How can I get a separate y, y' and y'' in equation
(y')^2 + y*y''+y'*y = 0
Best regards
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?
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.