Unformatted text preview: 2 x dx = e 1 2 log x = x 1 2 : x 1 2 dy dx + 1 2 x1 2 y = 1 2 x3 2 d dx ( yx 1 2 ) = 1 2 x3 2 yx 1 2 =x1 2 + C y =1 x + Cx1 2 Using the initial condition y (1) = 4, we get that 4 =1 + C (1). Hence C = 5 and y ( x ) =1 x + 5 x1 2 . 1...
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 Spring '07
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 Math, Derivative

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