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Σnn - 1cnx" + 2 nn - 1)x"-2 – 6 Σεx" = 0 n=2 n=2 n=0. 00 Σκα - 1cx*+ Σ« + 2k + 1)c+2x* – 6 Σax* = 0 k=2 A=0 A=0 2c₂ - 6c +(6c3-6c₁ )x+ Σ [(k² − k − 6)c₂ + (k+ 2)(k+1)C₁+2] x* = 0 k=2 Thus, 2cy – 6co = 0,6c3 – 6ci = 0 and (k - 3)(k + 2)ck + (k + 2)(k + 1)ck+2 = 0. Simplifying the above equations gives c2 = 3co and c3 = cı and ck+2 = - =- (k−3)ck for k = 2, 3, 4, ... (k+1) Substituting k = 2. 3. 4. ... into the last formula gives.

Σnn - 1cnx" + 2 nn - 1)x"-2 – 6 Σεx" = 0 n=2 n=2 n=0. 00 Σκα - 1cx*+ Σ« + 2k + 1)c+2x* – 6 Σax* = 0 k=2 A=0 A=0 2c₂ - 6c +(6c3-6c₁ )x+ Σ [(k² − k − 6)c₂ + (k+ 2)(k+1)C₁+2] x* = 0 k=2 Thus, 2cy – 6co = 0,6c3 – 6ci = 0 and (k - 3)(k + 2)ck + (k + 2)(k + 1)ck+2 = 0. Simplifying the above equations gives c2 = 3co and c3 = cı and ck+2 = - =- (k−3)ck for k = 2, 3, 4, ... (k+1) Substituting k = 2. 3. 4. ... into the last formula gives.

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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I'm not sure how they get this line from the line prior (marked in image). When I try it, I get zero c_0 and zero c_1 but here they have nonzero coefficients. 

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Given: The given differential equation is (x² + 1)y" - 6y = 0 and x = 0 is the ordinary point. Definition used: If x = xo is an ordinary point of the differential equation then there exist two linearly independent solution in the form of power series centered at xo. y = =Σ Cn(x-xo)" n=0 Calculation: Given differential equation is (x² + 1)y" - 6y=0 so consider power series y = n=2 Substitute y = nx", and the second derivative y" = Σcn(n-1)x"-² in the differential equation as, 00 00 Σn(n-1)cnx" + n(n − 1)c₁x²-² – 6 k=2 n=0 Σk(k-1)cx + C4 = 00 n=2 00 00 Σ cnx" = 0 n=0 (k+ 2)(k + 1)ck+2x² − 6 k=0 00 n=2 2₁₁x² = 0 k=0 00 2c₂ - 6€ + (6c3 - 6c₁ )x+ Σ [(k² − k − 6)ck + (k+2)(k+ 1)Ck+2] x² = 0 k=2 Thus, 2c₂6c0 = 0, 6c36c₁ = 0 and (k - 3)(k+ 2)ck + (k+2)(k+ 1)ck+2 = 0. Simplifying the above equations gives c₂= 3co and c3 = C₁ and C+2 = Substituting k = 2, 3, 4, ... into the last formula gives, = Co C5 = 0 C6 = 300 Similarly find other coefficients using the above formula. ΣC.X². n=0 Substituting all the coefficients in the power series y = cx" gives, ΣΩ n=0 (k-3) for k= 2, 3, 4, ... (k+1) y = co + c₁x + c₂x² + C3x²³ + C₁x² + c6x² + ... y = co + c₁x +3cox² + c₁x²³ + cox² = {{coxº +... y = co(1 + 3x² + x4 − ½ x6 + ...) + c₁(x + x³) The above solution is of the form y = co(vi) + ci(v2). Thus, the two power series solutions are y₁ = 1 + 3x² + x² - ²x² + ... and y₂ = x + x³
Transcribed Image Text:Given: The given differential equation is (x² + 1)y" - 6y = 0 and x = 0 is the ordinary point. Definition used: If x = xo is an ordinary point of the differential equation then there exist two linearly independent solution in the form of power series centered at xo. y = =Σ Cn(x-xo)" n=0 Calculation: Given differential equation is (x² + 1)y" - 6y=0 so consider power series y = n=2 Substitute y = nx", and the second derivative y" = Σcn(n-1)x"-² in the differential equation as, 00 00 Σn(n-1)cnx" + n(n − 1)c₁x²-² – 6 k=2 n=0 Σk(k-1)cx + C4 = 00 n=2 00 00 Σ cnx" = 0 n=0 (k+ 2)(k + 1)ck+2x² − 6 k=0 00 n=2 2₁₁x² = 0 k=0 00 2c₂ - 6€ + (6c3 - 6c₁ )x+ Σ [(k² − k − 6)ck + (k+2)(k+ 1)Ck+2] x² = 0 k=2 Thus, 2c₂6c0 = 0, 6c36c₁ = 0 and (k - 3)(k+ 2)ck + (k+2)(k+ 1)ck+2 = 0. Simplifying the above equations gives c₂= 3co and c3 = C₁ and C+2 = Substituting k = 2, 3, 4, ... into the last formula gives, = Co C5 = 0 C6 = 300 Similarly find other coefficients using the above formula. ΣC.X². n=0 Substituting all the coefficients in the power series y = cx" gives, ΣΩ n=0 (k-3) for k= 2, 3, 4, ... (k+1) y = co + c₁x + c₂x² + C3x²³ + C₁x² + c6x² + ... y = co + c₁x +3cox² + c₁x²³ + cox² = {{coxº +... y = co(1 + 3x² + x4 − ½ x6 + ...) + c₁(x + x³) The above solution is of the form y = co(vi) + ci(v2). Thus, the two power series solutions are y₁ = 1 + 3x² + x² - ²x² + ... and y₂ = x + x³
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