226_pdfsam_math 54 differential equation solutions odd

# 226_pdfsam_math 54 differential equation solutions odd -...

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

Chapter 4 Adding a general solution to the homogeneous equation, we get y ( t )= c 1 cos 2 t + c 2 sin 2 t 1 2 cos 2 t ln | sec 2 t +tan2 t |− 1 5 e t . 19. A general solution of the corresponding homogeneous equation is given by y h ( t )= c 1 e t + c 2 e t . We will try to Fnd a particular solution to the original nonhomogeneous equation of the form y p ( t )= v 1 ( t ) y 1 ( t )+ v 2 ( t ) y 2 ( t ), where y 1 ( t )= e t and y 2 ( t )= e t . We apply formulas (10) on page 195 in the text, but replace indeFnite integrals by deFnite integrals. Note that y 1 ( t ) y 0 2 ( t ) y 0 1 ( t ) y 2 ( t )= e x e x ( e x ) e x =2 . With g ( t )=1 /t and integration from 1 to t , formulas (10) yield v 1 ( t )= t Z 1 g ( x ) y 2 ( x ) 2 dx = 1 2 t Z 1 e x x dx , v 2 ( t )= t Z 1 g ( x ) y 1 ( x ) 2 dx = 1 2 t Z 1 e x x dx . (Notice that we have chosen the lower limit of integration to be equal to 1 because the initial conditions are given at 1. We could have chosen any other value for the lower limit, but the
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: choice of 1 will make the determination of the constants c 1 and c 2 easier.) Thus y p ( t ) = e t 2 t Z 1 e − x x dx − e − t 2 t Z 1 e x x dx , and so a general solution to the original di±erential equation is y ( t ) = c 1 e − t + c 2 e t + e t 2 t Z 1 e − x x dx − e − t 2 t Z 1 e x x dx . By plugging in the Frst initial condition (and using the fact that the integral of a function from a to a is zero which is why we have chosen the lower limit of integration to be the initial point, t = 1), we Fnd that y (1) = c 1 e − 1 + c 2 e 1 = 0 . 222...
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 Berkeley.

Ask a homework question - tutors are online