Internal problem ID [9274]
Internal file name [OUTPUT/8210_Monday_June_06_2022_02_17_52_AM_58227135/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 941.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "first_order_ode_lie_symmetry_calculated"
Maple gives the following as the ode type
[[_1st_order, _with_linear_symmetries], _rational, [_Abel, `2nd type`, `class C`]]
\[ \boxed {y^{\prime }-\frac {-32 x y-72 x^{3}+32 x^{2}-32 x +64 y^{3}+48 y^{2} x^{2}-192 y^{2} x +12 y x^{4}-96 x^{3} y+192 x^{2} y+x^{6}-12 x^{5}+48 x^{4}}{64 y+16 x^{2}-64 x +64}=0} \]
Writing the ode as \begin {align*} y^{\prime }&=\frac {x^{6}-12 x^{5}+12 y \,x^{4}+48 x^{4}-96 x^{3} y +48 x^{2} y^{2}-72 x^{3}+192 x^{2} y -192 x \,y^{2}+64 y^{3}+32 x^{2}-32 x y -32 x}{16 x^{2}-64 x +64 y +64}\\ 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 (x^{6}-12 x^{5}+12 y \,x^{4}+48 x^{4}-96 x^{3} y +48 x^{2} y^{2}-72 x^{3}+192 x^{2} y -192 x \,y^{2}+64 y^{3}+32 x^{2}-32 x y -32 x \right ) \left (b_{3}-a_{2}\right )}{16 x^{2}-64 x +64 y +64}-\frac {\left (x^{6}-12 x^{5}+12 y \,x^{4}+48 x^{4}-96 x^{3} y +48 x^{2} y^{2}-72 x^{3}+192 x^{2} y -192 x \,y^{2}+64 y^{3}+32 x^{2}-32 x y -32 x \right )^{2} a_{3}}{256 \left (x^{2}-4 x +4 y +4\right )^{2}}-\left (\frac {6 x^{5}-60 x^{4}+48 x^{3} y +192 x^{3}-288 x^{2} y +96 x \,y^{2}-216 x^{2}+384 x y -192 y^{2}+64 x -32 y -32}{16 x^{2}-64 x +64 y +64}-\frac {\left (x^{6}-12 x^{5}+12 y \,x^{4}+48 x^{4}-96 x^{3} y +48 x^{2} y^{2}-72 x^{3}+192 x^{2} y -192 x \,y^{2}+64 y^{3}+32 x^{2}-32 x y -32 x \right ) \left (2 x -4\right )}{16 \left (x^{2}-4 x +4 y +4\right )^{2}}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )-\left (\frac {12 x^{4}-96 x^{3}+96 x^{2} y +192 x^{2}-384 x y +192 y^{2}-32 x}{16 x^{2}-64 x +64 y +64}-\frac {x^{6}-12 x^{5}+12 y \,x^{4}+48 x^{4}-96 x^{3} y +48 x^{2} y^{2}-72 x^{3}+192 x^{2} y -192 x \,y^{2}+64 y^{3}+32 x^{2}-32 x y -32 x}{4 \left (x^{2}-4 x +4 y +4\right )^{2}}\right ) \left (x b_{2}+y b_{3}+b_{1}\right ) = 0 \end{equation} Putting the above in normal form gives \[ -\frac {x^{12} a_{3}-24 x^{11} a_{3}+24 x^{10} y a_{3}+240 x^{10} a_{3}-480 x^{9} y a_{3}+240 x^{8} y^{2} a_{3}-1296 x^{9} a_{3}+3840 x^{8} y a_{3}-3840 x^{7} y^{2} a_{3}+1280 x^{6} y^{3} a_{3}+80 x^{8} a_{2}+4096 x^{8} a_{3}-16 x^{8} b_{3}-15552 x^{7} y a_{3}+23040 