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    dw/dt = (a + bw) w^2

    I separated the variables and then used partial fractions but everything got too complicated and I end up with two 'log (w)' terms and a 'w' term on the LHS and t on the RHS. I can't rearrange that to w=[something in terms of t]. So may be I'm doing it wrong.

    (e.g. I got log (f(w)) + k(w) = t + c) where f and k are different linear functions of w.)

    Please help.

    Thanks
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    (Original post by VA11ZZY)
    dw/dt = (a + bw) w^2

    I separated the variables and then used partial fractions but everything got too complicated and I end up with two 'log (w)' terms and a 'w' term on the LHS and t on the RHS. I can't rearrange that to w=[something in terms of t]. So may be I'm doing it wrong.

    (e.g. I got log (f(w)) + k(w) = t + c) where f and k are different linear functions of w.)

    Please help.

    Thanks
    You won't be able to find the expression explicitly. Can you show us how you did the partial fraction?

    I end up with (checked with Wolfram for various values of a and b):
    Spoiler:
    Show


     \displaystyle -\frac{1}{aw} - \frac{b}{a^2}\log(w) + \frac{b}{a^2}\log(a+bw) = t+C

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    (Original post by Blazy)
    You won't be able to find the expression explicitly. Can you show us how you did the partial fraction?

    I end up with (checked with Wolfram for various values of a and b):
    Spoiler:
    Show















     \displaystyle -\frac{1}{aw} - \frac{b}{a^2}\log(w) + \frac{b}{a^2}\log(a+bw) = t+C














    Thanks very much for your answer.

    Yes - Wolfram Alpha gives that answer - the problem, as you pointed out, is rearranging the expression to give w in terms of t.

    Someone on stack exchange gave this answer (leaving out the constants for simplicity)

    \frac{dw}{dt} = -{w}^3 - {w}^2

    solves to

     log(w+1) - log(w) - \frac{1}{w}   = c - t

    (consistent with your answer)

    And then using the substitution of v=-\left(1+\frac{1}{w}\right)

    it rearranges to

    w=\frac{-1}{1+v}=\frac{1}{1+Lambert_W(-\exp(c-1-t))} because Lambert_W(−exp(c−1−t)) = v

    The Lambert function appears to be a numerical function which gives the solution to Y = Xe^X. Now I just need to figure out how to use it in this context!
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    (Original post by VA11ZZY)
    Thanks very much for your answer.

    Yes - Wolfram Alpha gives that answer - the problem, as you pointed out, is rearranging the expression to give w in terms of t.

    Someone on stack exchange gave this answer (leaving out the constants for simplicity)

    \frac{dw}{dt} = -{w}^3 - {w}^2

    solves to

     log(w+1) - log(w) - \frac{1}{w}   = c - t

    (consistent with your answer)

    And then using the substitution of v=-\left(1+\frac{1}{w}\right)

    it rearranges to

    w=\frac{-1}{1+v}=\frac{1}{1+Lambert_W(-\exp(c-1-t))} because Lambert_W(−exp(c−1−t)) = v

    The Lambert function appears to be a numerical function which gives the solution to Y = Xe^X. Now I just need to figure out how to use it in this context!
    I see. Assuming what you've written is true, we can pull off a similar trick then. It'll be hard to do this without hints but basically:

    1)  \log(a+bw) = \log(a(1+\frac{b}{aw})) = \log(a) + \log((1+\frac{b}{aw})

  and stick  \log(a) into our constant of integration.

    2) Make the substitution  v = -(1+\frac{a}{bw})

    Then work your way through for a similar answer, remembering the basic log rules - I think this will work.

    EDIT: Just realised where that substitution comes from. Making the substitution  v = -(b+\frac{a}{w}) directly works fine (without doing step 1).
 
 
 
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