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Unformatted text preview: y = x , so solutions exist and are unique everywhere except possibly along y = x . Alternatively, f ( x, y ) = ( xy ) 5 , ∂f ∂y =1 5 , so solutions exist and are unique everywhere. 3. f ( x, y ) = x 2 y1 , ∂f ∂y =x 2 y2 , which are both discontinuous when y = 0, so solutions exist and are unique everywhere except possibly along y = 0. 4. f ( x, y ) = ( x + y )2 , ∂f ∂y =2 ( x + y ) 3 , which are both discontinuous when y =x , so solutions exist and are unique everywhere except possibly along y =x ....
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 Winter '07
 brown
 dx, Nicke Andersson

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