Internal problem ID [9134]
Internal file name [OUTPUT/8069_Monday_June_06_2022_01_39_00_AM_13241258/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 800.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "abelFirstKind", "first_order_ode_lie_symmetry_calculated"
Maple gives the following as the ode type
[[_homogeneous, `class C`], _rational, _Abel]
\[ \boxed {y^{\prime }-\frac {-b^{3}+6 b^{2} x -12 b \,x^{2}+8 x^{3}-4 y^{2} b +8 x y^{2}+8 y^{3}}{\left (2 x -b \right )^{3}}=0} \]
Writing the ode as \begin {align*} y^{\prime }&=-\frac {-b^{3}+6 b^{2} x -12 b \,x^{2}-4 y^{2} b +8 x^{3}+8 y^{2} x +8 y^{3}}{b^{3}-6 b^{2} x +12 b \,x^{2}-8 x^{3}}\\ y^{\prime }&= \omega \left ( x,y\right ) \end {align*}
The condition of Lie symmetry is the linearized PDE given by \begin {align*} \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0\tag {A} \end {align*}
The type of this ode is not in the lookup table. To determine \(\xi ,\eta \) then (A) is solved using ansatz. Making bivariate polynomials of degree 1 to use as anstaz gives \begin{align*} \tag{1E} \xi &= x a_{2}+y a_{3}+a_{1} \\ \tag{2E} \eta &= x b_{2}+y b_{3}+b_{1} \\ \end{align*} Where the unknown coefficients are \[ \{a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}\} \] Substituting equations (1E,2E) and \(\omega \) into (A) gives \begin{equation} \tag{5E} b_{2}-\frac {\left (-b^{3}+6 b^{2} x -12 b \,x^{2}-4 y^{2} b +8 x^{3}+8 y^{2} x +8 y^{3}\right ) \left (b_{3}-a_{2}\right )}{b^{3}-6 b^{2} x +12 b \,x^{2}-8 x^{3}}-\frac {\left (-b^{3}+6 b^{2} x -12 b \,x^{2}-4 y^{2} b +8 x^{3}+8 y^{2} x +8 y^{3}\right )^{2} a_{3}}{\left (b^{3}-6 b^{2} x +12 b \,x^{2}-8 x^{3}\right )^{2}}-\left (-\frac {6 b^{2}-24 x b +24 x^{2}+8 y^{2}}{b^{3}-6 b^{2} x +12 b \,x^{2}-8 x^{3}}+\frac {\left (-b^{3}+6 b^{2} x -12 b \,x^{2}-4 y^{2} b +8 x^{3}+8 y^{2} x +8 y^{3}\right ) \left (-6 b^{2}+24 x b -24 x^{2}\right )}{\left (b^{3}-6 b^{2} x +12 b \,x^{2}-8 x^{3}\right )^{2}}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )+\frac {\left (-8 b y +16 y x +24 y^{2}\right ) \left (x b_{2}+y b_{3}+b_{1}\right )}{b^{3}-6 b^{2} x +12 b \,x^{2}-8 x^{3}} = 0 \end{equation} Putting the above in normal form gives \[ -\frac {b^{6} a_{2}+b^{6} a_{3}-b^{6} b_{2}-b^{6} b_{3}-12 b^{5} x a_{2}-12 b^{5} x a_{3}+12 b^{5} x b_{2}+12 b^{5} x b_{3}+60 b^{4} x^{2} a_{2}+60 b^{4} x^{2} a_{3}-60 b^{4} x^{2} b_{2}-60 b^{4} x^{2} b_{3}+8 b^{4} x y b_{2}+4 b^{4} y^{2} a_{2}+8 b^{4} y^{2} a_{3}+4 b^{4} y^{2} b_{3}-160 b^{3} x^{3} a_{2}-160 b^{3} x^{3} a_{3}+160 b^{3} x^{3} b_{2}+160 b^{3} x^{3} b_{3}-64 b^{3} x^{2} y b_{2}-16 b^{3} x \,y^{2} a_{2}-64 b^{3} x \,y^{2} a_{3}-24 b^{3} x \,y^{2} b_{2}-32 b^{3} x \,y^{2} b_{3}-8 b^{3} y^{3} a_{2}-16 b^{3} y^{3} b_{3}+240 b^{2} x^{4} a_{2}+240 b^{2} x^{4} a_{3}-240 b^{2} x^{4} b_{2}-240 b^{2} x^{4} b_{3}+192 b^{2} x^{3} y b_{2}+192 b^{2} x^{2} y^{2} a_{3}+144 b^{2} x^{2} y^{2} b_{2}+96 b^{2} x^{2} y^{2} b_{3}+96 b^{2} x \,y^{3} b_{3}-32 b^{2} y^{4} a_{3}-192 b \,x^{5} a_{2}-192 b \,x^{5} a_{3}+192 b \,x^{5} b_{2}+192 b \,x^{5} b_{3}-256 b \,x^{4} y b_{2}+64 b \,x^{3} y^{2} a_{2}-256 b \,x^{3} y^{2} a_{3}-288 b \,x^{3} y^{2} b_{2}-128 b \,x^{3} y^{2} b_{3}+96 b \,x^{2} y^{3} a_{2}-192 b \,x^{2} y^{3} b_{3}+128 b x \,y^{4} a_{3}-64 b \,y^{5} a_{3}+64 x^{6} a_{2}+64 x^{6} a_{3}-64 x^{6} b_{2}-64 x^{6} b_{3}+128 x^{5} y b_{2}-64 x^{4} y^{2} a_{2}+128 x^{4} y^{2} a_{3}+192 x^{4} y^{2} b_{2}+64 x^{4} y^{2} b_{3}-128 x^{3} y^{3} a_{2}+128 x^{3} y^{3} b_{3}-128 x^{2} y^{4} a_{3}+128 x \,y^{5} a_{3}+64 y^{6} a_{3}+8 b^{4} y b_{1}-64 b^{3} x y b_{1}+16 b^{3} y^{2} a_{1}-24 b^{3} y^{2} b_{1}+192 b^{2} x^{2} y b_{1}-96 b^{2} x \,y^{2} a_{1}+144 b^{2} x \,y^{2} b_{1}-48 b^{2} y^{3} a_{1}-256 b \,x^{3} y b_{1}+192 b \,x^{2} y^{2} a_{1}-288 b \,x^{2} y^{2} b_{1}+192 b x \,y^{3} a_{1}+128 x^{4} y b_{1}-128 x^{3} y^{2} a_{1}+192 x^{3} y^{2} b_{1}-192 x^{2} y^{3} a_{1}}{\left (b^{3}-6 b^{2} x +12 b \,x^{2}-8 x^{3}\right )^{2}} = 0 \] Setting the numerator to zero gives \begin{equation} \tag{6E} -b^{6} a_{2}-b^{6} a_{3}+b^{6} b_{2}+b^{6} b_{3}+12 b^{5} x a_{2}+12 b^{5} x a_{3}-12 b^{5} x b_{2}-12 b^{5} x b_{3}-60 b^{4} x^{2} a_{2}-60 b^{4} x^{2} a_{3}+60 b^{4} x^{2} b_{2}+60 b^{4} x^{2} b_{3}-8 b^{4} x y b_{2}-4 b^{4} y^{2} a_{2}-8 b^{4} y^{2} a_{3}-4 b^{4} y^{2} b_{3}+160 b^{3} x^{3} a_{2}+160 b^{3} x^{3} a_{3}-160 b^{3} x^{3} b_{2}-160 b^{3} x^{3} b_{3}+64 b^{3} x^{2} y b_{2}+16 b^{3} x \,y^{2} a_{2}+64 b^{3} x \,y^{2} a_{3}+24 b^{3} x \,y^{2} b_{2}+32 b^{3} x \,y^{2} b_{3}+8 b^{3} y^{3} a_{2}+16 b^{3} y^{3} b_{3}-240 b^{2} x^{4} a_{2}-240 b^{2} x^{4} a_{3}+240 b^{2} x^{4} b_{2}+240 b^{2} x^{4} b_{3}-192 b^{2} x^{3} y b_{2}-192 b^{2} x^{2} y^{2} a_{3}-144 b^{2} x^{2} y^{2} b_{2}-96 b^{2} x^{2} y^{2} b_{3}-96 b^{2} x \,y^{3} b_{3}+32 b^{2} y^{4} a_{3}+192 b \,x^{5} a_{2}+192 b \,x^{5} a_{3}-192 b \,x^{5} b_{2}-192 b \,x^{5} b_{3}+256 b \,x^{4} y b_{2}-64 b \,x^{3} y^{2} a_{2}+256 b \,x^{3} y^{2} a_{3}+288 b \,x^{3} y^{2} b_{2}+128 b \,x^{3} y^{2} b_{3}-96 b \,x^{2} y^{3} a_{2}+192 b \,x^{2} y^{3} b_{3}-128 b x \,y^{4} a_{3}+64 b \,y^{5} a_{3}-64 x^{6} a_{2}-64 x^{6} a_{3}+64 x^{6} b_{2}+64 x^{6} b_{3}-128 x^{5} y b_{2}+64 x^{4} y^{2} a_{2}-128 x^{4} y^{2} a_{3}-192 x^{4} y^{2} b_{2}-64 x^{4} y^{2} b_{3}+128 x^{3} y^{3} a_{2}-128 x^{3} y^{3} b_{3}+128 x^{2} y^{4} a_{3}-128 x \,y^{5} a_{3}-64 y^{6} a_{3}-8 b^{4} y b_{1}+64 b^{3} x y b_{1}-16 b^{3} y^{2} a_{1}+24 b^{3} y^{2} b_{1}-192 b^{2} x^{2} y b_{1}+96 b^{2} x \,y^{2} a_{1}-144 b^{2} x \,y^{2} b_{1}+48 b^{2} y^{3} a_{1}+256 b \,x^{3} y b_{1}-192 b \,x^{2} y^{2} a_{1}+288 b \,x^{2} y^{2} b_{1}-192 b x \,y^{3} a_{1}-128 x^{4} y b_{1}+128 x^{3} y^{2} a_{1}-192 x^{3} y^{2} b_{1}+192 x^{2} y^{3} a_{1} = 0 \end{equation} Looking at the above PDE shows the following are all the terms with \(\{x, y\}\) in them. \[ \{x, y\} \] The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them \[ \{x = v_{1}, y = v_{2}\} \] The above PDE (6E) now becomes \begin{equation} \tag{7E} -b^{6} a_{2}-b^{6} a_{3}+b^{6} b_{2}+b^{6} b_{3}+12 b^{5} a_{2} v_{1}+12 b^{5} a_{3} v_{1}-12 b^{5} b_{2} v_{1}-12 b^{5} b_{3} v_{1}-60 b^{4} a_{2} v_{1}^{2}-4 b^{4} a_{2} v_{2}^{2}-60 b^{4} a_{3} v_{1}^{2}-8 b^{4} a_{3} v_{2}^{2}+60 b^{4} b_{2} v_{1}^{2}-8 b^{4} b_{2} v_{1} v_{2}+60 b^{4} b_{3} v_{1}^{2}-4 b^{4} b_{3} v_{2}^{2}+160 b^{3} a_{2} v_{1}^{3}+16 b^{3} a_{2} v_{1} v_{2}^{2}+8 b^{3} a_{2} v_{2}^{3}+160 b^{3} a_{3} v_{1}^{3}+64 b^{3} a_{3} v_{1} v_{2}^{2}-160 b^{3} b_{2} v_{1}^{3}+64 b^{3} b_{2} v_{1}^{2} v_{2}+24 b^{3} b_{2} v_{1} v_{2}^{2}-160 b^{3} b_{3} v_{1}^{3}+32 b^{3} b_{3} v_{1} v_{2}^{2}+16 b^{3} b_{3} v_{2}^{3}-240 b^{2} a_{2} v_{1}^{4}-240 b^{2} a_{3} v_{1}^{4}-192 b^{2} a_{3} v_{1}^{2} v_{2}^{2}+32 b^{2} a_{3} v_{2}^{4}+240 b^{2} b_{2} v_{1}^{4}-192 b^{2} b_{2} v_{1}^{3} v_{2}-144 b^{2} b_{2} v_{1}^{2} v_{2}^{2}+240 b^{2} b_{3} v_{1}^{4}-96 b^{2} b_{3} v_{1}^{2} v_{2}^{2}-96 b^{2} b_{3} v_{1} v_{2}^{3}+192 b a_{2} v_{1}^{5}-64 b a_{2} v_{1}^{3} v_{2}^{2}-96 b a_{2} v_{1}^{2} v_{2}^{3}+192 b a_{3} v_{1}^{5}+256 b a_{3} v_{1}^{3} v_{2}^{2}-128 b a_{3} v_{1} v_{2}^{4}+64 b a_{3} v_{2}^{5}-192 b b_{2} v_{1}^{5}+256 b b_{2} v_{1}^{4} v_{2}+288 b b_{2} v_{1}^{3} v_{2}^{2}-192 b b_{3} v_{1}^{5}+128 b b_{3} v_{1}^{3} v_{2}^{2}+192 b b_{3} v_{1}^{2} v_{2}^{3}-64 a_{2} v_{1}^{6}+64 a_{2} v_{1}^{4} v_{2}^{2}+128 a_{2} v_{1}^{3} v_{2}^{3}-64 a_{3} v_{1}^{6}-128 a_{3} v_{1}^{4} v_{2}^{2}+128 a_{3} v_{1}^{2} v_{2}^{4}-128 a_{3} v_{1} v_{2}^{5}-64 a_{3} v_{2}^{6}+64 b_{2} v_{1}^{6}-128 b_{2} v_{1}^{5} v_{2}-192 b_{2} v_{1}^{4} v_{2}^{2}+64 b_{3} v_{1}^{6}-64 b_{3} v_{1}^{4} v_{2}^{2}-128 b_{3} v_{1}^{3} v_{2}^{3}-8 b^{4} b_{1} v_{2}-16 b^{3} a_{1} v_{2}^{2}+64 b^{3} b_{1} v_{1} v_{2}+24 b^{3} b_{1} v_{2}^{2}+96 b^{2} a_{1} v_{1} v_{2}^{2}+48 b^{2} a_{1} v_{2}^{3}-192 b^{2} b_{1} v_{1}^{2} v_{2}-144 b^{2} b_{1} v_{1} v_{2}^{2}-192 b a_{1} v_{1}^{2} v_{2}^{2}-192 b a_{1} v_{1} v_{2}^{3}+256 