Unformatted text preview: n 1 2 2 x x b x ln x Cy x y ; b ≠ Note: For this case we have a possibility that the second solution may have C = This would give 2 series solutions, but we have no way of knowing this before we have solved the ODE completely Case 3: Roots are equal 2 1 2 1 r r r r = ⇒ =Two solutions exist: ( 29 ( 29 ∑ ∞ = += n r n n 1 1 x x c x y ; c ≠ and ( 29 ( 29 ( 29 ∑ ∞ = ++ = n r n n 1 2 2 x x b x ln x y x y Note: The log term always occurs in this case. Finding the second solution: We can use Reduction of Order to find the second solution....
View
Full
Document
This note was uploaded on 04/11/2008 for the course CHNE 525 taught by Professor Prinja during the Fall '08 term at New Mexico.
 Fall '08
 Prinja

Click to edit the document details