x^{6} y^{2} a_{3}-15360 x^{5} y^{3} a_{3}+3840 x^{4} y^{4} a_{3}+64 x^{7} a_{1}-1152 x^{7} a_{2}-7744 x^{7} a_{3}+128 x^{7} b_{2}+256 x^{7} b_{3}+1024 x^{6} y a_{2}+32896 x^{6} y a_{3}-128 x^{6} y b_{3}-62208 x^{5} y^{2} a_{3}+61440 x^{4} y^{3} a_{3}-30720 x^{3} y^{4} a_{3}+6144 x^{2} y^{5} a_{3}-896 x^{6} a_{1}+6592 x^{6} a_{2}+9024 x^{6} a_{3}+128 x^{6} b_{1}-1536 x^{6} b_{2}-1600 x^{6} b_{3}+768 x^{5} y a_{1}-10752 x^{5} y a_{2}-32640 x^{5} y a_{3}+1536 x^{5} y b_{2}+1536 x^{5} y b_{3}+4608 x^{4} y^{2} a_{2}+66048 x^{4} y^{2} a_{3}-82944 x^{3} y^{3} a_{3}+61440 x^{2} y^{4} a_{3}-24576 x \,y^{5} a_{3}+4096 y^{6} a_{3}+4992 x^{5} a_{1}-19200 x^{5} a_{2}-7680 x^{5} a_{3}-1536 x^{5} b_{1}+6912 x^{5} b_{2}+4992 x^{5} b_{3}-7680 x^{4} y a_{1}+40704 x^{4} y a_{2}+8832 x^{4} y a_{3}+1536 x^{4} y b_{1}-12288 x^{4} y b_{2}-6144 x^{4} y b_{3}+3072 x^{3} y^{2} a_{1}-30720 x^{3} y^{2} a_{2}+6144 x^{3} y^{2} b_{2}+8192 x^{2} y^{3} a_{2}-2048 x^{2} y^{3} a_{3}+2048 x^{2} y^{3} b_{3}-14208 x^{4} a_{1}+29696 x^{4} a_{2}+5632 x^{4} a_{3}+6912 x^{4} b_{1}-14592 x^{4} b_{2}-8192 x^{4} b_{3}+27648 x^{3} y a_{1}-67584 x^{3} y a_{2}+7168 x^{3} y a_{3}-12288 x^{3} y b_{1}+30720 x^{3} y b_{2}+9216 x^{3} y b_{3}-18432 x^{2} y^{2} a_{1}+58368 x^{2} y^{2} a_{2}-30720 x^{2} y^{2} a_{3}+6144 x^{2} y^{2} b_{1}-24576 x^{2} y^{2} b_{2}+3072 x^{2} y^{2} b_{3}+4096 x \,y^{3} a_{1}-24576 x \,y^{3} a_{2}+26624 x \,y^{3} a_{3}+8192 x \,y^{3} b_{2}-8192 x \,y^{3} b_{3}+4096 y^{4} a_{2}-8192 y^{4} a_{3}+4096 y^{4} b_{3}+21504 x^{3} a_{1}-22528 x^{3} a_{2}-2048 x^{3} a_{3}-14336 x^{3} b_{1}+14336 x^{3} b_{2}+7168 x^{3} b_{3}-44032 x^{2} y a_{1}+45056 x^{2} y a_{2}-13312 x^{2} y a_{3}+30720 x^{2} y b_{1}-26624 x^{2} y b_{2}-4096 x^{2} y b_{3}+30720 x \,y^{2} a_{1}-28672 x \,y^{2} a_{2}+28672 x \,y^{2} a_{3}-24576 x \,y^{2} b_{1}+12288 x \,y^{2} b_{2}-10240 x \,y^{2} b_{3}-8192 y^{3} a_{1}+4096 y^{3} a_{2}-14336 y^{3} a_{3}+8192 y^{3} b_{1}+8192 y^{3} b_{3}-15360 x^{2} a_{1}+8192 x^{2} a_{2}+1024 x^{2} a_{3}+12288 x^{2} b_{1}-6144 x^{2} b_{2}-4096 x^{2} b_{3}+28672 x y a_{1}-8192 x y a_{2}+4096 x y a_{3}-24576 x y b_{1}+8192 x y b_{2}+4096 x y b_{3}-14336 y^{2} a_{1}-4096 y^{2} a_{3}+12288 y^{2} b_{1}-4096 y^{2} b_{2}+4096 x a_{1}-4096 x a_{2}+8192 x b_{2}+2048 x b_{3}-4096 y a_{1}-2048 y a_{3}-8192 y b_{2}-2048 a_{1}-4096 b_{2}}{256 \left (x^{2}-4 x +4 y +4\right )^{2}} = 0 \] Setting the numerator to zero gives \begin{equation} \tag{6E} -x^{12} a_{3}+24 x^{11} a_{3}-24 x^{10} y a_{3}-240 x^{10} a_{3}+480 x^{9} y a_{3}-240 x^{8} y^{2} a_{3}+1296 x^{9} a_{3}-3840 x^{8} y a_{3}+3840 x^{7} y^{2} a_{3}-1280 x^{6} y^{3} a_{3}-80 x^{8} a_{2}-4096 x^{8} a_{3}+16 x^{8} b_{3}+15552 x^{7} y a_{3}-23040 x^{6} y^{2} a_{3}+15360 x^{5} y^{3} a_{3}-3840 x^{4} y^{4} a_{3}-64 x^{7} a_{1}+1152 x^{7} a_{2}+7744 x^{7} a_{3}-128 x^{7} b_{2}-256 x^{7} b_{3}-1024 x^{6} y a_{2}-32896 x^{6} y a_{3}+128 x^{6} y b_{3}+62208 x^{5} y^{2} a_{3}-61440 x^{4} y^{3} a_{3}+30720 x^{3} y^{4} a_{3}-6144 x^{2} y^{5} a_{3}+896 x^{6} a_{1}-6592 x^{6} a_{2}-9024 x^{6} a_{3}-128 x^{6} b_{1}+1536 x^{6} b_{2}+1600 x^{6} b_{3}-768 x^{5} y a_{1}+10752 x^{5} y a_{2}+32640 x^{5} y a_{3}-1536 x^{5} y b_{2}-1536 x^{5} y b_{3}-4608 x^{4} y^{2} a_{2}-66048 x^{4} y^{2} a_{3}+82944 x^{3} y^{3} a_{3}-61440 x^{2} y^{4} a_{3}+24576 x \,y^{5} a_{3}-4096 y^{6} a_{3}-4992 x^{5} a_{1}+19200 x^{5} a_{2}+7680 x^{5} a_{3}+1536 x^{5} b_{1}-6912 x^{5} b_{2}-4992 x^{5} b_{3}+7680 x^{4} y a_{1}-40704 x^{4} y a_{2}-8832 x^{4} y a_{3}-1536 x^{4} y b_{1}+12288 x^{4} y b_{2}+6144 x^{4} y b_{3}-3072 x^{3} y^{2} a_{1}+30720 x^{3} y^{2} a_{2}-6144 x^{3} y^{2} b_{2}-8192 x^{2} y^{3} a_{2}+2048 x^{2} y^{3} a_{3}-2048 x^{2} y^{3} b_{3}+14208 x^{4} a_{1}-29696 x^{4} a_{2}-5632 x^{4} a_{3}-6912 x^{4} b_{1}+14592 x^{4} b_{2}+8192 x^{4} b_{3}-27648 x^{3} y a_{1}+67584 x^{3} y a_{2}-7168 x^{3} y a_{3}+12288 x^{3} y b_{1}-30720 x^{3} y b_{2}-9216 x^{3} y b_{3}+18432 x^{2} y^{2} a_{1}-58368 x^{2} y^{2} a_{2}+30720 x^{2} y^{2} a_{3}-6144 x^{2} y^{2} b_{1}+24576 x^{2} y^{2} b_{2}-3072 x^{2} y^{2} b_{3}-4096 x \,y^{3} a_{1}+24576 x \,y^{3} a_{2}-26624 x \,y^{3} a_{3}-8192 x \,y^{3} b_{2}+8192 x \,y^{3} b_{3}-4096 y^{4} a_{2}+8192 y^{4} a_{3}-4096 y^{4} b_{3}-21504 x^{3} a_{1}+22528 x^{3} a_{2}+2048 x^{3} a_{3}+14336 x^{3} b_{1}-14336 x^{3} b_{2}-7168 x^{3} b_{3}+44032 x^{2} y a_{1}-45056 x^{2} y a_{2}+13312 x^{2} y a_{3}-30720 x^{2} y b_{1}+26624 x^{2} y b_{2}+4096 x^{2} y b_{3}-30720 x \,y^{2} a_{1}+28672 x \,y^{2} a_{2}-28672 x \,y^{2} a_{3}+24576 x \,y^{2} b_{1}-12288 x \,y^{2} b_{2}+10240 x \,y^{2} b_{3}+8192 y^{3} a_{1}-4096 y^{3} a_{2}+14336 y^{3} a_{3}-8192 y^{3} b_{1}-8192 y^{3} b_{3}+15360 