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Math Help - solve

  1. #31
    Junior Member ceode's Avatar
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    yes.. bernoullis eqn...
    dy/dx +Py = Q y^x , where P and Q are functions of x alone
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  2. #32
    A Plied Mathematician
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    Well, technically,

    dy/dx +Py = Q y^n,

    but yes, it's a Bernoulli equation. Can you take it from here?
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  3. #33
    Junior Member ceode's Avatar
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    I think so....
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  4. #34
    A Plied Mathematician
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    Ok. Let me know if you get stuck.
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  5. #35
    Junior Member ceode's Avatar
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    sorry.. I took a break...

    I got I.F as 1/x

    Now answer would be

    1/ux = 1/(2.x^2) + c

    where u = logy

    Is that correct
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  6. #36
    A Plied Mathematician
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    You shouldn't regard delays in replying on this forum as unusual. I have just been doing other things for a while.

    I'm not sure I agree with your solution of the Bernoulli equation. Can you show your steps, please?
    Last edited by mr fantastic; October 15th 2010 at 01:59 PM.
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  7. #37
    Junior Member ceode's Avatar
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    multiply the whole eqn by 1/u^2

    u^-2 . du/dx + 1/(ux) = 1/x^2

    put z=u^-1

    now, eqn becomes

    -dz/dx + z/x = x^-2

    dz/dx - z/x = -x^-2

    P=-1/x , Q=-x^2

    IF = 1/x

    solution

    z . 1/x = inegral -x^-2 * 1/x dx + c
    Last edited by ceode; October 15th 2010 at 09:57 AM. Reason: error
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  8. #38
    Junior Member ceode's Avatar
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    Ok. Its time for me to go....
    Reply me when you are free
    Thank you very much
    love you
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  9. #39
    A Plied Mathematician
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    Ok, I see where you're going now. IF stands for "integrating factor", not "if" in caps. I agree with your post # 36. You should solve for u, and then substitute back in for y. y is the goal, remember.

    Good work!
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