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Math Help - Solve y'=y(1 + xy^4)

  1. #1
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    Solve y'=y(1 + xy^4)

    Last edited by mr fantastic; October 21st 2010 at 04:49 AM. Reason: Re-titled.
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  2. #2
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    It's Bernoulli, with standard form

    y'-y=xy^{5}
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  3. #3
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    ok
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  4. #4
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    bernulli solution problematic points

    y'-y=5xy^5\\
    \frac{y'}{y^5}-\frac{1}{y^4}=5x \, \, \, \, \, \,z=\frac{1}{y^4} \, \, \, \, \, \,\, \,dz=\frac{1}{y^5}dy
    how does from  dz=\frac{1}{y^5}dy we get z' and y' we only have dz and dy here???
    in order to get y' we need to have dy/dx we dont have it here.

    then
    z'-z=5x \, \, \, \, \, \, a=-1 \, \, \, \, \, A=-z \, \, \, \, \, \ e^{-z}(z'-z)=e^{-z} 5z \, \, \, \, \, (e^{-z}z)'=e^{-z} 5z \, \, \, \, \, e^{-z}z=\int ^z e^{-t} 5t +c after solving by parts i get
    e^{-z}z=-5ze^{-z}+5e^{-z} +c

    c=e^{-z}z+5ze^{-z}-5e^{-z}
    what to do now?
    Last edited by transgalactic; October 20th 2010 at 11:28 PM.
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  5. #5
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    First off, if you let z = y^{-4} then \displaystyle \frac{dz}{dy} = -4y^{-5}.
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  6. #6
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    ok i missed that -4 in the defferentiation
    what about my questions
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  7. #7
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    Well, you need to transform the DE which is in terms of \displaystyle \frac{dy}{dx} into a DE in terms of \displaystyle \frac{dz}{dx}.

    You have z = y^{-4}.


    By the chain rule, \displaystyle \frac{dz}{dx} = \frac{dz}{dy}\,\frac{dy}{dx}.

    So \displaystyle \frac{dz}{dx} = -4y^{-5}\,\frac{dy}{dx} and thus \displaystyle \frac{dy}{dx} = -\frac{1}{4}y^5\,\frac{dz}{dx}.


    Substituting everything into the DE

    \displaystyle \frac{dy}{dx} - y = 5x\,y^5

    \displaystyle -\frac{1}{4}y^5\,\frac{dz}{dx} - y = 5x\,y^5

    \displaystyle -\frac{1}{4}\,\frac{dz}{dx} - y^{-4} = 5x

    \displaystyle -\frac{1}{4}\,\frac{dz}{dx} - z = 5x

    \displaystyle \frac{dz}{dx} + 4z = -20x.


    This is now first order linear, which you can solve using an integrating factor.
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  8. #8
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    i got a similiar lenear differential equation
    and i solved it by multiplying in exponet etc
    and it doesnt come out
    ??
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  9. #9
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    Your integrating factor is e^{\int{4\,dx}} = e^{4x}.

    So multiplying through by the integrating factor gives

    \displaystyle e^{4x}\,\frac{dz}{dx} + 4e^{4x}z = -20x\,e^{4x}

    \displaystyle \frac{d}{dx}(e^{4x}z) = -20x\,e^{4x}

    \displaystyle e^{4x}z = \int{-20x\,e^{4x}\,dx}

    \displaystyle e^{4x}z = -20\int{x\,e^{4x}\,dx}

    \displaystyle e^{4x}z = -20\left(\frac{1}{4}x\,e^{4x} - \int{\frac{1}{4}e^{4x}\,dx}\right)

    \displaystyle e^{4x}z = -5x\,e^{4x} + 5\int{e^{4x}\,dx}

    \displaystyle e^{4x}z = -5x\,e^{4x} + \frac{5}{4}e^{4x} + C

    \displaystyle z = -5x + \frac{5}{4} + Ce^{-4x}

    \displaystyle y^{-4} = -5x + \frac{5}{4} + Ce^{-4x}

    \displaystyle y = \left(-5x + \frac{5}{4} + Ce^{-4x}\right)^{-\frac{1}{4}}.
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