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**Unformatted text preview: **2 x 3 + C 3 / 2 . (b) Using the initial condition, y (0) = 0, we Fnd that 0 = y (0) = 2(0) 3 + C 3 / 2 = C 3 / 2 C = 0 , and so y = (2 x/ 3 + 0) 3 / 2 = (2 x/ 3) 3 / 2 , x 0, is a solution to the initial value problem. (c) The function y ( x ) 0, clearly, satisFes both, the dierential equation dy/dx = y 1 / 3 and the initial condition y (0) = 0. (d) In notation of Theorem 1 on page 12, f ( x, y ) = y 1 / 3 and so f y = d dy ( y 1 / 3 ) = 1 3 y 2 / 3 = 1 3 y 2 / 3 . 33...

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