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Proof of Integration Factor for diff eq

  1. Jul 1, 2008 #1

    Ald

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    1. The problem statement, all variables and given/known data

    Looking for help on this Exact first order differential equation problem. Thanks

    Show that if ((partial M/partial t)-(partial N/partial y))/M=Q(y)

    then the differential equation M(t,y)+N(t,y)dy/dt=0 has an integrating factor

    μ(y)=exp(integral Q(y)dy).



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 1, 2008 #2

    Dick

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    If mu(y) is an integrating factor, then mu(y)*M(t,y)*dt+mu(y)*N(t,y)*dy=0 is exact. Is it? Try the usual test for exactness. BTW, be clear where the y dependence is in mu(y). It's not in the integration variable, that's just a dummy. y is one of the limits of integration.
     
  4. Jul 1, 2008 #3

    Ald

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    Thanks for your help, yes your recommendation makes sense.

    I assume these are the partial derivatives I take, there is no actual expression to take the partial of, how do I prove exactness?

    partial(mu(y)*M(t,y))/partial dy=partial (mu(y)*N(t,y))/partial dt
     
  5. Jul 1, 2008 #4

    Dick

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    Use your formula for mu(y). The partial derivative of mu(y) wrt to y can be written in a different way. The partial derivatives of M and N, you just write as partial derivatives. You are trying to show that mu is an integrating factor leads to the expression in the 'if' part of your statement.
     
  6. Jul 1, 2008 #5

    Ald

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    I'm sorry what does 'wrt' mean?
     
  7. Jul 1, 2008 #6

    Dick

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    wrt='with respect to'. What's the derivative of mu(y) with respect to y?
     
  8. Jul 1, 2008 #7

    Ald

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    The only other way I could find to write the mu(y) was in the form mu(t,y)=1/(xM-yN) is this what you had in mind?
    Thanks
     
  9. Jul 1, 2008 #8

    Ald

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    Thanks Dick for getting me in right direction, I think I have it proved.
    Thanks again
     
  10. Jul 1, 2008 #9

    Dick

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    Did it really work? You differentiated the exponential and used the chain rule, right? I'm kind of curious. I had to change [tex]e^{\int_0^y Q(t)dt}[/tex] by putting a negative sign inside the exp. You could also alter it by setting y as the lower limit, of course. Did you hit that snag?
     
  11. Jul 1, 2008 #10

    Ald

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    Attached is what I did, I haven't rewritten yet. Sometimes ignorance is bliss, I hope I did it right.
     

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  12. Jul 1, 2008 #11

    Dick

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    If ignorance is bliss, I'm experiencing it now. An attachment can take hours to clear approval.
     
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