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Unformatted text preview: , to get 2 x dx dy + 8 y = 0 . Upon substituting x = 2 2, y = 0, this yields 4 2 dx dy + 0 = 0 which does have a solution, dx dy = 0 . In this case, we can see the reason for our diculty by explicitly solving the original equation ( 3.12 ) for y in terms of x : near (2 , 1), y can be...
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- Fall '08