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Math Help - Separation in differential equation

  1. #1
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    Separation in differential equation

    How can I get a separate y, y' and y'' in equation

    (y')^2 + y*y''+y'*y = 0

    Best regards

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  2. #2
    A Plied Mathematician
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    I'm not entirely sure I understand your question. However, I would note that

    \dfrac{d}{dx}(y^{2})=2yy', and hence

    \dfrac{d^{2}}{dx^{2}}(y^{2})=2[(y')^{2}+yy''], and hence

    \dfrac{1}{2}\,\dfrac{d^{2}}{dx^{2}}(y^{2})=(y')^{2  }+yy''.

    The DE becomes

    \dfrac{(y^{2})''}{2}+yy'=0. We can now integrate directly, once, to obtain

    \dfrac{1}{2}\,(y^{2})'+\dfrac{y^{2}}{2}=C_{1}, or re-defining C_{1}, we have

    (y^{2})'+y^{2}=C_{1}.

    Can you continue?
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  3. #3
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    Thank you Mr Ackbeet. I can continue. Anyway, I want to separate y,y' and y'' in diff equation to get in form ay''+by'+cy+g(y'^2). How can we do this if we can? Thank a lot.
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  4. #4
    A Plied Mathematician
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    Hmm. That might be difficult. The substitution u=y^{2} yields the DE

    u''+u'=0,

    which is technically in the form you're looking for, with a=b=1, and c=g=0.

    The substitution u=yy' gives you the same DE, actually, just one order lower:

    u'+u=0.

    In this case, you have a=g=0, and b=c=1.

    I don't know if that helps you or not. That's a strange form to try to get the DE into - at least I'm not familiar with it. It looks semi-Ricatti, but in a second-order form. I don't know.

    Simply dividing through by yy' yields the immediately integrable expression

    \dfrac{y'}{y}+\dfrac{y''}{y'}+1=0.

    Other than this, I'm about out of ideas.
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