x^{2} a_{1}-8192 x^{2} a_{2}-1024 x^{2} a_{3}-12288 x^{2} b_{1}+6144 x^{2} b_{2}+4096 x^{2} b_{3}-28672 x y a_{1}+8192 x y a_{2}-4096 x y a_{3}+24576 x y b_{1}-8192 x y b_{2}-4096 x y b_{3}+14336 y^{2} a_{1}+4096 y^{2} a_{3}-12288 y^{2} b_{1}+4096 y^{2} b_{2}-4096 x a_{1}+4096 x a_{2}-8192 x b_{2}-2048 x b_{3}+4096 y a_{1}+2048 y a_{3}+8192 y b_{2}+2048 a_{1}+4096 b_{2} = 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} \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} 2048 a_{1}+4096 b_{2}-a_{3} v_{1}^{12}+24 a_{3} v_{1}^{11}-240 a_{3} v_{1}^{10}+1296 a_{3} v_{1}^{9}-4096 a_{3} v_{2}^{6}-24 a_{3} v_{1}^{10} v_{2}+480 a_{3} v_{1}^{9} v_{2}-240 a_{3} v_{1}^{8} v_{2}^{2}-3840 a_{3} v_{1}^{8} v_{2}+3840 a_{3} v_{1}^{7} v_{2}^{2}-1280 a_{3} v_{1}^{6} v_{2}^{3}+15552 a_{3} v_{1}^{7} v_{2}-23040 a_{3} v_{1}^{6} v_{2}^{2}+15360 a_{3} v_{1}^{5} v_{2}^{3}-3840 a_{3} v_{1}^{4} v_{2}^{4}+62208 a_{3} v_{1}^{5} v_{2}^{2}-61440 a_{3} v_{1}^{4} v_{2}^{3}+30720 a_{3} v_{1}^{3} v_{2}^{4}-6144 a_{3} v_{1}^{2} v_{2}^{5}+82944 a_{3} v_{1}^{3} v_{2}^{3}-61440 a_{3} v_{1}^{2} v_{2}^{4}+24576 a_{3} v_{1} v_{2}^{5}+\left (-30720 a_{1}+28672 a_{2}-28672 a_{3}+24576 b_{1}-12288 b_{2}+10240 b_{3}\right ) v_{1} v_{2}^{2}+\left (-28672 a_{1}+8192 a_{2}-4096 a_{3}+24576 b_{1}-8192 b_{2}-4096 b_{3}\right ) v_{1} v_{2}+\left (-1024 a_{2}-32896 a_{3}+128 b_{3}\right ) v_{1}^{6} v_{2}+\left (-768 a_{1}+10752 a_{2}+32640 a_{3}-1536 b_{2}-1536 b_{3}\right ) v_{1}^{5} v_{2}+\left (-4608 a_{2}-66048 a_{3}\right ) v_{1}^{4} v_{2}^{2}+\left (7680 a_{1}-40704 a_{2}-8832 a_{3}-1536 b_{1}+12288 b_{2}+6144 b_{3}\right ) v_{1}^{4} v_{2}+\left (-3072 a_{1}+30720 a_{2}-6144 b_{2}\right ) v_{1}^{3} v_{2}^{2}+\left (-27648 a_{1}+67584 a_{2}-7168 a_{3}+12288 b_{1}-30720 b_{2}-9216 b_{3}\right ) v_{1}^{3} v_{2}+\left (-8192 a_{2}+2048 a_{3}-2048 b_{3}\right ) v_{1}^{2} v_{2}^{3}+\left (18432 a_{1}-58368 a_{2}+30720 a_{3}-6144 b_{1}+24576 b_{2}-3072 b_{3}\right ) v_{1}^{2} v_{2}^{2}+\left (44032 a_{1}-45056 a_{2}+13312 a_{3}-30720 b_{1}+26624 b_{2}+4096 b_{3}\right ) v_{1}^{2} v_{2}+\left (-4096 a_{1}+24576 a_{2}-26624 a_{3}-8192 b_{2}+8192 b_{3}\right ) v_{1} v_{2}^{3}+\left (-4992 a_{1}+19200 a_{2}+7680 a_{3}+1536 b_{1}-6912 b_{2}-4992 b_{3}\right ) v_{1}^{5}+\left (14208 a_{1}-29696 a_{2}-5632 a_{3}-6912 b_{1}+14592 