Differential Equations Lecture Work Solutions 279

Differential - x − ct and the last term is a function of x ct exactly as we expect from D’Alembert solution 279

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
The limits of integration become x 0 and x 0 +2 ct .Thu stheso lu t ion u ( x 0 ,t )= Z x 0 +2 ct x 0 1 2 c V ( z ) dz + K ( x 0 ) But x 0 = x ct u ( x, t )= Z x + ct x ct 1 2 c V ( z ) dz + K ( x ct ) Now break the integral using the point zero. u ( x, t )= K ( x ct ) Z x ct 0 1 2 c V ( z ) dz + Z x + ct 0 1 2 c V (
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: x − ct and the last term is a function of x + ct , exactly as we expect from D’Alembert solution. 279...
View Full Document

This note was uploaded on 12/22/2011 for the course MAP 2302 taught by Professor Bell,d during the Fall '08 term at UNF.

Ask a homework question - tutors are online