Question

Problem #14: Consider the following differential equation. [6 marks] (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 Σc, Cnxh n=0 then the recurrence formula for the coefficients would be given by Ck+2 = g(k) Ck, k > 2. Enter the function gk) into the answer box below. (b) Find the solution to the above differential equation with initial conditions y(0) = 0 and y'(0) = 9. (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 y(O) = 7 and y'(0) = 0. (-5*k*(k-1)+8*+6)/((k+2)*(k+1)) Enter your answer as a symbolic function of k, as in these examples Problem #14(a): -5k(k – 1) + 8 + 6 (k + 2)(k + 1) 9*x+36*x^3 Enter your answer as a symbolic function of x, as in these examples Problem #14(b): 9x + 3683 7+7*(6/2)*x^2+7 Enter your answer as a symbolic function of X, as in these examples Problem #14(c): 7+7Qx2 +7

Consider the following differential equation. (1 + 5x2) y′′ − 8xy′ − 6y  =  0

(a)	If you were to look for a power series solution about x0  =  0, i.e., of the form ∞ Σ n=0 cn xn then the recurrence formula for the coefficients would be given by ck+2  =  g(k) ck , k  ≥  2. Enter the function g(k) into the answer box below.

(b)	Find the solution to the above differential equation with initial conditions y(0)  =  0 and y′(0)  =  9. (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 y(0)  =  7 and y′(0)  =  0.

Answer 1

o ->0 Given differential equation (itsely! gay! by=0 At x=0 is ordinary point Y= not axta, x² tez x3 +94x4 + topx" & 7 be Solution of equ. a, +20,*+ 30gx2+404++ - +manent 29, + Gaze+ 1204+*+--thiriyaninze Substituting y, y' fy" in eae) (452£29476032 +124707 +12942+ ---} + n(n-1)anen-> ---} a, +20,*+ 30; x++42_xt- In thanente --- -68 ao ta, eto x² taz x3 +9429 + 7 fa...y=0 -8296

coefficient of x- 1. Mt2) Cnt 1) antz + 5 nch-han- - 8 n. an-bano M+2) Cnti) antz +5n-1) an - snan-banzo -an 64-2)(n+1) Ant2 = [56(n-1)] + sonantan = -an&50²50-sn-oh -an [ 50²-130-6 -(5n_13n-6) +) Cnt) anta an -30 g) = -(50²-130-6) (nt) (n+2) -(-6) Que na2 2 (20-26-6) A2 = 390 3.4 n=1 +12 12 -(5-13-6) 2.3 =a2 au = 390 14. 7/91

Substituting a193,94 in ea g=aota, x + 390) at +291) x² + (300) et e- y = 26C1+34 +374 + -->) +0,62+--) Required solution y' = a0(Gat 12x²+ --) + a, ( 14 The --) Ye=0 -> 0= ao C1+0+0) +0,(0) 000 y'=9 - 9- QoCoto) + a, 9,29 complete solution with condition goro & yolg y = q (appit --)) condition y(o)=7& y'(0) 70 90=7 &a, co solution y = 1 (1+32*+ 324 --) ▼