352_pdfsam_math 54 differential equation solutions odd

# 352_pdfsam_math 54 differential equation solutions odd -...

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

Chapter 6 On ( 1 , 1) (even on ( −∞ , 1) ) we have f 1 ( x ) ≡− f 2 ( x ) or, equivalently, f 1 ( x )+ f 2 ( x ) 0 and so these functions are linearly dependent on ( 1 , 1). However, their linear combina- tion c 1 f 1 ( x )+ c 2 f 2 ( x )= ( ( c 2 c 1 )( x 1) ,x 1; ( c 1 + c 2 )( x 1) ,x> 1 cannot vanish identically on ( −∞ , ) unless c 1 c 2 =0and c 1 + c 2 = 0, which implies c 1 = c 2 =0 . 31. (a) Linearity of diFerentiation and the product rule yield y 0 ( x )=( v ( x ) e x ) 0 = v 0 ( x ) e x + v ( x )( e x ) 0 =[ v 0 ( x )+ v ( x )] e x , y 0 ( x )=[ v 0 ( x )+ v ( x )] 0 e x +[ v 0 ( x )+ v ( x )] ( e x ) 0 =[ v 0 ( x )+2 v 0 ( x )+ v ( x )] e x , y 0 ( x )=[ v 0 ( x )+2 v 0 ( x )+ v ( x )] 0 e x +[ v 0 ( x )+2 v 0 ( x )+ v ( x )] ( e x ) 0 =[ v 0 ( x )+3 v 0 ( x )+3 v 0 ( x )+ v ( x )] e x . (b) Substituting y , y 0 , y 0 ,and y 0 into the diFerential equation (32), we obtain [ v 0 +3 v 0 +3 v 0 + v ] e x 2[ v 0 +2 v 0 + v ] e x 5[ v 0 + v ] e x +6 ve x =0 [( v 0 +3 v 0 +3 v 0 + v ) 2( v 0 +2 v 0 + v ) 5( v 0 + v )+6 v ] e x =0 v 0 + v 0 6 v 0 =0 , wh e r ew ehav eu s edth efa c ttha tth efun c t ion e x is never zero. Let v 0 =:
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 Berkeley.

Ask a homework question - tutors are online