first order and higher degree differential equation Can someone please solve this differential equation with detailed solution
$$y= 2x\frac{dy}{dx} + y^3\bigg(\frac{dy}{dx}\bigg)^3$$
Thanks in advance
 A: Hint: This ODE does have a Lie point symmetry. The infinitesimal generator is given by
$$X=\dfrac{5}{3}x\partial_x+y\partial y$$ 
Use the method of canonical coordinate 
$$Xr(x,y)=0 \qquad Xs(x,y)=1$$
to obtain the substitution (we choose the most simple solution)
$$y(r,s)=r\exp(s)$$
$$x(r,s)=\exp\left[\dfrac{5}{3}s\right].$$
Then determine the derivative in the new coordinates:
$$\dfrac{dy(r,s)}{dx(r,s)}=\dfrac{\dfrac{\partial y}{\partial r}dr+\dfrac{\partial y}{\partial s}ds}{\dfrac{\partial x}{\partial r}dr+\dfrac{\partial x}{\partial s}ds}=\dfrac{\exp(s)+r\exp(s)s'}{\dfrac{5}{3}\exp(\dfrac{5}{3}s)s'}$$
Substitute the same expression on the right-hand side of the ODE and you will obtain an ODE $s'(r)=g(r)$, which boils down to a "simple" integration. I wrote "simple" because it might happen that you cannot find an explicit solution for the integral.
Maple returns after simplification:
$$27\, \left( {\frac {\rm d}{{\rm d}r}}s \left( r \right)  \right) ^{3}{
r}^{6}+81\, \left( {\frac {\rm d}{{\rm d}r}}s \left( r \right) 
 \right) ^{2}{r}^{5}+81\, \left( {\frac {\rm d}{{\rm d}r}}s \left( r
 \right)  \right) {r}^{4}+25\,r \left( {\frac {\rm d}{{\rm d}r}}s
 \left( r \right)  \right) ^{3}+27\,{r}^{3}+150\, \left( {\frac 
{\rm d}{{\rm d}r}}s \left( r \right)  \right) ^{2}=0
$$
This is a polynomial in $ds/dr$ of degree 3. Solve the cubic equation (by Cardano's formula) to obtain three first order separable ODEs for each solution of the form $s'(r)=g(r)$.
