# Find a solution in the form of a power series for an ODE

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Find a solution in the form of a power series for the equation

y" - 2*x*y' = 0

(ie find 2 linearly independent solutions y1(x) and y2(x)).

After doing that, note that the equation can also be solved directly by integration:

y"/y' = 2x

ln(y') = x^2 + c1

y' = ke^(x^2) k=e^c1

y = k* integral((e^(t^2))dt + C) from 0 to x

Thus, one of your power series solutions gives and explicit form for the integral:

integral(e^(t^2)dt) from 0 to x

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##### Solution Summary

A solution is found in the form of a power series for an ODE. Independent solutions are solved by using integration.

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Solution. Assume that a solution of the ODE (in form of power series) is as follows.

y(x)=a(0)+a(1)x+a(2) +a(3) +...+a(n) +... (1)

where a(i), i=0,1,2,... are coefficients.

Then by (1) we have,

y'(x)=a(1)+2*a(2)x+3*a(3) +...+n*a(n) +(n+1)*a(n+1) +... (2)

and

y''(x)=2*a(2)+6*a(3)x+...+n(n-1)*a(n) +(n+1)n* a(n+1) +(n+2)(n+1)*a(n+2) +... ...

###### 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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