b_{2}+8192 b_{3}\right ) v_{1}^{4}+\left (-21504 a_{1}+22528 a_{2}+2048 a_{3}+14336 b_{1}-14336 b_{2}-7168 b_{3}\right ) v_{1}^{3}+\left (15360 a_{1}-8192 a_{2}-1024 a_{3}-12288 b_{1}+6144 b_{2}+4096 b_{3}\right ) v_{1}^{2}+\left (-4096 a_{1}+4096 a_{2}-8192 b_{2}-2048 b_{3}\right ) v_{1}+\left (-4096 a_{2}+8192 a_{3}-4096 b_{3}\right ) v_{2}^{4}+\left (8192 a_{1}-4096 a_{2}+14336 a_{3}-8192 b_{1}-8192 b_{3}\right ) v_{2}^{3}+\left (14336 a_{1}+4096 a_{3}-12288 b_{1}+4096 b_{2}\right ) v_{2}^{2}+\left (4096 a_{1}+2048 a_{3}+8192 b_{2}\right ) v_{2}+\left (-80 a_{2}-4096 a_{3}+16 b_{3}\right ) v_{1}^{8}+\left (-64 a_{1}+1152 a_{2}+7744 a_{3}-128 b_{2}-256 b_{3}\right ) v_{1}^{7}+\left (896 a_{1}-6592 a_{2}-9024 a_{3}-128 b_{1}+1536 b_{2}+1600 b_{3}\right ) v_{1}^{6} = 0 \end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve \begin {align*} -61440 a_{3}&=0\\ -23040 a_{3}&=0\\ -6144 a_{3}&=0\\ -4096 a_{3}&=0\\ -3840 a_{3}&=0\\ -1280 a_{3}&=0\\ -240 a_{3}&=0\\ -24 a_{3}&=0\\ -a_{3}&=0\\ 24 a_{3}&=0\\ 480 a_{3}&=0\\ 1296 a_{3}&=0\\ 3840 a_{3}&=0\\ 15360 a_{3}&=0\\ 15552 a_{3}&=0\\ 24576 a_{3}&=0\\ 30720 a_{3}&=0\\ 62208 a_{3}&=0\\ 82944 a_{3}&=0\\ 2048 a_{1}+4096 b_{2}&=0\\ -4608 a_{2}-66048 a_{3}&=0\\ -3072 a_{1}+30720 a_{2}-6144 b_{2}&=0\\ 4096 a_{1}+2048 a_{3}+8192 b_{2}&=0\\ -8192 a_{2}+2048 a_{3}-2048 b_{3}&=0\\ -4096 a_{2}+8192 a_{3}-4096 b_{3}&=0\\ -1024 a_{2}-32896 a_{3}+128 b_{3}&=0\\ -80 a_{2}-4096 a_{3}+16 b_{3}&=0\\ -4096 a_{1}+4096 a_{2}-8192 b_{2}-2048 b_{3}&=0\\ 14336 a_{1}+4096 a_{3}-12288 b_{1}+4096 b_{2}&=0\\ -4096 a_{1}+24576 a_{2}-26624 a_{3}-8192 b_{2}+8192 b_{3}&=0\\ -768 a_{1}+10752 a_{2}+32640 a_{3}-1536 b_{2}-1536 b_{3}&=0\\ -64 a_{1}+1152 a_{2}+7744 a_{3}-128 b_{2}-256 b_{3}&=0\\ 8192 a_{1}-4096 a_{2}+14336 a_{3}-8192 b_{1}-8192 b_{3}&=0\\ -30720 a_{1}+28672 a_{2}-28672 a_{3}+24576 b_{1}-12288 b_{2}+10240 b_{3}&=0\\ -28672 a_{1}+8192 a_{2}-4096 a_{3}+24576 b_{1}-8192 b_{2}-4096 b_{3}&=0\\ -27648 a_{1}+67584 a_{2}-7168 a_{3}+12288 b_{1}-30720 b_{2}-9216 b_{3}&=0\\ -21504 a_{1}+22528 a_{2}+2048 a_{3}+14336 b_{1}-14336 b_{2}-7168 b_{3}&=0\\ -4992 a_{1}+19200 a_{2}+7680 a_{3}+1536 b_{1}-6912 b_{2}-4992 b_{3}&=0\\ 896 a_{1}-6592 a_{2}-9024 a_{3}-128 b_{1}+1536 b_{2}+1600 b_{3}&=0\\ 7680 a_{1}-40704 a_{2}-8832 a_{3}-1536 b_{1}+12288 b_{2}+6144 b_{3}&=0\\ 14208 a_{1}-29696 a_{2}-5632 a_{3}-6912 b_{1}+14592 b_{2}+8192 b_{3}&=0\\ 15360 a_{1}-8192 a_{2}-1024 a_{3}-12288 b_{1}+6144 b_{2}+4096 b_{3}&=0\\ 18432 a_{1}-58368 a_{2}+30720 a_{3}-6144 b_{1}+24576 b_{2}-3072 b_{3}&=0\\ 44032 a_{1}-45056 a_{2}+13312 a_{3}-30720 b_{1}+26624 b_{2}+4096 b_{3}&=0 \end {align*}
