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Math Help - Solve the Bernoulli equation.

  1. #1
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    Solve the Bernoulli equation.

    \frac{dy}{dx}+xy=xy^3

    I have no idea how to do these. And we *have* to do it as a Bernoulli equation.

    Initial values: (0, 1/3)
    Last edited by A Beautiful Mind; September 24th 2010 at 01:35 PM.
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  2. #2
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    Quote Originally Posted by A Beautiful Mind View Post
    \frac{dy}{dx}+xy=xy^3

    I have no idea how to do these. And we *have* to do it as a Bernoulli equation.

    Initial values: (1/3, 0)
    The procedure is given here (and no doubt it's in your classnotes and textbook too): Bernoulli differential equation - Wikipedia, the free encyclopedia

    What part of applying this procedure are you stuck on?
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    Well, my answer I get before using the initial conditions is:

    y = \sqrt{1+Ce^\frac{-x^2}{2}}


    w'+xw=x, where w = \frac{1}{y^2}
    w' = -3y^{-4}
    w'=\frac{-3}{y^{4}}

    e^{\frac{x^2}{2} is the integrating factor...the rest I tried to put on here, but the stuff keeps getting messed up in the latex.
    Last edited by A Beautiful Mind; September 22nd 2010 at 11:13 PM.
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    Quote Originally Posted by A Beautiful Mind View Post
    Well, my answer I get before using the initial conditions is:

    y = \sqrt{1+Ce^\frac{-x^2}{2}}
    http://www.wolframalpha.com/input/?i...+xy+%3D+xy%5E3

    The DE is seperable in fact. Are you required to do it as a Bernoulli?
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    Yes, we're required to do it as a Bernoulli. I noticed it was separable too, but she's very strict about this.
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    I still need help with this...I'm getting nowhere. And my teachers office hours are cancelled. And the tutors can't do above Calc III. :/
    Last edited by mr fantastic; September 24th 2010 at 03:09 PM. Reason: Restored original post.
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    Quote Originally Posted by A Beautiful Mind View Post
    Well, my answer I get before using the initial conditions is:

    y = \sqrt{1+Ce^\frac{-x^2}{2}}


    w'+xw=x, Mr F says: This is wrong.

    where w = \frac{1}{y^2} Mr F says: Yes, this is the correct substitution.

    w' = -3y^{-4} Mr F says: This is wrong.

    w'=\frac{-3}{y^{4}}

    e^{\frac{x^2}{2} is the integrating factor...the rest I tried to put on here, but the stuff keeps getting messed up in the latex.
    There are a number of things you are not doing correctly. Have you read any references on the required technique eg. http://en.wikipedia.org/wiki/Bernoul...ntial_equation.

    Please show all your work, every step, so that it can be reviewed.
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    \frac{dy}{dx}+xy=xy^3

    n = 3

    u = y^{1-3}

    u = y^-2<br />
    \frac{du}{dx}

    \frac{du}{dx} = -2y^{-3}

    \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}

    \frac{1}{-2y^{-3}}\frac{du}{dx}+xy=xy^3

    \frac{du}{dx}-2y^{-2}=-2x

    Getting integrating factor.

    P(x) = -2x

    e^{\int-2x} = e^{-x^2}

    e^{-x^2}\frac{du}{dx}+(-2y^{-2}xe^{-x^2}) = -2xe^{-x^2}

    Did the integration in my head. You integrate [e^-x^/2(u)}]' and -2xe^{-x^2}. It's easy to see on the right side the -2x term will cancel out by u-substitution.

    e^{-x^2}u = e^-x^2+C

    Sub in the u.

    \frac{e^{-x^2}}{y^2} = e^-x^2+C

    \frac{1}{y^2}=1+Ce^{x^2}

    Plug in initial values to this equation. Y(0) = 1/3

    \frac{1}{(1/3)^2} = 1+Ce^0
    9 = 1+C
    C= 8

    \frac{1}{y^2}= 1+8e^{x^2}

    And yes, I read your first reference to wikipedia, but I got the first answer I posted yesterday as that weird square root. Then, I tried looking for examples.
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  9. #9
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    Quote Originally Posted by A Beautiful Mind View Post
    \frac{dy}{dx}+xy=xy^3

    n = 3

    u = y^{1-3}

    u = y^-2<br />
    \frac{du}{dx}

    \frac{du}{dx} = -2y^{-3}

    \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}

    \frac{1}{-2y^{-3}}\frac{du}{dx}+xy=xy^3 Mr F says: All OK to here (although there are parts of your working I find difficult to understand)

    \frac{du}{dx}-2y^{-2}=-2x Mr F says: This is wrong. There is a missing x. It should be \displaystyle \frac{du}{dx} - 2xy^{-2} = -2x. From which you get \displaystyle \frac{du}{dx} - 2ux = -2x.

    Getting integrating factor.

    P(x) = -2x

    e^{\int-2x} = e^{-x^2}

    e^{-x^2}\frac{du}{dx}+(-2y^{-2}xe^{-x^2}) = -2xe^{-x^2}

    Did the integration in my head. You integrate [e^-x^/2(u)}]' and -2xe^{-x^2}. It's easy to see on the right side the -2x term will cancel out by u-substitution.

    e^{-x^2}u = e^-x^2+C

    Sub in the u.

    \frac{e^{-x^2}}{y^2} = e^-x^2+C

    \frac{1}{y^2}=1+Ce^{x^2}

    Plug in initial values to this equation. Y(0) = 1/3

    \frac{1}{(1/3)^2} = 1+Ce^0
    9 = 1+C
    C= 8

    \frac{1}{y^2}= 1+8e^{x^2}

    And yes, I read your first reference to wikipedia, but I got the first answer I posted yesterday as that weird square root. Then, I tried looking for examples.
    OK, this final answer is correct. But a lot of your working presents as wrong even though you finally get the correct answer. You will lose marks for that.
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    Sorry, the x reappears in the next step, though. I subbed u =y^-2 though in that part.
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