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Thread: Bernoulli equation

  1. #1
    Senior Member Vinod's Avatar
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    Bernoulli equation

    Hello,
    I want to solve this bernoulli differential equation.But my answer is different.

    $y'+2xy=xy^2$

    Multiplying both sides of equation by $y^{-2}$ and -1, we get

    $-y^{-2}y'-xy^{-1}=-x$ Let $z=y^{-1}$ Then,the equation became z'-xz=-x. Now the equation is linear in z.Using P(x)=-x produces $\int P(x)dx=\int -xdx=\frac{-x^2}{2}$ which implies that $e^{-x^2/2}$is an integrating factor.Multiplying the linear equation by this factor we get

    $z'e^{-x^2/2}-xze^{-x^2/2}=-xe^{-x^2/2}$

    $\frac{d[ze^{-x^2/2}]}{dx}=-xe^{-x^2/2}$

    $ze^{-x^2/2}=\int -xe^{-x^2/2}dx$

    $ze^{-x^2/2}=e^{-x^2/2}+c$
    $y^{-1}=1+ce^{x^2/2}$
    $y=\frac{1}{1+ce^{x^2/2}}$

    But answer is $\frac{-2}{Ce^{x^2}-1}$ Where is i am wrong in my answer, would any forumite explain me?
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  2. #2
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    Re: Bernoulli equation

    Hey Vinod.

    Hint - Think about when you integrate e^[cu] du and the constant c.
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  3. #3
    Senior Member Vinod's Avatar
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    Re: Bernoulli equation

    Quote Originally Posted by chiro View Post
    Hey Vinod.

    Hint - Think about when you integrate e^[cu] du and the constant c.
    Hello chiro,
    Integrating $e^{cu}du$, we get $e^{cu}/c$. that's it.
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  4. #4
    Senior Member Vinod's Avatar
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    Re: Bernoulli equation

    Quote Originally Posted by chiro View Post
    Hey Vinod.

    Hint - Think about when you integrate e^[cu] du and the constant c.
    Hello,
    After rectification of my wrong answer in post#1, i got the answer $\frac{2}{1+2ce^{x^2}}$ but it is also different.
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  5. #5
    MHF Contributor MarkFL's Avatar
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    Re: Bernoulli equation

    2 times some arbitrary constant is still just an arbitrary constant, right?
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  6. #6
    Senior Member Vinod's Avatar
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    Re: Bernoulli equation

    Quote Originally Posted by MarkFL View Post
    2 times some arbitrary constant is still just an arbitrary constant, right?
    Hello, Yes, You are right.Answer i got is not different from the answer provided.
    Thanks from MarkFL
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