 = " An- in - Can4)+1 A = 3 An-2 (40-1)(n+1)+1 A2 = 300 - 340 Az = 0 22 / 7x3+1 3 A2 15x5+1 9 Ao 1672 "2 (11= 244 [ Ao + 3 A0 x² + 900 xt - I 22 1672 Hence 76x) = c. 7, (x) teyn) is general solution of differential Equation.

Answer 2

3. Find a Frobenius senies solution 30. use the small index. 404" tary +91 - 1) = 0 - 1 0 443"-tang-(2x++)40 - Theorem a at least one nontoive where & is a definte (real Hypothesis the point to is reqular singular point o the differential equation. conclusion - The differential equation has of the solution f one non trivial form (x-xo) en (x-xo". a definte real or complex) Constant which may be determined, and to is valid in some deleted interva wheat (R70) about to -solution - since x=0 is a regular singulas point of the D.E and we seek solution for ocuCR, we assume non = I en xtr where cofo then my! (040) en 17).- ntr-2 I (nto) (ntri cn l determined, and this solution deleted interval oc/x-ko KR HO and ry no substituting the series (1) , (i) & (ii) weger. am L ()(n+x-1) en 3**8219*(n*) ema" * 3*°) 3 n o enant 4 (tr) (ntr-1) Cn x atryanto o 45 g (ntrenze no 200+8+2 + I enorner - - no nto. 9 Š [46n+x) *78-1) 4 (nt) + ]en - 3en10

Honce the use small index I 40 (8-1) 4gr +17662er con + (7-1+9(ne) "11] Chung the lowest o the coerhonnt of Equating to zero the coepient of t poner of a (ie the coefficien Therefore the led to the quadrat Coefficient olxal. to the quadratic ear, whese the +7(7-1).499+1 = 0 coto This is the indicia doeffenifat equation we get indicial equation of 482 124987120 2) 4724584150 and observe that its roots are 4r +474841-0 -). 48(741) +1(8+1)-0 (11 (4841) - 2) 72-1, y = -1 where og in the solution they are real a un equal. 14 (647) (n+7-1) 49 (64x) + 1] .cm + 30- 4 0 m Letting vir [4 (-4) (-4). 974)+1] cm - 3 en-yo Conssponoleing to the larger root of the indicial egna (m) [4 (na) + 9)+1] CS-361-220 + f(n).( 49-59) +1] -360120 3 (-) (40+4) 1) C-360-120

* |(141-16 417 em - 36m2.30 - nu v-XX] cn - 36.220 - 46049 - 1) (-3 C. 2o. 2) [ 4n 13 cm - 3 (n.220 = c = 3 (n. 2 na (42137 21 7 9 = 300 7 Oy = 34, 30 154 15 10 Letting 8 = -1) and using the value, a la ga y colatt + is, * ----) 3) we now let a=-1 to obtain the recurrence formular. [4(0-1) (-2) 49 (0-1) + 1] C - 360-120 =) [(n-1) ( 40-8 + 9) + 1] cm - 3 co-150 3 (my (4041) 41 ) en = 3 Cny 2) (4 n(n-1) + n-yax) Car 3 cm i 4- S .. O (40-3) - 2012 .36 Y = Co [ Therefor + 3 + 4x + tox+---]-( the two (3), (4) Frobenius senies.