44_pdfsam_math 54 differential equation solutions odd

44_pdfsam_math 54 differential equation solutions odd -...

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

View Full Document Right Arrow Icon
Chapter 2 where v B is driver B’s constant velocity, and k> 0 is a positive constant. Separating variables we get dv A v 2 A = kdt Z dv A v 2 A = Z 1 v A ( t ) = kt + C. From the initial condition we fnd 1 v B = 1 v A (0) = k · 0+ C = C C = 1 v B . Thus v A ( t )= 1 +1 /v B = v B v B . The ±unction s ( t ) there±ore satisfes ds dt = v B v B ,s (0) = 0 . Integrating we obtain s ( t Z v B v B dt = 1 k ln ( v B +1)+ C 1 . To fnd C 1 we use the initial condition: 0= s (0) = 1 k ln ( v B k · 0+1)+ C 1 = C 1 C 1 =0 . So, s ( t 1 k ln ( v B +1) . At the moment t = t 1 when driver A’s speed was halved, i.e.,
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.

Ask a homework question - tutors are online