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2.7 Homogeneous Diff Eq. 2

# 2.7 Homogeneous Diff Eq. 2 - 2.7 So U1 Un(u = U1 Ui-1 Ui 1...

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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 ) = 0 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 n -dimensional, 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 PNU-MATH

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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 But R k ( t ) e ( c k - c 1 ) t = 0 implies R k = 0 It’s a contradiction to P k
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2.7 Homogeneous Diff Eq. 2 - 2.7 So U1 Un(u = U1 Ui-1 Ui 1...

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