359_pdfsam_math 54 differential equation solutions odd

359_pdfsam_math 54 differential equation solutions odd - c...

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

View Full Document Right Arrow Icon
Exercises 6.2 Substituting these constants into the general solution, we get the answer y ( x )= e 2 x 2 e x sin 2 x. 23. Rewriting the system in operator form yields ( D 3 1) [ x ]+( D +1)[ y ]=0 , ( D 1)[ x ]+ y =0 . Multiplying the second equation in this sytem by ( D + 1) and subtracting the result from the Frst equation, we get ±( D 3 1 ) ( D +1)( D 1) ² [ x ]= D 2 ( D 1)[ x ]=0 . Since the roots of the characteristic equation, r 2 ( r 1) = 0 are r = 0 of multiplicity two and r = 1, a general solution x ( t )isg ivenby x ( t )= c 1 + c 2 t + c 3 e t . ±rom the second equation in the original system, we obtain y ( t )= x ( t ) x 0 ( t )= ( c 1 + c 2 t + c 3 e t ) ( c 1 + c 2 t + c 3 e t ) 0 =( c 1 c 2 )+ c 2 t. 25. A linear combination of the given functions c 0 e rx + c 1 xe rx + c 2 x 2 e rx + ··· + c m 1 x m 1 e rx = ( c 0 + c 1 x + c 2 x 2 + ··· + c m x m ) e rx (6.3)
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: c + c 1 x + c 2 x 2 + ··· + c m − 1 x m − 1 , vanishes on this interval (the exponential factor, e rx , is never zero). But, as we have proved in Problem 27, Section 6.1, the system of monomials { 1 , x, . . . , x n } is linearly independent on any interval. Thus, the linear combination (6.3) vanishes on an inteval if and only if it has all zero coefficients, i.e., c = c 1 = . . . = c m − 1 = 0. Therefore, the system { e rx , xe rx , . . . , x m − 1 e rx } is linearly independent on any interval, in particular, on ( −∞ , ∞ ). 355...
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