If we want to find the solution of the differential equation y- 2.xy = 0 with the help of power series around x, = 0 then Which of the following is the Recurrence Relation to be obtained? 2a, n20 (n+2) 2а, n20 (n+2) 2a n20 (n+1) 2a, a +2 n20 (n+1) 2a,-1 an+1 n20 %3D (n+2)

If we want to find the solution of the differential equation y- 2.xy = 0 with the help of power series around x, = 0 then Which of the following is the Recurrence Relation to be obtained? 2a, n20 (n+2) 2а, n20 (n+2) 2a n20 (n+1) 2a, a +2 n20 (n+1) 2a,-1 an+1 n20 %3D (n+2)

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. Seek a power series solution of the form y(r) =ana" to the differential equation n=0 y"(r) – 2ry (x) – 4y(x) = 0, xo = 0. (a) Set-up a recurrence for the cofficients, an. (b) Find the first four terms of two independent power series solutions yı and y2 in such a way that you can write y(r) = a0y1(x) + a1y2(x). (c) Determine the Taylor series of e about the point ro = 0. %3D (Hint: Use the Taylor series for e".) (d) By using Part(c), find the explicit form of y2(x).
Find a power series expansion about x = 0 for a general solution to the given differential equation. Your answer should include a general formula for the coefficients. -z" -x²z' - xz = 0 What is the power series solution to the differential equation? A. Z(x) = a 1 + B. Z(X) = a1 + ∞ Σ (-1) - n=1 (1.4.7... (3n - 2))² (3n)! ∞ Σ (-1). n=1 ∞ OC. Z(x) = a 1+ Σ n=1 D. Z(X) =a 1+ Σ n=1 (3.6.9...(3n)) (3n - 2)! (1.4.7... (3n - 2))² (3n)! -x³n (3.6.9...(3n)) (3n - 2)! 3n 3n + a X + 2,³^) - ₁,[(x + 2, + x+ Σ n=1 X + x+ Σ n=1 ∞ n=1 (-1)^. ∞ Σ (-1). n=1 (2.5.8...(3n-1))² (3n + 1)! (1.4.7...(3n-2)) +1 (3n + 2)! -x³n+ (2.5.8--- (3n-1))² (3n+ 1)! (1.4.7...(3n-2)) (3n+2)! -x³n+1 -x³n+1 -x³n+ 1
I need detailed explanation solving this problem (c) from Engineering Mathematics, please.
B1. Advance maths
Can you help me with the third part?
Find singular points of the differential equation 2x2y′′ − xy′ + (1 + x)y = 0, and solve it around that singular points using power series solutions. Find the first three nonzero terms in each of two solutions about the singular point. Solve the recurrence relation you find (i.e express an in terms of n) and finally show that solutions you find is converging. (Hint: Find the roots of indicial equation and construct the recurrence relation).
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 +...
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
Solve the given ODE using the power series method. y'' + xy' + 2y = 0 4. Obtain the recurrence relation. as Oas+2 = s + 1 as O as+2 s + 2 as as+2 = 8 + 2 as O as+2 = - s + 1 5. Determine the general solution. Use up to the 5th-degree term of the solution. 1 1 + a1 x 3 1 ,5 + 15 Y = ao[ 1 - - (-- 1 1 x° + 2 y = ao 1 + a1 - ... 1 y = ao[ 1 + 2 1 + -x* + 8 1 x° + 15 + ai x + 3 ... ... 1 4 + 1 + a1( x + 1 y = ao(1+x² + + 3 8 - |00