2.24.2 Problem 2

Internal problem ID [13566]
Book : Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 1, section 1.3. Abel Equations of the Second Kind. subsection 1.3.3-2.
Problem number : 2
Date solved : Sunday, January 18, 2026 at 08:42:40 PM
CAS classification : [_rational, [_Abel, `2nd type`, `class A`]]

Entering first order ode abel second kind solver

\begin{align*} y y^{\prime }&=\left (3 a x +b \right ) y-a^{2} x^{3}-b a \,x^{2}+c x \\ \end{align*}

2.24.2.1 Solved using first_order_ode_abel_second_kind_solved_by_converting_to_first_kind

32.688 (sec)

This is Abel second kind ODE, it has the form

\[ \left (y+g\right )y^{\prime }= f_0(x)+f_1(x) y +f_2(x)y^{2}+f_3(x)y^{3} \]

Comparing the above to given ODE which is

\begin{align*}y y^{\prime } = \left (3 a x +b \right ) y-a^{2} x^{3}-b a \,x^{2}+c x\tag {1} \end{align*}

Shows that

\begin{align*} g &= 0\\ f_0 &= -a^{2} x^{3}-b a \,x^{2}+c x\\ f_1 &= 3 a x +b\\ f_2 &= 0\\ f_3 &= 0 \end{align*}

Applying transformation

\begin{align*} y&=\frac {1}{u(x)}-g \end{align*}

Results in the new ode which is Abel first kind

\begin{align*} u^{\prime }\left (x \right ) = \left (a^{2} x^{3}+b a \,x^{2}-c x \right ) u \left (x \right )^{3}+\left (-3 a x -b \right ) u \left (x \right )^{2} \end{align*}

Which is now solved. Entering first order ode abel first kind solverThis is Abel first kind ODE, it has the form

\[ u^{\prime }\left (x \right )= f_0(x)+f_1(x) u \left (x \right ) +f_2(x)u \left (x \right )^{2}+f_3(x)u \left (x \right )^{3} \]

Comparing the above to given ODE which is

\begin{align*}u^{\prime }\left (x \right )&=\left (a^{2} x^{3}+b a \,x^{2}-c x \right ) u \left (x \right )^{3}+\left (-3 a x -b \right ) u \left (x \right )^{2}\tag {1} \end{align*}

Therefore

\begin{align*} f_0 &= 0\\ f_1 &= 0\\ f_2 &= -3 a x -b\\ f_3 &= a^{2} x^{3}+b a \,x^{2}-c x \end{align*}

Hence

\begin{align*} f'_{0} &= 0\\ f'_{3} &= 3 a^{2} x^{2}+2 x b a -c \end{align*}

Since \(f_2(x)=-3 a x -b\) is not zero, then the followingtransformation is used to remove \(f_2\). Let \(u \left (x \right ) = u(x) - \frac {f_2}{3 f_3}\) or

\begin{align*} u \left (x \right ) &= u(x) - \left ( \frac {-3 a x -b}{3 a^{2} x^{3}+3 b a \,x^{2}-3 c x} \right ) \\ &= u \left (x \right )+\frac {3 a x +b}{3 x \left (a^{2} x^{2}+x b a -c \right )} \end{align*}

The above transformation applied to (1) gives a new ODE as

\begin{align*} u^{\prime }\left (x \right ) = \frac {\left (27 a^{6} x^{9}+81 a^{5} b \,x^{8}+81 a^{4} b^{2} x^{7}-81 a^{4} c \,x^{7}+27 a^{3} b^{3} x^{6}-162 a^{3} b c \,x^{6}-81 a^{2} b^{2} c \,x^{5}+81 a^{2} c^{2} x^{5}+81 a b \,c^{2} x^{4}-27 c^{3} x^{3}\right ) u \left (x \right )^{3}}{27 x^{2} \left (a^{2} x^{2}+x b a -c \right )^{2}}+\frac {\left (-81 a^{4} x^{5}-135 a^{3} b \,x^{4}-63 a^{2} b^{2} x^{3}+81 a^{2} c \,x^{3}-9 a \,b^{3} x^{2}+54 a b c \,x^{2}+9 b^{2} c x \right ) u \left (x \right )}{27 x^{2} \left (a^{2} x^{2}+x b a -c \right )^{2}}+\frac {-2 b^{3}-9 b c}{27 x^{2} \left (a^{2} x^{2}+x b a -c \right )^{2}}\tag {2} \end{align*}

The above ODE (2) can now be solved.

Entering first order ode LIE solverWriting the ode as

\begin{align*} u^{\prime }\left (x \right )&=\frac {27 x^{9} u^{3} a^{6}+81 x^{8} u^{3} a^{5} b +81 x^{7} u^{3} a^{4} b^{2}-81 x^{7} u^{3} a^{4} c +27 x^{6} u^{3} a^{3} b^{3}-162 x^{6} u^{3} a^{3} b c -81 x^{5} u^{3} a^{2} b^{2} c +81 x^{5} u^{3} a^{2} c^{2}+81 x^{4} u^{3} a b \,c^{2}-81 x^{5} u \,a^{4}-135 x^{4} u \,a^{3} b -27 x^{3} u^{3} c^{3}-63 x^{3} u \,a^{2} b^{2}+81 x^{3} u \,a^{2} c -9 u a \,b^{3} x^{2}+54 u a b c \,x^{2}+9 u \,b^{2} c x -2 b^{3}-9 b c}{27 x^{2} \left (a^{2} x^{2}+x b a -c \right )^{2}}\\ u^{\prime }\left (x \right )&= \omega \left ( x,u \left (x \right )\right ) \end{align*}

The condition of Lie symmetry is the linearized PDE given by

\begin{align*} \eta _{x}+\omega \left ( \eta _{u \left (x \right )}-\xi _{x}\right ) -\omega ^{2}\xi _{u \left (x \right )}-\omega _{x}\xi -\omega _{u \left (x \right )}\eta =0\tag {A} \end{align*}

To determine \(\xi ,\eta \) then (A) is solved using ansatz. Making bivariate polynomials of degree 3 to use as anstaz gives

\begin{align*} \tag{1E} \xi &= u^{3} a_{10}+u^{2} x a_{9}+u \,x^{2} a_{8}+x^{3} a_{7}+u^{2} a_{6}+u x a_{5}+x^{2} a_{4}+u a_{3}+x a_{2}+a_{1} \\ \tag{2E} \eta &= u^{3} b_{10}+u^{2} x b_{9}+u \,x^{2} b_{8}+x^{3} b_{7}+u^{2} b_{6}+u x b_{5}+x^{2} b_{4}+u b_{3}+x b_{2}+b_{1} \\ \end{align*}

Where the unknown coefficients are

\[ \{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}, a_{8}, a_{9}, a_{10}, b_{1}, b_{2}, b_{3}, b_{4}, b_{5}, b_{6}, b_{7}, b_{8}, b_{9}, b_{10}\} \]

Substituting equations (1E,2E) and \(\omega \) into (A) gives

\begin{equation} \tag{5E} \text {Expression too large to display} \end{equation}

Putting the above in normal form gives

\[ \text {Expression too large to display} \]

Setting the numerator to zero gives

\begin{equation} \tag{6E} \text {Expression too large to display} \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} \text {Expression too large to display} \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} \text {Expression too large to display} \end{equation}

