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Math Help - Tricky First Order D.E

  1. #1
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    Tricky First Order D.E

    I'm asked to solve the following D.E using an integrating factor:



    The problem is the equation is not only dependant on y but also ln(y), so I can't see how the usual method will work. How would I go about solving this?
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  2. #2
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    Let u = \ln y so your ODE becomes \left(x^2+u\right)dx + xdu = 0. See how that goes.
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    \dfrac{1}{M}\left(\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}\right)=\ldots=-\dfrac{1}{y}

    This means that the equation has an integrating factor that depends on y


    Edited: Sorry, I didn't see Danny's post.
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  4. #4
    A Plied Mathematician
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    When I see a problem like this, and I want to find out what integrating factor there might be, I, with the help of Danny there, have worked out a decently general method. It won't solve every first-order, but it will do many. It works like this: assume an integrating factor of the form

    h(x^{m}y^{n}),

    where h is an unknown function. The idea is that you hope to be able to obtain a differential equation that governs the behavior of h. You multiply the DE by this integrating factor thus:

    h(x^{m}y^{n})\,y(x^{2}+\ln(y))\,dx+h(x^{m}y^{n})\,  x\,dy=0.

    You then utilize the conditions for exactness:

    \frac{\partial }{\partial y} \left[h(x^{m}y^{n})\,y(x^{2}+\ln(y))\right]=\frac{\partial }{\partial x} \left[h(x^{m}y^{n})\,x\right].

    So you take your derivatives and boil it all down. The hope is to get a separable DE for h. I get the following:

    \frac{h'(x^{m}y^{n})}{h(x^{m}y^{n})}=-\frac{x^{2}+\ln(y)}{x^{m}y^{n}(nx^{2}+n\,\ln(y)-m)}.

    You have some freedoms here with m and n. I can see that if I choose m = 0, the equation simplifies drastically. You should be able to finish from here.

    So here's a decently general method for solving quite a few integrating factor-type nonlinear problems.
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  5. #5
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    Thanks a lot, I've got this solved now.
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