b b_{1} v_{1}^{3} v_{2}+288 b b_{1} v_{1}^{2} v_{2}^{2}+128 a_{1} v_{1}^{3} v_{2}^{2}+192 a_{1} v_{1}^{2} v_{2}^{3}-128 b_{1} v_{1}^{4} v_{2}-192 b_{1} v_{1}^{3} v_{2}^{2} = 0 \end{equation} Collecting the above on the terms \(v_i\) introduced, and these are \[ \{v_{1}, v_{2}\} \] Equation (7E) now becomes \begin{equation} \tag{8E} -128 b a_{3} v_{1} v_{2}^{4}-b^{6} a_{2}-b^{6} a_{3}+b^{6} b_{2}+b^{6} b_{3}+32 b^{2} a_{3} v_{2}^{4}+64 b a_{3} v_{2}^{5}+128 a_{3} v_{1}^{2} v_{2}^{4}-128 a_{3} v_{1} v_{2}^{5}-128 b_{2} v_{1}^{5} v_{2}-8 b^{4} b_{1} v_{2}+\left (-64 a_{2}-64 a_{3}+64 b_{2}+64 b_{3}\right ) v_{1}^{6}+\left (192 b a_{2}+192 b a_{3}-192 b b_{2}-192 b b_{3}\right ) v_{1}^{5}+\left (-240 b^{2} a_{2}-240 b^{2} a_{3}+240 b^{2} b_{2}+240 b^{2} b_{3}\right ) v_{1}^{4}+\left (160 b^{3} a_{2}+160 b^{3} a_{3}-160 b^{3} b_{2}-160 b^{3} b_{3}\right ) v_{1}^{3}-64 a_{3} v_{2}^{6}+\left (-60 b^{4} a_{2}-60 b^{4} a_{3}+60 b^{4} b_{2}+60 b^{4} b_{3}\right ) v_{1}^{2}+\left (12 b^{5} a_{2}+12 b^{5} a_{3}-12 b^{5} b_{2}-12 b^{5} b_{3}\right ) v_{1}+\left (8 b^{3} a_{2}+16 b^{3} b_{3}+48 b^{2} a_{1}\right ) v_{2}^{3}+\left (-4 b^{4} a_{2}-8 b^{4} a_{3}-4 b^{4} b_{3}-16 b^{3} a_{1}+24 b^{3} b_{1}\right ) v_{2}^{2}+\left (64 a_{2}-128 a_{3}-192 b_{2}-64 b_{3}\right ) v_{1}^{4} v_{2}^{2}+\left (256 b b_{2}-128 b_{1}\right ) v_{1}^{4} v_{2}+\left (128 a_{2}-128 b_{3}\right ) v_{1}^{3} v_{2}^{3}+\left (-64 b a_{2}+256 b a_{3}+288 b b_{2}+128 b b_{3}+128 a_{1}-192 b_{1}\right ) v_{1}^{3} v_{2}^{2}+\left (-192 b^{2} b_{2}+256 b b_{1}\right ) v_{1}^{3} v_{2}+\left (-96 b a_{2}+192 b b_{3}+192 a_{1}\right ) v_{1}^{2} v_{2}^{3}+\left (-192 b^{2} a_{3}-144 b^{2} b_{2}-96 b^{2} b_{3}-192 b a_{1}+288 b b_{1}\right ) v_{1}^{2} v_{2}^{2}+\left (64 b^{3} b_{2}-192 b^{2} b_{1}\right ) v_{1}^{2} v_{2}+\left (-96 b^{2} b_{3}-192 b a_{1}\right ) v_{1} v_{2}^{3}+\left (16 b^{3} a_{2}+64 b^{3} a_{3}+24 b^{3} b_{2}+32 b^{3} b_{3}+96 b^{2} a_{1}-144 b^{2} b_{1}\right ) v_{1} v_{2}^{2}+\left (-8 b^{4} b_{2}+64 b^{3} b_{1}\right ) v_{1} v_{2} = 0 \end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve \begin {align*} -128 a_{3}&=0\\ -64 a_{3}&=0\\ 128 a_{3}&=0\\ -128 b_{2}&=0\\ -128 b a_{3}&=0\\ 64 b a_{3}&=0\\ 32 b^{2} a_{3}&=0\\ -8 b^{4} b_{1}&=0\\ 128 a_{2}-128 b_{3}&=0\\ -64 a_{2}-64 a_{3}+64 b_{2}+64 b_{3}&=0\\ 64 a_{2}-128 a_{3}-192 b_{2}-64 b_{3}&=0\\ -96 b^{2} b_{3}-192 b a_{1}&=0\\ 256 b b_{2}-128 b_{1}&=0\\ -192 b^{2} b_{2}+256 b b_{1}&=0\\ 64 b^{3} b_{2}-192 b^{2} b_{1}&=0\\ -8 b^{4} b_{2}+64 b^{3} b_{1}&=0\\ -96 b a_{2}+192 b b_{3}+192 a_{1}&=0\\ 8 b^{3} a_{2}+16 b^{3} b_{3}+48 b^{2} a_{1}&=0\\ 192 b a_{2}+192 b a_{3}-192 b b_{2}-192 b b_{3}&=0\\ -240 b^{2} a_{2}-240 b^{2} a_{3}+240 b^{2} b_{2}+240 b^{2} b_{3}&=0\\ 160 b^{3} a_{2}+160 b^{3} a_{3}-160 b^{3} b_{2}-160 b^{3} b_{3}&=0\\ -60 b^{4} a_{2}-60 b^{4} a_{3}+60 b^{4} b_{2}+60 b^{4} b_{3}&=0\\ 12 b^{5} a_{2}+12 b^{5} a_{3}-12 b^{5} b_{2}-12 b^{5} b_{3}&=0\\ -b^{6} a_{2}-b^{6} a_{3}+b^{6} b_{2}+b^{6} b_{3}&=0\\ -4 b^{4} a_{2}-8 b^{4} a_{3}-4 b^{4} b_{3}-16 b^{3} a_{1}+24 b^{3} b_{1}&=0\\ -192 b^{2} a_{3}-144 b^{2} b_{2}-96 b^{2} b_{3}-192 b a_{1}+288 b b_{1}&=0\\ -64 b a_{2}+256 b a_{3}+288 b b_{2}+128 b b_{3}+128 a_{1}-192 b_{1}&=0\\ 16 b^{3} a_{2}+64 b^{3} a_{3}+24 b^{3} b_{2}+32 b^{3} b_{3}+96 b^{2} a_{1}-144 b^{2} b_{1}&=0 \end {align*}
Solving the above equations for the unknowns gives \begin {align*} a_{1}&=-\frac {b b_{3}}{2}\\ a_{2}&=b_{3}\\ a_{3}&=0\\ b_{1}&=0\\ b_{2}&=0\\ b_{3}&=b_{3} \end {align*}
Substituting the above solution in the anstaz (1E,2E) (using \(1\) as arbitrary value for any unknown in the RHS) gives \begin{align*} \xi &= -\frac {b}{2}+x \\ \eta &= y \\ \end{align*} The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( x,y\right ) \to \left ( R,S \right )\) where \(\left ( R,S \right )\) are the canonical coordinates which make the original ode become a quadrature and hence solved by integration.
