{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

M257-316Notes_Lecture7

# M257-316Notes_Lecture7 - Chapter 6 Lecture 7 Frobenius...

This preview shows pages 1–3. Sign up to view the full content.

Chapter 6 Lecture 7 - Frobenius Series about Regular Singular Points Example 1: Ly = 2 x 2 y xy + (1 + x ) y = 0 x = 0 is a RSP. y = n =0 a n x n + r (6.1) Ly = 2 x 2 n =0 a n ( n + r )( n + r 1) x n + r 2 x n =0 a n ( n + r ) x n + r 1 + (1 + x ) n =0 a n x n + r = 0 n =0 a n { 2( n + r )( n + r 1) ( n + r ) + 1 } x n + r + n =0 a n x n + r +1 = 0 (6.2) m = n + 1 n = 0 m = 1 n = m 1 Therefore a 0 { 2 r ( r 1) r + 1 } x r + n =1 [ a n { 2( n + r )( n + r 1) ( n + r ) + 1 } + a n 1 ] x n + r = 0 . 37

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

View Full Document
Lecture 7 - Frobenius Series about Regular Singular Points x r > Indicial Equation: 2 r 2 3 r + 1 = (2 r 1)( r 1) = 0 r = 1 2 , r = 1. a 0 arbitrary Recursion a n = a n 1 (2 n + 2 r 3)( n + r ) + 1 (6.3) Let r = 1 / 2: a n = a n 1 (2 n 2)( n + 1 / 2) + 1 = a n 1 ( n 1)(2 n + 1) + 1 = a n 1 n (2 n 1) n = 1 : a 1 = a 0 1 ; n = 2 : a 2 = a 1 2 . 3 = + a 0 2 . 3 a 3 = a 2 3 . 5 = a 0 1 . (2 . 3)(3 . 5) ; a 4 = a 3 4 . 7 = + a 0 1(2 . 3)(3 . 5)(4 . 7) (6.4) a n = ( 1) n n !1 . 3 . 5 . (2 n 1) y 1 ( x ) = x 1 / 2 n =0 ( 1) n n ! x n 1 . 3 . 5 . (2 n 1)= x 1 / 2 n =0 ( 1) n 2 ( n 1) n !
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern