This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: March 26, 2010 The differential equation of interest is: ln x y 00 + 1 2 y + y = 0 . Determine the first three nonzero terms in the series ∑ ∞ n =0 a n ( x 1) n + r First verify that x = 1 is indeed a singular point (show this) : lim x → 1 1 2 ( x 1) ln x = 1 2 < ∞ lim x → 1 " ( x 1) 2 ln x # = 0 < ∞ ∴ ( x = 1) is a singular point. Also, there exists at least one solution of the form y = ( x 1) r ∞ X n =0 a n ( x 1) n = ∞ X n =0 a n ( x 1) n + r . (1) Aside: You can use a change of variables t = x 1 to rewrite the differential equation in terms of t and move the singular point to t = 0 : let t = x 1 ⇒ dy dx = dy dt dt dx = dy dt · 1 = y ( t ) ⇒ ln xy 00 ( x ) + 1 2 y ( x ) + y ( x ) = ln ( t + 1) y 00 ( t ) + 1 2 y ( t ) + y ( t ) = 0 Then the solution appears in form of y ( t ) = t r ∞ X n =0 ˆ a n t n = ∞ X n =0 ˆ a n t n + r . This will make the expressions appear more compact and is definitely a good idea to do. In fact, do this. But I’ll keepfact, do this....
View
Full
Document
This note was uploaded on 05/02/2011 for the course GE 207K taught by Professor None during the Spring '10 term at University of Texas.
 Spring '10
 None

Click to edit the document details