Elementary Differential Equations : Power Series Methods
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Power Series Methods - Introduction and Review of Power Series
14. Find two linearly independent power series solutions of the given differential equation. Determine the radius of the convergence of each series, and identify the general solution in terms of familiar elementary functions.
y" + y = x
Answer:
y(x) = x + c0 (1 - x2/2! + x4/4! + x6/6! + ...) + (c1-1) (x - x3/3! + x5/5! + x7/7! + ...) = x + c0cosx + (c1-1) sinx ;  = 
24. Establish the binomial series in (12) by means of the following steps. (a) Show that y = (1 +x) satisfies the initial value problem (1 + x)y' = y, y(0) = 1. (b) Show that the power series method gives the binomial series in (12) as the solution of the initial value problem in part (a), and that this series converges if x < 1. (x) Explain why the validity of the binomial series given in (12) follows from parts (a) and (b).
(1 +x) = 1 + x + [( - 1)x2/2!] + [( - 1)( - 2)x3/3!] + ... (12)
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Elementary differential equations and power series methods are investigated. The solution is detailed and well presented. The response received a rating of "5/5" from the student who originally posted the question.
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Power Series Methods - Introduction and Review of Power Series
14. Find two linearly independent power series solutions of the given differential equation. Determine the radius of the convergence of each series, and identify the general solution in terms of familiar elementary functions.
y" + y = x .................................(1)
Answer:
y(x) = x + c0 (1 - x2/2! + x4/4! + x6/6! + ...) + (c1-1) (x - x3/3! + x5/5! + x7/7! + ...) = x + c0cosx + (c1-1) sinx ; =
Solution. We can assume that (1) has a power series solution as follows.
.................(2)
where are constants.
Then we take derivatives on both sides of (2) and we have
and
...............(3)
We substitute (2) and (3) to (1) to get
i.e.,
So,
...
Education
- BSc , Wuhan Univ. China
- MA, Shandong Univ.
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- "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
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- "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
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