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Thread: Checking a Bernoulli Equation

  1. #1
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    Checking a Bernoulli Equation

    The Equation is:

    $\displaystyle t^2y'+2ty-y^3=0$

    I have the solution, which thanks to Mathematica, I know to be correct, it is:

    $\displaystyle \pm \sqrt{\frac{5t}{2+5t^5c}}$

    I have to do a solution check, and keep getting terribly lost in the algebra. I am wondering if anyone has any suggestions or tricks on how to plug this mess back in without it turning into the never-ending algebra problem. Thanks.
    Last edited by Lyov; Feb 21st 2011 at 06:15 PM. Reason: I typed the solution incorrectly.
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  2. #2
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    Quote Originally Posted by Lyov View Post
    The Equation is:

    $\displaystyle t^2y'+2ty-y^3=0$

    I have the solution, which thanks to Mathematica, I know to be correct, it is:

    $\displaystyle \pm \sqrt{\frac{5t}{2+5t^2c}}$

    I have to do a solution check, and keep getting terribly lost in the algebra. I am wondering if anyone has any suggestions or tricks on how to plug this mess back in without it turning into the never-ending algebra problem. Thanks.
    My only thought is to solve for $\displaystyle \displaystyle 2 + 5ct^2$ in terms of y:
    $\displaystyle \displaystyle 2 + 5ct^2 = \frac{5t}{y}$

    That helps a little anyway, since you will have a $\displaystyle \displaystyle (2 + 5ct^2)^2$ in your derivative.

    -Dan
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  3. #3
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    $\displaystyle \displaystyle t^2\,\frac{dy}{dt} + 2t\,y = y^3$

    Make the substitution $\displaystyle \displaystyle v = y^{1-3} = y^{-2} \implies y = v^{-\frac{1}{2}} \implies \frac{dy}{dt} = -\frac{1}{2}v^{-\frac{3}{2}}\,\frac{dv}{dt}$ and the integral becomes

    $\displaystyle \displaystyle -\frac{1}{2}t^2v^{-\frac{3}{2}}\,\frac{dv}{dt} + 2t\,v^{-\frac{1}{2}} = v^{-\frac{3}{2}}$

    $\displaystyle \displaystyle -\frac{1}{2}t^2\,\frac{dv}{dt} + 2t\,v = 1$

    $\displaystyle \displaystyle -\frac{1}{2}\,\frac{dv}{dt} + 2t^{-1}v = t^{-2}$

    $\displaystyle \displaystyle \frac{dv}{dt} - 4t^{-1}v = -2t^{-2}$.

    This is now first-order linear, so mulitplying both sides by the integrating factor $\displaystyle \displaystyle e^{\int{-4t^{-1}\,dt}} = e^{-4\ln{t}} = e^{\ln{\left(t^{-4}\right)}} = t^{-4}$ gives

    $\displaystyle \displaystyle t^{-4}\,\frac{dv}{dt} - 4t^{-5}v = -2t^{-6}$

    $\displaystyle \displaystyle \frac{d}{dt}\left(t^{-4}v\right) = -2t^{-6}$

    $\displaystyle \displaystyle t^{-4}v = \int{-2t^{-6}\,dt}$

    $\displaystyle \displaystyle t^{-4}v = \frac{2}{5}t^{-5} + C$

    $\displaystyle \displaystyle v = \frac{2}{5}t^{-1} + Ct^4$

    $\displaystyle \displaystyle y^{-2} = \frac{2}{5}t^{-1} + Ct^4$

    $\displaystyle \displaystyle \frac{1}{y^2} = \frac{2}{5t} + Ct^4$

    $\displaystyle \displaystyle \frac{1}{y^2} = \frac{2 + 5Ct^5}{5t}$

    $\displaystyle \displaystyle y^2 = \frac{5t}{2 + 5Ct^5}$

    $\displaystyle \displaystyle y = \pm \sqrt{\frac{5t}{2 + 5Ct}}$.

    I agree with your final answer.
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