3f-fall2010-exam_3_solutions

3f-fall2010-exam_3_solutions - Math 3F — Exam#3 Solutions...

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
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math 3F — Exam #3 Solutions Laney College, Fall 2010 Fred Bourgoin 1. Use power series to solve the initial-value problem: y ′′- xy ′- y = 0 , y (0) = 1 , y ′ (0) = 1 . Since the initial conditions are given at x = 0, it makes sense to center your series at x = 0. Write your answer in sigma notation. (22 points) Solution. Let y = ∞ ∑ n =0 a n x n . Then y ′ = ∞ ∑ n =1 na n x n − 1 , y ′′ = ∞ ∑ n =2 n ( n- 1) a n x n − 2 , and y ′′- xy ′- y = ∞ ∑ n =2 n ( n- 1) a n x n − 2- x ∞ ∑ n =1 na n x n − 1- ∞ ∑ n =0 a n x n = ∞ ∑ n =0 ( n + 2)( n + 1) a n +2 x n- ∞ ∑ n =1 na n x n- ∞ ∑ n =0 a n x n = ∞ ∑ n =0 ( n + 2)( n + 1) a n +2 x n- ∞ ∑ n =0 na n x n- ∞ ∑ n =0 a n x n = ∞ ∑ n =0 [ ( n + 2)( n + 1) a n +2- na n- a n ] x n = 0 , so ( n + 2)( n + 1) a n +2- na n- a n = 0 for all n , and we have the recurrence relation a n +2 = a n n + 2 ....
View Full Document

{[ snackBarMessage ]}

Page1 / 3

3f-fall2010-exam_3_solutions - Math 3F — Exam#3 Solutions...

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

View Full Document
Ask a homework question - tutors are online