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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 ) = ( x-y ) 5 , ∂f ∂y =-1 5 , so solutions exist and are unique everywhere. 3. f ( x, y ) = x 2 y-1 , ∂f ∂y =-x 2 y-2 , 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