First order differential equation unresolved over derivative First order differential equation unresolved over derivative
I really dont know what to do whit this equation, i have tried to integrate it, but i can do it.
$$y´^3-yy´^2-x^2y´+x^2y=0$$
 A: $$y´^3-yy´^2-x^2y´+x^2y=0$$
Factoring,
$$y'^2(y'-y)-x^2(y'-y)=0$$
$$(y'-y)(y'^2-x^2)=0$$
$$(y'-y)(y'+x)(y'-x)=0$$
This will give you three solutions.
A: One thing which can be done with the equation
$y´^3-yy´^2-x^2y´+x^2y=0 \tag{1}$
is to observe that
$y´^3-yy´^2-x^2y´+x^2y = y'^2(y' - y) -x^2(y' - y) = (y'^2 - x^2)(y' -y)$
$= (y' + x)(y' - x)(y' - y), \tag{2}$
so that (1) becomes
$(y' + x)(y' - x)(y' - y) = 0. \tag{3}$
It now appears that (1) is really three ODEs in one, as it were, since a solution to
$y' + x = 0, \tag{4}$
or
$y' - x = 0 \tag{5}$
or
$y' - y = 0 \tag{6}$
will satisfy (1).  Taking $y(0) = y_0$ as the initial condition (for example; there are other possibilities $y_0 = y(x_0)$ for $x_0 \in \Bbb R$), we see that the solutions to (4)-(6) are
$y = -\dfrac{1}{2}x^2 + y_0, \tag{7}$
$y = \dfrac{1}{2}x^2 + y_0, \tag{8}$
and
$y = y_0e^x, \tag{9}$
respectively, and exactly one of these holds.  It seems to me that establishing this point rigorously  will involve invoking smoothness and continuation arguments, but I'm leaving such an in-depth study to my readers.  From an algebraic point of view, we are choosing a branch of solutions to the cubic equation in $y'$ (1); there seems to be an interesting if elementary interplay between ideas from algebraic geometry and analysis taking place here, but I haven't the time right now to be more thorough; I merely present intuitions I have attained by groping in the dark in the hope that someone can take things further.  
But anyway, thare's (at least) one thing which can done with equation (1).
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
