Consider the differential equation 2x2y" + 3xy' + (2x - 1)y=0.The indicial equation is 2r²+r-1=0.The recurrence relation is ck[2(k + r) + (k+r-1) + 3(k+r)-1]+2Ck-1=0.A series solution corresponding to the indicial root r = - 1 isy=x-¹[1 + Σk=1kxk], whereOack=(-2)k/[k!(-1).1.3...(2k-3)]Obck=(-2)k/[k!(-1)(2k-3)!]Ock=(-2)/[k!1.3...(2k-5)]O d.ck = -2k/[k! 1.3...(2k-3)]Oeck=(-2)/[k!(-1).1.3...(2k-1)]

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Answered: Consider the differential equation…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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2) Suppose you have a differential equation that satisfies the conditions of Theorem 6.2.1- Existence of Power Series Solutions with center x, = 0. It is then guaranteed that a solution exists of the form y(x) = E-0 Cnx". The recurrence relation for the coefficients was calculated to be k 3 Ck+2 k = 1,2,3, .. 2(k + 2) k+1 2(k + 2)(k + 1) 3 C2 = -7C0 Find the solution y(x) to the initial-value problem if y(0) = 0 and y'(0) = 1. Give at least the first 3 nonzero terms of the solution.
In solving a differential equation by Frobenius method of series solution about x = 0, we obtain the following indicial equation and recurrence relation: 7(2r–1)=0 and 1 Cn+1 = %3D 6n +r+3n where n = 0,1,2,3,... Find the solution corresponding to the larger root of the indicial equation. Include the first three nonzero terms, use Co=I O A. Y = 1+x 4 x+... 133 O B. Y =x-1/2/ 4 3 +.. 133 O c.y =xl/2| 12 4 3 +... c. 4 2 O D. y = x1/2 1+x+ 133 ++.. O E.Y =x1/2 1+x+ 19 4 2 +...
I know the answer is x+(1/2)(x^2)+(1/2)(x^3)+(13/24)(x^4)+(11/15)(x^5) but can you help me on verifying the answer? Somehow it matches the original equation is what I'm looking for.
The second order equation 81a?y" + 81ry' + (81x? – 16) y = 0 has a regular singular point at a = 0, and has series solutions of the form 00 (1) Insert the formal power series into the differential equation, we derive an equation Cn+ n-2 So we have the indicial equation and a recurrence relation Cn_2 for n = 2, 3, . (2) From the indicial equation we can solve the indicial roots (enter your results as a comma separated list). For any one of the indicial roots, we have c = (3) Let r, be the smaller indicial root. Then r and the recurrence relation becomes Cn 2 for n = 2, 3, - . . Let co =1 From the recurrence relation, we have a solution I+1+ z+2+ T+3+ z+4+ +... Y1 = z+ (4) Let r2 be the larger indicial root Then T2 = and the recurrence relation becomes Cn= Cn 2 for n = 2, 3, -.. Let co = 1 From the recurrence relation, we have another solution. 2 = 1+ T+2+ Ta+0 +... (5) The general solution is given by y = Ay + By, with arbitrary constants A, B.
2- Consider the differential equation (²+3)y" - 4y' = 0. The recurrence relation for the coefficients an of the power series solution an" about the ordinary point o=0 is given by b) an+2= an+2= an+2= none of these an+2= 12-0 4nan+1 + (n² = n)an 3n² +9n+6 n22 (4n+4)an+1-(n² - n)an 3n² +9n+6 (n + 4)an+1 +n²an 3n² +9n+6 an+1 + (n²-n)an 3n² +9n+6 1 " n>2 n≥ 2 n22
Find two power series solutions of the given differential equation about the ordinary point x = 0. (x2+1)y" 6y = 0 Oy₁ = 1 + 3x² + x²-to 1,6 Oy₁ = 1 + 3x² + 5x4 + 7x6 + Oy₁ = 1 + 3x² + x4 + +6 5 + + and y₂ = x + x³ and Y2 = x + 2x³ + 3x5 + 4x7. and y₂ = x - x³ 34 ₁=1+²+x+6+ ... and y₂ = x+3x² Oy₁ = 1 - 3x² + 5x4 – 7x6 + and y₂ = x - 2x³ + 3x5 - 4x² + .
Find the recurrence relation of the coefficients for the two linearly independent power series solutions for the differential equation: y" + 2xy' + 4y = 0
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