Setting each coefficients in (8E) to zero gives the following equations to solve

\begin{align*} 4374 a^{8} b_{7}&=0\\ 5832 a^{10} a_{9}&=0\\ -2187 a^{10} b_{7}&=0\\ -729 a^{12} a_{8}&=0\\ -1458 a^{12} a_{9}&=0\\ -2187 a^{12} a_{10}&=0\\ -1458 c^{6} a_{6}&=0\\ -2187 c^{6} a_{10}&=0\\ -13122 a^{11} b a_{10}&=0\\ 13122 a b \,c^{5} a_{10}&=0\\ 4374 a^{8} a_{7}+1458 a^{8} b_{8}&=0\\ -4374 a^{10} a_{7}-1458 a^{10} b_{8}&=0\\ 729 a^{10} a_{8}-729 a^{10} b_{9}&=0\\ 1458 b^{2} c^{4} a_{10}+729 c^{5} a_{10}&=0\\ -324 b^{3} c^{3} a_{10}-1458 b \,c^{4} a_{10}&=0\\ -243 b^{4} c^{2} a_{10}+243 b^{2} c^{3} a_{10}&=0\\ -5832 a \,b^{3} c^{3} a_{10}+4374 a b \,c^{4} a_{10}&=0\\ -32805 a^{2} b^{2} c^{4} a_{10}+13122 a^{2} c^{5} a_{10}&=0\\ 43740 a^{3} b^{3} c^{3} a_{10}-65610 a^{3} b \,c^{4} a_{10}&=0\\ -43740 a^{9} b^{3} a_{10}+65610 a^{9} b c a_{10}&=0\\ -32805 a^{10} b^{2} a_{10}+13122 a^{10} c a_{10}&=0\\ -729 a^{12} a_{5}-4374 a^{11} b a_{8}&=0\\ 6561 a^{10} a_{6}+26973 a^{9} b a_{9}&=0\\ -1458 a^{12} a_{6}-8748 a^{11} b a_{9}&=0\\ 3645 a^{8} b_{4}+16767 a^{7} b b_{7}&=0\\ -2187 a^{10} b_{4}-10935 a^{9} b b_{7}&=0\\ 8748 a b \,c^{5} a_{6}-1458 c^{6} a_{9}&=0\\ -4 b^{6} a_{3}-36 b^{4} c a_{3}-81 b^{2} c^{2} a_{3}&=0\\ 72 b^{5} c a_{6}+216 b^{3} c^{2} a_{6}-486 b \,c^{3} a_{6}&=0\\ -8 b^{6} a_{6}-72 b^{4} c a_{6}-162 b^{2} c^{2} a_{6}&=0\\ 108 b^{5} c a_{10}+378 b^{3} c^{2} a_{10}-486 b \,c^{3} a_{10}&=0\\ -12 b^{6} a_{10}-108 b^{4} c a_{10}-243 b^{2} c^{2} a_{10}&=0\\ -32805 a^{4} b^{4} c^{2} a_{10}+131220 a^{4} b^{2} c^{3} a_{10}-32805 a^{4} c^{4} a_{10}&=0\\ 13122 a^{5} b^{5} c a_{10}-131220 a^{5} b^{3} c^{2} a_{10}+131220 a^{5} b \,c^{3} a_{10}&=0\\ -13122 a^{7} b^{5} a_{10}+131220 a^{7} b^{3} c a_{10}-131220 a^{7} b \,c^{2} a_{10}&=0\\ -32805 a^{8} b^{4} a_{10}+131220 a^{8} b^{2} c a_{10}-32805 a^{8} c^{2} a_{10}&=0\\ -21870 a^{2} b^{2} c^{4} a_{6}+8748 a^{2} c^{5} a_{6}+8748 a b \,c^{5} a_{9}&=0\\ 30618 a^{9} b a_{6}+50544 a^{8} b^{2} a_{9}-21870 a^{8} c a_{9}&=0\\ -8748 a^{11} b a_{6}-21870 a^{10} b^{2} a_{9}+8748 a^{10} c a_{9}&=0\\ -2187 a^{6} b^{6} a_{10}+65610 a^{6} b^{4} c a_{10}-196830 a^{6} b^{2} c^{2} a_{10}+43740 a^{6} c^{3} a_{10}&=0\\ 8748 a^{2} b^{4} c^{2} a_{10}-30618 a^{2} b^{2} c^{3} a_{10}+8019 a^{2} c^{4} a_{10}-729 c^{6} a_{3}&=0\\ 29160 a^{3} b^{3} c^{3} a_{6}-43740 a^{3} b \,c^{4} a_{6}-21870 a^{2} b^{2} c^{4} a_{9}+8748 a^{2} c^{5} a_{9}&=0\\ 57834 a^{8} b^{2} a_{6}+49086 a^{7} b^{3} a_{9}-25515 a^{8} c a_{6}-78732 a^{7} b c a_{9}&=0\\ -21870 a^{10} b^{2} a_{6}-29160 a^{9} b^{3} a_{9}+8748 a^{10} c a_{6}+43740 a^{9} b c a_{9}&=0\\ 972 a \,b^{4} c^{2} a_{10}+4374 a \,b^{2} c^{3} a_{10}+972 b^{2} c^{4} a_{6}+729 c^{5} a_{6}&=0\\ 486 a \,b^{5} c a_{10}-3888 a \,b^{3} c^{2} a_{10}-216 b^{3} c^{3} a_{6}-972 b \,c^{4} a_{6}&=0\\ -729 a^{12} a_{3}-4374 a^{11} b a_{5}-10935 a^{10} b^{2} a_{8}+4374 a^{10} c a_{8}&=0\\ 2916 a^{8} b_{2}+13851 a^{7} b b_{4}+24300 a^{6} b^{2} b_{7}-15309 a^{6} c b_{7}&=0\\ -2187 a^{10} b_{2}-10935 a^{9} b b_{4}-21870 a^{8} b^{2} b_{7}+10935 a^{8} c b_{7}&=0\\ 2187 a^{8} a_{4}+729 a^{8} b_{5}+16767 a^{7} b a_{7}+5832 a^{7} b b_{8}&=0\\ -3645 a^{10} a_{4}-1458 a^{10} b_{5}-21141 a^{9} b a_{7}-7290 a^{9} b b_{8}&=0\\ 1458 a^{10} a_{5}-729 a^{10} b_{6}+2916 a^{9} b a_{8}-3645 a^{9} b b_{9}&=0\\ -4 b^{6} a_{5}-36 b^{4} c a_{5}-108 b^{3} c^{2} a_{1}-81 b^{2} c^{2} a_{5}-486 b \,c^{3} a_{1}&=0\\ -21870 a^{4} b^{4} c^{2} a_{6}+87480 a^{4} b^{2} c^{3} a_{6}+29160 a^{3} b^{3} c^{3} a_{9}-21870 a^{4} c^{4} a_{6}-43740 a^{3} b \,c^{4} a_{9}&=0\\ 56376 a^{7} b^{3} a_{6}+26244 a^{6} b^{4} a_{9}-93312 a^{7} b c a_{6}-108864 a^{6} b^{2} c a_{9}+29160 a^{6} c^{2} a_{9}&=0\\ -29160 a^{9} b^{3} a_{6}-21870 a^{8} b^{4} a_{9}+43740 a^{9} b c a_{6}+87480 a^{8} b^{2} c a_{9}-21870 a^{8} c^{2} a_{9}&=0\\ -108 a \,b^{6} a_{10}+486 a \,b^{4} c a_{10}-162 b^{4} c^{2} a_{6}+4374 a \,b^{2} c^{2} a_{10}+243 b^{2} c^{3} a_{6}&=0\\ -4374 a^{11} b a_{3}-10935 a^{10} b^{2} a_{5}-14580 a^{9} b^{3} a_{8}+4374 a^{10} c a_{5}+21870 a^{9} b c a_{8}&=0\\ -5832 a^{3} b^{5} c a_{10}+58320 a^{3} b^{3} c^{2} a_{10}-64152 a^{3} b \,c^{3} a_{10}+4374 a b \,c^{5} a_{3}-729 c^{6} a_{5}&=0\\ -8 b^{6} a_{9}+36 b^{5} c a_{3}-72 b^{4} c a_{9}+54 b^{3} c^{2} a_{3}-162 b^{2} c^{2} a_{9}-486 b \,c^{3} a_{3}&=0\\ 29889 a^{6} b^{4} a_{6}+7533 a^{5} b^{5} a_{9}-130734 a^{6} b^{2} c a_{6}-72900 a^{5} b^{3} c a_{9}+36450 a^{6} c^{2} a_{6}+74358 a^{5} b \,c^{2} a_{9}&=0\\ 8748 a^{5} b^{5} c a_{6}-87480 a^{5} b^{3} c^{2} a_{6}-21870 a^{4} b^{4} c^{2} a_{9}+87480 a^{5} b \,c^{3} a_{6}+87480 a^{4} b^{2} c^{3} a_{9}-21870 a^{4} c^{4} a_{9}&=0\\ -21870 a^{8} b^{4} a_{6}-8748 a^{7} b^{5} a_{9}+87480 a^{8} b^{2} c a_{6}+87480 a^{7} b^{3} c a_{9}-21870 a^{8} c^{2} a_{6}-87480 a^{7} b \,c^{2} a_{9}&=0\\ 8748 a^{7} b a_{4}+2916 a^{7} b b_{5}+23328 a^{6} b^{2} a_{7}+8748 a^{6} b^{2} b_{8}-17496 a^{6} c a_{7}-5832 a^{6} c b_{8}&=0\\ 2187 a^{8} b_{1}+10935 a^{7} b b_{2}+19926 a^{6} b^{2} b_{4}+16038 a^{5} b^{3} b_{7}-12393 a^{6} c b_{4}-43740 a^{5} b c b_{7}&=0\\ -2187 a^{10} b_{1}-10935 a^{9} b b_{2}-21870 a^{8} b^{2} b_{4}-21870 a^{7} b^{3} b_{7}+10935 a^{8} c b_{4}+43740 a^{7} b c b_{7}&=0\\ -1458 a^{6} b^{6} a_{6}+43740 a^{6} b^{4} c a_{6}+8748 a^{5} b^{5} c a_{9}-131220 a^{6} b^{2} c^{2} a_{6}-87480 a^{5} b^{3} c^{2} a_{9}+29160 a^{6} c^{3} a_{6}+87480 a^{5} b \,c^{3} a_{9}&=0\\ -8748 a^{7} b^{5} a_{6}-1458 a^{6} b^{6} a_{9}+87480 a^{7} b^{3} c a_{6}+43740 a^{6} b^{4} c a_{9}-87480 a^{7} b \,c^{2} a_{6}-131220 a^{6} b^{2} c^{2} a_{9}+29160 a^{6} c^{3} a_{9}&=0\\ -10935 a^{10} b^{2} a_{3}-14580 a^{9} b^{3} a_{5}-10935 a^{8} b^{4} a_{8}+4374 a^{10} c a_{3}+21870 a^{9} b c a_{5}+43740 a^{8} b^{2} c a_{8}-10935 a^{8} c^{2} a_{8}&=0\\ -972 a^{2} b^{5} c a_{10}-3402 a^{2} b^{3} c^{2} a_{10}-3888 a \,b^{3} c^{3} a_{6}+4374 a^{2} b \,c^{3} a_{10}+1458 a b \,c^{4} a_{6}+972 b^{2} c^{4} a_{9}+1458 c^{5} a_{9}&=0\\ 2187 a^{10} a_{3}+6561 a^{9} b a_{5}-3645 a^{9} b b_{6}+4860 a^{8} b^{2} a_{8}-7290 a^{8} b^{2} b_{9}-729 a^{8} c a_{8}+3645 a^{8} c b_{9}&=0\\ -14580 a^{9} b^{3} a_{3}-10935 a^{8} b^{4} a_{5}-4374 a^{7} b^{5} a_{8}+21870 a^{9} b c a_{3}+43740 a^{8} b^{2} c a_{5}+43740 a^{7} b^{3} c a_{8}-10935 a^{8} c^{2} a_{5}-43740 a^{7} b \,c^{2} a_{8}&=0\\ 324 a^{3} b^{6} a_{10}-486 a^{3} b^{4} c a_{10}+5832 a^{2} b^{4} c^{2} a_{6}-8748 a^{3} b^{2} c^{2} a_{10}-17010 a^{2} b^{2} c^{3} a_{6}-3888 a \,b^{3} c^{3} a_{9}+3645 a^{2} c^{4} a_{6}-2187 a b \,c^{4} a_{9}&=0\\ 972 a^{4} b^{5} a_{10}-3888 a^{3} b^{5} c a_{6}+3402 a^{4} b^{3} c a_{10}+34992 a^{3} b^{3} c^{2} a_{6}+5832 a^{2} b^{4} c^{2} a_{9}-4374 a^{4} b \,c^{2} a_{10}-34992 a^{3} b \,c^{3} a_{6}-9720 a^{2} b^{2} c^{3} a_{9}&=0\\ 1458 a^{4} b^{6} a_{10}-45927 a^{4} b^{4} c a_{10}+153090 a^{4} b^{2} c^{2} a_{10}-10935 a^{2} b^{2} c^{4} a_{3}-39366 a^{4} c^{3} a_{10}+4374 a^{2} c^{5} a_{3}+4374 a b \,c^{5} a_{5}-729 c^{6} a_{8}&=0\\ 13122 a^{5} b^{5} a_{10}-145800 a^{5} b^{3} c a_{10}+14580 a^{3} b^{3} c^{3} a_{3}+166212 a^{5} b \,c^{2} a_{10}-21870 a^{3} b \,c^{4} a_{3}-10935 a^{2} b^{2} c^{4} a_{5}+4374 a^{2} c^{5} a_{5}+4374 a b \,c^{5} a_{8}&=0\\ 8019 a^{7} b b_{1}+15552 a^{6} b^{2} b_{2}+13122 a^{5} b^{3} b_{4}+4374 a^{4} b^{4} b_{7}-9477 a^{6} c b_{2}-34992 a^{5} b c b_{4}-42282 a^{4} b^{2} c b_{7}+19683 a^{4} c^{2} b_{7}&=0\\ -10935 a^{9} b b_{1}-21870 a^{8} b^{2} b_{2}-21870 a^{7} b^{3} b_{4}-10935 a^{6} b^{4} b_{7}+10935 a^{8} c b_{2}+43740 a^{7} b c b_{4}+65610 a^{6} b^{2} c b_{7}-21870 a^{6} c^{2} b_{7}&=0\\ 10206 a^{9} b a_{3}+12150 a^{8} b^{2} a_{5}-7290 a^{8} b^{2} b_{6}+4860 a^{7} b^{3} a_{8}-7290 a^{7} b^{3} b_{9}-4374 a^{8} c a_{5}+3645 a^{8} c b_{6}+14580 a^{7} b c