{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

351_pdfsam_math 54 differential equation solutions odd

# 351_pdfsam_math 54 differential equation solutions odd -...

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

Exercises 6.1 25. Clearly, it is suﬃcient to prove (9) just for two functions, y 1 and y 2 . Using the linear property of differentiation, we have L [ y 1 + y 2 ] = [ y 1 + y 2 ] ( n ) + p 1 [ y 1 + y 2 ] ( n 1) + · · · + p n [ y 1 + y 2 ] = y ( n ) 1 + y ( n ) 2 + p 1 y ( n 1) 1 + y ( n 1) 2 + · · · + p n [ y 1 + y 2 ] = y ( n ) 1 + p 1 y ( n 1) 1 + · · · + p n y 1 + y ( n ) 2 + p 1 y ( n 1) 2 + · · · + p n y 2 = L [ y 1 ] + L [ y 1 ] . Next, we verify (10). L [ cy ] = [ cy ] ( n ) + p 1 [ cy ] ( n 1) + · · · + p n [ cy ] = cy ( n ) + p 1 cy ( n 1) + · · · + p n cy = c y ( n ) + p 1 y ( n 1) + · · · + p n y = cL [ y ] . 27. A linear combination c 0 + c 1 x + c 2 x 2 + · · · + c n x n of the functions from the given set is a polynomial of degree at most n and so, by the funda- mental theorem of algebra, it cannot have more than n zeros unless it is the zero polynomial, i.e., it has all zero coeﬃcients. Thus, if this linear combination vanishes on a whole interval ( a, b ), then it follows that c 0 = c 1 = c 2 = . . . = c n = 0. Therefore, the set of functions { 1 , x, x
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}