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Unformatted text preview: 3 e 5 x 3 x 2 + C . Since y (0) = 1 , we have 3 e 5(0) 3 2 + C = 1 3 + C = 1 C = 4 Hence y (1) = 3 e 5(1) 3 1 2 + 4 = 3 e 5 + 1 . 20. We see that 3 log 6 4 6 3 6 log 3 3 3 4 = 3 log 6 6 3 1 / 4 6 log 3 3 4 1 / 3 = 3 log 6 6 3 / 4 6 log 3 3 4 / 3 = 3 3 4 log 6 6 6 4 3 log 3 3 = 9 4 log 6 6 8 log 3 3 = 9 4 8 = 23 4 ....
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 Fall '09
 RAdin

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