Solving the above equations for the unknowns gives \begin {align*} a_{1}&=a_{1}\\ a_{2}&=0\\ a_{3}&=0\\ b_{1}&=a_{1}\\ b_{2}&=-\frac {a_{1}}{2}\\ b_{3}&=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 &= 1 \\ \eta &= -\frac {x}{2}+1 \\ \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 {-\frac {x}{2}+1}{1}\\ &= -\frac {x}{2}+1 \end {align*}
This is easily solved to give \begin {align*} y = -\frac {1}{4} x^{2}+x +c_{1} \end {align*}
Where now the coordinate \(R\) is taken as the constant of integration. Hence \begin {align*} R &= \frac {1}{4} x^{2}-x +y \end {align*}
And \(S\) is found from \begin {align*} dS &= \frac {dx}{\xi } \\ &= \frac {dx}{1} \end {align*}
Integrating gives \begin {align*} S &= \int { \frac {dx}{T}}\\ &= x \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 {x^{6}-12 x^{5}+12 y \,x^{4}+48 x^{4}-96 x^{3} y +48 x^{2} y^{2}-72 x^{3}+192 x^{2} y -192 x \,y^{2}+64 y^{3}+32 x^{2}-32 x y -32 x}{16 x^{2}-64 x +64 y +64} \end {align*}
Evaluating all the partial derivatives gives \begin {align*} R_{x} &= \frac {x}{2}-1\\ R_{y} &= 1\\ S_{x} &= 1\\ 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 {16 x^{2}-64 x +64 y +64}{-64+x^{6}-12 x^{5}+12 \left (y +4\right ) x^{4}+32 \left (-3 y -2\right ) x^{3}+16 \left (3 y^{2}+12 y -1\right ) x^{2}+64 \left (-3 y^{2}+1\right ) x +64 y^{3}-64 y}\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 {R +1}{R^{3}-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 {R +1}{R^{3}-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*} x = \int _{}^{y}\frac {\frac {1}{4} x^{2}-x +\textit {\_a} +1}{\left (\frac {1}{4} x^{2}-x +\textit {\_a} \right )^{3}-\frac {x^{2}}{4}+x -\textit {\_a} -1}d \textit {\_a} +c_{1} \end {align*}
Which simplifies to \begin {align*} x = \int _{}^{y}\frac {\frac {1}{4} x^{2}-x +\textit {\_a} +1}{\left (\frac {1}{4} x^{2}-x +\textit {\_a} \right )^{3}-\frac {x^{2}}{4}+x -\textit {\_a} -1}d \textit {\_a} +c_{1} \end {align*}
