Unformatted text preview: a = 0 and a 1 = 3. So a 2 = a 4 = Â·Â·Â· = 0 and, from the recurrence relation with m = 1, a 3 = (11 1 + 1 a = 0 . So a 5 = a 7 = Â·Â·Â· = 0 and hence y = 3 x is the solution. 17. Put the equation (1x 2 ) y Â±Â±2 xy Â± + 2 y = 0 in the form y Â±Â±2 x 1x 2 + 2 1x 2 y = 0 ....
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 Fall '11
 StuartChalk
 Zagreb, Highways in Croatia, Recurrence relation, power series method

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