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Unformatted text preview: GE 207K Not Exact Equations – Problem Set September 15, 2010 Solve the following firstorder differential equation. 1. y dx + ( x 2 y x ) dy = 0 , μ = μ ( x ) (1) Writing the equation in standard form we find y + ( x 2 y x ) dy dx = 0 M = y , N = x 2 y x M y = 1 , N x = 2 xy 1 Comparing M y and N x , we note that M y 6 = N x ⇒ Not exact. We are told that there exists an integrating factor which is a function of x . Therefore, M y N x N = 1 2 xy + 1 x 2 y x = 2(1 xy ) x (1 xy ) = 2 x ⇒ μ ( x ) = e R My Nx N dx = e R 2 x dx = x 2 . Next, we multiply through by the integrating factor μ = x 2 to obtain. yx 2 + y 1 x dy dx = 0 . Make sure from now on you work with this equation! ! We can now either integrate M ( x,y ) with respect to x OR integrate N ( x,y ) with respect to y . Integrating M ( x,y ) with respect to x , we find, ψ ( x,y ) = Z M ( x,y ) dx = Z yx 2 dx = y x + h ( y ) Note that we didn’t forget to include the arbitrary function (not a constant)! ! Recall that ψ x = M and ψ y = N . Therefore, differentiating previous equation with respect to y and equating it to N ( x,y ), we find 1 x + dh dy  {z } ψ x = 1 x + y  {z } N ⇒ h = y 2 2 1 GE 207K Not Exact Equations – Problem Set September 15, 2010 And now we’ve found ψ to its entirety! ψ ( x,y ) = y x + y 2 2 Recall that the solution to an exact differential equation is ψ ( x,y ) = C . Therefore, ∴ y x + y 2 2 = C . (2) 2 GE 207K Not Exact Equations – Problem Set September 15, 2010 2. Show that the following differential equation is not exact. Then find an integrating factorShow that the following differential equation is not exact....
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This note was uploaded on 10/08/2010 for the course GE 207K taught by Professor None during the Fall '10 term at University of Texas at Austin.
 Fall '10
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