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A Bernoulli DE

  1. Oct 6, 2009 #1
    1. The problem statement, all variables and given/known data

    Solve the given differential equation by using appropriate substitution.

    [tex]3(1+t^{2})\frac{dy}{dx} = 2ty(y^{3} - 1)[/tex]

    2. Relevant equations

    [tex]y = u^{\frac{1}{1-n}}[/tex]

    3. The attempt at a solution

    [tex]\frac{dy}{dt} = \frac{2ty^{4} - 2ty}{3 + 3t^{2}}[/tex]

    [tex]\frac{dy}{dt} + \frac{2ty}{3 + 3t^{2}} = \frac{2ty^{4}}{3 + 3t^{2}}[/tex]

    [tex]-\frac{1}{3}u^{-\frac{4}{3}}\frac{du}{dt} + (\frac{2ty}{3 + 3t^{2}})u^{-\frac{1}{3}} = (\frac{2ty^{4}}{3 + 3t^{2}})(u^{-\frac{1}{3}})^{4}[/tex]

    [tex]-\frac{1}{3}\frac{du}{dt} + (\frac{2ty}{3 + 3t^{2}})\frac{u^{-\frac{1}{3}}}{u^{-\frac{4}{3}}} = \frac{2ty^{4}}{3 + 3t^{2}}[/tex]

    [tex]-\frac{1}{3}\frac{du}{dt} + (\frac{2ty}{3 + 3t^{2}})u = \frac{2ty^{4}}{3 + 3t^{2}}[/tex]

    [tex]\frac{du}{dt} - (\frac{2ty}{1 + t^{2}})u = -\frac{2ty^{4}}{1 + t^{2}}[/tex]

    Integrating factor:

    [tex]I(t) = e^{-\int\frac{2t}{1+t^{2}}dt}[/tex]

    use substitution:

    [tex]e^{ln|u|^{-1}}[/tex]

    [tex]I(t) = \frac{1}{1+t^{2}}[/tex]

    [tex]y(t) = \frac{1}{\frac{1}{1+t^{2}}}[-\int\frac{2t}{(1+t^{2})(1+t^{2})}dt + c][/tex]

    [tex]y(t) = 1+t^{2}[\frac{1}{1+t^{2}} + c][/tex]

    [tex]y(t) = 1 + c + ct^{2}[/tex]
     
  2. jcsd
  3. Oct 6, 2009 #2

    LCKurtz

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    For one thing, you shouldn't have t, u, and y all in the same equation. Here's an easier way to get started. Rewrite your equation:

    3(1 + t2)y' + 2ty = 2ty4

    Now multiply through by y-4:

    3(1 + t2)y-4y' + 2ty-3 = 2t

    Now let u = y-3, u' = -3y-4y'

    This gets you to the equation

    -(1 + t2)u' + 2tu = 2t

    without all the fractional exponents and resulting errors.
     
  4. Oct 6, 2009 #3
    I made some typos in there here I fixed it:

    1. The problem statement, all variables and given/known data

    Solve the given differential equation by using appropriate substitution.

    [tex]3(1+t^{2})\frac{dy}{dx} = 2ty(y^{3} - 1)[/tex]

    2. Relevant equations

    [tex]y = u^{\frac{1}{1-n}}[/tex]

    3. The attempt at a solution

    [tex]\frac{dy}{dt} = \frac{2ty^{4} - 2ty}{3 + 3t^{2}}[/tex]

    [tex]\frac{dy}{dt} + \frac{2ty}{3 + 3t^{2}} = \frac{2ty^{4}}{3 + 3t^{2}}[/tex]

    [tex]-\frac{1}{3}u^{-\frac{4}{3}}\frac{du}{dt} + (\frac{2t}{3 + 3t^{2}})u^{-\frac{1}{3}} = (\frac{2t}{3 + 3t^{2}})(u^{-\frac{1}{3}})^{4}[/tex]

    [tex]-\frac{1}{3}\frac{du}{dt} + (\frac{2t}{3 + 3t^{2}})\frac{u^{-\frac{1}{3}}}{u^{-\frac{4}{3}}} = \frac{2t}{3 + 3t^{2}}[/tex]

    [tex]-\frac{1}{3}\frac{du}{dt} + (\frac{2t}{3 + 3t^{2}})u = \frac{2t}{3 + 3t^{2}}[/tex]

    [tex]\frac{du}{dt} - (\frac{2t}{1 + t^{2}})u = -\frac{2t}{1 + t^{2}}[/tex]

    Integrating factor:

    [tex]I(t) = e^{-\int\frac{2t}{1+t^{2}}dt}[/tex]

    use substitution:

    [tex]e^{ln|u|^{-1}}[/tex]

    [tex]I(t) = \frac{1}{1+t^{2}}[/tex]

    [tex]y(t) = \frac{1}{\frac{1}{1+t^{2}}}[-\int\frac{2t}{(1+t^{2})(1+t^{2})}dt + c][/tex]

    [tex]y(t) = 1+t^{2}[\frac{1}{1+t^{2}} + c][/tex]

    [tex]y(t) = 1 + c + ct^{2}[/tex]
     
  5. Oct 6, 2009 #4

    LCKurtz

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    Interesting step that last one. Where did the denominator go?
     
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