The characteristic pde which is used to find the canonical coordinates is \begin {align*} \frac {d x}{\xi } &= \frac {d y}{\eta } = dS \tag {1} \end {align*}
The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial y}\right ) S(x,y) = 1\). Starting with the first pair of ode’s in (1) gives an ode to solve for the independent variable \(R\) in the canonical coordinates, where \(S(R)\). Therefore \begin {align*} \frac {dy}{dx} &= \frac {\eta }{\xi }\\ &= \frac {y}{-\frac {b}{2}+x}\\ &= -\frac {2 y}{-2 x +b} \end {align*}
This is easily solved to give \begin {align*} y = c_{1} \left (-2 x +b \right ) \end {align*}
Where now the coordinate \(R\) is taken as the constant of integration. Hence \begin {align*} R &= \frac {y}{-2 x +b} \end {align*}
And \(S\) is found from \begin {align*} dS &= \frac {dx}{\xi } \\ &= \frac {dx}{-\frac {b}{2}+x} \end {align*}
Integrating gives \begin {align*} S &= \int { \frac {dx}{T}}\\ &= \ln \left (2 x -b \right ) \end {align*}
Where the constant of integration is set to zero as we just need one solution. Now that \(R,S\) are found, we need to setup the ode in these coordinates. This is done by evaluating \begin {align*} \frac {dS}{dR} &= \frac { S_{x} + \omega (x,y) S_{y} }{ R_{x} + \omega (x,y) R_{y} }\tag {2} \end {align*}
Where in the above \(R_{x},R_{y},S_{x},S_{y}\) are all partial derivatives and \(\omega (x,y)\) is the right hand side of the original ode given by \begin {align*} \omega (x,y) &= -\frac {-b^{3}+6 b^{2} x -12 b \,x^{2}-4 y^{2} b +8 x^{3}+8 y^{2} x +8 y^{3}}{b^{3}-6 b^{2} x +12 b \,x^{2}-8 x^{3}} \end {align*}
Evaluating all the partial derivatives gives \begin {align*} R_{x} &= \frac {2 y}{\left (-2 x +b \right )^{2}}\\ R_{y} &= \frac {1}{-2 x +b}\\ S_{x} &= -\frac {2}{-2 x +b}\\ S_{y} &= 0 \end {align*}
Substituting all the above in (2) and simplifying gives the ode in canonical coordinates. \begin {align*} \frac {dS}{dR} &= -\frac {2 \left (-2 x +b \right )^{3}}{b^{3}+2 \left (-3 x +y \right ) b^{2}+4 \left (3 x^{2}-2 y x +y^{2}\right ) b -8 x^{3}+8 x^{2} y -8 y^{2} x -8 y^{3}}\tag {2A} \end {align*}
We now need to express the RHS as function of \(R\) only. This is done by solving for \(x,y\) in terms of \(R,S\) from the result obtained earlier and simplifying. This gives \begin {align*} \frac {dS}{dR} &= \frac {2}{8 R^{3}-4 R^{2}-2 R -1} \end {align*}
The above is a quadrature ode. This is the whole point of Lie symmetry method. It converts an ode, no matter how complicated it is, to one that can be solved by integration when the ode is in the canonical coordiates \(R,S\). Integrating the above gives \begin {align*} S \left (R \right ) = \int \frac {2}{8 R^{3}-4 R^{2}-2 R -1}d R +c_{1}\tag {4} \end {align*}
To complete the solution, we just need to transform (4) back to \(x,y\) coordinates. This results in \begin {align*} \ln \left (2 x -b \right ) = \int _{}^{\frac {y}{-2 x +b}}\frac {2}{8 \textit {\_a}^{3}-4 \textit {\_a}^{2}-2 \textit {\_a} -1}d \textit {\_a} +c_{1} \end {align*}
Which simplifies to \begin {align*} \ln \left (2 x -b \right ) = \int _{}^{\frac {y}{-2 x +b}}\frac {2}{8 \textit {\_a}^{3}-4 \textit {\_a}^{2}-2 \textit {\_a} -1}d \textit {\_a} +c_{1} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \ln \left (2 x -b \right ) &= \int _{}^{\frac {y}{-2 x +b}}\frac {2}{8 \textit {\_a}^{3}-4 \textit {\_a}^{2}-2 \textit {\_a} -1}d \textit {\_a} +c_{1} \\ \end{align*}
Verification of solutions
\[ \ln \left (2 x -b \right ) = \int _{}^{\frac {y}{-2 x +b}}\frac {2}{8 \textit {\_a}^{3}-4 \textit {\_a}^{2}-2 \textit {\_a} -1}d \textit {\_a} +c_{1} \] Verified OK.
