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M257-316Notes_Lecture6

M257-316Notes_Lecture6 - Chapter 5 Lecture 6 Singular...

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Chapter 5 Lecture 6 - Singular points 5.1 Radius of Convergence and Nearest Singular Points EG. 1: (1 + x 2 ) y + 2 xy + 4 x 2 y = 0 . (1) If we were given y (0) = 0 and y (0) = 1 then we would want a power series expansion of the form y = n =0 c n x n about x 0 = 0 . (5.1) Roots of 1+ x 2 = 0 are x = ± i , so we expect the radius of convergence of the TS for 1 1+ x 2 to be 1 since 1 1+ x 2 = 1 x 2 + x 4 1 lim a n +2 a n = 1 ρ = 1 . (5.2) (2) If we were given y (1) = 1, y (1) = 0 then a power series expansion of the form c n ( x 1) n is required. In this case ρ = 2. EG. 2: ( x 1)(2 x 1) y + 2 xy 2 y = 0. x = 0 is an ordinary point. x = 1 and x = 1 2 are singular points. One solution of this equation is y ( x ) = 1 x 1 = (1 + x + x 2 + · · · ) ρ = 1 . (5.3) 31

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Lecture 6 - Singular points This TS solution about the ordinary point x = 0 converges beyond the singular point x = 1 2 . EG: ( x 2 2 x ) y + 5( x 1) y + 3 y = 0 y (1) = 7 y (1) = 3. x = 1 is an ordinary point. x = 0 is a singular point ( x 1) 2 1 y + 5( x 1) y + 3 y = 0. Let t = x 1 so that d dt = d dx and the equation is transformed to ( t 2 1) ˙ ˙ y + 5 t ˙ y + 3 y = 0 y = n =0 c n t n , y = n =1 c n nt n 1 , y = n =2 c n n ( n 1) t n 2 n =2 n ( n 1) c n t n n =2 n ( n 1) c n t n 2 + 5
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