5-2. Power series - Here is the idea substitute y = ∑ n =...

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Power series, solution about ordinary points. If L <1 then the power series converges at x. Radius of convergence: R=lim|c n+1 /c n | as n tents to infinity.
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Suppouse the linear second order differential equation a 2 ( x ) y ' ' + a 1 ( x ) y ' + a 0 ( x ) y = 0 (5) is put into standard form y '' + P ( x ) y ' + Q ( x ) y = 0 (6) by dividing by the leading coefficient a 2 ( x ) .
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Definition. A point x 0 is said to be an ordinary point of the differential equation (5) if both P ( x ) Q ( x ) in the standard form (6) are analytic at x 0 . A point that is not an ordinary point is said to be a singular point of the equation. Every finite value of x is an ordinary point of the differential equation y '' + e x y ' + sinx y = 0 x = x 0 is an ordinary point of (5) if a 2 ( x ) 0 , whereas x = x 0 is singular point of (5). Theorem. A solution of the form y = n = 0 c n ( x x 0 ) n is said to be a solution about the ordinary point x 0 . Finding a power series solution. The power series method of the solving a linear DE with variable coefficients is often described as “ the method of undetermined series coefficients”.
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Unformatted text preview: Here is the idea: substitute y = ∑ n = ∞ c n x n into the differential equation and then equate all coefficients to the right-hand side of the equation to determine the coefficients c n . Example 1. Solve Slution. Let y = ∑ n = ∞ c n x n . Then y' = ∑ n = 1 ∞ nc n x n − 1 and y' ' = ∑ n = 2 ∞ n ( n − 1 ) c n x n − 2 . Substituting y ∧ y ' ' into differential equation gives ¿ 2 c 2 + ∑ n = 3 ∞ n ( n − 1 ) c n x n − 2 + ∑ n = ∞ c n x n + 1 Let k = n − 2 ∧ k = n + 1 . Then c k − 1 x k = ¿ 2 c 2 + ∑ k = 1 ∞ ( k + 2 )( k + 1 ) c k + 2 x k + ∑ k = 1 ∞ ¿ This relation generates consecutive coefficients of the assumed solution one at the time as we let k take on the successive integers Example 2. Use power series method to solve given differential equation about ordinary point x = 0, y '' +( cosx ) y = . Solution....
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