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Math Help - Find the general solution

  1. #1
    MHF Contributor alexmahone's Avatar
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    Find the general solution

    Find the general solution of the differential equation 3y+x^3y^4+3xy'=0.

    y'+\frac{y}{x}=\frac{-x^2y^4}{3}

    xy'+y=\frac{-x^3y^4}{3}

    \frac{d}{dx}(xy)=\frac{-x^3y^4}{3}

    xy=\frac{-x^4y^4}{12}+C

    But the answer given in the book is y(x)=x^{-1}(C+\ln x)^{-1/3}. Where have I gone wrong?
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: Find the general solution

    Quote Originally Posted by alexmahone View Post
    Find the general solution of the differential equation 3y+x^3y^4+3xy'=0.

    y'+\frac{y}{x}=\frac{-x^2y^4}{3}

    xy'+y=\frac{-x^3y^4}{3}

    \frac{d}{dx}(xy)=\frac{-x^3y^4}{3}

    xy=\frac{-x^4y^4}{12}+C

    But the answer given in the book is y(x)=x^{-1}(C+\ln x)^{-1/3}. Where have I gone wrong?
    Setting z=x\ y Your DE is...

    z^{'}= -\frac{z^{4}}{3 x} (1)

    ... and the integration is immediate...

    z= x y= -\frac{1}{(\ln x+c)^{\frac{1}{3}}} (2)

    Kind regards

    \chi \sigma
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  3. #3
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    Re: Find the general solution

    Quote Originally Posted by alexmahone View Post
    Find the general solution of the differential equation 3y+x^3y^4+3xy'=0.

    y'+\frac{y}{x}=\frac{-x^2y^4}{3}

    xy'+y=\frac{-x^3y^4}{3}

    \frac{d}{dx}(xy)=\frac{-x^3y^4}{3}

    xy=\frac{-x^4y^4}{12}+C

    But the answer given in the book is y(x)=x^{-1}(C+\ln x)^{-1/3}. Where have I gone wrong?
    You have gone wrong by treating y as a constant and not as a function of x.

    You should recognise that this is actually a Bernoulli Equation.
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  4. #4
    MHF Contributor alexmahone's Avatar
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    Re: Find the general solution

    Quote Originally Posted by chisigma View Post
    Setting z=x\ y Your DE is...

    z^{'}= -\frac{z^{4}}{3 x} (1)

    ... and the integration is immediate...

    z= x y= -\frac{1}{(\ln x+c)^{\frac{1}{3}}} (2)

    Kind regards

    \chi \sigma
    But what's wrong with my solution?
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  5. #5
    MHF Contributor alexmahone's Avatar
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    Re: Find the general solution

    Quote Originally Posted by Prove It View Post
    You have gone wrong by treating y as a constant and not as a function of x.

    You should recognise that this is actually a Bernoulli Equation.
    I know that this is a Bernoulli equation; that is why I multiplied by the integrating factor x. Which step of my solution is actually invalid?
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  6. #6
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    Re: Find the general solution

    Quote Originally Posted by alexmahone View Post
    I know that this is a Bernoulli equation; that is why I multiplied by the integrating factor x. Which step of my solution is actually invalid?
    Which steps are invalid?

    1. Not making an appropriate substitution.

    2. Performing integration treating a function of the independent variable like a constant...
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  7. #7
    MHF Contributor chisigma's Avatar
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    Re: Find the general solution

    Quote Originally Posted by alexmahone View Post
    But what's wrong with my solution?
    You arrived [correctly...] to write...

    \frac{d}{dx} (x y)= - \frac{x^{3}y^{4}}{3} (1)

    ... and then You [incorrectly...] operated as in case of separate variables... but in (1) x and y aren't separate...

    Kind regards

    \chi \sigma
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