Question

Consider the following differential equation. (1 + 5x2)y" – 8xY' – 6y = 0 (a) If you were to look for a power series solution about xo = 0, i.e., of the form Σ τη x2 n=0 then the recurrence formula for the coefficients would be given by C +2 g(k) Cx. k > 2. Enter the function g(k) into the answer box below. (b) Find the solution to the above differential equation with initial conditions (0) = 0 and y'O) = 6. (Note that this solution is a terminating power series.) (C) Find the first three nonzero terms in the solution to the above differential equation with initial conditions v0 = 4 and y'O) = 0.

Answer 1

(Holly" - goxy-by=0 m2 nzo dy dă - n=1 puttina this values we Na2 y= Å enan get - 8x ² annan putting (17582) Ě enn(n-1) & ) nr 2 No2 nal -6 Į Ensen zo nzo non I en n(n-1) x + 5 Ź Cnncny) a n=2 n 22 0 -8I Cnnri 6 I ensi n=1 no the coefficient of an we we get Eque nating Centa (ont2)(n+1) + 5 en non-1)-8 nen-ben=0 Conta (272) (+1) = em [8n76 - 5 n(n-1)] -en [598-130-6] =-5 en [ 13 =-5 en (n + 2/5) (n-3)

Cut 2 - - 5 « (12) Ск-3) (k+) (k+1) 14 77 0 ) Now, с. - — 5eo Pis) (-3) — зе, 2:1 со. — — 54 (+ 1 ) C(-3) (1+1)(1+1) — -5 (-2) Тя е е 3x7 4 = - 5 о (2+25) (2-3) (2) (2+1) — 5x12- (-) е, 413 е, - Зco. C5 -5 (+215) (3-5) (2+1) 3+1) С 2 O ee — 54 (4t s ) (3) (4+7-9 сет) — 5х20 x 1 14 - 110 — си с LYS

Cy = - 5 CF ( 6+%) (5-3) (5+2)(611) 0 Caicou 913) o - 5% (6+6) (6-3) (6+2)(6+1) CG X 31 243 CG = 8X7 대 82 = 11X12 sen Cou ㅋ and so on. Hence the salution is Y = Co+&++€3x?+ Expo Cg Mest C432L = @[2+33*+ 32"+ 45 46 4 Lyon ] + a[xt a +7/30 440) = Co [62+122?+66467] 9[1+: t 722 12

neral solk Now, y (0)=0 givers. eo=o. 4'(o)=96 gives 9=6 Hence the general ered solution is y=[1+7%2] e Naw, 4(0)=4 givers Co=4. 44(0)=0 gives. 9=0 is y =6[1+3324 3247 11 206+ --..] The