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Unformatted text preview: 2.7. So U 1 U n ( u ) = U 1 U i 1 U i +1 U n U i ( u ) = U 1 U i 1 U i +1 U n (0) = 0 u N ( U 1 U n ) 10. Suppose that b 1 e c 1 t + + b n e c n t = 0 ,c i s are distinct Apply the mathematical induction on n , If n = 1, then b 1 e c 1 t = 0 b 1 = 0 Assume that this assertion is true for n 1 We are going to prove that this is also true for n ( D c n I )( b 1 e c 1 t + + b n e c n t ) = 0 b 1 c 1 e c 1 t + + b n 1 c n 1 e c n 1 t + b n c n e c n t ( b 1 c n e c 1 t + + b n 1 c n e c n 1 t + b n c n e c n t ) = b 1 ( c 1 c n ) e c 1 t + b 2 ( c 2 c n ) e c 2 t + + b n 1 ( c n 1 c n ) e c n 1 t = 0 By the induction hypothesis and c i c n 6 = 0 , i = 1 , ,n 1 ,b i = 0 { e c 1 t , ,e c n t } is linearly independent Since the solution space is ndimensional, the given set is a basis for the solution space of the differential equation 11. Suppose that k i =1 n i 1 j =0 c ij t j e c i t = 0 146 PNUMATH 2.7. Let P i ( t ) = n i 1 j =0 c ij t j Then we have P 1 ( t ) e c 1 t + P 2 ( t ) e c 2 t + + P k ( t ) e c k t = 0 Assume that not all c ij are zero, then P i 6 = 0 Say, P k Divide the equation by e c 1 t P 1 ( t ) + P 2 ( t ) e ( c 2 c 1 ) t + + P k ( t ) e ( c k c 1 ) t = 0 (1) Upon differentiating (1) sufficiently many times we can reduces P 1 ( t ) to 0 Q 2 ( t ) e ( c 2 c 1 ) t + + Q k ( t ) e ( c k c 1 ) t = 0, and deg Q i =deg P i and Q k does not vanish identically Continuing this process, R k ( t ) e ( c k c 1 ) t = 0, and deg R k =deg P k and R k does not vanish identically...
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This note was uploaded on 02/14/2012 for the course MAS 4105 taught by Professor Rudyak during the Spring '09 term at University of Florida.
 Spring '09
 RUDYAK

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