It follows that y x 1 5 x ln x Cx The initial condition y 1 7 gives 7 1 C or C

# It follows that y x 1 5 x ln x cx the initial

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It follows that y ( x ) = 1 5 x ln( x ) + Cx . The initial condition, y (1) = - 7, gives - 7 = 1 C or C = - 1 7 . The solution is y ( x ) = 1 5 x ln( x ) - x/ 7 = 7 35 x ln( x ) - x . 9. Consider the Bernoulli Eqn: dy dx - 4 y x = y 4 x 2 . Use the substitution, u = y 1 - 4 = y - 3 , then du dx = - 3 y - 4 dy dx . Multiple the equation above by - 3 y - 4 , which gives - 3 y - 4 dy dx + 12 xy 3 = - 3 x 2 or du dx + 12 u x = - 3 x 2 .

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This linear equation in u ( x ) has the integrating factor μ ( x ) = exp 12 R dx x = x 12 . We have d dx x 12 u = - 3 x 10 , so x 12 u ( x ) = - 3 11 x 11 + C = x 12 y 3 ( x ) . It follows that y 3 ( x ) = x 12 C - 3 11 x 11 . The initial condition, y (1) = 1, gives 1 = 1 C - 3 11 or C = 14 11 . The solution is y 3 ( x ) = x 12 14 11 - 3 11 x 11 = 11 x 12 14 - 3 x 11 or y ( x ) = x 4 11 14 - 3 x 11 1 / 3 .
• Fall '08
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