This is Abel first kind ODE, it has the form \[ y^{\prime }= f_0(x)+f_1(x) y +f_2(x)y^{2}+f_3(x)y^{3} \] Comparing the above to given ODE which is \begin {align*} y^{\prime }&=-\frac {8 y^{3}}{b^{3}-6 b^{2} x +12 b \,x^{2}-8 x^{3}}-\frac {\left (-4 b +8 x \right ) y^{2}}{b^{3}-6 b^{2} x +12 b \,x^{2}-8 x^{3}}-\frac {-b^{3}+6 b^{2} x -12 b \,x^{2}+8 x^{3}}{b^{3}-6 b^{2} x +12 b \,x^{2}-8 x^{3}}\tag {1} \end {align*}
Therefore \begin {align*} f_0(x) &= \frac {b^{3}}{b^{3}-6 b^{2} x +12 b \,x^{2}-8 x^{3}}-\frac {6 b^{2} x}{b^{3}-6 b^{2} x +12 b \,x^{2}-8 x^{3}}+\frac {12 b \,x^{2}}{b^{3}-6 b^{2} x +12 b \,x^{2}-8 x^{3}}-\frac {8 x^{3}}{b^{3}-6 b^{2} x +12 b \,x^{2}-8 x^{3}}\\ f_1(x) &= 0\\ f_2(x) &= \frac {4 b}{b^{3}-6 b^{2} x +12 b \,x^{2}-8 x^{3}}-\frac {8 x}{b^{3}-6 b^{2} x +12 b \,x^{2}-8 x^{3}}\\ f_3(x) &= -\frac {8}{b^{3}-6 b^{2} x +12 b \,x^{2}-8 x^{3}} \end {align*}
Since \(f_2(x)=\frac {4 b}{b^{3}-6 b^{2} x +12 b \,x^{2}-8 x^{3}}-\frac {8 x}{b^{3}-6 b^{2} x +12 b \,x^{2}-8 x^{3}}\) is not zero, then the first step is to apply the following transformation to remove \(f_2\). Let \(y = u(x) - \frac {f_2}{3 f_3}\) or \begin {align*} y &= u(x) - \left ( \frac {\frac {4 b}{b^{3}-6 b^{2} x +12 b \,x^{2}-8 x^{3}}-\frac {8 x}{b^{3}-6 b^{2} x +12 b \,x^{2}-8 x^{3}}}{-\frac {24}{b^{3}-6 b^{2} x +12 b \,x^{2}-8 x^{3}}} \right ) \\ &= u \left (x \right )-\frac {x}{3}+\frac {b}{6} \end {align*}
The above transformation applied to (1) gives a new ODE as \begin {align*} u^{\prime }\left (x \right ) = -\frac {8 u \left (x \right )^{3}}{b^{3}-6 b^{2} x +12 b \,x^{2}-8 x^{3}}+\frac {2 u \left (x \right ) b^{2}}{3 \left (b^{3}-6 b^{2} x +12 b \,x^{2}-8 x^{3}\right )}-\frac {8 u \left (x \right ) b x}{3 \left (b^{3}-6 b^{2} x +12 b \,x^{2}-8 x^{3}\right )}+\frac {8 u \left (x \right ) x^{2}}{3 \left (b^{3}-6 b^{2} x +12 b \,x^{2}-8 x^{3}\right )}+\frac {38 b^{3}}{27 \left (b^{3}-6 b^{2} x +12 b \,x^{2}-8 x^{3}\right )}-\frac {76 b^{2} x}{9 \left (b^{3}-6 b^{2} x +12 b \,x^{2}-8 x^{3}\right )}+\frac {152 b \,x^{2}}{9 \left (b^{3}-6 b^{2} x +12 b \,x^{2}-8 x^{3}\right )}-\frac {304 x^{3}}{27 \left (b^{3}-6 b^{2} x +12 b \,x^{2}-8 x^{3}\right )}\tag {2} \end {align*}
The above ODE (2) can now be solved as separable.
Writing the ode as \begin {align*} u^{\prime }\left (x \right )&=-\frac {2 \left (-19 b^{3}-9 u \,b^{2}+114 b^{2} x +36 u b x -228 b \,x^{2}+108 u^{3}-36 u \,x^{2}+152 x^{3}\right )}{27 \left (b^{3}-6 b^{2} x +12 b \,x^{2}-8 x^{3}\right )}\\ u^{\prime }\left (x \right )&= \omega \left ( x,u\right ) \end {align*}
The condition of Lie symmetry is the linearized PDE given by \begin {align*} \eta _{x}+\omega \left ( \eta _{u}-\xi _{x}\right ) -\omega ^{2}\xi _{u}-\omega _{x}\xi -\omega _{u}\eta =0\tag {A} \end {align*}
The type of this ode is not in the lookup table. To determine \(\xi ,\eta \) then (A) is solved using ansatz. Making bivariate polynomials of degree 1 to use as anstaz gives \begin{align*} \tag{1E} \xi &= u a_{3}+x a_{2}+a_{1} \\ \tag{2E} \eta &= u b_{3}+x b_{2}+b_{1} \\ \end{align*} Where the unknown coefficients are \[ \{a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}\} \] Substituting equations (1E,2E) and \(\omega \) into (A) gives \begin{equation} \tag{5E} b_{2}-\frac {2 \left (-19 b^{3}-9 u \,b^{2}+114 b^{2} x +36 u b x -228 b \,x^{2}+108 u^{3}-36 u \,x^{2}+152 x^{3}\right ) \left (b_{3}-a_{2}\right )}{27 \left (b^{3}-6 b^{2} x +12 b \,x^{2}-8 x^{3}\right )}-\frac {4 \left (-19 b^{3}-9 u \,b^{2}+114 b^{2} x +36 u b x -228 b \,x^{2}+108 u^{3}-36 u \,x^{2}+152 x^{3}\right )^{2} a_{3}}{729 \left (b^{3}-6 b^{2} x +12 b \,x^{2}-8 x^{3}\right )^{2}}-\left (-\frac {2 \left (114 b^{2}+36 b u -456 x b -72 x u +456 x^{2}\right )}{27 \left (b^{3}-6 b^{2} x +12 b \,x^{2}-8 x^{3}\right )}+\frac {2 \left (-19 b^{3}-9 u \,b^{2}+114 b^{2} x +36 u b x -228 b \,x^{2}+108 u^{3}-36 u \,x^{2}+152 x^{3}\right ) \left (-6 b^{2}+24 x b -24 x^{2}\right )}{27 \left (b^{3}-6 b^{2} x +12 b \,x^{2}-8 x^{3}\right )^{2}}\right ) \left (u a_{3}+x a_{2}+a_{1}\right )+\frac {2 \left (-9 b^{2}+36 x b +324 u^{2}-36 x^{2}\right ) \left (u b_{3}+x b_{2}+b_{1}\right )}{27 \left (b^{3}-6 b^{2} x +12 b \,x^{2}-8 x^{3}\right )} = 0 \end{equation} Putting the above in normal form gives \[ -\frac {1026 b^{6} a_{2}+1444 b^{6} a_{3}-729 