Unformatted text preview: since ! ! y 4xy + 3y 2 " x ( ) = 4x + 6y # ! ! x x 2 + 2xy ( ) = 2x + 2y , it is nonexact If we multiply the equation by the expression p(x,y) = x 2 , we get x 2 (4xy + 3y 2 – x) dx + x 3 (x + 2y) dy = 0 then ! ! y 4x 3 y + 3x 2 y 2 " x 3 ( ) = 4x 3 + 6x 2 y = ! ! x x 4 + 2x 3 y ( ) this new equation is exact. So, it is the essential equivalent exact equation and p(x,y) = x 2 is an integrating factor. Remark : The oneparameter family of solutions of the essential equivalent exact equation is also a solution of the nonexact equation. However, the multiplication by the integrating factor may result in either: a) The lost of (one or more) solutions of the nonexact equation. b) The gain of (one or more) solutions of the nonexact equation. c) both cases. Later on we will see examples of these case and moreover that it is not always possible to find an integrating factor....
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 Fall '08
 STAFF
 Algebra, Multiplication

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