**Unformatted text preview: **y = ( x 2 + 1) 2 d dx [ y ( x 2 + 1) 2 ] = x 4 + 2 x 2 + 1 y ( x 2 + 1) 2 = x 5 5 + 2 x 3 3 + x + C y = 1 ( x 2 + 1) 2 ( x 5 5 + 2 x 3 3 + x + C ) Using the initial condition y (0) =-2, we have-2 = C , so y ( x ) = 1 ( x 2 + 1) 2 ( x 5 5 + 2 x 3 3 + x-2) . 1...

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- Spring '07
- staff
- Math, UCLA ID, -2. dx