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Math Help - using the homogeneous equation...

  1. #1
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    using the homogeneous equation...

    ok so here is the problem...

    (x^2+ xy)dx - (2xy + y^2)dy = 0 let y = ux

    ok the problem seems easy enough... if y = ux then dy = udx + xdu

    so now im ready to substitute...

    when i solved it i got something like

    ln|x| = (2u^2 + u)/(-u^3-2u^2+u+1)

    so where did i go wrong...

    (x^2 +ux^2)dx - [(2ux^2 + (ux)^2)](udx + xdu) from here i used algebra to simplify?

    thanks in advance...
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  2. #2
    A Plied Mathematician
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    You started out just fine. With both substitutions y=ux and x=vy, you get a horrendous integration that looks like

    \displaystyle-\int\frac{u(2+u)}{1+u-2u^{2}-u^{3}}\,du or

    \displaystyle\int\frac{v(v+1)}{v^{3}+v^{2}-2v-1}\,dv.

    Both integrals are pretty nasty. Not really sure what to do from here. Ideas, anyone else?
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by slapmaxwell1 View Post
    ln|x| = (2u^2 + u)/(-u^3-2u^2+u+1)

    Check the numerator. It should be:

    \ln \left |{x}\right |=\displaystyle\int \dfrac{u^2+2u}{1+u-2u^2-u^3}\;du


    Fernando Revilla


    Edited: Sorry, I didn't see Ackbeet's post
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  4. #4
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    i tried i couldnt finish the problem on my exam, so thats what i left..i even ran it through maxima, and maxima wouldnt integrate it...its the only problem that gave me fits...thanks for checking my work.
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  5. #5
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    Mathematica Version 4 and WolframAlpha won't integrate it, either. You're not alone! You're probably good if you reduce the DE to integrals - but that's about as far as you can go with this one, I think, unless there's a clever substitution or an integrating factor to get the equation exact. It's not linear or Bernoulli in x or y.
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  6. #6
    A Plied Mathematician
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    Maybe General can help us out. He's elected not to allow visitor messages or PM's. We'll just have to hope he sees this thread. That fairly likely, though.
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  7. #7
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    i will redo the problem and try to make it exact...that just might work. thanks for the tip..
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