In Exercise 17.4, problem 4, in Stewarts Early Transcendentals Calculus, the problem says to use the power series to solve the differential equation: (x-3)y' + 2y = 0. The answer establishes a value for C1 as 2/3 Co. It then substitutes this value in the C2 equation which is ((3C1)/(3 * 2)). The answer is that C2 is equal to 3Co/32. How do they get this equation? I would have thought it would be 2Co/32. Substitute 0 for n in equation (6),(0+2)coCo+13(0+1)2c0C13(1)2c0(7)3C1Substitute 1 for n in equation (6),(1+2)e3(1+1)C1+1ЗсуC2=3(2)2coSubstituteforCi2033(2)3(co)3(3)с23co(8)32C2

Answered: Substitute 0 for n in equation (6),…

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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In Exercise 17.4, problem 4, in Stewarts Early Transcendentals Calculus, the problem says to use the power series to solve the differential equation: (x-3)y' + 2y = 0.

The answer establishes a value for C₁ as 2/3 C_o. It then substitutes this value in the C₂ equation which is ((3C₁)/(3 * 2)). The answer is that C₂ is equal to 3C_o/3². How do they get this equation? I would have thought it would be 2C_o/3².

Substitute 0 for n in equation (6),
(0+2)co
Co+1
3(0+1)
2c0
C1
3(1)
2c0
(7)
3
C1
Substitute 1 for n in equation (6),
(1+2)e
3(1+1)
C1+1
Зсу
C2=
3(2)
2co
Substitute
for
Ci
20
3
3(2)
3(co)
3(3)
с2
3co
(8)
32
C2

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