ExactEquation

ExactEquation - N and solve for g '( y ): 3 3 2 2 x g y x y...

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EXACT EQUATION Solving an Exact first order differential equation from section 1.6 p62. See the document FirstOrderDiffEq.pdf for a comparison between this and other methods. THE PROBLEM 31. p62 ( ) ( ) 2 3 3 2 0 x y dx x y dy + + + = Find the general solution. The equation is already written in the form M dx N dy + = 0 where M x y N x y = + = + 2 3 3 2 In order to be an Exact equation, the partial derivative of M with respect to y must equal the partial derivative of N with respect to x : M y N x = To verify that we have an Exact equation, we find: d dy d dx x y x y ( ) ( ) 2 3 3 3 2 3 + = + = THE SOLUTION To solve, we first integrate M with respect to x and use g ( y ) for the constant of integration: M dx x y dx x xy g y = + = + + ( ) ( ) 2 3 3 2 Now we take this result and differentiate with respect to y : d dy x xy g y x g y [ ( )] ( ) 2 3 3 + + = + We set this equal to
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Unformatted text preview: N and solve for g '( y ): 3 3 2 2 x g y x y g y y + = + = ( ) ( ) We integrate to find g ( y ): g y g y y dy y C ( ) ( ) = = = + 2 2 1 Substitute this into the first expression containing g ( y ) to obtain F ( x , y ): F x y x xy g y x xy y C ( , ) ( ) = + + = + + + 2 2 2 1 3 3 If an initial condition is given, the value of C 1 can be found, yielding a particular solution. In this case, an initial condition was not given. THE ANSWER The general solution is then written in this form, absorbing the value of C 1 into C . x xy y C 2 2 3 + + = Tom Penick tomzap@eden.com www.teicontrols.com/notes October 24, 1997...
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This note was uploaded on 03/29/2008 for the course MATH 308 taught by Professor Comech during the Spring '08 term at Texas A&M.

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