Any help, please? How can I start to solve them? I tried to use $y'=p$. Also I tried $x=e^x$ and so many methods, but I couldn't reach them to the end. I always got blocked in the middle. Thanks in advance.
First equation: $$6x^2y-6y'^2+(12x^2-3x^3)y'+x^5-6x^4=0.$$ Second equation: $$x=\frac{y\ln y}{y'}-\frac{y'^2}{y^2}$$ with $y>0$.
On
FIRST EQUATION : $$6x^2y-6y'^2+(12x^2-3x^3)y'+x^5-6x^4=0.$$ By inspection one can see a particular solution : $$y=\frac13 x^3\quad\implies\quad y'=x^2$$ $6x^2(\frac13 x^3)-6(x^2)^2+(12x^2-3x^3)(x^2)+x^5-6x^4=$
$=2x^5-6x^4+12x^4-3x^5+x^5-6x^4=0.$
This draw us to the change of function : $$y(x)=\frac13 x^3+u(x)\quad\implies\quad y'=x^2+u'$$ $$6x^2(\frac13 x^3)-6(x^2+u')^2+(12x^2-3x^3)(x^2+u')+x^5-6x^4=0.$$ After simplification : $$6x^2u-3x^3u'-6u'^2=0$$ The change of $x$ into $-x$ doesn't change the equation. This draw us to the change of variable $X=x^2$ which leads to : $$u(X)=X\frac{du}{dX}+4\left(\frac{du}{dX}\right)^2$$ This is a Clairault's differential equation.
SECOND EQUATION : $$x=\frac{y\ln y}{y'}-\frac{y'^2}{y^2}$$ Since Aryadeva already answered there is no need to reapeat the method of solving.
For the second differential equation: $$x=\frac{y\ln y}{y'}-\frac{y'^2}{y^2}$$ $$x=\frac{\ln y}{(\ln y)'}-{((\ln y)')^2}$$ Substitute $u=\ln y$: $$x=\frac{u}{u'}-{u'^2}$$ Multiply by $u'$: $$xu'=u-u'^3$$ This DE is of the form: $$u=xu'+f(u')$$ This is Clairaut's differential equation.