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Unformatted text preview: since ! ! y 4xy + 3y 2 " x ( ) = 4x + 6y # ! ! x x 2 + 2xy ( ) = 2x + 2y , it is non-exact 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 one-parameter family of solutions of the essential equivalent exact equation is also a solution of the non-exact equation. However, the multiplication by the integrating factor may result in either: a) The lost of (one or more) solutions of the non-exact equation. b) The gain of (one or more) solutions of the non-exact 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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This note was uploaded on 07/23/2011 for the course MAP 2302 taught by Professor Staff during the Fall '08 term at FIU.
- Fall '08