Unformatted text preview: as p ( x ) q ( y ). We also see that the equation is linear with y as the dependent variable. 3. Here M ( x, y ) = ye xy + 2 x , N ( x, y ) = xe xy − 2 y . Thus ∂M ∂y = ∂ ∂y ( ye xy + 2 x ) = e xy + y ∂ ∂y ( e xy ) = e xy + ye xy x = e xy (1 + yx ) , ∂N ∂x = ∂ ∂x ( xe xy − 2 y ) = e xy + x ∂ ∂x ( e xy ) = e xy + xe xy y = e xy (1 + xy ) , ∂M/∂y = ∂N/∂x , and the equation is exact. We write the equation in the form dy dx = − ye xy + 2 x xe xy − 2 y 59...
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations, Derivative, ∂y ∂y ∂N, xy dx xe

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