Integrating factors of exact equations

# Integrating factors of exact equations - example So this...

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So, exact equations will be of the form: M(x,y)dx+N(x,y)dy=0 To test if the equation is exact, check if If the equation is not exact, you need to multiply by integrating factor μ(x,y) so, now you need to expand the equation Exanding this gives: Separate the terms: Now, you need to make an assumption that is a function of only x or only y To do this, first evaluate the term This term will simplify to a form that is very similar to either the original M term or the original N term If it simplifies to the M term, assume is a function of y only. Then, = 0, and the bold equation simplifies to: And you can solve for μ. If the is similar to the orignal N term, you will get: Then, your original equation will become:

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μ (x,y)*M(x,y)dx+ μ (x,y)*N(x,y)dy=0 And you solve it as you would for a simple exact equation.
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Unformatted text preview: example: So, this equation is not exact, so use IF => => so, the term appears to be similar to the original M term. So, assume that the IF is a function of y only, and becomes 0 Now, which simplifies to which comes out to: Now, the differential equation becomes: (check on your own that this is now exact) Now, so integrate M with respect to x (I found out after class from a very helpful greek student in the class that the symbol I've been drawing on the board and calling phi is most likely to actually be a psi Ψ) Integrating M gives: Now, N= , so take the derivative of the Ψ we just found to solve for the f(y) term: so, f'(y)= f(y)= So, the final answer is: and since Ψ=C, Question I will take up the beginning of the next tutorial (please try it on your own): Y(x+y+1)dx + x(x+3y+2)dy = 0...
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