272_pdfsam_math 54 differential equation solutions odd

272_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 5 We use initial conditions to determine constants C 1 , C 2 , C 3 ,and C 4 . Diferentiating (5.7) and (5.8), we get 3= x (0) = C 1 e 0 + C 2 e 0 + C 3 cos 0 + C 4 sin 0 = C 1 + C 2 + C 3 , 1= x 0 (0) = C 1 e 0 C 2 e 0 C 3 sin 0 + C 4 cos 0 = C 1 C 2 + C 4 , 1= y (0) = C 1 e 0 + C 2 e 0 C 3 cos 0 C 4 sin 0 = C 1 + C 2 C 3 , y 0 (0) = C 1 e 0 C 2 e 0 + C 3 sin 0 C 4 cos 0 = C 1 C 2 C 4 C 1 + C 2 + C 3 =3 , C 1 C 2 + C 4 =1 , C 1 + C 2 C 3 , C 1 C 2 C 4 = 1 . Solving we obtain C 1 = C 2 = C 3 = C 4 = 1. So, the desired solution is x ( t )= e t + e t +cos t +sin t, y ( t e t + e t cos t sin t. 23. We will attempt to solve this system by Frst eliminating the ±unction y ( t ). Thus, we multiply the Frst equation by ( D + 2) and the second by ( D 1). There±ore, we obtain ( D +2)( D 1)[ x ]+( D D 1) D [ y ]=( D +2) ± 3 e 2 t ² =6 e 2 t 6 e 2 t =0 , ( D 1)( D +2)[ x ] ( D 1)( D y ]= ( D 1) ± 3 e t ² = 3 e t +3 e t . Adding these two equations yields
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