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Thread: 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

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

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

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

    The DE becomes

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

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

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

    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 $\displaystyle u=y^{2}$ yields the DE

    $\displaystyle u''+u'=0,$

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

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

    $\displaystyle u'+u=0.$

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

    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 $\displaystyle yy'$ yields the immediately integrable expression

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

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