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Unformatted text preview: Series sol. near ordinary points w/ given initial conds. Let’s revisit the 2 problems that we have solved in class – but let’s solve them with some initial conditions. We will solve them two using two different methods. 1. Using the results we obtained in ordinary_ps.pdf . 2. Using a Taylor series expansion of the solution about x . Before doing so, recall that the Talor series expansion of a function about point x is given by y ( x ) = y ( x ) + y ( x )( x x ) + y 00 ( x ) 2! ( x x ) 2 + y 000 ( x ) 3! ( x x ) 3 + y (4) ( x ) 4! ( x x ) 4 + ··· . (1) For a given second order initial value problem (IVP), we already know y ( x ) and y ( x ) . What we need to find are y 00 ( x ) ,y 000 ( x ) , ··· . We will find these using the given differential equation, as illustrated in the problems below. 1 c HF, 2011 Series sol. near ordinary points w/ given initial conds. Problem 1 (1 + x ) y 00 + y = 0 , x = 0 . y (0) = 1 ,y (0) = 1 . (2) Solution – Method A : Recall that we previous found the general solution to the problem to be (see ordinary_ps.pdf ) y = a 1 1 2 x 2 + 1 6 x 3 1 24 x 4 + ··· + a 1 x 1 6 x 3 + 1 12 x 4 + ··· , (3) where a and a 1 are two arbitrary constants. Using the first initial conditions, we find y (0) = 1 = a 1 1 2 2 + 1 6 3 1 24 4 + ··· + a 1 1 6 3 + 1 12 4 + ··· ⇒ a = 1 ....
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 Spring '10
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 Derivative, Taylor Series, Constant of integration, Boundary value problem

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