exam2_sol - Exam 2 Solutions 1] (a) If a solution is in the...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Exam 2 Solutions 1] (a) If a solution is in the form y x = n = 0 A n x n , then substituting into the equation gives us 1 x 2 n = 2 n n 1 A n x n 2 2x n = 0 n A n x n 1 2 n = 0 A n x n = 0 , or n = 2 n n 1 A n x n 2 n = 0 2 2n n n 1  A n x n = 0 , or n = 0 n 2  n 1 A n 2 x n  2 2n n n 1  A n x n = 0 . So the condition on the coefficients A n ( n from 0 to ) is n 2  n 1 A n 2  2 2n n n 1  A n = 0 , or n 2  n 1 A n 2 − n 2  n 1 A n = 0 . This, A 0 and A 1 are arbitrary, all even A n are determined by A 0 , and all odd A n are determined by A 1 . The first independent solution is: A 0 1 x 2 x 4 / 3 x 6 / 5 x 8 / 7 ... . The second independent solution is just
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 08/30/2011 for the course CGN 2200 taught by Professor Glagola during the Spring '11 term at University of Florida.

Page1 / 2

exam2_sol - Exam 2 Solutions 1] (a) If a solution is in the...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online