b^{6} b_{2}-1026 b^{6} b_{3}+486 b^{5} u a_{2}+1368 b^{5} u a_{3}-12312 b^{5} x a_{2}-17328 b^{5} x a_{3}+9234 b^{5} x b_{2}+12312 b^{5} x b_{3}+1296 b^{4} u^{2} a_{3}-3888 b^{4} u x a_{2}-13680 b^{4} u x a_{3}+61560 b^{4} x^{2} a_{2}+86640 b^{4} x^{2} a_{3}-48600 b^{4} x^{2} b_{2}-61560 b^{4} x^{2} b_{3}-5832 b^{3} u^{3} a_{2}-16416 b^{3} u^{3} a_{3}-11664 b^{3} u^{3} b_{3}-10368 b^{3} u^{2} x a_{3}-17496 b^{3} u^{2} x b_{2}+11664 b^{3} u \,x^{2} a_{2}+54720 b^{3} u \,x^{2} a_{3}-164160 b^{3} x^{3} a_{2}-231040 b^{3} x^{3} a_{3}+136080 b^{3} x^{3} b_{2}+164160 b^{3} x^{3} b_{3}-42768 b^{2} u^{4} a_{3}+98496 b^{2} u^{3} x a_{3}+69984 b^{2} u^{3} x b_{3}+31104 b^{2} u^{2} x^{2} a_{3}+104976 b^{2} u^{2} x^{2} b_{2}-15552 b^{2} u \,x^{3} a_{2}-109440 b^{2} u \,x^{3} a_{3}+246240 b^{2} x^{4} a_{2}+346560 b^{2} x^{4} a_{3}-213840 b^{2} x^{4} b_{2}-246240 b^{2} x^{4} b_{3}+171072 b \,u^{4} x a_{3}+69984 b \,u^{3} x^{2} a_{2}-196992 b \,u^{3} x^{2} a_{3}-139968 b \,u^{3} x^{2} b_{3}-41472 b \,u^{2} x^{3} a_{3}-209952 b \,u^{2} x^{3} b_{2}+7776 b u \,x^{4} a_{2}+109440 b u \,x^{4} a_{3}-196992 b \,x^{5} a_{2}-277248 b \,x^{5} a_{3}+178848 b \,x^{5} b_{2}+196992 b \,x^{5} b_{3}+46656 u^{6} a_{3}-171072 u^{4} x^{2} a_{3}-93312 u^{3} x^{3} a_{2}+131328 u^{3} x^{3} a_{3}+93312 u^{3} x^{3} b_{3}+20736 u^{2} x^{4} a_{3}+139968 u^{2} x^{4} b_{2}-43776 u \,x^{5} a_{3}+65664 x^{6} a_{2}+92416 x^{6} a_{3}-62208 x^{6} b_{2}-65664 x^{6} b_{3}+486 b^{5} b_{1}+972 b^{4} u a_{1}-4860 b^{4} x b_{1}-17496 b^{3} u^{2} b_{1}-7776 b^{3} u x a_{1}+19440 b^{3} x^{2} b_{1}-34992 b^{2} u^{3} a_{1}+104976 b^{2} u^{2} x b_{1}+23328 b^{2} u \,x^{2} a_{1}-38880 b^{2} x^{3} b_{1}+139968 b \,u^{3} x a_{1}-209952 b \,u^{2} x^{2} b_{1}-31104 b u \,x^{3} a_{1}+38880 b \,x^{4} b_{1}-139968 u^{3} x^{2} a_{1}+139968 u^{2} x^{3} b_{1}+15552 u \,x^{4} a_{1}-15552 x^{5} b_{1}}{729 \left (b^{3}-6 b^{2} x +12 b \,x^{2}-8 x^{3}\right )^{2}} = 0 \] Setting the numerator to zero gives \begin{equation} \tag{6E} -1026 b^{6} a_{2}-1444 b^{6} a_{3}+729 b^{6} b_{2}+1026 b^{6} b_{3}-486 b^{5} u a_{2}-1368 b^{5} u a_{3}+12312 b^{5} x a_{2}+17328 b^{5} x a_{3}-9234 b^{5} x b_{2}-12312 b^{5} x b_{3}-1296 b^{4} u^{2} a_{3}+3888 b^{4} u x a_{2}+13680 b^{4} u x a_{3}-61560 b^{4} x^{2} a_{2}-86640 b^{4} x^{2} a_{3}+48600 b^{4} x^{2} b_{2}+61560 b^{4} x^{2} b_{3}+5832 b^{3} u^{3} a_{2}+16416 b^{3} u^{3} a_{3}+11664 b^{3} u^{3} b_{3}+10368 b^{3} u^{2} x a_{3}+17496 b^{3} u^{2} x b_{2}-11664 b^{3} u \,x^{2} a_{2}-54720 b^{3} u \,x^{2} a_{3}+164160 b^{3} x^{3} a_{2}+231040 b^{3} x^{3} a_{3}-136080 b^{3} x^{3} b_{2}-164160 b^{3} x^{3} b_{3}+42768 b^{2} u^{4} a_{3}-98496 b^{2} u^{3} x a_{3}-69984 b^{2} u^{3} x b_{3}-31104 b^{2} u^{2} x^{2} a_{3}-104976 b^{2} u^{2} x^{2} b_{2}+15552 b^{2} u \,x^{3} a_{2}+109440 b^{2} u \,x^{3} a_{3}-246240 b^{2} x^{4} a_{2}-346560 b^{2} x^{4} a_{3}+213840 b^{2} x^{4} b_{2}+246240 b^{2} x^{4} b_{3}-171072 b \,u^{4} x a_{3}-69984 b \,u^{3} x^{2} a_{2}+196992 b \,u^{3} x^{2} a_{3}+139968 b \,u^{3} x^{2} b_{3}+41472 b \,u^{2} x^{3} a_{3}+209952 b \,u^{2} x^{3} b_{2}-7776 b u \,x^{4} a_{2}-109440 b u \,x^{4} a_{3}+196992 b \,x^{5} a_{2}+277248 b \,x^{5} a_{3}-178848 b \,x^{5} b_{2}-196992 b \,x^{5} b_{3}-46656 u^{6} a_{3}+171072 u^{4} x^{2} a_{3}+93312 u^{3} x^{3} a_{2}-131328 u^{3} x^{3} a_{3}-93312 u^{3} x^{3} b_{3}-20736 u^{2} x^{4} a_{3}-139968 u^{2} x^{4} b_{2}+43776 u \,x^{5} a_{3}-65664 x^{6} a_{2}-92416 x^{6} a_{3}+62208 x^{6} b_{2}+65664 x^{6} b_{3}-486 b^{5} b_{1}-972 b^{4} u a_{1}+4860 b^{4} x b_{1}+17496 b^{3} u^{2} b_{1}+7776 b^{3} u x a_{1}-19440 b^{3} x^{2} b_{1}+34992 b^{2} u^{3} a_{1}-104976 b^{2} u^{2} x b_{1}-23328 b^{2} u \,x^{2} a_{1}+38880 b^{2} x^{3} b_{1}-139968 b \,u^{3} x a_{1}+209952 b \,u^{2} x^{2} b_{1}+31104 b u \,x^{3} a_{1}-38880 b \,x^{4} b_{1}+139968 u^{3} x^{2} a_{1}-139968 u^{2} x^{3} b_{1}-15552 u \,x^{4} a_{1}+15552 x^{5} b_{1} = 0 \end{equation} Looking at the above PDE shows the following are all the terms with \(\{u, x\}\) in them. \[ \{u, x\} \] The following substitution is now made to be able to collect on all terms with \(\{u, x\}\) in them \[ \{u = v_{1}, x = v_{2}\} \] The above PDE (6E) now becomes \begin{equation} \tag{7E} -1026 b^{6} a_{2}-1444 b^{6} a_{3}+729 b^{6} b_{2}+1026 b^{6} b_{3}-486 