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Math Help - solve

  1. #16
    A Plied Mathematician
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    Hmm. Well, that tells me that you have a rather difficult course load right now, and that it seems like you spend what you think is a disproportionate amount of time on math. It tells me that you are trying hard to learn.

    But it does not tell me that you want to learn. How badly do you want to learn math? How willing are you to do whatever it takes?
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  2. #17
    Junior Member ceode's Avatar
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    Hmmm.. Iam doing CS engg... Its my first year... I never study it. There is no need to. I learn everything from class itself. Iam proud to say iam the topper in my class in computer.
    And yes i will do no matter what it takes to master maths.
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  3. #18
    A Plied Mathematician
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    Very well. Then I will do my part to help you master what you need. However, that also means that I will take you at your word. You will need to do the major part of the work, or you will never master math. Math is not a spectator sport - you have to do it.

    What you need to do right now is to substitute u into the original differential equation such that the only variables are u and x. Post # 6 tells you how to do that. You have all the expressions that you need. So what do you get?
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  4. #19
    Junior Member ceode's Avatar
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    by putting u=logy

    dy/dx + (yu)/x = (y.u^2)/x^2
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  5. #20
    A Plied Mathematician
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    I would agree with what you have so far except that you should have

    dy/dx + (yu)/x = (y u^2)/x^2.

    I still see some y's in there. How can you get rid of those?
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  6. #21
    Junior Member ceode's Avatar
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    sorry about that x

    dy/dx + (e^u u)/x = (e^u . u^2)/x^2

    Now is it okay
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  7. #22
    A Plied Mathematician
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    Better. But you still have the y' (or, dy/dx) on the far left. How does that substitute?
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  8. #23
    Junior Member ceode's Avatar
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    du/dx = 1/y . dy/dx

    dy/dx = y . du/dx
    =e^u . du/dx
    then

    e^u . du/dx + (e^u u)/x = (e^u . u^2)/x^2
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  9. #24
    A Plied Mathematician
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    Excellent. Now, stare at this equation for a while. Do you see any way forward?
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  10. #25
    Junior Member ceode's Avatar
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    should i eliminate e^u
    equation becomes

    du/dx + (u)/x = (u^2)/x^2
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  11. #26
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    I would agree that that is the next step. But you have to justify it. You are dividing by something. Any time you divide by something, you must be sure that what you are dividing by is not zero. So, can e^u ever be zero?
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  12. #27
    Junior Member ceode's Avatar
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    I multiplied the whole eqn by 1/e^u , e^u cannot be 0
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  13. #28
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    Right. The exponential function is never zero. Ok. So now what? Take a look at

    du/dx + (u)/x = (u^2)/x^2.

    Do you notice anything?
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  14. #29
    Junior Member ceode's Avatar
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    Now should i multiply the whole eqn by x^2 ,
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  15. #30
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    No, I don't think I would do that. Let me rewrite slightly.

    \displaystyle \frac{du}{dx} + \frac{1}{x}\, u = \frac{1}{x^{2}}\,u^{2}.

    Do you recognize anything?
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