Unformatted text preview: . We are in Case II. The solutions are of the form y 1 = ∞ ³ m =0 a m x m and y 2 = y 1 ln x + ∞ ³ m =0 b m x m , with a ± = 0. Let us determine y 1 . We use y instead of y to simplify the notation. We have y = ∞ ³ m =0 a m x m ; y ± = ∞ ³ m =0 ma m x m1 ; y ±± = ∞ ³ m =0 m ( m1) a m x m2 ....
View
Full
Document
This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.
 Fall '11
 StuartChalk

Click to edit the document details