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Math Help - bernoulli

  1. #16
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    Quote Originally Posted by Prove It View Post
    Step 1: Solve for \displaystyle y^{-2}.

    Step 2: Take both sides to the power of [tex]\displaystyle -\frac{1}{2}[/math].
    i have never seen this before... i would only have thought of multiplying by y^3 but that wouldnt help

    what do you do with a -1/2 power? 1/2 is root but what do you do with minus half? and do you apply it to thE WHOLE rhs or just the individual terms>?

    ah, is it (1/y^2)^-1/2 = 1/(1/(sqrt(y^2)) = 1/(1/y) = y

    same thing to the rhs?

    so it becomes 1/(sqrt(rhs))?
    Last edited by mr fantastic; February 22nd 2011 at 03:19 AM. Reason: Merged posts.
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  2. #17
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Prove It View Post
    Except \displaystyle \frac{dv}{dx} - \frac{3v}{x} = \frac{x + 1}{x} is not a homogeneous equation, and is not separable.

    Look to my answer #6. I said solve the homogeneous equation. For a lineal equation always the homogeneous is separable. Solving this one, you can use (obviously that was my proposal) the variation of constants method.


    Since it's first-order linear, the Integrating Factor method is needed...

    It is not necessary.


    Fernando Revilla
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  3. #18
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    it worked just fine and the i.f. method is easier anyway. plus as i said, this question does not require knowledge of homogeneous equations so that was not the intended method
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  4. #19
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by mathcore View Post
    it worked just fine and the i.f. method is easier anyway. plus as i said, this question does not require knowledge of homogeneous equations so that was not the intended method

    All right, but up untill my answer #6 nobody could know which was the intended method.


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