hw_4_7_09

hw_4_7_09 - MA 221 Homework Solutions Due date: April 7,...

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

View Full Document Right Arrow Icon
MA 221 Homework Solutions Due date : April 7 , 2009 Section 10.6 Problems 5 ̄ ,9 ̄ ,1 ̄ 1 ̄ (Underlined problems are to be handed in) 5 ̄ .) f x h 0 x / a ,0 x a , f x h 0 L x / L a , a x L and g 0 0 The problem is consistent because g 0 0 g L and f 0 0 f L The formal solution is given in equation (5) on page 625 of the text with the coefficients given in equations (6) and (7) on page 626. By equation (7) g x 0 n 1 b n n  L sin n x L . Thus, each term in this infinite series must be zero and so b n 0 for all n ’s. Therefore, the formal solution given in equation (5) on page 625 becomes u x , t n 1 a n cos n  t L sin n x L Using equation (7) on page 609 for n 1,2,3,. ..wehave a n 2 L 0 L f x sin n x L dx 2 L h 0 a 0 a x sin n x L dx h 0 a L L x L a sin n x L dx 2 h 0 L 1 a 0 a x sin n x L dx L L a a L sin n x L dx 1 L a a L x sin n x L dx By using integration by parts x sin n x L dx xL n cos n x L L 2 n 2 2 sin n x L For n .. a n 2 h 0 L 1 a aL n cos n a L L 2 n 2 2 sin n a L  L 2 n L a cos n cos n a L  1 L a L 2 n cos n aL n cos n a L  L 2 n 2 2 sin n a n 2 h 0 L 2 n 2 2 a L a sin n a L , n u x , t 2 h 0 L 2
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.

Page1 / 4

hw_4_7_09 - MA 221 Homework Solutions Due date: April 7,...

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