b_{9}&=0\\ -2916 a^{10} a_{2}-1458 a^{10} b_{3}-17496 a^{9} b a_{4}-7290 a^{9} b b_{5}-40824 a^{8} b^{2} a_{7}-14580 a^{8} b^{2} b_{8}+20412 a^{8} c a_{7}+7290 a^{8} c b_{8}&=0\\ 972 a^{4} b^{6} a_{6}+972 a^{5} b^{4} a_{10}-28431 a^{4} b^{4} c a_{6}-3888 a^{3} b^{5} c a_{9}+4374 a^{5} b^{2} c a_{10}+88938 a^{4} b^{2} c^{2} a_{6}+27702 a^{3} b^{3} c^{2} a_{9}-21870 a^{4} c^{3} a_{6}-20412 a^{3} b \,c^{3} a_{9}&=0\\ 8262 a^{5} b^{5} a_{6}+972 a^{4} b^{6} a_{9}+324 a^{6} b^{3} a_{10}-87480 a^{5} b^{3} c a_{6}-24786 a^{4} b^{4} c a_{9}+1458 a^{6} b c a_{10}+96228 a^{5} b \,c^{2} a_{6}+67068 a^{4} b^{2} c^{2} a_{9}-14580 a^{4} c^{3} a_{9}&=0\\ 324 a \,b^{4} c a_{1}-4 b^{6} a_{8}+1458 a \,b^{2} c^{2} a_{1}-36 b^{4} c a_{8}-54 b^{3} c^{2} a_{2}-54 b^{3} c^{2} b_{3}-81 b^{2} c^{2} a_{8}-243 b \,c^{3} a_{2}-243 b \,c^{3} b_{3}&=0\\ -36 a \,b^{6} a_{3}+378 a \,b^{4} c a_{3}+36 b^{5} c a_{5}+2430 a \,b^{2} c^{2} a_{3}+108 b^{3} c^{2} a_{5}-108 b^{3} c^{2} b_{6}+243 b^{2} c^{3} a_{1}-243 b \,c^{3} a_{5}-486 b \,c^{3} b_{6}&=0\\ -972 a^{2} b^{5} a_{10}+324 a \,b^{5} c a_{6}-2970 a^{2} b^{3} c a_{10}-2916 a \,b^{3} c^{2} a_{6}-162 b^{4} c^{2} a_{9}-108 b^{3} c^{3} a_{3}+6318 a^{2} b \,c^{2} a_{10}+486 b^{2} c^{3} b_{10}-486 b \,c^{4} a_{3}&=0\\ 48843 a^{6} b^{4} a_{10}-10935 a^{4} b^{4} c^{2} a_{3}-220158 a^{6} b^{2} c a_{10}+43740 a^{4} b^{2} c^{3} a_{3}+14580 a^{3} b^{3} c^{3} a_{5}+62694 a^{6} c^{2} a_{10}-10935 a^{4} c^{4} a_{3}-21870 a^{3} b \,c^{4} a_{5}-10935 a^{2} b^{2} c^{4} a_{8}+4374 a^{2} c^{5} a_{8}&=0\\ 4374 a^{5} b^{5} c a_{3}+93312 a^{7} b^{3} a_{10}-43740 a^{5} b^{3} c^{2} a_{3}-10935 a^{4} b^{4} c^{2} a_{5}-157464 a^{7} b c a_{10}+43740 a^{5} b \,c^{3} a_{3}+43740 a^{4} b^{2} c^{3} a_{5}+14580 a^{3} b^{3} c^{3} a_{8}-10935 a^{4} c^{4} a_{5}-21870 a^{3} b \,c^{4} a_{8}&=0\\ -243 a^{2} b^{6} a_{10}+7533 a^{2} b^{4} c a_{10}+648 a \,b^{4} c^{2} a_{6}-15795 a^{2} b^{2} c^{2} a_{10}+2916 a \,b^{2} c^{3} a_{6}-216 b^{3} c^{3} a_{9}+486 b^{2} c^{4} a_{3}-2187 a^{2} c^{3} a_{10}-972 b \,c^{4} a_{9}+729 c^{5} a_{3}&=0\\ -72 a \,b^{6} a_{6}+432 a \,b^{4} c a_{6}+72 b^{5} c a_{9}-81 b^{4} c^{2} a_{3}+3402 a \,b^{2} c^{2} a_{6}+270 b^{3} c^{2} a_{9}-162 b^{3} c^{2} b_{10}+243 b^{2} c^{3} a_{3}-243 b \,c^{3} a_{9}-729 b \,c^{3} b_{10}&=0\\ 11178 a^{6} b^{2} b_{1}+10206 a^{5} b^{3} b_{2}+3645 a^{4} b^{4} b_{4}+243 a^{3} b^{5} b_{7}-6561 a^{6} c b_{1}-26244 a^{5} b c b_{2}-33534 a^{4} b^{2} c b_{4}-14580 a^{3} b^{3} c b_{7}+15309 a^{4} c^{2} b_{4}+37179 a^{3} b \,c^{2} b_{7}&=0\\ -21870 a^{8} b^{2} b_{1}-21870 a^{7} b^{3} b_{2}-10935 a^{6} b^{4} b_{4}-2187 a^{5} b^{5} b_{7}+10935 a^{8} c b_{1}+43740 a^{7} b c b_{2}+65610 a^{6} b^{2} c b_{4}+43740 a^{5} b^{3} c b_{7}-21870 a^{6} c^{2} b_{4}-65610 a^{5} b \,c^{2} b_{7}&=0\\ -216 a^{2} b^{5} a_{1}-540 a^{2} b^{3} c a_{1}+216 a \,b^{4} c a_{2}+108 a \,b^{4} c b_{3}+1944 a^{2} b \,c^{2} a_{1}+972 a \,b^{2} c^{2} a_{2}+486 a \,b^{2} c^{2} b_{3}-54 b^{3} c^{2} b_{5}-243 b^{2} c^{3} b_{1}-243 b \,c^{3} b_{5}&=0\\ -2187 a^{8} a_{1}+729 a^{7} b a_{2}+12150 a^{6} b^{2} a_{4}+4374 a^{6} b^{2} b_{5}+14094 a^{5} b^{3} a_{7}+5832 a^{5} b^{3} b_{8}-10935 a^{6} c a_{4}-2916 a^{6} c b_{5}-46656 a^{5} b c a_{7}-17496 a^{5} b c b_{8}&=0\\ -4374 a^{7} b^{5} a_{3}-729 a^{6} b^{6} a_{5}+43740 a^{7} b^{3} c a_{3}+21870 a^{6} b^{4} c a_{5}+4374 a^{5} b^{5} c a_{8}+51030 a^{9} b a_{10}-43740 a^{7} b \,c^{2} a_{3}-65610 a^{6} b^{2} c^{2} a_{5}-43740 a^{5} b^{3} c^{2} a_{8}+14580 a^{6} c^{3} a_{5}+43740 a^{5} b \,c^{3} a_{8}&=0\\ -10935 a^{8} b^{4} a_{3}-4374 a^{7} b^{5} a_{5}-729 a^{6} b^{6} a_{8}+43740 a^{8} b^{2} c a_{3}+43740 a^{7} b^{3} c a_{5}+21870 a^{6} b^{4} c a_{8}+10935 a^{10} a_{10}-10935 a^{8} c^{2} a_{3}-43740 a^{7} b \,c^{2} a_{5}-65610 a^{6} b^{2} c^{2} a_{8}+14580 a^{6} c^{3} a_{8}&=0\\ -2187 a^{10} a_{1}-13851 a^{9} b a_{2}-7290 a^{9} b b_{3}-33534 a^{8} b^{2} a_{4}-14580 a^{8} b^{2} b_{5}-39366 a^{7} b^{3} a_{7}-14580 a^{7} b^{3} b_{8}+16767 a^{8} c a_{4}+7290 a^{8} c b_{5}+78732 a^{7} b c a_{7}+29160 a^{7} b c b_{8}&=0\\ -729 a^{6} b^{6} a_{3}+21870 a^{6} b^{4} c a_{3}+4374 a^{5} b^{5} c a_{5}+96228 a^{8} b^{2} a_{10}-65610 a^{6} b^{2} c^{2} a_{3}-43740 a^{5} b^{3} c^{2} a_{5}-10935 a^{4} b^{4} c^{2} a_{8}-43011 a^{8} c a_{10}+14580 a^{6} c^{3} a_{3}+43740 a^{5} b \,c^{3} a_{5}+43740 a^{4} b^{2} c^{3} a_{8}-10935 a^{4} c^{4} a_{8}&=0\\ 19440 a^{8} b^{2} a_{3}+12150 a^{7} b^{3} a_{5}-7290 a^{7} b^{3} b_{6}+3645 a^{6} b^{4} a_{8}-3645 a^{6} b^{4} b_{9}-8019 a^{8} c a_{3}-14580 a^{7} b c a_{5}+14580 a^{7} b c b_{6}+2430 a^{6} b^{2} c a_{8}+21870 a^{6} b^{2} c b_{9}-4374 a^{6} c^{2} a_{8}-7290 a^{6} c^{2} b_{9}&=0\\ -720 a^{2} b^{5} a_{6}-72 a \,b^{6} a_{9}+162 a \,b^{5} c a_{3}-2160 a^{2} b^{3} c a_{6}+324 a \,b^{4} c a_{9}+324 a \,b^{4} c b_{10}-1944 a \,b^{3} c^{2} a_{3}-81 b^{4} c^{2} a_{5}+4860 a^{2} b \,c^{2} a_{6}+2916 a \,b^{2} c^{2} a_{9}+1458 a \,b^{2} c^{2} b_{10}+243 b^{2} c^{3} b_{6}&=0\\ 19440 a^{7} b^{3} a_{3}+7290 a^{6} b^{4} a_{5}-3645 a^{6} b^{4} b_{6}+1944 a^{5} b^{5} a_{8}-729 a^{5} b^{5} b_{9}-29160 a^{7} b c a_{3}-19440 a^{6} b^{2} c a_{5}+21870 a^{6} b^{2} c b_{6}+14580 a^{5} b^{3} c b_{9}+2916 a^{6} c^{2} a_{5}-7290 a^{6} c^{2} b_{6}-17496 a^{5} b \,c^{2} a_{8}-21870 a^{5} b \,c^{2} b_{9}&=0\\ -7290 a^{7} b a_{1}+972 a^{6} b^{2} a_{2}+6804 a^{5} b^{3} a_{4}+2916 a^{5} b^{3} b_{5}+3402 a^{4} b^{4} a_{7}+1458 a^{4} b^{4} b_{8}-4374 a^{6} c a_{2}-29160 a^{5} b c a_{4}-8748 a^{5} b c b_{5}-39852 a^{4} b^{2} c a_{7}-17496 a^{4} b^{2} c b_{8}+21870 a^{4} c^{2} a_{7}+8748 a^{4} c^{2} b_{8}&=0\\ -468 a^{2} b^{5} a_{3}-36 a \,b^{6} a_{5}-1350 a^{2} b^{3} c a_{3}+270 a \,b^{4} c a_{5}+216 a \,b^{4} c b_{6}-972 a \,b^{3} c^{2} a_{1}+36 b^{5} c a_{8}+3402 a^{2} b \,c^{2} a_{3}+1944 a \,b^{2} c^{2} a_{5}+972 a \,b^{2} c^{2} b_{6}+162 b^{3} c^{2} a_{8}-108 b^{3} c^{2} b_{9}-486 b \,c^{3} b_{9}&=0\\ -3888 a^{3} b^{5} a_{10}-648 a^{2} b^{5} c a_{6}+37908 a^{3} b^{3} c a_{10}-2268 a^{2} b^{3} c^{2} a_{6}+648 a \,b^{4} c^{2} a_{9}-1944 a \,b^{3} c^{3} a_{3}-24786 a^{3} b \,c^{2} a_{10}+2916 a^{2} b \,c^{3} a_{6}+2916 a \,b^{2} c^{3} a_{9}-1458 a b \,c^{4} a_{3}+486 b^{2} c^{4} a_{5}+1458 c^{5} a_{5}+729 c^{5} b_{6}&=0\\ -21870 a^{7} b^{3} b_{1}-10935 a^{6} b^{4} b_{2}-2187 a^{5} b^{5} b_{4}+43740 a^{7} b c b_{1}+65610 a^{6} b^{2} c b_{2}+43740 a^{5} b^{3} c b_{4}+10935 a^{4} b^{4} c b_{7}-4374 a^{8} a_{8}-1458 a^{8} b_{9}-21870 a^{6} c^{2} b_{2}-65610 a^{5} b \,c^{2} b_{4}-65610 a^{4} b^{2} c^{2} b_{7}+21870 a^{4} c^{3} b_{7}&=0\\ -10206 a^{6} b^{2} a_{1}-486 a^{5} b^{3} a_{2}+1215 a^{4} b^{4} a_{4}+729 a^{4} b^{4} b_{5}+243 a^{3} b^{5} a_{7}+2187 a^{6} c a_{1}-11664 a^{5} b c a_{2}-23814 a^{4} b^{2} c a_{4}-8748 a^{4} b^{2} c b_{5}-11664 a^{3} b^{3} c a_{7}-5832 a^{3} b^{3} c b_{8}+15309 a^{4} c^{2} a_{4}+4374 a^{4} c^{2} b_{5}+34263 a^{3} b \,c^{2} a_{7}+17496 a^{3} b \,c^{2} b_{8}&=0\\ 7290 a^{5} b^{3} b_{1}+2916 a^{4} b^{4} b_{2}+243 a^{3} b^{5} b_{4}-17496 a^{5} b c b_{1}-162 a^{4} b^{3} a_{7}-54 a^{4} b^{3} b_{8}-24786 a^{4} b^{2} c b_{2}-11664 a^{3} b^{3} c b_{4}-729 a^{2} b^{4} c b_{7}-729 a^{4} b c a_{7}-243 a^{4} b c b_{8}+10935 a^{4} c^{2} b_{2}+28431 a^{3} b \,c^{2} b_{4}+18225 a^{2} b^{2} c^{2} b_{7}-10935 a^{2} c^{3} b_{7}&=0\\ -10206 a^{9} b a_{1}-26244 a^{8} b^{2} a_{2}-14580 a^{8} b^{2} b_{3}-32076 a^{7} b^{3} a_{4}-14580 a^{7} b^{3} b_{5}-18954 a^{6} b^{4} a_{7}-7290 a^{6} b^{4} b_{8}+13122 a^{8} c a_{2}+7290 a^{8} c b_{3}+64152 a^{7} b c a_{4}+29160 a^{7} b c b_{5}+113724 a^{6} b^{2} c a_{7}+43740 a^{6} b^{2} c b_{8}-37908 a^{6} c^{2} a_{7}-14580 a^{6} c^{2} b_{8}&=0\\ -10935 a^{6} b^{4} b_{1}-2187 a^{5} b^{5} b_{2}+65610 a^{6} b^{2} c b_{1}+43740 