Results 1 to 3 of 3

Math Help - Checking a Bernoulli Equation

  1. #1
    Newbie
    Joined
    Nov 2010
    Posts
    3

    Checking a Bernoulli Equation

    The Equation is:

    t^2y'+2ty-y^3=0

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

    \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; February 21st 2011 at 06:15 PM. Reason: I typed the solution incorrectly.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,933
    Thanks
    336
    Awards
    1
    Quote Originally Posted by Lyov View Post
    The Equation is:

    t^2y'+2ty-y^3=0

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

    \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 2 + 5ct^2 in terms of y:
    \displaystyle 2 + 5ct^2 = \frac{5t}{y}

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

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,549
    Thanks
    1418
    \displaystyle t^2\,\frac{dy}{dt} + 2t\,y = y^3

    Make the substitution \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 -\frac{1}{2}t^2v^{-\frac{3}{2}}\,\frac{dv}{dt} + 2t\,v^{-\frac{1}{2}} = v^{-\frac{3}{2}}

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

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

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

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

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

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

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

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

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

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

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

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

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

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

    I agree with your final answer.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Bernoulli's Equation
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: September 6th 2011, 04:44 AM
  2. bernoulli's equation
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: July 25th 2011, 06:00 PM
  3. Bernoulli equation
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: April 22nd 2009, 08:39 AM
  4. bernoulli equation
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: February 6th 2009, 06:44 AM
  5. Bernoulli's equation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 9th 2008, 05:30 AM

Search Tags


/mathhelpforum @mathhelpforum