216_pdfsam_math 54 differential equation solutions odd

# 216_pdfsam_math 54 differential equation solutions odd - |...

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

Chapter 4 The system (9) on page 195 in the text becomes v 0 1 ( t )cos2 t + v 0 2 ( t )sin2 t =0 2 v 0 1 ( t )sin2 t +2 v 0 2 ( t )cos2 t =tan2 t. (4.5) Multiplying the frst equation in (4.5) by sin 2 t , the second equation by (1 / 2) cos 2 t , and adding the resulting equations together, we get v 0 2 ( t )= 1 2 sin 2 t v 2 = 1 2 Z sin 2 tdt = 1 4 cos 2 t + c 3 . From the frst equation in (4.5) we also obtain v 0 1 ( t )= v 0 2 ( t )tan2 t = 1 2 sin 2 2 t cos 2 t = 1 2 1 cos 2 2 t cos 2 t = 1 2 (cos 2 t sec 2 t ) v 1 ( t )= 1 2 Z (cos 2 t sec 2 t ) dt = 1 4 (sin 2 t ln | sec 2 t +tan2 t | )+ c 4 . We take c 3 = c 4 = 0 since we need just one particular solution. Thus y p ( t )= 1 4 (sin 2 t ln | sec 2 t +tan2 t | )cos2 t 1 4 cos 2 t sin 2 t = 1 4 cos 2 t ln | sec 2 t +tan2 t | and a general solution to the given equation is y ( t )= y h ( t )+ y p ( t )= c 1 cos 2 t + c 2 sin 2 t 1 4 cos 2 t ln |
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: | sec 2 t + tan 2 t | . 2. From Example 1 on page 196 in the text, we know that unctions y 1 ( t ) = cos t and y 2 ( t ) = sin t are two linearly independent solutions to the corresponding homogeneous equation, and so its general solution is given by y h ( t ) = c 1 cos t + c 2 sin t. Now we apply the method o variation o parameters to fnd a particular solution to the original equation. By the ormula (3) on page 194 in the text, y p ( t ) has the orm y p ( t ) = v 1 ( t ) y 1 ( t ) + v 2 ( t ) y 2 ( t ) . 212...
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