a^{5} b^{3} c b_{2}+10935 a^{4} b^{4} c b_{4}-6561 a^{8} a_{5}-2187 a^{8} b_{6}-13122 a^{7} b a_{8}-5103 a^{7} b b_{9}-21870 a^{6} c^{2} b_{1}-65610 a^{5} b \,c^{2} b_{2}-65610 a^{4} b^{2} c^{2} b_{4}-21870 a^{3} b^{3} c^{2} b_{7}+21870 a^{4} c^{3} b_{4}+43740 a^{3} b \,c^{3} b_{7}&=0\\ -162 a^{2} b^{6} a_{6}-2160 a^{3} b^{4} a_{10}+5427 a^{2} b^{4} c a_{6}+324 a \,b^{5} c a_{9}+324 a \,b^{4} c^{2} a_{3}-9720 a^{3} b^{2} c a_{10}-11421 a^{2} b^{2} c^{2} a_{6}-2187 a \,b^{3} c^{2} a_{9}-1458 a \,b^{3} c^{2} b_{10}+1458 a \,b^{2} c^{3} a_{3}-108 b^{3} c^{3} a_{5}-2187 a^{2} c^{3} a_{6}-1458 a b \,c^{3} a_{9}+2916 a b \,c^{3} b_{10}-486 b \,c^{4} a_{5}+729 c^{5} a_{1}&=0\\ 10935 a^{4} b^{4} c b_{1}-29160 a^{7} b a_{3}-27459 a^{6} b^{2} a_{5}-11178 a^{6} b^{2} b_{6}-11664 a^{5} b^{3} a_{8}-4374 a^{5} b^{3} b_{9}-65610 a^{4} b^{2} c^{2} b_{1}-21870 a^{3} b^{3} c^{2} b_{2}+8748 a^{6} c a_{5}+6561 a^{6} c b_{6}+1458 a^{5} b c a_{8}+8748 a^{5} b c b_{9}+21870 a^{4} c^{3} b_{1}+43740 a^{3} b \,c^{3} b_{2}+21870 a^{2} b^{2} c^{3} b_{4}-10935 a^{2} c^{4} b_{4}-10935 a b \,c^{4} b_{7}&=0\\ -2187 a^{5} b^{5} b_{1}+43740 a^{5} b^{3} c b_{1}+10935 a^{4} b^{4} c b_{2}-8748 a^{8} a_{3}-21141 a^{7} b a_{5}-8019 a^{7} b b_{6}-16281 a^{6} b^{2} a_{8}-6804 a^{6} b^{2} b_{9}-65610 a^{5} b \,c^{2} b_{1}-65610 a^{4} b^{2} c^{2} b_{2}-21870 a^{3} b^{3} c^{2} b_{4}+2187 a^{6} c a_{8}+3645 a^{6} c b_{9}+21870 a^{4} c^{3} b_{2}+43740 a^{3} b \,c^{3} b_{4}+21870 a^{2} b^{2} c^{3} b_{7}-10935 a^{2} c^{4} b_{7}&=0\\ -18954 a^{8} b^{2} a_{1}-24786 a^{7} b^{3} a_{2}-14580 a^{7} b^{3} b_{3}-15309 a^{6} b^{4} a_{4}-7290 a^{6} b^{4} b_{5}-3645 a^{5} b^{5} a_{7}-1458 a^{5} b^{5} b_{8}+9477 a^{8} c a_{1}+49572 a^{7} b c a_{2}+29160 a^{7} b c b_{3}+91854 a^{6} b^{2} c a_{4}+43740 a^{6} b^{2} c b_{5}+72900 a^{5} b^{3} c a_{7}+29160 a^{5} b^{3} c b_{8}-30618 a^{6} c^{2} a_{4}-14580 a^{6} c^{2} b_{5}-109350 a^{5} b \,c^{2} a_{7}-43740 a^{5} b \,c^{2} b_{8}&=0\\ -7776 a^{5} b^{3} a_{1}-972 a^{4} b^{4} a_{2}+5832 a^{5} b c a_{1}-540 a^{4} b^{3} a_{8}-108 a^{4} b^{3} b_{9}-7776 a^{4} b^{2} c a_{2}-5832 a^{3} b^{3} c a_{4}-2916 a^{3} b^{3} c b_{5}-972 a^{2} b^{4} c a_{7}-2430 a^{4} b c a_{8}-486 a^{4} b c b_{9}+8748 a^{4} c^{2} a_{2}+23328 a^{3} b \,c^{2} a_{4}+8748 a^{3} b \,c^{2} b_{5}+12636 a^{2} b^{2} c^{2} a_{7}+8748 a^{2} b^{2} c^{2} b_{8}-8748 a^{2} c^{3} a_{7}-5832 a^{2} c^{3} b_{8}&=0\\ 216 a^{3} b^{6} a_{6}-22356 a^{4} b^{4} a_{10}-324 a^{3} b^{4} c a_{6}-648 a^{2} b^{5} c a_{9}+2916 a^{2} b^{4} c^{2} a_{3}+86994 a^{4} b^{2} c a_{10}-5832 a^{3} b^{2} c^{2} a_{6}-2268 a^{2} b^{3} c^{2} a_{9}-3402 a^{2} b^{2} c^{3} a_{3}-1944 a \,b^{3} c^{3} a_{5}-17496 a^{4} c^{2} a_{10}+2916 a^{2} b \,c^{3} a_{9}-729 a^{2} c^{4} a_{3}-5103 a b \,c^{4} a_{5}-3645 a b \,c^{4} b_{6}+486 b^{2} c^{4} a_{8}+2187 c^{5} a_{8}+729 c^{5} b_{9}&=0\\ -26244 a^{5} b^{3} a_{3}-7371 a^{4} b^{4} a_{5}-2187 a^{4} b^{4} b_{6}-1134 a^{3} b^{5} a_{8}-243 a^{3} b^{5} b_{9}+36450 a^{5} b c a_{3}+18468 a^{4} b^{2} c a_{5}+16038 a^{4} b^{2} c b_{6}+4860 a^{3} b^{3} c a_{8}+2916 a^{3} b^{3} c b_{9}+21870 a^{2} b^{2} c^{3} b_{1}+2187 a^{4} c^{2} a_{5}-6561 a^{4} c^{2} b_{6}+14580 a^{3} b \,c^{2} a_{8}-2187 a^{3} b \,c^{2} b_{9}-10935 a^{2} c^{4} b_{1}-10935 a b \,c^{4} b_{2}+2187 c^{5} b_{4}&=0\\ -38637 a^{6} b^{2} a_{3}-18954 a^{5} b^{3} a_{5}-7290 a^{5} b^{3} b_{6}-5184 a^{4} b^{4} a_{8}-1458 a^{4} b^{4} b_{9}-21870 a^{3} b^{3} c^{2} b_{1}+15309 a^{6} c a_{3}+18954 a^{5} b c a_{5}+17496 a^{5} b c b_{6}+2430 a^{4} b^{2} c a_{8}+7290 a^{4} b^{2} c b_{9}+43740 a^{3} b \,c^{3} b_{1}+21870 a^{2} b^{2} c^{3} b_{2}+8748 a^{4} c^{2} a_{8}-2187 a^{4} c^{2} b_{9}-10935 a^{2} c^{4} b_{2}-10935 a b \,c^{4} b_{4}+2187 c^{5} b_{7}&=0\\ 10935 a^{6} b^{4} a_{3}+2673 a^{5} b^{5} a_{5}-729 a^{5} b^{5} b_{6}+486 a^{4} b^{6} a_{8}+216 a^{6} b^{3} a_{9}-41310 a^{6} b^{2} c a_{3}-14580 a^{5} b^{3} c a_{5}+14580 a^{5} b^{3} c b_{6}-3645 a^{4} b^{4} c a_{8}+3645 a^{4} b^{4} c b_{9}-21870 a^{8} a_{10}+972 a^{6} b c a_{9}+10206 a^{6} c^{2} a_{3}+4374 a^{5} b \,c^{2} a_{5}-21870 a^{5} b \,c^{2} b_{6}-18954 a^{4} b^{2} c^{2} a_{8}-21870 a^{4} b^{2} c^{2} b_{9}+10206 a^{4} c^{3} a_{8}+7290 a^{4} c^{3} b_{9}&=0\\ 648 a^{4} b^{5} a_{6}+216 a^{3} b^{6} a_{9}-1944 a^{3} b^{5} c a_{3}-63180 a^{5} b^{3} a_{10}+2268 a^{4} b^{3} c a_{6}-324 a^{3} b^{4} c a_{9}+11664 a^{3} b^{3} c^{2} a_{3}+2916 a^{2} b^{4} c^{2} a_{5}+97686 a^{5} b c a_{10}-2916 a^{4} b \,c^{2} a_{6}-5832 a^{3} b^{2} c^{2} a_{9}-5832 a^{3} b \,c^{3} a_{3}+3888 a^{2} b^{2} c^{3} a_{5}+7290 a^{2} b^{2} c^{3} b_{6}-1944 a \,b^{3} c^{3} a_{8}-4374 a^{2} c^{4} a_{5}-3645 a^{2} c^{4} b_{6}-8748 a b \,c^{4} a_{8}-3645 a b \,c^{4} b_{9}&=0\\ 2187 a^{4} b^{4} b_{1}+243 a^{3} b^{5} b_{2}-216 a^{4} b^{3} a_{4}-54 a^{4} b^{3} b_{5}-16038 a^{4} b^{2} c b_{1}-216 a^{3} b^{4} a_{7}-108 a^{3} b^{4} b_{8}-8748 a^{3} b^{3} c b_{2}-729 a^{2} b^{4} c b_{4}-972 a^{4} b c a_{4}-243 a^{4} b c b_{5}+6561 a^{4} c^{2} b_{1}-972 a^{3} b^{2} c a_{7}-486 a^{3} b^{2} c b_{8}+19683 a^{3} b \,c^{2} b_{2}+13851 a^{2} b^{2} c^{2} b_{4}+729 a \,b^{3} c^{2} b_{7}-8019 a^{2} c^{3} b_{4}-10206 a b \,c^{3} b_{7}&=0\\ 3402 a^{5} b^{5} a_{3}+486 a^{4} b^{6} a_{5}+216 a^{6} b^{3} a_{6}+648 a^{5} b^{4} a_{9}-29160 a^{5} b^{3} c a_{3}-7290 a^{4} b^{4} c a_{5}+3645 a^{4} b^{4} c b_{6}-1944 a^{3} b^{5} c a_{8}-72900 a^{7} b a_{10}+972 a^{6} b c a_{6}+2916 a^{5} b^{2} c a_{9}+26244 a^{5} b \,c^{2} a_{3}+2916 a^{4} b^{2} c^{2} a_{5}-21870 a^{4} b^{2} c^{2} b_{6}-2916 a^{3} b^{3} c^{2} a_{8}-7290 a^{3} b^{3} c^{2} b_{9}+2916 a^{4} c^{3} a_{5}+7290 a^{4} c^{3} b_{6}+23328 a^{3} b \,c^{3} a_{8}+14580 a^{3} b \,c^{3} b_{9}&=0\\ -1080 a^{3} b^{4} a_{3}-414 a^{2} b^{5} a_{5}-108 a^{2} b^{5} b_{6}+1215 a^{2} b^{4} c a_{1}-36 a \,b^{6} a_{8}-4860 a^{3} b^{2} c a_{3}-1215 a^{2} b^{3} c a_{5}-270 a^{2} b^{3} c b_{6}-2673 a^{2} b^{2} c^{2} a_{1}+162 a \,b^{4} c a_{8}+216 a \,b^{4} c b_{9}-243 a \,b^{3} c^{2} a_{2}+2916 a^{2} b \,c^{2} a_{5}+972 a^{2} b \,c^{2} b_{6}-2187 a^{2} c^{3} a_{1}+1458 a \,b^{2} c^{2} a_{8}+972 a \,b^{2} c^{2} b_{9}-1458 a b \,c^{3} a_{2}-243 b^{2} c^{3} a_{4}+729 c^{4} b_{5}&=0\\ -540 a^{3} b^{4} a_{1}-162 a^{2} b^{5} a_{2}-54 a^{2} b^{5} b_{3}-2430 a^{3} b^{2} c a_{1}-405 a^{2} b^{3} c a_{2}-135 a^{2} b^{3} c b_{3}+108 a \,b^{4} c a_{4}+108 a \,b^{4} c b_{5}+729 a \,b^{3} c^{2} b_{1}+1458 a^{2} b \,c^{2} a_{2}+486 a^{2} b \,c^{2} b_{3}+486 a \,b^{2} c^{2} a_{4}+486 a \,b^{2} c^{2} b_{5}-1458 a b \,c^{3} b_{1}+54 b^{3} c^{2} a_{7}-54 b^{3} c^{2} b_{8}-243 b^{2} c^{3} b_{2}+243 b \,c^{3} a_{7}-243 b \,c^{3} b_{8}+729 c^{4} b_{2}&=0\\ 486 a^{4} b^{6} a_{3}+648 a^{5} b^{4} a_{6}+648 a^{4} b^{5} a_{9}-10935 a^{4} b^{4} c a_{3}-1944 a^{3} b^{5} c a_{5}-95499 a^{6} b^{2} a_{10}+2916 a^{5} b^{2} c a_{6}+2268 a^{4} b^{3} c a_{9}+24786 a^{4} b^{2} c^{2} a_{3}+4374 a^{3} b^{3} c^{2} a_{5}-7290 a^{3} b^{3} c^{2} b_{6}+2916 a^{2} b^{4} c^{2} a_{8}+41553 a^{6} c a_{10}-2916 a^{4} b \,c^{2} a_{9}-4374 a^{4} c^{3} a_{3}+8748 a^{3} b \,c^{3} a_{5}+14580 a^{3} b \,c^{3} b_{6}+11178 a^{2} b^{2} c^{3} a_{8}+7290 a^{2} b^{2} c^{3} b_{9}-8019 a^{2} c^{4} a_{8}-3645 a^{2} c^{4} b_{9}&=0\\ -81 a^{2} b^{6} a_{3}-1620 a^{3} b^{4} a_{6}-666 a^{2} b^{5} a_{9}-162 a^{2} b^{5} b_{10}+3321 a^{2} b^{4} c a_{3}+162 a \,b^{5} c a_{5}-7290 a^{3} b^{2} c a_{6}-2025 a^{2} b^{3} c a_{9}-405 a^{2} b^{3} c b_{10}-7047 a^{2} b^{2} c^{2} a_{3}-1215 a \,b^{3} c^{2} a_{5}-729 a \,b^{3} c^{2} b_{6}-81 b^{4} c^{2} a_{8}+4374 a^{2} b \,c^{2} a_{9}+1458 a^{2} b \,c^{2} b_{10}-2187 a^{2} c^{3} a_{3}-1458 a b \,c^{3} a_{5}+1458 a b \,c^{3} b_{6}-243 b^{2} c^{3} a_{8}+243 b^{2} c^{3} b_{9}+729 c^{4} b_{9}&=0\\ -324 