351_pdfsam_math 54 differential equation solutions odd

351_pdfsam_math 54 differential equation solutions odd -...

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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 diFerentiation, 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 ] = h y ( n ) 1 + y ( n ) 2 i + p 1 h y ( n 1) 1 + y ( n 1) 2 i + + p n [ y 1 + y 2 ] = h y ( n ) 1 + p 1 y ( n 1) 1 + + p n y 1 i + h y ( n ) 2 + p 1 y ( n 1) 2 + + p n y 2 i = 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 ² = 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 2 ,...,x n } is linearly independent on any interval (
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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.

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