**Unformatted text preview: **2 x dx = e 1 2 log x = x 1 2 : x 1 2 dy dx + 1 2 x-1 2 y = 1 2 x-3 2 d dx ( yx 1 2 ) = 1 2 x-3 2 yx 1 2 =-x-1 2 + C y =-1 x + Cx-1 2 Using the initial condition y (1) = 4, we get that 4 =-1 + C (1). Hence C = 5 and y ( x ) =-1 x + 5 x-1 2 . 1...

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- Spring '07
- staff
- Math, Derivative