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Unformatted text preview:  x  ) = x − 2 . To solve the equation, we multiply it by the integrating factor x − 2 to obtain (3 + yx − 2 ) dx + ( y − x − 1 ) dy = 0 . This is now exact. Thus, we want to Fnd F ( x, y ). To do this, we integrate M ( x, y ) = 3+ yx − 2 with respect to x to get F ( x, y ) = Z ( 3 + yx − 2 ) dx = 3 x − yx − 1 + g ( y ) ⇒ F y ( x, y ) = − x − 1 + g ( y ) = N ( x, y ) = y − x − 1 ⇒ g ( y ) = y ⇒ g ( y ) = y 2 2 . Therefore, F ( x, y ) = 3 x − yx − 1 + y 2 2 . And so we see that an implicit solution is y 2 2 − y x + 3 x = C. 74...
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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