b^{5} a_{2} v_{1}+12312 b^{5} a_{2} v_{2}-1368 b^{5} a_{3} v_{1}+17328 b^{5} a_{3} v_{2}-9234 b^{5} b_{2} v_{2}-12312 b^{5} b_{3} v_{2}+3888 b^{4} a_{2} v_{1} v_{2}-61560 b^{4} a_{2} v_{2}^{2}-1296 b^{4} a_{3} v_{1}^{2}+13680 b^{4} a_{3} v_{1} v_{2}-86640 b^{4} a_{3} v_{2}^{2}+48600 b^{4} b_{2} v_{2}^{2}+61560 b^{4} b_{3} v_{2}^{2}+5832 b^{3} a_{2} v_{1}^{3}-11664 b^{3} a_{2} v_{1} v_{2}^{2}+164160 b^{3} a_{2} v_{2}^{3}+16416 b^{3} a_{3} v_{1}^{3}+10368 b^{3} a_{3} v_{1}^{2} v_{2}-54720 b^{3} a_{3} v_{1} v_{2}^{2}+231040 b^{3} a_{3} v_{2}^{3}+17496 b^{3} b_{2} v_{1}^{2} v_{2}-136080 b^{3} b_{2} v_{2}^{3}+11664 b^{3} b_{3} v_{1}^{3}-164160 b^{3} b_{3} v_{2}^{3}+15552 b^{2} a_{2} v_{1} v_{2}^{3}-246240 b^{2} a_{2} v_{2}^{4}+42768 b^{2} a_{3} v_{1}^{4}-98496 b^{2} a_{3} v_{1}^{3} v_{2}-31104 b^{2} a_{3} v_{1}^{2} v_{2}^{2}+109440 b^{2} a_{3} v_{1} v_{2}^{3}-346560 b^{2} a_{3} v_{2}^{4}-104976 b^{2} b_{2} v_{1}^{2} v_{2}^{2}+213840 b^{2} b_{2} v_{2}^{4}-69984 b^{2} b_{3} v_{1}^{3} v_{2}+246240 b^{2} b_{3} v_{2}^{4}-69984 b a_{2} v_{1}^{3} v_{2}^{2}-7776 b a_{2} v_{1} v_{2}^{4}+196992 b a_{2} v_{2}^{5}-171072 b a_{3} v_{1}^{4} v_{2}+196992 b a_{3} v_{1}^{3} v_{2}^{2}+41472 b a_{3} v_{1}^{2} v_{2}^{3}-109440 b a_{3} v_{1} v_{2}^{4}+277248 b a_{3} v_{2}^{5}+209952 b b_{2} v_{1}^{2} v_{2}^{3}-178848 b b_{2} v_{2}^{5}+139968 b b_{3} v_{1}^{3} v_{2}^{2}-196992 b b_{3} v_{2}^{5}+93312 a_{2} v_{1}^{3} v_{2}^{3}-65664 a_{2} v_{2}^{6}-46656 a_{3} v_{1}^{6}+171072 a_{3} v_{1}^{4} v_{2}^{2}-131328 a_{3} v_{1}^{3} v_{2}^{3}-20736 a_{3} v_{1}^{2} v_{2}^{4}+43776 a_{3} v_{1} v_{2}^{5}-92416 a_{3} v_{2}^{6}-139968 b_{2} v_{1}^{2} v_{2}^{4}+62208 b_{2} v_{2}^{6}-93312 b_{3} v_{1}^{3} v_{2}^{3}+65664 b_{3} v_{2}^{6}-486 b^{5} b_{1}-972 b^{4} a_{1} v_{1}+4860 b^{4} b_{1} v_{2}+7776 b^{3} a_{1} v_{1} v_{2}+17496 b^{3} b_{1} v_{1}^{2}-19440 b^{3} b_{1} v_{2}^{2}+34992 b^{2} a_{1} v_{1}^{3}-23328 b^{2} a_{1} v_{1} v_{2}^{2}-104976 b^{2} b_{1} v_{1}^{2} v_{2}+38880 b^{2} b_{1} v_{2}^{3}-139968 b a_{1} v_{1}^{3} v_{2}+31104 b a_{1} v_{1} v_{2}^{3}+209952 b b_{1} v_{1}^{2} v_{2}^{2}-38880 b b_{1} v_{2}^{4}+139968 a_{1} v_{1}^{3} v_{2}^{2}-15552 a_{1} v_{1} v_{2}^{4}-139968 b_{1} v_{1}^{2} v_{2}^{3}+15552 b_{1} v_{2}^{5} = 0 \end{equation} Collecting the above on the terms \(v_i\) introduced, and these are \[ \{v_{1}, v_{2}\} \] Equation (7E) now becomes \begin{equation} \tag{8E} \left (93312 a_{2}-131328 a_{3}-93312 b_{3}\right ) v_{1}^{3} v_{2}^{3}+\left (-69984 b a_{2}+196992 b a_{3}+139968 b b_{3}+139968 a_{1}\right ) v_{1}^{3} v_{2}^{2}+\left (-98496 b^{2} a_{3}-69984 b^{2} b_{3}-139968 b a_{1}\right ) v_{1}^{3} v_{2}+\left (-20736 a_{3}-139968 b_{2}\right ) v_{1}^{2} v_{2}^{4}+\left (41472 b a_{3}+209952 b b_{2}-139968 b_{1}\right ) v_{1}^{2} v_{2}^{3}+\left (-31104 b^{2} a_{3}-104976 b^{2} b_{2}+209952 b b_{1}\right ) v_{1}^{2} v_{2}^{2}+\left (10368 b^{3} a_{3}+17496 b^{3} b_{2}-104976 b^{2} b_{1}\right ) v_{1}^{2} v_{2}+\left (-7776 b a_{2}-109440 b a_{3}-15552 a_{1}\right ) v_{1} v_{2}^{4}+\left (15552 b^{2} a_{2}+109440 b^{2} a_{3}+31104 b a_{1}\right ) v_{1} v_{2}^{3}+\left (-11664 b^{3} a_{2}-54720 b^{3} a_{3}-23328 b^{2} a_{1}\right ) v_{1} v_{2}^{2}+\left (3888 b^{4} a_{2}+13680 b^{4} a_{3}+7776 b^{3} a_{1}\right ) v_{1} v_{2}-46656 a_{3} v_{1}^{6}-171072 b a_{3} v_{1}^{4} v_{2}-1026 b^{6} a_{2}+1026 b^{6} b_{3}-1444 b^{6} a_{3}-486 b^{5} b_{1}+729 b^{6} b_{2}+\left (5832 b^{3} a_{2}+16416 b^{3} a_{3}+11664 b^{3} b_{3}+34992 b^{2} a_{1}\right ) v_{1}^{3}+\left (-1296 b^{4} a_{3}+17496 b^{3} b_{1}\right ) v_{1}^{2}+\left (-486 b^{5} a_{2}-1368 b^{5} a_{3}-972 b^{4} a_{1}\right ) v_{1}+\left (-65664 a_{2}-92416 a_{3}+62208 b_{2}+65664 b_{3}\right ) v_{2}^{6}+\left (196992 b a_{2}+277248 b a_{3}-178848 b b_{2}-196992 b b_{3}+15552 b_{1}\right ) v_{2}^{5}+\left (-246240 b^{2} a_{2}-346560 b^{2} a_{3}+213840 b^{2} b_{2}+246240 b^{2} b_{3}-38880 b b_{1}\right ) v_{2}^{4}+\left (164160 b^{3} a_{2}+231040 b^{3} a_{3}-136080 b^{3} b_{2}-164160 b^{3} b_{3}+38880 b^{2} b_{1}\right ) v_{2}^{3}+\left (-61560 b^{4} a_{2}-86640 b^{4} a_{3}+48600 b^{4} b_{2}+61560 b^{4} b_{3}-19440 b^{3} b_{1}\right ) v_{2}^{2}+\left (12312 b^{5} a_{2}+17328 b^{5} a_{3}-9234 b^{5} b_{2}-12312 b^{5} b_{3}+4860 b^{4} b_{1}\right ) v_{2}+42768 b^{2} a_{3} v_{1}^{4}+171072 