a^{4} b^{3} a_{1}-432 a^{3} b^{4} a_{2}-108 a^{3} b^{4} b_{3}-108 a^{2} b^{5} a_{4}-54 a^{2} b^{5} b_{5}-729 a^{2} b^{4} c b_{1}-1458 a^{4} b c a_{1}-1944 a^{3} b^{2} c a_{2}-486 a^{3} b^{2} c b_{3}-270 a^{2} b^{3} c a_{4}-135 a^{2} b^{3} c b_{5}+5103 a^{2} b^{2} c^{2} b_{1}+108 a \,b^{4} c b_{8}+729 a \,b^{3} c^{2} b_{2}+972 a^{2} b \,c^{2} a_{4}+486 a^{2} b \,c^{2} b_{5}-2187 a^{2} c^{3} b_{1}+486 a \,b^{2} c^{2} b_{8}-4374 a b \,c^{3} b_{2}-243 b^{2} c^{3} b_{4}+1458 c^{4} b_{4}&=0\\ -3159 a^{4} b^{4} a_{1}-243 a^{3} b^{5} a_{2}-594 a^{4} b^{3} a_{5}-108 a^{4} b^{3} b_{6}+8262 a^{4} b^{2} c a_{1}-864 a^{3} b^{4} a_{8}-216 a^{3} b^{4} b_{9}-243 a^{2} b^{4} c a_{4}-2673 a^{4} b c a_{5}-486 a^{4} b c b_{6}+2187 a^{4} c^{2} a_{1}-3888 a^{3} b^{2} c a_{8}-972 a^{3} b^{2} c b_{9}+12393 a^{3} b \,c^{2} a_{2}+7533 a^{2} b^{2} c^{2} a_{4}+4374 a^{2} b^{2} c^{2} b_{5}+1215 a \,b^{3} c^{2} a_{7}-6561 a^{2} c^{3} a_{4}-2916 a^{2} c^{3} b_{5}-4374 a b \,c^{3} a_{7}-5832 a b \,c^{3} b_{8}&=0\\ -1620 a^{3} b^{5} a_{3}-81 a^{2} b^{6} a_{5}-972 a^{4} b^{3} a_{6}-1512 a^{3} b^{4} a_{9}-324 a^{3} b^{4} b_{10}+16524 a^{3} b^{3} c a_{3}+2592 a^{2} b^{4} c a_{5}+729 a^{2} b^{4} c b_{6}+162 a \,b^{5} c a_{8}-4374 a^{4} b c a_{6}-6804 a^{3} b^{2} c a_{9}-1458 a^{3} b^{2} c b_{10}-7290 a^{3} b \,c^{2} a_{3}-1944 a^{2} b^{2} c^{2} a_{5}-5103 a^{2} b^{2} c^{2} b_{6}-486 a \,b^{3} c^{2} a_{8}-729 a \,b^{3} c^{2} b_{9}-4374 a^{2} c^{3} a_{5}+2187 a^{2} c^{3} b_{6}-2916 a b \,c^{3} a_{8}-1458 a b \,c^{3} b_{9}+2187 c^{5} b_{1}&=0\\ -9558 a^{4} b^{4} a_{3}-1377 a^{3} b^{5} a_{5}-243 a^{3} b^{5} b_{6}-81 a^{2} b^{6} a_{8}-918 a^{4} b^{3} a_{9}-162 a^{4} b^{3} b_{10}+34506 a^{4} b^{2} c a_{3}+10692 a^{3} b^{3} c a_{5}+5832 a^{3} b^{3} c b_{6}+1863 a^{2} b^{4} c a_{8}+729 a^{2} b^{4} c b_{9}-4131 a^{4} b c a_{9}-729 a^{4} b c b_{10}-4374 a^{4} c^{2} a_{3}+3645 a^{3} b \,c^{2} a_{5}-10935 a^{3} b \,c^{2} b_{6}+3159 a^{2} b^{2} c^{2} a_{8}-729 a^{2} b^{2} c^{2} b_{9}-10935 a b \,c^{4} b_{1}-6561 a^{2} c^{3} a_{8}-729 a^{2} c^{3} b_{9}+2187 c^{5} b_{2}&=0\\ -2754 a^{3} b^{5} a_{6}-162 a^{2} b^{6} a_{9}-324 a^{2} b^{5} c a_{3}-1296 a^{4} b^{3} a_{10}+27216 a^{3} b^{3} c a_{6}+4698 a^{2} b^{4} c a_{9}+1458 a^{2} b^{4} c b_{10}-1134 a^{2} b^{3} c^{2} a_{3}+324 a \,b^{4} c^{2} a_{5}-5832 a^{4} b c a_{10}-16038 a^{3} b \,c^{2} a_{6}-6318 a^{2} b^{2} c^{2} a_{9}-10206 a^{2} b^{2} c^{2} b_{10}+1458 a^{2} b \,c^{3} a_{3}+1458 a \,b^{2} c^{3} a_{5}-4374 a b \,c^{4} a_{1}-108 b^{3} c^{3} a_{8}-4374 a^{2} c^{3} a_{9}+4374 a^{2} c^{3} b_{10}-486 b \,c^{4} a_{8}+1458 c^{5} a_{2}+1458 c^{5} b_{3}&=0\\ 108 a^{3} b^{6} a_{3}-15957 a^{4} b^{4} a_{6}-2511 a^{3} b^{5} a_{9}-486 a^{3} b^{5} b_{10}-162 a^{3} b^{4} c a_{3}-324 a^{2} b^{5} c a_{5}+60750 a^{4} b^{2} c a_{6}+21384 a^{3} b^{3} c a_{9}+11664 a^{3} b^{3} c b_{10}-2916 a^{3} b^{2} c^{2} a_{3}-1134 a^{2} b^{3} c^{2} a_{5}+10206 a^{2} b^{2} c^{3} a_{1}+324 a \,b^{4} c^{2} a_{8}-10935 a^{4} c^{2} a_{6}-5103 a^{3} b \,c^{2} a_{9}-21870 a^{3} b \,c^{2} b_{10}+1458 a^{2} b \,c^{3} a_{5}-5103 a^{2} c^{4} a_{1}+1458 a \,b^{2} c^{3} a_{8}-8019 a b \,c^{4} a_{2}-7290 a b \,c^{4} b_{3}+2187 c^{5} a_{4}+1458 c^{5} b_{5}&=0\\ -486 a^{3} b^{5} a_{1}-648 a^{4} b^{3} a_{3}-972 a^{3} b^{4} a_{5}-216 a^{3} b^{4} b_{6}+5832 a^{3} b^{3} c a_{1}-360 a^{2} b^{5} a_{8}-108 a^{2} b^{5} b_{9}+486 a^{2} b^{4} c a_{2}-2916 a^{4} b c a_{3}-4374 a^{3} b^{2} c a_{5}-972 a^{3} b^{2} c b_{6}+1458 a^{3} b \,c^{2} a_{1}-1080 a^{2} b^{3} c a_{8}-270 a^{2} b^{3} c b_{9}+2430 a^{2} b^{2} c^{2} a_{2}+486 a \,b^{3} c^{2} a_{4}+2430 a^{2} b \,c^{2} a_{8}+972 a^{2} b \,c^{2} b_{9}-4374 a^{2} c^{3} a_{2}-2916 a b \,c^{3} a_{4}-2916 a b \,c^{3} b_{5}-486 b^{2} c^{3} a_{7}+1458 c^{4} b_{8}&=0\\ -17496 a^{7} b^{3} a_{1}-11664 a^{6} b^{4} a_{2}-7290 a^{6} b^{4} b_{3}-2916 a^{5} b^{5} a_{4}-1458 a^{5} b^{5} b_{5}+34992 a^{7} b c a_{1}+108 a^{6} b^{3} a_{8}+69984 a^{6} b^{2} c a_{2}+43740 a^{6} b^{2} c b_{3}+58320 a^{5} b^{3} c a_{4}+29160 a^{5} b^{3} c b_{5}+17496 a^{4} b^{4} c a_{7}+7290 a^{4} b^{4} c b_{8}-13122 a^{8} a_{9}-4374 a^{8} b_{10}+486 a^{6} b c a_{8}-23328 a^{6} c^{2} a_{2}-14580 a^{6} c^{2} b_{3}-87480 a^{5} b \,c^{2} a_{4}-43740 a^{5} b \,c^{2} b_{5}-104976 a^{4} b^{2} c^{2} a_{7}-43740 a^{4} b^{2} c^{2} b_{8}+34992 a^{4} c^{3} a_{7}+14580 a^{4} c^{3} b_{8}&=0\\ 243 a^{3} b^{5} b_{1}-270 a^{4} b^{3} a_{2}-54 a^{4} b^{3} b_{3}-324 a^{3} b^{4} a_{4}-108 a^{3} b^{4} b_{5}-5832 a^{3} b^{3} c b_{1}-54 a^{2} b^{5} a_{7}-54 a^{2} b^{5} b_{8}-729 a^{2} b^{4} c b_{2}-1215 a^{4} b c a_{2}-243 a^{4} b c b_{3}-1458 a^{3} b^{2} c a_{4}-486 a^{3} b^{2} c b_{5}+10935 a^{3} b \,c^{2} b_{1}-135 a^{2} b^{3} c a_{7}-135 a^{2} b^{3} c b_{8}+9477 a^{2} b^{2} c^{2} b_{2}+729 a \,b^{3} c^{2} b_{4}+486 a^{2} b \,c^{2} a_{7}+486 a^{2} b \,c^{2} b_{8}-5103 a^{2} c^{3} b_{2}-7290 a b \,c^{3} b_{4}-243 b^{2} c^{3} b_{7}+2187 c^{4} b_{7}&=0\\ -8019 a^{6} b^{4} a_{1}-2187 a^{5} b^{5} a_{2}-1458 a^{5} b^{5} b_{3}+108 a^{6} b^{3} a_{5}+48114 a^{6} b^{2} c a_{1}+324 a^{5} b^{4} a_{8}+43740 a^{5} b^{3} c a_{2}+29160 a^{5} b^{3} c b_{3}+13851 a^{4} b^{4} c a_{4}+7290 a^{4} b^{4} c b_{5}-15309 a^{8} a_{6}-43011 a^{7} b a_{9}-16038 a^{7} b b_{10}+486 a^{6} b c a_{5}-16038 a^{6} c^{2} a_{1}+1458 a^{5} b^{2} c a_{8}-65610 a^{5} b \,c^{2} a_{2}-43740 a^{5} b \,c^{2} b_{3}-83106 a^{4} b^{2} c^{2} a_{4}-43740 a^{4} b^{2} c^{2} b_{5}-33534 a^{3} b^{3} c^{2} a_{7}-14580 a^{3} b^{3} c^{2} b_{8}+27702 a^{4} c^{3} a_{4}+14580 a^{4} c^{3} b_{5}+67068 a^{3} b \,c^{3} a_{7}+29160 a^{3} b \,c^{3} b_{8}&=0\\ 324 a^{4} b^{5} a_{3}+108 a^{3} b^{6} a_{5}-44712 a^{5} b^{3} a_{6}-13770 a^{4} b^{4} a_{9}-4374 a^{4} b^{4} b_{10}+1134 a^{4} b^{3} c a_{3}-162 a^{3} b^{4} c a_{5}-11664 a^{3} b^{3} c^{2} a_{1}-324 a^{2} b^{5} c a_{8}+67068 a^{5} b c a_{6}+44712 a^{4} b^{2} c a_{9}+32076 a^{4} b^{2} c b_{10}-1458 a^{4} b \,c^{2} a_{3}-2916 a^{3} b^{2} c^{2} a_{5}+23328 a^{3} b \,c^{3} a_{1}-1134 a^{2} b^{3} c^{2} a_{8}+17496 a^{2} b^{2} c^{3} a_{2}+14580 a^{2} b^{2} c^{3} b_{3}-4374 a^{4} c^{2} a_{9}-13122 a^{4} c^{2} b_{10}+1458 a^{2} b \,c^{3} a_{8}-8748 a^{2} c^{4} a_{2}-7290 a^{2} c^{4} b_{3}-11664 a b \,c^{4} a_{4}-7290 a b \,c^{4} b_{5}+2916 c^{5} a_{7}+1458 c^{5} b_{8}&=0\\ 324 a^{5} b^{4} a_{3}+324 a^{4} b^{5} a_{5}+6561 a^{4} b^{4} c a_{1}+108 a^{3} b^{6} a_{8}-67068 a^{6} b^{2} a_{6}-37422 a^{5} b^{3} a_{9}-14580 a^{5} b^{3} b_{10}+1458 a^{5} b^{2} c a_{3}+1134 a^{4} b^{3} c a_{5}-39366 a^{4} b^{2} c^{2} a_{1}-162 a^{3} b^{4} c a_{8}-18954 a^{3} b^{3} c^{2} a_{2}-14580 a^{3} b^{3} c^{2} b_{3}+28431 a^{6} c a_{6}+49572 a^{5} b c a_{9}+34992 a^{5} b c b_{10}-1458 a^{4} b \,c^{2} a_{5}+13122 a^{4} c^{3} a_{1}-2916 a^{3} b^{2} c^{2} a_{8}+37908 a^{3} b \,c^{3} a_{2}+29160 a^{3} b \,c^{3} b_{3}+24786 a^{2} b^{2} c^{3} a_{4}+14580 a^{2} b^{2} c^{3} b_{5}-12393 a^{2} c^{4} a_{4}-7290 a^{2} c^{4} b_{5}-15309 a b \,c^{4} a_{7}-7290 a b \,c^{4} b_{8}&=0\\ -1458 a^{5} b^{5} a_{1}+108 a^{6} b^{3} a_{3}+324 a^{5} b^{4} a_{5}+29160 a^{5} b^{3} c a_{1}+324 a^{4} b^{5} a_{8}+10206 a^{4} b^{4} c a_{2}+7290 a^{4} b^{4} c b_{3}-51030 a^{7} b a_{6}-55890 a^{6} b^{2} a_{9}-22356 a^{6} b^{2} b_{10}+486 a^{6} b c a_{3}+1458 a^{5} b^{2} c a_{5}-43740 a^{5} b \,c^{2} a_{1}+1134 a^{4} b^{3} c a_{8}-61236 a^{4} b^{2} c^{2} a_{2}-43740 a^{4} b^{2} c^{2} b_{3}-26244 a^{3} b^{3} c^{2} a_{4}-14580 a^{3} b^{3} c^{2} b_{5}+21870 a^{6} c a_{9}+13122 a^{6} c b_{10}-1458 a^{4} b \,c^{2} a_{8}+20412 a^{4} c^{3} a_{2}+14580 a^{4} c^{3} b_{3}+52488 a^{3} b \,c^{3} a_{4}+29160 a^{3} b \,c^{3} b_{5}+32076 a^{2} b^{2} c^{3} a_{7}+14580 a^{2} b^{2} c^{3} b_{8}-16038 a^{2} c^{4} a_{7}-7290 a^{2} c^{4} b_{8}&=0 \end{align*}

