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Math Help - Bernoulli equation

  1. #1
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    Bernoulli equation

    I'm trying to figure out how to do this problem and we've just covered bernoulli equations in class so I know I have to use that. This is the problem:

    Solve the equation
    y' = (X cos(t) + T)y - y^3
    where X and T are constants.

    So far I have this:

    dy/dt - (X cos(t) + T)y = -y^3

    then using integrating factor:

    e^\int(Xcos(t) + T)) = e^(Xsin(t) + Tt + C) = e^(Tt)* e^(xsin(t)) * e^C

    But where do I go from here?
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  2. #2
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    Re: Bernoulli equation

    You can't go straight to the integrating factor, because you haven't done the substitution required to turn the DE into something first order linear.

    Let \displaystyle \begin{align*} v = y^{1 - 3} = y^{-2} \end{align*}. Then we have \displaystyle \begin{align*} y = v^{-\frac{1}{2}} \implies \frac{dy}{dt} = -\frac{1}{2}v^{-\frac{3}{2}}\, \frac{dv}{dt} \end{align*}. Substituting into the DE gives

    \displaystyle \begin{align*} \frac{dy}{dt} - \left( X \cos{(t)} + T \right) y &= -y^3 \\ -\frac{1}{2}v^{-\frac{3}{2}}\,\frac{dv}{dt} - \left( X \cos{(t)} + T \right) v^{-\frac{1}{2}} &= -v^{-\frac{3}{2}} \\ \frac{dv}{dt} + 2 \left( X \cos{(t)} + T \right) v &= 2 \end{align*}

    This is now first order linear, so now you can apply an integrating factor.
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  3. #3
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    Re: Bernoulli equation

    Alright, thanks a lot. One questions though: How did you get -v^{-\frac{3}{2}} on the right side in the 2nd line?
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  4. #4
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    Re: Bernoulli equation

    We know \displaystyle \begin{align*} y = v^{-\frac{1}{2}} \end{align*} so \displaystyle \begin{align*} y^3 = \left( v^{-\frac{1}{2}} \right)^3 = v^{-\frac{3}{2}} \end{align*}.
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  5. #5
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    Re: Bernoulli equation

    So for the integrating factor I get e^{2Xsin(t) + Tt + C}, is that correct? Do I then multiply the entire equation by this? I've kind of forgotten how to do the rest of the steps .
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  6. #6
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    Re: Bernoulli equation

    Not quite. Your integrating factor is \displaystyle \begin{align*} e^{\int{2 \left[ X\cos{(t)} + T \right] dt}} = e^{2 \left[ X\sin{(t)} + Tt \right]} \end{align*}. Note that any of the family of solutions to this integral will work, so for simplicity we disregard the +C. Anyway, yes we multiply both sides of the DE by this integrating factor to give

    \displaystyle \begin{align*} e^{2\left[ X\sin{(t)} + Tt \right]}\,\frac{dv}{dt} + 2e^{2\left[ X\cos{(t)} + Tt \right]} \left[X \cos{(t)} + T \right] v &= 2e^{2\left[ X\sin{(t)} + Tt \right]} \\ \frac{d}{dt} \left[ e^{2\left[ X\sin{(t)} + Tt \right] }\, v \right] &= 2e^{2\left[ X\sin{(t)} + Tt \right]} \\ e^{2\left[ X\sin{(t)} + Tt \right]} &= \int{2e^{2\left[ X\sin{(t)} + Tt \right]}\,dt} \end{align*}

    Can you go from here?
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