232_pdfsam_math 54 differential equation solutions odd

# 232_pdfsam_math 54 differential equation solutions odd - [...

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Chapter 4 or, equivalently, y = 1 t c , where c is an arbitrary constant. (b) A linear combination of y 1 ( t ):=1 / ( t c 1 )and y 2 ( t ):=1 / ( t c 2 ), C 1 y 1 ( t )+ C 2 y 2 ( t )= C 1 t c 1 + C 2 t c 2 = ( C 1 + C 2 ) t ( C 1 c 2 + C 2 c 1 ) ( t c 1 )( t c 2 ) , is identically zero in a neighborhood of t =0ifandon lyif( C 1 + C 2 ) t ( C 1 c 2 + C 2 c 1 ) 0. Thus the numerator must be the zero polynomial, i.e., C 1 and C 2 must satisfy C 1 + C 2 =0 , C 1 c 2 + C 2 c 1 =0 C 2 = C 1 , C 1 ( c 2 c 1 )=0 . Since c 1 6 = c 2 , the second equation implies that C 1 =0 ,andthen C 2 = 0 from the Frst equation. Thus, only the trivial linear combination of y 1 ( t )and y 2 ( t ) vanishes identically around the origin, and so these functions are linearly independent. (c) ±or any function of the form y c ( t ):=1 / ( t c ), the equality y 0 c ( t )= 1 ( t c ) 2
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Unformatted text preview: [ y c ( t )] 2 holds for all t 6 = c . In particular, at t = 0, y c (0) = [ y c (0)] 2 . (We assume that c 6 = 0; otherwise, t = 0 is not in the domain.) Obviously, this equality fails for any positive initial velocity y (0), in particular, it is false for given data, y (0) = 1 and y (0) = 2. 6. Rewriting given equation in the equivalent form y = ( k/m ) y , we see that the function f ( y ) in the energy integral lemma is ( k/m ) y . So, F ( y ) = Z k m y dy = k 2 m y 2 + C. 228...
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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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