Partial Differential Equation (Solution needed, by usage of Symmetries)

Partial Differential Equation
Method to be used: Symmetries
Equation: $U_t*U_{x}^{2}-U_{xx}=0=F $        is         $U_t*U_{x}^{2}=U_{xx} $

Has an infinite dimensional algebra of infinitesimal point symmetries. Let M ⊂ R and let F : M × R 1+p → R define the differential equation.

Standard formulation for the second prolongation:
$Pr^{(2)} X=ξ \frac{∂}{∂x} +τ \frac{∂}{∂t}+η \frac{∂}{∂u}+η_{x} \frac{∂}{∂ux}+η_{t} \frac{∂}{∂ut} +η_{xx} \frac{∂}{∂uxx} $

First the ETA_x, ETA_t and ETA_xx must be defined:
$η_{t}=D_{t}η-u_{x}D_{t}ξ- u_{t}D_{t}τ $
$η_{x}=D_{x}η-u_{x}D_{x}ξ- u_{t}D_{x}τ $
$η_{xx}=D_{x}η-u_{xx}D_{x}ξ- u_{xt}D_{x}τ $

Adding these definitions to the second prolongation gives:

$Pr^{(2)} X=0+0+0+η_{x}2u_{t}u_{x} +η_{t}u_{x}^{2}  -η_{xx} $

When worked out and substitution of $U_t*U_{x}^{2}=U_{xx} $ in the above mentioned equation after implementing the different forms of ETA give the second prolongation.

Based on the second prolongation and the there out following sub-equations I determined the definition of ETA, XI and TAU:
$ξ(x,u,t)=(\frac{1}{4}C_1 u^2-\frac{1}{2} D_1 u-\frac{1}{2} C_1 t+H_1 )x+F(u,t)$
$η(u,t)=(C_1 t+C_2 )u+D_1 t+D_2$
$τ(t)=C_1 t^2+2C_2 t+C_3$

Substituting these definitions into the "standard" formula for a vector field X results in:

$X=ξ \frac{∂}{∂x} +τ \frac{∂}{∂t}+η \frac{∂}{∂u}$
$X=((-\frac{1}{4} C_1 u^2-\frac{1}{2} D_1 u-\frac{1}{2} C_1 t+H_1 )x+F(u,t)) \frac{∂}{∂x} +(C_1 t^2+2C_2 t+C_3 ) \frac{∂}{∂t}+((C_1 t+C_2 )u+D_1 t+D_2 ) \frac{∂}{∂u}$

This results in the following infinitesimal symmetry generators:
$X_1=((-\frac{1}{4} u^2-\frac{1}{2} t)x) \frac{∂}{∂x} +t^2 \frac{∂}{∂t}+ut \frac{∂}{∂u} $             Where C_1 = 1
$X_2=2t \frac{∂}{∂t} +u \frac{∂}{∂u} $                                                                      Where C_2 = 1
$X_3=\frac{∂}{∂t} $                                                                                             Where C_3 = 1
$X_4=-\frac{1}{2}ux\frac{∂}{∂x} $                                                                            Where D_1 = 1
$X_5=\frac{∂}{∂u} $                                                                                            Where D_2 = 1
$X_6=x\frac{∂}{∂x} $                                                                                         Where H_1 = 1

Up till this point it has been checked and concluded to be correct. What to do? Using these symmetry generators I have to find solution(s) to the on the top stated partial differential equation. The goal is to find for every generator (where possible) the solution that this symmetry is giving to the PDE.  I do not know how to do this, internet has not yet been helpful to me