61_pdfsam_math 54 differential equation solutions odd

61_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
Exercises 2.3 The last integral can be found by integrating by parts twice which leads back to an integral similar to the original. Combining these two similar integrals and simplifying, we obtain Z e 2 t cos ± πt 12 ² dt = e 2 t ³ 2cos ( πt 12 ) + π 12 sin ( πt 12 4+( π 12 ) 2 + C. Thus we see that x ( t )= 1 2 2cos ( πt 12 ) + π 12 sin ( πt 12 ) 4+( π 12 ) 2 + Ce 2 t . Using the initial condition, t =0and x = 10, to solve for C ,weobta in C = 19 2 + 2 4+( π 12 ) 2 . Therefore, the desired solution is x ( t )= 1 2 2cos ( πt 12 ) + π 12 sin ( πt 12 ) 4+( π 12 ) 2 + µ 19 2 + 2 4+( π 12 ) 2 e 2 t . 39. Let T j ( t ), j =0 , 1 , 2 ,... , denote the temperature in the classroom for 9 + j t< 10 + j , where t = 13 denotes 1 : 00 p.m. , t = 14 denotes 2 : 00
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: , etc. Then T (9) = 0 , (2.12) and the continuity of the temperature implies that lim t 10+ j = T j +1 (10 + j ) , j = 0 , 1 , 2 , . . . . (2.13) According to the work of the heating unit, the temperature satisFes the equation dT j dt = 1 T j , if j = 2 k T j , if j = 2 k + 1 , 9 + j < t < 10 + j k = 0 , 1 , . . . . The general solutions of these equations are: for j even dT j dt = 1 T j dT j 1 T j = dt ln | 1 T j | = t + c j T j ( t ) = 1 C j e t ; 57...
View Full 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