3_1 - Solutions 3.1-Page 204 Problem 5 Find a power series...

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Solutions 3.1-Page 204 Problem 5 Find a power series solution of the given differential equation. Determine the radius of convergence of the resulting series, and use the series in Eqs.(5) through (12) to identify the series solution in terms of familiar elementary functions. y x y 2 = The differential equation can be rewritten as . Substituting and into the differential equation yields 0 2 = y x y = = 1 n n n x c y = = 1 1 n n n x nc y ∑∑ = = = 10 2 1 nn n n n n x c x x nc Simplifying further yields = = + = 2 1 n n n n x c x nc In the first summation, the index can be shifted to start at n = -3. = = + + + = + 30 2 2 3 ) 3 ( n n n n x c x c n The identity principle can be used to equate coefficients. ) 3 ( ) 3 ( 3 3 + = = + + + n c c c c n n n n n and c if the two summations are to be equal. 0 2 1 = = c Substituting values for n in order to discern a pattern gives:
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18 ) 6 ( 3 6 : 3 0 5 : 2 0 4 : 1 3 : 0 0 0 3 6 2 5 1 4 0 3 c c c c n c c n c c n c c n = = = = = = = = = = = = The pattern that can be fit to these coefficients is that only the coefficients with an index that is a multiple of 3 is nonzero. Rewriting this with k as the new index gives: ! 3 0 3 k c c k k = Therefore + + = + + = = = ... ! 2 1 3 3 1 ... ) ! 2 ( 3 ) ! 1 ( 3 ! 3 2 3 3 0 6 2 0 3 0 0 0 3 0 x x c x c x c c x k c y k k k This summation matches the form of if the variable is x e 3 3 x instead of x . So 3 / 0 3 x e c y = The radius of convergence is ( ) () = + = + = = + + ! )! 1 ( 3 lim )! 1 ( 3 / ! 3 / lim lim 1 0 0 1 k k k c k c c c n k k n n n n ρ =
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Problem 7 Find a power series solution of the given differential equation. Determine the radius of convergence of the resulting series, and use the series in Eqs.(5) through (12) to identify the series solution in terms of familiar elementary functions. 0
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3_1 - Solutions 3.1-Page 204 Problem 5 Find a power series...

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