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Ae x Be 5x with derivative y Ae x 5Be 5x Evaluating at x 0, we have 4 A B 1 A 5B Solving this pair of equations, we get A 19 4 and B 3 4, so our solution is (1213) y 19 4 e x 3 4 e 5x Example 124 A function x x t satisfies the differential equation (1214) x 2x 15x 0. ݼ t w School Damascus College Ballarat;. AE = (1/2)Y 325 Y = 600, so AE = 300 325 = 625 We also know that AE = Y inventory and when Y = 600, we have 625 = 600 inventory inventory = $25 Solving through, we find that $25 worth of inventory was sold in 16 to make up for the difference between AE and output 2.

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