This results in \begin {align*} x = \int _{}^{y}\frac {\frac {1}{4} x^{2}-x +\textit {\_a} +1}{\left (\frac {1}{4} x^{2}-x +\textit {\_a} \right )^{3}-\frac {x^{2}}{4}+x -\textit {\_a} -1}d \textit {\_a} +c_{1} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} x &= \int _{}^{y}\frac {\frac {1}{4} x^{2}-x +\textit {\_a} +1}{\left (\frac {1}{4} x^{2}-x +\textit {\_a} \right )^{3}-\frac {x^{2}}{4}+x -\textit {\_a} -1}d \textit {\_a} +c_{1} \\ \end{align*}
Verification of solutions
\[ x = \int _{}^{y}\frac {\frac {1}{4} x^{2}-x +\textit {\_a} +1}{\left (\frac {1}{4} x^{2}-x +\textit {\_a} \right )^{3}-\frac {x^{2}}{4}+x -\textit {\_a} -1}d \textit {\_a} +c_{1} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{6}+12 y x^{4}-12 x^{5}+48 y^{2} x^{2}-96 x^{3} y+48 x^{4}+64 y^{3}-192 y^{2} x +192 x^{2} y-16 x^{2} y^{\prime }-72 x^{3}-64 y^{\prime } y-32 x y+64 y^{\prime } x +32 x^{2}-64 y^{\prime }-32 x =0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {32 x y+72 x^{3}-32 x^{2}+32 x -64 y^{3}-48 y^{2} x^{2}+192 y^{2} x -12 y x^{4}+96 x^{3} y-192 x^{2} y-x^{6}+12 x^{5}-48 x^{4}}{-16 x^{2}-64 y+64 x -64} \end {array} \]
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 Chini differential order: 1; looking for linear symmetries differential order: 1; found: 1 linear symmetries. Trying reduction of order 1st order, trying the canonical coordinates of the invariance group <- 1st order, canonical coordinates successful`
✓ Solution by Maple
Time used: 0.032 (sec). Leaf size: 35
dsolve(diff(y(x),x) = (-32*x*y(x)-72*x^3+32*x^2-32*x+64*y(x)^3+48*x^2*y(x)^2-192*x*y(x)^2+12*y(x)*x^4-96*x^3*y(x)+192*x^2*y(x)+x^6-12*x^5+48*x^4)/(64*y(x)+16*x^2-64*x+64),y(x), singsol=all)
\[ y \left (x \right ) = -\frac {x^{2}}{4}+x +\operatorname {RootOf}\left (-x +\int _{}^{\textit {\_Z}}\frac {\textit {\_a} +1}{\textit {\_a}^{3}-\textit {\_a} -1}d \textit {\_a} +c_{1} \right ) \]
✓ Solution by Mathematica
Time used: 0.274 (sec). Leaf size: 53
DSolve[y'[x] == (-32*x + 32*x^2 - 72*x^3 + 48*x^4 - 12*x^5 + x^6 - 32*x*y[x] + 192*x^2*y[x] - 96*x^3*y[x] + 12*x^4*y[x] - 192*x*y[x]^2 + 48*x^2*y[x]^2 + 64*y[x]^3)/(64 - 64*x + 16*x^2 + 64*y[x]),y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [x-8 \text {RootSum}\left [11776 \text {$\#$1}^3-40 \text {$\#$1}-1\&,\text {$\#$1} \log \left (17664 \text {$\#$1}^2-1472 \text {$\#$1}+11 x^2+44 y(x)-44 x-40\right )\&\right ]=c_1,y(x)\right ] \]