Solving the above equations for the unknowns gives

\begin{align*} a_{1}&=0\\ a_{2}&=-b_{3}\\ a_{3}&=0\\ a_{4}&=\frac {a b b_{3}}{c}\\ a_{5}&=0\\ a_{6}&=0\\ a_{7}&=\frac {a^{2} b_{3}}{c}\\ a_{8}&=0\\ a_{9}&=0\\ a_{10}&=0\\ b_{1}&=0\\ b_{2}&=0\\ b_{3}&=b_{3}\\ b_{4}&=0\\ b_{5}&=-\frac {2 a b b_{3}}{c}\\ b_{6}&=0\\ b_{7}&=0\\ b_{8}&=-\frac {3 a^{2} b_{3}}{c}\\ b_{9}&=0\\ b_{10}&=0 \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 {x \left (a^{2} x^{2}+x b a -c \right )}{c} \\ \eta &= -\frac {u \left (3 a^{2} x^{2}+2 x b a -c \right )}{c} \\ \end{align*}

Shifting is now applied to make \(\xi =0\) in order to simplify the rest of the computation

\begin{align*} \eta &= \eta - \omega \left (x,u\right ) \xi \\ &= -\frac {u \left (3 a^{2} x^{2}+2 x b a -c \right )}{c} - \left (\frac {27 x^{9} u^{3} a^{6}+81 x^{8} u^{3} a^{5} b +81 x^{7} u^{3} a^{4} b^{2}-81 x^{7} u^{3} a^{4} c +27 x^{6} u^{3} a^{3} b^{3}-162 x^{6} u^{3} a^{3} b c -81 x^{5} u^{3} a^{2} b^{2} c +81 x^{5} u^{3} a^{2} c^{2}+81 x^{4} u^{3} a b \,c^{2}-81 x^{5} u \,a^{4}-135 x^{4} u \,a^{3} b -27 x^{3} u^{3} c^{3}-63 x^{3} u \,a^{2} b^{2}+81 x^{3} u \,a^{2} c -9 u a \,b^{3} x^{2}+54 u a b c \,x^{2}+9 u \,b^{2} c x -2 b^{3}-9 b c}{27 x^{2} \left (a^{2} x^{2}+x b a -c \right )^{2}}\right ) \left (\frac {x \left (a^{2} x^{2}+x b a -c \right )}{c}\right ) \\ &= -\frac {\left (a^{2} u \,x^{3}+a b u \,x^{2}-c u x +\frac {1}{3} b \right ) \left (a^{4} x^{6} u^{2}+2 a^{3} b \,u^{2} x^{5}+a^{2} u^{2} \left (b^{2}-2 c \right ) x^{4}-\frac {a b u \left (6 c u +a \right ) x^{3}}{3}+\left (-\frac {1}{3} u \,b^{2} a +c^{2} u^{2}\right ) x^{2}+\frac {b c u x}{3}-\frac {2 b^{2}}{9}-c \right )}{x c \left (a^{2} x^{2}+x b a -c \right )}\\ \xi &= 0 \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 \(S(R)\). Since \(\xi =0\) then in this special case

\begin{align*} R = x \end{align*}

\(S\) is found from

\begin{align*} S &= \int { \frac {1}{\eta }} dy\\ &= \int { \frac {1}{-\frac {\left (a^{2} u \,x^{3}+a b u \,x^{2}-c u x +\frac {1}{3} b \right ) \left (a^{4} x^{6} u^{2}+2 a^{3} b \,u^{2} x^{5}+a^{2} u^{2} \left (b^{2}-2 c \right ) x^{4}-\frac {a b u \left (6 c u +a \right ) x^{3}}{3}+\left (-\frac {1}{3} u \,b^{2} a +c^{2} u^{2}\right ) x^{2}+\frac {b c u x}{3}-\frac {2 b^{2}}{9}-c \right )}{x c \left (a^{2} x^{2}+x b a -c \right )}}} dy \end{align*}

Which results in

\begin{align*} S&= -27 x \left (a^{2} x^{2}+x b a -c \right ) c \left (\frac {\frac {\left (3 a^{2} x^{3}+3 b a \,x^{2}-3 c x \right ) \ln \left (9 a^{4} x^{6} u^{2}+18 a^{3} b \,u^{2} x^{5}+9 a^{2} b^{2} u^{2} x^{4}-18 a^{2} c \,u^{2} x^{4}-18 a b c \,u^{2} x^{3}-3 a^{2} b u \,x^{3}-3 a \,b^{2} u \,x^{2}+9 c^{2} u^{2} x^{2}+3 b c u x -2 b^{2}-9 c \right )}{18 a^{4} x^{6}+36 a^{3} b \,x^{5}+18 a^{2} b^{2} x^{4}-36 a^{2} c \,x^{4}-36 b c \,x^{3} a +18 c^{2} x^{2}}-\frac {2 \left (-2 b -\frac {\left (3 a^{2} x^{3}+3 b a \,x^{2}-3 c x \right ) \left (-3 a^{2} b \,x^{3}-3 x^{2} b^{2} a +3 b c x \right )}{2 \left (9 a^{4} x^{6}+18 a^{3} b \,x^{5}+9 a^{2} b^{2} x^{4}-18 a^{2} c \,x^{4}-18 b c \,x^{3} a +9 c^{2} x^{2}\right )}\right ) \operatorname {arctanh}\left (\frac {2 \left (9 a^{4} x^{6}+18 a^{3} b \,x^{5}+9 a^{2} b^{2} x^{4}-18 a^{2} c \,x^{4}-18 b c \,x^{3} a +9 c^{2} x^{2}\right ) u -3 a^{2} b \,x^{3}-3 x^{2} b^{2} a +3 b c x}{9 \sqrt {a^{4} b^{2} x^{6}+4 a^{4} c \,x^{6}+2 a^{3} b^{3} x^{5}+8 a^{3} b c \,x^{5}+a^{2} b^{4} x^{4}+2 a^{2} b^{2} c \,x^{4}-8 a^{2} c^{2} x^{4}-2 a \,b^{3} c \,x^{3}-8 a b \,c^{2} x^{3}+b^{2} c^{2} x^{2}+4 c^{3} x^{2}}}\right )}{9 \sqrt {a^{4} b^{2} x^{6}+4 a^{4} c \,x^{6}+2 a^{3} b^{3} x^{5}+8 a^{3} b c \,x^{5}+a^{2} b^{4} x^{4}+2 a^{2} b^{2} c \,x^{4}-8 a^{2} c^{2} x^{4}-2 a \,b^{3} c \,x^{3}-8 a b \,c^{2} x^{3}+b^{2} c^{2} x^{2}+4 c^{3} x^{2}}}}{9 c}-\frac {\ln \left (3 a^{2} u \,x^{3}+3 a b u \,x^{2}-3 c u x +b \right )}{9 c \left (3 a^{2} x^{3}+3 b a \,x^{2}-3 c x \right )}\right ) \end{align*}

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,u) S_{u} }{ R_{x} + \omega (x,u) R_{u} }\tag {2} \end{align*}

Where in the above \(R_{x},R_{u},S_{x},S_{u}\) are all partial derivatives and \(\omega (x,u)\) is the right hand side of the original ode given by

\begin{align*} \omega (x,u) &= \frac {27 x^{9} u^{3} a^{6}+81 x^{8} u^{3} a^{5} b +81 x^{7} u^{3} a^{4} b^{2}-81 x^{7} u^{3} a^{4} c +27 x^{6} u^{3} a^{3} b^{3}-162 x^{6} u^{3} a^{3} b c -81 x^{5} u^{3} a^{2} b^{2} c +81 x^{5} u^{3} a^{2} c^{2}+81 x^{4} u^{3} a b \,c^{2}-81 x^{5} u \,a^{4}-135 x^{4} u \,a^{3} b -27 x^{3} u^{3} c^{3}-63 x^{3} u \,a^{2} b^{2}+81 x^{3} u \,a^{2} c -9 u a \,b^{3} x^{2}+54 u a b c \,x^{2}+9 u \,b^{2} c x -2 b^{3}-9 b c}{27 x^{2} \left (a^{2} x^{2}+x b a -c \right )^{2}} \end{align*}

Evaluating all the partial derivatives gives

\begin{align*} R_{x} &= 1\\ R_{u} &= 0\\ S_{x} &= -\frac {u c \left (3 a^{2} x^{2}+2 x b a -c \right )}{\left (u x \left (a^{2} x^{2}+x b a -c \right )+\frac {b}{3}\right ) \left (x^{2} \left (a^{2} x^{2}+x b a -c \right )^{2} u^{2}-\frac {b x \left (a^{2} x^{2}+x b a -c \right ) u}{3}-\frac {2 b^{2}}{9}-c \right )}\\ S_{u} &= -\frac {x c \left (a^{2} x^{2}+x b a -c \right )}{\left (u x \left (a^{2} x^{2}+x b a -c \right )+\frac {b}{3}\right ) \left (x^{2} \left (a^{2} x^{2}+x b a -c \right )^{2} u^{2}-\frac {b x \left (a^{2} x^{2}+x b a -c \right ) u}{3}-\frac {2 b^{2}}{9}-c \right )} \end{align*}

Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.

\begin{align*} \frac {dS}{dR} &= -\frac {c}{x \left (a^{2} x^{2}+x b a -c \right )}\tag {2A} \end{align*}

We now need to express the RHS as function of \(R\) only. This is done by solving for \(x,u\) in terms of \(R,S\) from the result obtained earlier and simplifying. This gives

\begin{align*} \frac {dS}{dR} &= -\frac {c}{R \left (R^{2} a^{2}+R a b -c \right )} \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 \(R,S\).