a_{3} v_{1}^{4} v_{2}^{2}+43776 a_{3} v_{1} v_{2}^{5} = 0 \end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve \begin {align*} -46656 a_{3}&=0\\ 43776 a_{3}&=0\\ 171072 a_{3}&=0\\ -171072 b a_{3}&=0\\ 42768 b^{2} a_{3}&=0\\ -20736 a_{3}-139968 b_{2}&=0\\ 93312 a_{2}-131328 a_{3}-93312 b_{3}&=0\\ -65664 a_{2}-92416 a_{3}+62208 b_{2}+65664 b_{3}&=0\\ -1296 b^{4} a_{3}+17496 b^{3} b_{1}&=0\\ -7776 b a_{2}-109440 b a_{3}-15552 a_{1}&=0\\ 15552 b^{2} a_{2}+109440 b^{2} a_{3}+31104 b a_{1}&=0\\ -11664 b^{3} a_{2}-54720 b^{3} a_{3}-23328 b^{2} a_{1}&=0\\ 3888 b^{4} a_{2}+13680 b^{4} a_{3}+7776 b^{3} a_{1}&=0\\ -486 b^{5} a_{2}-1368 b^{5} a_{3}-972 b^{4} a_{1}&=0\\ -98496 b^{2} a_{3}-69984 b^{2} b_{3}-139968 b a_{1}&=0\\ 41472 b a_{3}+209952 b b_{2}-139968 b_{1}&=0\\ -31104 b^{2} a_{3}-104976 b^{2} b_{2}+209952 b b_{1}&=0\\ 10368 b^{3} a_{3}+17496 b^{3} b_{2}-104976 b^{2} b_{1}&=0\\ -69984 b a_{2}+196992 b a_{3}+139968 b b_{3}+139968 a_{1}&=0\\ 5832 b^{3} a_{2}+16416 b^{3} a_{3}+11664 b^{3} b_{3}+34992 b^{2} a_{1}&=0\\ 196992 b a_{2}+277248 b a_{3}-178848 b b_{2}-196992 b b_{3}+15552 b_{1}&=0\\ -246240 b^{2} a_{2}-346560 b^{2} a_{3}+213840 b^{2} b_{2}+246240 b^{2} b_{3}-38880 b b_{1}&=0\\ 164160 b^{3} a_{2}+231040 b^{3} a_{3}-136080 b^{3} b_{2}-164160 b^{3} b_{3}+38880 b^{2} b_{1}&=0\\ -61560 b^{4} a_{2}-86640 b^{4} a_{3}+48600 b^{4} b_{2}+61560 b^{4} b_{3}-19440 b^{3} b_{1}&=0\\ 12312 b^{5} a_{2}+17328 b^{5} a_{3}-9234 b^{5} b_{2}-12312 b^{5} b_{3}+4860 b^{4} b_{1}&=0\\ -1026 b^{6} a_{2}-1444 b^{6} a_{3}+729 b^{6} b_{2}+1026 b^{6} b_{3}-486 b^{5} b_{1}&=0 \end {align*}
Solving the above equations for the unknowns gives \begin {align*} a_{1}&=-\frac {b b_{3}}{2}\\ a_{2}&=b_{3}\\ a_{3}&=0\\ b_{1}&=0\\ b_{2}&=0\\ b_{3}&=b_{3} \end {align*}
Substituting the above solution in the anstaz (1E,2E) (using \(1\) as arbitrary value for any unknown in the RHS) gives \begin{align*} \xi &= -\frac {b}{2}+x \\ \eta &= u \\ \end{align*} The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( x,u\right ) \to \left ( R,S \right )\) where \(\left ( R,S \right )\) are the canonical coordinates which make the original ode become a quadrature and hence solved by integration.
The characteristic pde which is used to find the canonical coordinates is \begin {align*} \frac {d x}{\xi } &= \frac {d u}{\eta } = dS \tag {1} \end {align*}
The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial u}\right ) S(x,u) = 1\). Starting with the first pair of ode’s in (1) gives an ode to solve for the independent variable \(R\) in the canonical coordinates, where \(S(R)\). Unable to determine \(R\). Terminating
Since unable to solve for \(u \left (x \right )\), will terminate solution.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime } b^{3}-6 y^{\prime } b^{2} x +12 y^{\prime } b \,x^{2}-8 y^{\prime } x^{3}+8 y^{3}-4 y^{2} b +8 x y^{2}-b^{3}+6 b^{2} x -12 b \,x^{2}+8 x^{3}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {b^{3}-6 b^{2} x +12 b \,x^{2}+4 y^{2} b -8 x^{3}-8 x y^{2}-8 y^{3}}{b^{3}-6 b^{2} x +12 b \,x^{2}-8 x^{3}} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli trying separable trying inverse linear trying homogeneous types: trying homogeneous C trying homogeneous types: trying homogeneous D <- homogeneous successful <- homogeneous successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 41
dsolve(diff(y(x),x) = (-b^3+6*b^2*x-12*b*x^2+8*x^3-4*y(x)^2*b+8*x*y(x)^2+8*y(x)^3)/(2*x-b)^3,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}\frac {1}{\textit {\_a}^{3}-\textit {\_a}^{2}-\textit {\_a} -1}d \textit {\_a} \right )+\ln \left (-2 x +b \right )+c_{1} \right ) \left (-2 x +b \right )}{2} \]
✓ Solution by Mathematica
Time used: 0.321 (sec). Leaf size: 128
DSolve[y'[x] == (-b^3 + 6*b^2*x - 12*b*x^2 + 8*x^3 - 4*b*y[x]^2 + 8*x*y[x]^2 + 8*y[x]^3)/(-b + 2*x)^3,y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [-\frac {19}{3} \text {RootSum}\left [-19 \text {$\#$1}^3+6 \sqrt [3]{38} \text {$\#$1}-19\&,\frac {\log \left (\frac {\frac {4}{(b-2 x)^2}-\frac {24 y(x)}{(b-2 x)^3}}{4 \sqrt [3]{38} \sqrt [3]{\frac {1}{(b-2 x)^6}}}-\text {$\#$1}\right )}{2 \sqrt [3]{38}-19 \text {$\#$1}^2}\&\right ]=\frac {1}{9} 38^{2/3} \left (\frac {1}{(b-2 x)^6}\right )^{2/3} (b-2 x)^4 \log (b-2 x)+c_1,y(x)\right ] \]