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232_pdfsam_math 54 differential equation solutions odd

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

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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 = 0 if and only if ( 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 = c 2 , the second equation implies that C 1 = 0, and then C 2 = 0 from the first 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) For any function of the form y c ( t ) := 1 / ( t c ), the equality y c ( t ) = 1
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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 ² dy = − k 2 m y 2 + C. 228...
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