Since the ode has the form \(\frac {d}{d R}S \left (R \right )=f(R)\), then we only need to integrate \(f(R)\).

\begin{align*} \int {dS} &= \int {-\frac {c}{R \left (R^{2} a^{2}+R a b -c \right )}\, dR}\\ S \left (R \right ) &= -\frac {\ln \left (R^{2} a^{2}+R a b -c \right )}{2}+\frac {a b \,\operatorname {arctanh}\left (\frac {2 R \,a^{2}+a b}{\sqrt {a^{2} b^{2}+4 a^{2} c}}\right )}{\sqrt {a^{2} b^{2}+4 a^{2} c}}+\ln \left (R \right ) + c_2 \end{align*}

\begin{align*} S \left (R \right )&= -\frac {\ln \left (R^{2} a^{2}+R a b -c \right )}{2}+\frac {a b \,\operatorname {arctanh}\left (\frac {a \left (2 R a +b \right )}{\sqrt {a^{2} \left (b^{2}+4 c \right )}}\right )}{\sqrt {a^{2} \left (b^{2}+4 c \right )}}+\ln \left (R \right )+c_2 \end{align*}

To complete the solution, we just need to transform the above back to \(x,u\) coordinates. This results in

\begin{align*} -\frac {\ln \left (9 x^{2} \left (a^{2} x^{2}+x b a -c \right )^{2} u \left (x \right )^{2}-3 b x \left (a^{2} x^{2}+x b a -c \right ) u \left (x \right )-2 b^{2}-9 c \right ) \sqrt {b^{2}+4 c}+2 b \,\operatorname {arctanh}\left (\frac {2 u \left (x \right ) x \left (a^{2} x^{2}+x b a -c \right )-\frac {b}{3}}{\sqrt {b^{2}+4 c}}\right )-2 \ln \left (3 u \left (x \right ) x \left (a^{2} x^{2}+x b a -c \right )+b \right ) \sqrt {b^{2}+4 c}}{2 \sqrt {b^{2}+4 c}} = -\frac {\ln \left (a^{2} x^{2}+x b a -c \right )}{2}+\frac {a b \,\operatorname {arctanh}\left (\frac {a \left (2 a x +b \right )}{\sqrt {a^{2} \left (b^{2}+4 c \right )}}\right )}{\sqrt {a^{2} \left (b^{2}+4 c \right )}}+\ln \left (x \right )+c_2 \end{align*}

Substituting \(u=u \left (x \right )+\frac {-3 a x -b}{3 a^{2} x^{3}+3 b a \,x^{2}-3 c x}\) in the above solution gives

\begin{align*} -\frac {\ln \left (9 x^{2} \left (a^{2} x^{2}+x b a -c \right )^{2} \left (u \left (x \right )+\frac {-3 a x -b}{3 a^{2} x^{3}+3 b a \,x^{2}-3 c x}\right )^{2}-3 b x \left (a^{2} x^{2}+x b a -c \right ) \left (u \left (x \right )+\frac {-3 a x -b}{3 a^{2} x^{3}+3 b a \,x^{2}-3 c x}\right )-2 b^{2}-9 c \right ) \sqrt {b^{2}+4 c}+2 b \,\operatorname {arctanh}\left (\frac {2 \left (u \left (x \right )+\frac {-3 a x -b}{3 a^{2} x^{3}+3 b a \,x^{2}-3 c x}\right ) x \left (a^{2} x^{2}+x b a -c \right )-\frac {b}{3}}{\sqrt {b^{2}+4 c}}\right )-2 \ln \left (3 \left (u \left (x \right )+\frac {-3 a x -b}{3 a^{2} x^{3}+3 b a \,x^{2}-3 c x}\right ) x \left (a^{2} x^{2}+x b a -c \right )+b \right ) \sqrt {b^{2}+4 c}}{2 \sqrt {b^{2}+4 c}} = -\frac {\ln \left (a^{2} x^{2}+x b a -c \right )}{2}+\frac {a b \,\operatorname {arctanh}\left (\frac {a \left (2 a x +b \right )}{\sqrt {a^{2} \left (b^{2}+4 c \right )}}\right )}{\sqrt {a^{2} \left (b^{2}+4 c \right )}}+\ln \left (x \right )+c_2 \end{align*}

Simplifying the above gives

\begin{align*} -\frac {\sqrt {b^{2}+4 c}\, \ln \left (\left (1+x^{2} \left (a^{2} x^{2}+x b a -c \right ) u \left (x \right )^{2}+\left (-2 a \,x^{2}-b x \right ) u \left (x \right )\right ) \left (a^{2} x^{2}+x b a -c \right )\right )-2 \sqrt {b^{2}+4 c}\, \ln \left (\left (\left (a^{2} x^{2}+x b a -c \right ) u \left (x \right )-a \right ) x \right )+2 b \,\operatorname {arctanh}\left (\frac {2 a^{2} u \left (x \right ) x^{3}+2 a b u \left (x \right ) x^{2}-2 c u \left (x \right ) x -2 a x -b}{\sqrt {b^{2}+4 c}}\right )}{2 \sqrt {b^{2}+4 c}} &= -\frac {\ln \left (a^{2} x^{2}+x b a -c \right )}{2}+\frac {a b \,\operatorname {arctanh}\left (\frac {a \left (2 a x +b \right )}{\sqrt {a^{2} \left (b^{2}+4 c \right )}}\right )}{\sqrt {a^{2} \left (b^{2}+4 c \right )}}+\ln \left (x \right )+c_2 \\ \end{align*}

Substituting \(u \left (x \right )=\frac {1}{y}\) in the above solution gives

\[ -\frac {\sqrt {b^{2}+4 c}\, \ln \left (\left (1+\frac {x^{2} \left (a^{2} x^{2}+x b a -c \right )}{y^{2}}+\frac {-2 a \,x^{2}-b x}{y}\right ) \left (a^{2} x^{2}+x b a -c \right )\right )-2 \sqrt {b^{2}+4 c}\, \ln \left (\left (\frac {a^{2} x^{2}+x b a -c}{y}-a \right ) x \right )+2 b \,\operatorname {arctanh}\left (\frac {\frac {2 a^{2} x^{3}}{y}+\frac {2 b a \,x^{2}}{y}-\frac {2 c x}{y}-2 a x -b}{\sqrt {b^{2}+4 c}}\right )}{2 \sqrt {b^{2}+4 c}} = -\frac {\ln \left (a^{2} x^{2}+x b a -c \right )}{2}+\frac {a b \,\operatorname {arctanh}\left (\frac {a \left (2 a x +b \right )}{\sqrt {a^{2} \left (b^{2}+4 c \right )}}\right )}{\sqrt {a^{2} \left (b^{2}+4 c \right )}}+\ln \left (x \right )+c_2 \]

Simplifying the above gives

\begin{align*} -\frac {\sqrt {b^{2}+4 c}\, \ln \left (\frac {\left (y^{2}+x^{2} \left (a^{2} x^{2}+x b a -c \right )+\left (-2 a \,x^{2}-b x \right ) y\right ) \left (a^{2} x^{2}+x b a -c \right )}{y^{2}}\right )-2 \sqrt {b^{2}+4 c}\, \ln \left (\frac {\left (a^{2} x^{2}+x b a -c -a y\right ) x}{y}\right )-2 b \,\operatorname {arctanh}\left (\frac {-2 a^{2} x^{3}-2 b a \,x^{2}+2 y a x +b y+2 c x}{y \sqrt {b^{2}+4 c}}\right )}{2 \sqrt {b^{2}+4 c}} &= -\frac {\ln \left (a^{2} x^{2}+x b a -c \right )}{2}+\frac {a b \,\operatorname {arctanh}\left (\frac {a \left (2 a x +b \right )}{\sqrt {a^{2} \left (b^{2}+4 c \right )}}\right )}{\sqrt {a^{2} \left (b^{2}+4 c \right )}}+\ln \left (x \right )+c_2 \\ \end{align*}

Summary of solutions found

\begin{align*} -\frac {\sqrt {b^{2}+4 c}\, \ln \left (\frac {\left (y^{2}+x^{2} \left (a^{2} x^{2}+x b a -c \right )+\left (-2 a \,x^{2}-b x \right ) y\right ) \left (a^{2} x^{2}+x b a -c \right )}{y^{2}}\right )-2 \sqrt {b^{2}+4 c}\, \ln \left (\frac {\left (a^{2} x^{2}+x b a -c -a y\right ) x}{y}\right )-2 b \,\operatorname {arctanh}\left (\frac {-2 a^{2} x^{3}-2 b a \,x^{2}+2 y a x +b y+2 c x}{y \sqrt {b^{2}+4 c}}\right )}{2 \sqrt {b^{2}+4 c}} &= -\frac {\ln \left (a^{2} x^{2}+x b a -c \right )}{2}+\frac {a b \,\operatorname {arctanh}\left (\frac {a \left (2 a x +b \right )}{\sqrt {a^{2} \left (b^{2}+4 c \right )}}\right )}{\sqrt {a^{2} \left (b^{2}+4 c \right )}}+\ln \left (x \right )+c_2 \\ \end{align*}

2.24.2.2 Solved using first_order_ode_LIE

10.224 (sec)

Entering first order ode LIE solver

\begin{align*} y y^{\prime }&=\left (3 a x +b \right ) y-a^{2} x^{3}-b a \,x^{2}+c x \\ \end{align*}

Writing the ode as

\begin{align*} y^{\prime }&=\frac {-a^{2} x^{3}-b a \,x^{2}+3 y a x +b y +c x}{y}\\ 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*}

To determine \(\xi ,\eta \) then (A) is solved using ansatz. Making bivariate polynomials of degree 3 to use as anstaz gives

\begin{align*} \tag{1E} \xi &= x^{3} a_{7}+x^{2} y a_{8}+x \,y^{2} a_{9}+y^{3} a_{10}+x^{2} a_{4}+x y a_{5}+y^{2} a_{6}+x a_{2}+y a_{3}+a_{1} \\ \tag{2E} \eta &= x^{3} b_{7}+x^{2} y b_{8}+x \,y^{2} b_{9}+y^{3} b_{10}+x^{2} b_{4}+x y b_{5}+y^{2} b_{6}+x b_{2}+y b_{3}+b_{1} \\ \end{align*}

Where the unknown coefficients are

\[ \{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}, a_{8}, a_{9}, a_{10}, b_{1}, b_{2}, b_{3}, b_{4}, b_{5}, b_{6}, b_{7}, b_{8}, b_{9}, b_{10}\} \]

Substituting equations (1E,2E) and \(\omega \) into (A) gives

\begin{equation} \tag{5E} 3 x^{2} b_{7}+2 x y b_{8}+y^{2} b_{9}+2 x b_{4}+y b_{5}+b_{2}+\frac {\left (-a^{2} x^{3}-b a \,x^{2}+3 y a x +b y +c x \right ) \left (-3 x^{2} a_{7}+x^{2} b_{8}-2 x y a_{8}+2 x y b_{9}-y^{2} a_{9}+3 y^{2} b_{10}-2 x a_{4}+x b_{5}-y a_{5}+2 y b_{6}-a_{2}+b_{3}\right )}{y}-\frac {\left (-a^{2} x^{3}-b a \,x^{2}+3 y a x +b y +c x \right )^{2} \left (x^{2} a_{8}+2 x y a_{9}+3 y^{2} a_{10}+x a_{5}+2 y a_{6}+a_{3}\right )}{y^{2}}-\frac {\left (-3 a^{2} x^{2}-2 x b a +3 y a +c \right ) \left (x^{3} a_{7}+x^{2} y a_{8}+x \,y^{2} a_{9}+y^{3} a_{10}+x^{2} a_{4}+x y a_{5}+y^{2} a_{6}+x a_{2}+y a_{3}+a_{1}\right )}{y}-\left (\frac {3 a x +b}{y}-\frac {-a^{2} x^{3}-b a \,x^{2}+3 y a x +b y +c x}{y^{2}}\right ) \left (x^{3} b_{7}+x^{2} y b_{8}+x \,y^{2} b_{9}+y^{3} b_{10}+x^{2} b_{4}+x y b_{5}+y^{2} b_{6}+x b_{2}+y b_{3}+b_{1}\right ) = 0 \end{equation}

Putting the above in normal form gives

\[ \text {Expression too large to display} \]

Setting the numerator to zero gives

\begin{equation} \tag{6E} \text {Expression too large to display} \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} \text {Expression too large to display} \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} \text {Expression too large to display} \end{equation}

