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Unformatted text preview: 642:527 SOLUTIONS: ASSIGNMENT 2 FALL 2009 Some of these solutions were written by Professor Dan Ocone. 4.3.1 (a) The answer in the text seems sufficient. (b) Rewrite the equation as y ′′ − [(cos x ) /x ] y ′ +[5 /x ] y = 0. Then p ( x ) and q ( x ) are analytic at all points except x = 0. Since the functions xp ( x ) = cos x and x 2 q ( x ) = 5 x are analytic at 0, x = 0 is a regular singular point. Now the power series for cos( x ) about x = 0 has an infinite radius of convergence; the function which is equal to the constant 5 x is given as a power series all of whose terms except the first are zero, so its power series representation also has an infinite radius of convergence. Therefore, when the Frobenius method is applied it will produce series solutions, centered at x = 0, with infinite radii of convergence . (c) The singular points are the roots of ( x 2 − 3), i.e., ± √ 3. We have p ( x ) = 0 and q ( x ) = 1 / ( x 2 − 3); since ( x − √ 3) 2 q ( x ) = ( x − √ 3)( x + √ 3) is analytic at x = √ 3, x = √ 3 is a regular singular point; by the same reasoning so is x = − √ 3. Each series must have radius of convergence at least the distance to the other singular point nearest, i.e., 2 √ 3. (g) Write the equation as y ′′ + 1 ( x − 1)( x + 3) 2 y ′ + 1 ( x − 1)( x + 3) 2 y = 0 . Here p and q (which happen to be the same) are analytic at all points except 1 and − 3. Since ( x +3) p ( x ) = 1 / ( x − 1)( x +3) is not analytic at x = − 3, − 3 is an irregular singular point of the equation. However ( x − 1) p ( x ) = 1 / ( x +3) 2 and ( x − 1) 2 q ( x ) = ( x − 1) / ( x +3) 2 are analytic at x = 1 and so x = 1 is a regular singular point. The power series for 1 / ( x + 3) and ( x − 1) / ( x + 3) 2 with center x = 1 must have radius of convergence R = 4, since that is the distance from x to x = − 3, the nearest point of nonanalyticity. Thus the series solutions centered at x = 1 produced by the Frobenius method will have radii of convergence at least as large as R = 4. (n) Carrying out the differentiation gives the equation x 3 ( x − 1) y ′′ +(4 x 3 − 3 x 2 ) y ′ +2 y = 0, so p ( x ) = (4 x 3 − 3 x 2 ) / [ x 3 ( x − 1)] and q ( x ) = 2 / [ x 3 ( x − 1)]. These are analytic except at x = 0 and x = 1, so these are the singular points. Since x 2 p ( x ) == (4 x 3 − 3 x 2 ) / [ x 2 ( x − 1)] is not analytic at x = 0, x = 0 is an irregular singular point, but ( x − 1) p ( x ) = (4 x 3 − 3 x 2 ) /x 3 and ( x − 1) 2 q ( x ) = 2( x − 1) /x 3 are analytic at x = 1, so this is a regular singular point....
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This note was uploaded on 12/15/2009 for the course MATH 527 taught by Professor Staff during the Fall '08 term at Rutgers.
 Fall '08
 Staff
 Math, Cone

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