notexact_PS

# Notexact_PS - GE 207K Not Exact Equations – Problem Set Solve the following first-order differential equation 1 y dx x 2 y x dy = 0 μ = μ x(1

This preview shows pages 1–4. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: GE 207K Not Exact Equations – Problem Set September 15, 2010 Solve the following first-order 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....
View Full Document

## 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.

### Page1 / 8

Notexact_PS - GE 207K Not Exact Equations – Problem Set Solve the following first-order differential equation 1 y dx x 2 y x dy = 0 μ = μ x(1

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online