Setting each coefficients in (8E) to zero gives the following equations to solve

\begin{align*} c b_{1}&=0\\ -3 a a_{10}&=0\\ -24 a^{2} a_{10}&=0\\ 18 a^{3} a_{10}&=0\\ -a^{4} a_{8}&=0\\ -2 a^{4} a_{9}&=0\\ -3 a^{4} a_{10}&=0\\ -c a_{1}&=0\\ -a^{4} a_{5}-2 a^{3} b a_{8}&=0\\ -6 a^{3} b a_{10}+12 a^{3} a_{9}&=0\\ -2 a^{4} a_{6}-4 a^{3} b a_{9}+6 a^{3} a_{8}&=0\\ -16 a b a_{10}-6 a a_{9}+9 a b_{10}&=0\\ 24 a^{2} b a_{10}-14 a^{2} a_{9}-4 a^{2} b_{10}&=0\\ -a b b_{1}-c^{2} a_{3}+c b_{2}&=0\\ 2 a b a_{1}-2 b c a_{3}-2 c a_{2}+2 c b_{3}&=0\\ -a^{4} a_{3}-2 a^{3} b a_{5}-a^{2} b^{2} a_{8}+2 a^{2} c a_{8}-a^{2} b_{7}&=0\\ 2 a b c a_{3}-a^{2} b_{1}-a b b_{2}-c^{2} a_{5}+c b_{4}&=0\\ -b^{2} a_{3}-3 a a_{1}-b a_{2}+b b_{3}-c a_{3}+b_{2}&=0\\ -2 b^{2} a_{6}-3 a a_{3}-b a_{5}+2 b b_{6}-c a_{6}+b_{5}&=0\\ -2 a^{3} b a_{3}-a^{2} b^{2} a_{5}+2 a^{2} c a_{5}+2 a b c a_{8}-a^{2} b_{4}-a b b_{7}&=0\\ -3 a^{2} b^{2} a_{10}+12 a^{3} a_{6}+16 a^{2} b a_{9}+6 a^{2} c a_{10}-4 a^{2} a_{8}-3 a^{2} b_{9}&=0\\ -3 b^{2} a_{10}-3 a a_{6}-b a_{9}+3 b b_{10}-c a_{10}+b_{9}&=0\\ -a^{2} b^{2} a_{3}+2 a^{2} c a_{3}+2 a b c a_{5}-a^{2} b_{2}-a b b_{4}-c^{2} a_{8}+c b_{7}&=0\\ -4 a^{3} b a_{6}-2 a^{2} b^{2} a_{9}+6 a^{3} a_{5}+8 a^{2} b a_{8}+4 a^{2} c a_{9}+6 a^{2} a_{7}-2 a^{2} b_{8}&=0\\ 6 a \,b^{2} a_{10}-15 a^{2} a_{6}-9 a b a_{9}-4 a b b_{10}-18 a c a_{10}-9 a a_{8}+6 a b_{9}&=0\\ 2 a \,b^{2} a_{3}+3 a^{2} a_{1}+3 a b a_{2}-2 a b b_{3}-6 a c a_{3}-2 b c a_{5}-2 c^{2} a_{6}-3 c a_{4}+2 c b_{5}&=0\\ -4 a b a_{3}-b^{2} a_{5}-4 b c a_{6}-6 a a_{2}+3 a b_{3}-2 b a_{4}+b b_{5}-2 c a_{5}+3 c b_{6}+2 b_{4}&=0\\ 16 a^{2} b a_{6}+4 a \,b^{2} a_{9}+6 a b c a_{10}-5 a^{2} a_{5}-3 a^{2} b_{6}-2 a b a_{8}-3 a b b_{9}-12 a c a_{9}-12 a a_{7}+3 a b_{8}&=0\\ -10 a b a_{6}-2 b^{2} a_{9}-6 b c a_{10}-6 a a_{5}+6 a b_{6}-2 b a_{8}+2 b b_{9}-2 c a_{9}+4 c b_{10}+2 b_{8}&=0\\ -2 a^{2} b^{2} a_{6}+6 a^{3} a_{3}+8 a^{2} b a_{5}+4 a^{2} c a_{6}+2 a \,b^{2} a_{8}+4 a b c a_{9}+5 a^{2} a_{4}-2 a^{2} b_{5}+5 a b a_{7}-2 a b b_{8}-6 a c a_{8}&=0\\ 8 a^{2} b a_{3}+2 a \,b^{2} a_{5}+4 a b c a_{6}+4 a^{2} a_{2}-2 a^{2} b_{3}+4 a b a_{4}-2 a b b_{5}-6 a c a_{5}-2 b c a_{8}-2 c^{2} a_{9}-4 c a_{7}+2 c b_{8}&=0\\ 4 a \,b^{2} a_{6}-6 a^{2} a_{3}-3 a b a_{5}-3 a b b_{6}-12 a c a_{6}-b^{2} a_{8}-4 b c a_{9}-3 c^{2} a_{10}-9 a a_{4}+3 a b_{5}-3 b a_{7}+b b_{8}-3 c a_{8}+3 c b_{9}+3 b_{7}&=0 \end{align*}

Solving the above equations for the unknowns gives

\begin{align*} a_{1}&=0\\ a_{2}&=\frac {c b_{6}}{a}\\ a_{3}&=a_{3}\\ a_{4}&=-b b_{6}\\ a_{5}&=0\\ a_{6}&=0\\ a_{7}&=-a b_{6}\\ a_{8}&=0\\ a_{9}&=0\\ a_{10}&=0\\ b_{1}&=0\\ b_{2}&=c a_{3}\\ b_{3}&=\frac {a b a_{3}+c b_{6}}{a}\\ b_{4}&=-a b a_{3}\\ b_{5}&=3 a a_{3}-2 b b_{6}\\ b_{6}&=b_{6}\\ b_{7}&=-a^{2} a_{3}\\ b_{8}&=-3 a b_{6}\\ b_{9}&=0\\ b_{10}&=0 \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 {x \left (a^{2} x^{2}+x b a -c \right )}{a} \\ \eta &= -\frac {y \left (3 a^{2} x^{2}+2 x b a -y a -c \right )}{a} \\ \end{align*}

Shifting is now applied to make \(\xi =0\) in order to simplify the rest of the computation

\begin{align*} \eta &= \eta - \omega \left (x,y\right ) \xi \\ &= -\frac {y \left (3 a^{2} x^{2}+2 x b a -y a -c \right )}{a} - \left (\frac {-a^{2} x^{3}-b a \,x^{2}+3 y a x +b y +c x}{y}\right ) \left (-\frac {x \left (a^{2} x^{2}+x b a -c \right )}{a}\right ) \\ &= -\frac {\left (a^{2} x^{2}+\left (b x -y \right ) a -c \right ) \left (a^{2} x^{4}+x^{2} \left (b x -2 y \right ) a -b x y -c \,x^{2}+y^{2}\right )}{y a}\\ \xi &= 0 \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 \(S(R)\). Since \(\xi =0\) then in this special case

\begin{align*} R = x \end{align*}

\(S\) is found from

\begin{align*} S &= \int { \frac {1}{\eta }} dy\\ &= \int { \frac {1}{-\frac {\left (a^{2} x^{2}+\left (b x -y \right ) a -c \right ) \left (a^{2} x^{4}+x^{2} \left (b x -2 y \right ) a -b x y -c \,x^{2}+y^{2}\right )}{y a}}} dy \end{align*}

Which results in

\begin{align*} S&= \frac {a \ln \left (a^{2} x^{4}+a b \,x^{3}-2 a \,x^{2} y -b x y -c \,x^{2}+y^{2}\right )}{2 c}-\frac {a b x \,\operatorname {arctanh}\left (\frac {-2 a \,x^{2}-b x +2 y}{\sqrt {b^{2} x^{2}+4 c \,x^{2}}}\right )}{c \sqrt {b^{2} x^{2}+4 c \,x^{2}}}-\frac {a \ln \left (-a^{2} x^{2}-x b a +y a +c \right )}{c} \end{align*}

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 {-a^{2} x^{3}-b a \,x^{2}+3 y a x +b y +c x}{y} \end{align*}

Evaluating all the partial derivatives gives

\begin{align*} R_{x} &= 1\\ R_{y} &= 0\\ S_{x} &= -\frac {\left (a^{2} x^{3}+b a \,x^{2}+\left (-3 y a -c \right ) x -b y \right ) a}{\left (a^{2} x^{2}+x b a -y a -c \right ) \left (a^{2} x^{4}+a b \,x^{3}+\left (-2 y a -c \right ) x^{2}-b x y +y^{2}\right )}\\ S_{y} &= -\frac {y a}{\left (a^{2} x^{2}+x b a -y a -c \right ) \left (a^{2} x^{4}+a b \,x^{3}+\left (-2 y a -c \right ) x^{2}-b x y +y^{2}\right )} \end{align*}

Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.

\begin{align*} \frac {dS}{dR} &= 0\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} &= 0 \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 \(R,S\).

Since the ode has the form \(\frac {d}{d R}S \left (R \right )=f(R)\), then we only need to integrate \(f(R)\).

\begin{align*} \int {dS} &= \int {0\, dR} + c_2 \\ S \left (R \right ) &= c_2 \end{align*}

To complete the solution, we just need to transform the above back to \(x,y\) coordinates. This results in

\begin{align*} \frac {a \left (\ln \left (a^{2} x^{4}+a b \,x^{3}+\left (-2 a y-c \right ) x^{2}-b x y+y^{2}\right ) \sqrt {b^{2}+4 c}+2 b \,\operatorname {arctanh}\left (\frac {2 a \,x^{2}+b x -2 y}{\sqrt {b^{2}+4 c}\, x}\right )-2 \ln \left (-a^{2} x^{2}+\left (-b x +y\right ) a +c \right ) \sqrt {b^{2}+4 c}\right )}{2 \sqrt {b^{2}+4 c}\, c} = c_2 \end{align*}

Simplifying the above gives

\begin{align*} \frac {\left (b \,\operatorname {arctanh}\left (\frac {2 a \,x^{2}+b x -2 y}{\sqrt {b^{2}+4 c}\, x}\right )+\frac {\ln \left (a^{2} x^{4}+a b \,x^{3}-2 a \,x^{2} y-b x y-c \,x^{2}+y^{2}\right ) \sqrt {b^{2}+4 c}}{2}-\ln \left (-a^{2} x^{2}-x b a +a y+c \right ) \sqrt {b^{2}+4 c}\right ) a}{\sqrt {b^{2}+4 c}\, c} &= c_2 \\ \end{align*}

Summary of solutions found

\begin{align*} \frac {\left (b \,\operatorname {arctanh}\left (\frac {2 a \,x^{2}+b x -2 y}{\sqrt {b^{2}+4 c}\, x}\right )+\frac {\ln \left (a^{2} x^{4}+a b \,x^{3}-2 a \,x^{2} y-b x y-c \,x^{2}+y^{2}\right ) \sqrt {b^{2}+4 c}}{2}-\ln \left (-a^{2} x^{2}-x b a +a y+c \right ) \sqrt {b^{2}+4 c}\right ) a}{\sqrt {b^{2}+4 c}\, c} &= c_2 \\ \end{align*}

2.24.2.3 ✓ Maple. Time used: 0.003 (sec). Leaf size: 826

ode:=y(x)*diff(y(x),x) = (3*a*x+b)*y(x)-a^2*x^3-a*b*x^2+c*x; 
dsolve(ode,y(x), singsol=all);
 

\begin{align*} \text {Solution too large to show}\end{align*}

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 Chini 
differential order: 1; looking for linear symmetries 
trying exact 
trying Abel 
<- Abel successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y \left (x \right ) \left (\frac {d}{d x}y \left (x \right )\right )=\left (3 a x +b \right ) y \left (x \right )-a^{2} x^{3}-a b \,x^{2}+c x \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {\left (3 a x +b \right ) y \left (x \right )-a^{2} x^{3}-a b \,x^{2}+c x}{y \left (x \right )} \end {array} \]

2.24.2.4 ✓ Mathematica. Time used: 2.95 (sec). Leaf size: 194

ode=y[x]*D[y[x],x]==(3*a*x+b)*y[x]-a^2*x^3-a*b*x^2+c*x; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\[ \text {Solve}\left [\frac {2 a b \left (\text {RootSum}\left [\text {$\#$1}^4 a^2+\text {$\#$1}^3 a b-2 \text {$\#$1}^2 a y(x)-\text {$\#$1}^2 c-\text {$\#$1} b y(x)+y(x)^2\&,\frac {-2 \text {$\#$1}^3 a^2 \log (x-\text {$\#$1})-\text {$\#$1}^2 a b \log (x-\text {$\#$1})+2 \text {$\#$1} a y(x) \log (x-\text {$\#$1})+b y(x) \log (x-\text {$\#$1})+\text {$\#$1} c \log (x-\text {$\#$1})}{-4 \text {$\#$1}^3 a^2-3 \text {$\#$1}^2 a b+4 \text {$\#$1} a y(x)+2 \text {$\#$1} c+b y(x)}\&\right ]-\log \left (a \left (-a x^2-b x+y(x)\right )+c\right )\right )}{c (3 a+b+c-1)}=c_1,y(x)\right ] \]

2.24.2.5 ✗ Sympy

from sympy import * 
x = symbols("x") 
a = symbols("a") 
b = symbols("b") 
c = symbols("c") 
y = Function("y") 
ode = Eq(a**2*x**3 + a*b*x**2 - c*x - (3*a*x + b)*y(x) + y(x)*Derivative(y(x), x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

Timed Out
 

Python version: 3.12.3 (main, Aug 14 2025, 17:47:21) [GCC 13.3.0] 
Sympy version 1.14.0