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Unformatted text preview: Midterm Exam Solutions Problem 1 (8+3 6 Points) (a) (8 points) (i) (2 points) (1) y'' + xy' + 2 x 3 y = Irregular (Essential) Singular point at x=0. (ii) (2 points) (2) y'' x H x 1 L H x 2 L y' + x 2 H x 1 L H x 2 L y = Regular singular points at x=1 and x=2. (1 point each) (iii) (2 points) (3) y'' + 2 y' + 2 y = No singular points, all ordinary points. (iv) (2 points) (4) y'' + x x 1 y = Regular Singular points occur for x=2n p (1 point) except for x=0 which is an ordinary point since the singularity is removable(1 point): (5) lim x x x 1 = 1 (b) (6 points) (i) (3 points) Since there are singular points at x=1 and x=2, the nearest singular point to the origin is x=1. (1 point) By Fuch's theorem, the radius of convergence of the series is at least the distance to the nearest singularity, so the radius of convergence is at least 1 (1 point). Hence the series is guaranteed to converge in (1 points) (6) x < 1 (ii) (3 points) Since there are singular points at x=2n p , the nearest singular points to the origin are x=2 p (1 points). By Fuch's theo rem, the radius of convergence of the series is at least the distance to the nearest singularity, so the radius of convergence is at least 2 p (1 point). Hence the series is guaranteed to converge in (1 point) (7) x < 2 p (c) (6 points) (i) (2 points) As stated in class, an equation of the form Printed by Mathematica for Students (8) y'' + p H x L H x x L y' + q H x L H x x L 2 y = Where p and q are analytic at x leads to an indicial equation given by (9) n 2 + H p 1 L n + q = Where (10) p H x L = n = p n H x x L n q H x L = n = q n H x x L n In this problem the equation is (11) y'' x H x 1 L H x 2 L y' + x 2 H x 1 L H x 2 L y = so (12) p = x x 2 q = x 2 H x 1 L x 2 This gives p =1 and q =0. Plugging these in gives the indicial equation (1 point) (13) n 2 = So there is a double root of n =0 (1 point). (ii) (2 points) No, it can not. By Fuch's theorem, only one of the solutions will be in the form of a Taylor series. (iii) (2 points) Method 1 Fuch's theorem (or they might site theorem 25 from page 27 of the class notes) tells us that one solution will be of the form of a power series (14) y 1 = n = a n H x 1 L n and a second linearly independent solution will be of the form (15) y 2 = ln x 1 y 1 H x L + n = b n...
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This note was uploaded on 01/08/2011 for the course ACM 95b taught by Professor Nilesa.pierce during the Winter '09 term at Caltech.
 Winter '09
 NilesA.Pierce

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