Solve the differential equation $\left(y'\right)^3+\left(3x-6\right)\cdot \:y'=3x$ This equation, which does not respect the original loosed. Lagrange Equation and Claire.
$$\left(y'\right)^3+\left(3x-6\right)\cdot \:y'=3x$$
I did not want to address here, but I had to, because I can not solve it. Even WolframAlpfa knows incremental solutions :(
 A: Apply the method in http://www.ae.illinois.edu/lndvl/Publications/2002_IJND.pdf#page=2:
Let $F(x,y,t)=t^3+(3x-6)t-3x$ ,
Then $\dfrac{dx}{dt}=-\dfrac{\dfrac{\partial F}{\partial t}}{\dfrac{\partial F}{\partial x}+t\dfrac{\partial F}{\partial y}}=-\dfrac{3t^2+3x-6}{3t-3}=-\dfrac{x}{t-1}-\dfrac{t^2-2}{t-1}$
$\dfrac{dx}{dt}+\dfrac{x}{t-1}=-\dfrac{t^2-2}{t-1}$
I.F.$=e^{\int\frac{1}{t-1}dt}=e^{\ln(t-1)}=t-1$
$\therefore\dfrac{d}{dt}((t-1)x)=2-t^2$
$(t-1)x=\int(2-t^2)~dt$
$(t-1)x=2t-\dfrac{t^3}{3}+C_1$
$x=\dfrac{t(6-t^2)}{3(t-1)}+\dfrac{C_1}{t-1}$
$\therefore\dfrac{dy}{dt}=tx=\dfrac{t^2(6-t^2)}{3(t-1)}+\dfrac{C_1t}{t-1}$
$y=\int\left(\dfrac{t^2(6-t^2)}{3(t-1)}+\dfrac{C_1t}{t-1}\right)dt$
$y=\dfrac{(3C_1+5)\ln(t-1)}{3}+\dfrac{(3C_1+8)(t-1)}{3}-\dfrac{4(t-1)^3}{9}-\dfrac{(t-1)^4}{12}+C_2$
Hence $\begin{cases}x=\dfrac{t(6-t^2)}{3(t-1)}+\dfrac{C_1}{t-1}\\y=\dfrac{(3C_1+5)\ln(t-1)}{3}+\dfrac{(3C_1+8)(t-1)}{3}-\dfrac{4(t-1)^3}{9}-\dfrac{(t-1)^4}{12}+C_2\end{cases}$
A: Maple gives the following result (one of three solutions)
$$y(x)=\frac{1}{2}\int\frac{(12x+4\sqrt{4x^3-15x^2+48x-32})^{2/3}-4x+8}{(12x+4\sqrt{4x^3-15x^2+48x-32})^{1/3}}dx+c$$
EDIT:
You can substitute $z=y'$. Then your problem is just a polynomial of third degree. Find its solutions by Cardanos formula. Then integrate the resulting functions. Actually this problem is just a problem of integration.
