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EnzoF61
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Homework Statement
y' = [tex]\sqrt{(1-y^2)
}[/tex]
Initial condition y(0) = 0
a) Show y = sinx is a solution of the initial value problem.
b) Look for a solution of the initial value problem in the form of a power series about x = 0. Find coefficients up to the term in x^3 in this series.
Homework Equations
part a) was Ok.
The Attempt at a Solution
This is for my part b attempt.
(y')^2 + (y)^2 = 1
(cosx)^2 + (sinx)^2 = 1
[tex]\sum(-1)^ {2n} * (x^ {4n} )}/(2n!)^2[/tex] + [tex]\sum (-1)^{2n} * (x^ {4n+2})/((2n+1)!)^2[/tex] = 1
I had also tried the general solution series for y and y'. y = a0 + a1x + ... + an*x^n
y' = a1 + ... + n*an x^n
Please note the values such as a0, a1, ...an have the nth term as a subscript.
Maybe I should not be using part a where y=sinx and y'=cosx because I will have even powers when squared always...?
What series will allow me to evaluate at the x^3 term?
Thanks,
-Adam
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