Internal problem ID [9269]
Internal file name [OUTPUT/8205_Monday_June_06_2022_02_16_30_AM_63816909/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 936.
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
[[_1st_order, _with_linear_symmetries], _Abel]
\[ \boxed {y^{\prime }-y^{2}-\frac {7 x^{2} y}{16}+\frac {x y}{2}-y^{3}-\frac {3 y^{2} x^{2}}{8}+\frac {3 y^{2} x}{4}-\frac {3 y x^{4}}{64}+\frac {3 x^{3} y}{16}=-\frac {1}{4} x +1+\frac {5}{128} x^{4}-\frac {5}{64} x^{3}+\frac {1}{16} x^{2}+\frac {1}{512} x^{6}-\frac {3}{256} x^{5}} \]
Writing the ode as \begin {align*} y^{\prime }&=-\frac {1}{4} x +1+y^{2}+\frac {7}{16} x^{2} y -\frac {1}{2} x y +\frac {5}{128} x^{4}-\frac {5}{64} x^{3}+\frac {1}{16} x^{2}+y^{3}+\frac {3}{8} x^{2} y^{2}-\frac {3}{4} x \,y^{2}+\frac {3}{64} y \,x^{4}-\frac {3}{16} x^{3} y +\frac {1}{512} x^{6}-\frac {3}{256} x^{5}\\ 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}+\left (-\frac {1}{4} x +1+y^{2}+\frac {7}{16} x^{2} y -\frac {1}{2} x y +\frac {5}{128} x^{4}-\frac {5}{64} x^{3}+\frac {1}{16} x^{2}+y^{3}+\frac {3}{8} x^{2} y^{2}-\frac {3}{4} x \,y^{2}+\frac {3}{64} y \,x^{4}-\frac {3}{16} x^{3} y +\frac {1}{512} x^{6}-\frac {3}{256} x^{5}\right ) \left (b_{3}-a_{2}\right )-\left (-\frac {1}{4} x +1+y^{2}+\frac {7}{16} x^{2} y -\frac {1}{2} x y +\frac {5}{128} x^{4}-\frac {5}{64} x^{3}+\frac {1}{16} x^{2}+y^{3}+\frac {3}{8} x^{2} y^{2}-\frac {3}{4} x \,y^{2}+\frac {3}{64} y \,x^{4}-\frac {3}{16} x^{3} y +\frac {1}{512} x^{6}-\frac {3}{256} x^{5}\right )^{2} a_{3}-\left (-\frac {1}{4}+\frac {7}{8} x y -\frac {1}{2} y +\frac {5}{32} x^{3}-\frac {15}{64} x^{2}+\frac {1}{8} x +\frac {3}{4} x \,y^{2}-\frac {3}{4} y^{2}+\frac {3}{16} x^{3} y -\frac {9}{16} x^{2} y +\frac {3}{256} x^{5}-\frac {15}{256} x^{4}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )-\left (2 y +\frac {7}{16} x^{2}-\frac {1}{2} x +3 y^{2}+\frac {3}{4} x^{2} y -\frac {3}{2} x y +\frac {3}{64} x^{4}-\frac {3}{16} x^{3}\right ) \left (x b_{2}+y b_{3}+b_{1}\right ) = 0 \end{equation} Putting the above in normal form gives \[ \frac {15}{512} a_{3} x^{7} y^{2}+\frac {5}{2} a_{3} y^{4} x -\frac {15}{64} a_{3} y^{4} x^{4}-\frac {15}{4096} a_{3} y^{2} x^{8}+\frac {15}{8192} a_{3} y \,x^{9}+\frac {35}{1024} a_{3} y \,x^{7}-\frac {83}{1024} a_{3} x^{6} y +\frac {15}{16} a_{3} y^{4} x^{3}+\frac {15}{64} a_{3} y^{3} x^{5}-\frac {3}{16384} a_{3} y \,x^{10}-\frac {7}{8} x y a_{1}-\frac {3}{4} x \,y^{2} a_{1}-\frac {3}{16} x^{3} y a_{1}+\frac {9}{16} x^{2} y a_{1}-2 x y b_{2}-3 x \,y^{2} b_{2}-\frac {3}{4} x^{3} y b_{2}-\frac {3}{4} x^{2} y b_{1}+\frac {3}{2} x^{2} y b_{2}+\frac {3}{2} x y b_{1}-\frac {3}{64} x^{5} b_{2}-\frac {3}{64} x^{4} b_{1}+\frac {3}{16} x^{4} b_{2}+\frac {3}{16} x^{3} b_{1}+\frac {1}{2} a_{3} x -\frac {85}{4096} a_{3} x^{6}-a_{3} y^{6}-\frac {1}{262144} a_{3} x^{12}+\frac {5}{4096} a_{3} x^{9}-\frac {59}{16384} a_{3} x^{8}-\frac {31}{256} a_{3} x^{4}-\frac {19}{65536} a_{3} x^{10}+\frac {35}{4096} a_{3} x^{7}-a_{3} y^{4}-\frac {3}{2} a_{3} y^{2}-\frac {3}{16} a_{3} x^{2}+\frac {3}{16} a_{3} x^{3}+\frac {27}{512} a_{3} x^{5}-2 a_{3} y^{5}-\frac {5}{4} a_{3} y^{3}+\frac {3}{65536} a_{3} x^{11}+\frac {1}{2} y a_{1}-\frac {5}{32} x^{3} a_{1}+\frac {15}{64} x^{2} a_{1}-\frac {1}{8} x a_{1}+\frac {3}{4} y^{2} a_{1}-\frac {3}{256} x^{5} a_{1}+\frac {15}{256} x^{4} a_{1}-2 y b_{1}-\frac {7}{16} x^{3} b_{2}-\frac {7}{16} x^{2} b_{1}+\frac {1}{2} x^{2} b_{2}+\frac {1}{2} x b_{1}-3 y^{2} b_{1}-\frac {1}{4} x b_{3}-y^{2} b_{3}-y^{2} a_{2}+\frac {5}{128} x^{4} b_{3}-\frac {25}{128} x^{4} a_{2}-\frac {5}{64} x^{3} b_{3}+\frac {5}{16} x^{3} a_{2}+\frac {1}{16} x^{2} b_{3}-\frac {3}{16} x^{2} a_{2}-2 y^{3} b_{3}-y^{3} a_{2}+\frac {1}{512} x^{6} b_{3}-\frac {7}{512} x^{6} a_{2}-\frac {3}{256} x^{5} b_{3}+\frac {9}{128} x^{5} a_{2}+\frac {1}{2} x a_{2}+\frac {1}{4} y a_{3}+\frac {1}{4} a_{1}-a_{2}-a_{3}+b_{2}+b_{3}-\frac {21}{16} x^{2} y a_{2}+x y a_{2}-\frac {3}{8} x^{2} y^{2} b_{3}-\frac {9}{8} x^{2} y^{2} a_{2}+\frac {3}{4} x \,y^{2} b_{3}+\frac {3}{2} x \,y^{2} a_{2}-\frac {15}{64} y \,x^{4} a_{2}+\frac {3}{4} x^{3} y a_{2}+\frac {7}{8} a_{3} x y -\frac {159}{256} a_{3} y^{2} x^{4}-\frac {7}{4} a_{3} x^{2} y^{3}+\frac {3}{4} a_{3} x \,y^{3}+\frac {11}{16} a_{3} x^{3} y^{2}+\frac {9}{8} a_{3} x \,y^{2}-\frac {35}{16} a_{3} y^{4} x^{2}-\frac {57}{64} a_{3} x^{2} y -\frac {5}{512} a_{3} x^{8} y -\frac {15}{16} a_{3} x^{2} y^{2}-\frac {3}{4} a_{3} x^{2} y^{5}+\frac {1}{2} a_{3} x^{3} y -\frac {67}{256} a_{3} y \,x^{4}+\frac {3}{2} a_{3} x \,y^{5}-\frac {65}{512} a_{3} x^{6} y^{2}+\frac {25}{16} a_{3} x^{3} y^{3}+\frac {73}{512} a_{3} x^{5} y -\frac {25}{32} a_{3} x^{4} y^{3}+\frac {45}{128} a_{3} x^{5} y^{2}-\frac {5}{128} a_{3} x^{6} y^{3} = 0 \] Setting the numerator to zero gives \begin{equation} \tag{6E} -a_{3} x^{12}+12 a_{3} x^{11}-48 a_{3} y \,x^{10}-76 a_{3} x^{10}+480 a_{3} y \,x^{9}-960 a_{3} y^{2} x^{8}+320 a_{3} x^{9}-2560 a_{3} x^{8} y +7680 a_{3} x^{7} y^{2}-10240 a_{3} x^{6} y^{3}-944 a_{3} x^{8}+8960 a_{3} y \,x^{7}-33280 a_{3} x^{6} y^{2}+61440 a_{3} y^{3} x^{5}-61440 a_{3} y^{4} x^{4}+2240 a_{3} x^{7}-21248 a_{3} x^{6} y +92160 a_{3} x^{5} y^{2}-204800 a_{3} x^{4} y^{3}+245760 a_{3} y^{4} x^{3}-196608 a_{3} x^{2} y^{5}-3584 x^{6} a_{2}-5440 a_{3} x^{6}+512 x^{6} b_{3}+37376 a_{3} x^{5} y -162816 a_{3} y^{2} x^{4}+409600 a_{3} x^{3} y^{3}-573440 a_{3} y^{4} x^{2}+393216 a_{3} x \,y^{5}-262144 a_{3} y^{6}-3072 x^{5} a_{1}+18432 x^{5} a_{2}+13824 a_{3} x^{5}-12288 x^{5} b_{2}-3072 x^{5} b_{3}-61440 y \,x^{4} a_{2}-68608 a_{3} y \,x^{4}+180224 a_{3} x^{3} y^{2}-458752 a_{3} x^{2} y^{3}+655360 a_{3} y^{4} x -524288 a_{3} y^{5}+15360 x^{4} a_{1}-51200 x^{4} a_{2}-31744 a_{3} x^{4}-12288 x^{4} b_{1}+49152 x^{4} b_{2}+10240 x^{4} b_{3}-49152 x^{3} y a_{1}+196608 x^{3} y a_{2}+131072 a_{3} x^{3} y -196608 x^{3} y b_{2}-294912 x^{2} y^{2} a_{2}-245760 a_{3} x^{2} y^{2}-98304 x^{2} y^{2} b_{3}+196608 a_{3} x \,y^{3}-262144 a_{3} y^{4}-40960 x^{3} a_{1}+81920 x^{3} a_{2}+49152 a_{3} x^{3}+49152 x^{3} b_{1}-114688 x^{3} b_{2}-20480 x^{3} b_{3}+147456 x^{2} y a_{1}-344064 x^{2} y a_{2}-233472 a_{3} x^{2} y -196608 x^{2} y b_{1}+393216 x^{2} y b_{2}-196608 x \,y^{2} a_{1}+393216 x \,y^{2} a_{2}+294912 a_{3} x \,y^{2}-786432 x \,y^{2} b_{2}+196608 x \,y^{2} b_{3}-262144 y^{3} a_{2}-327680 a_{3} y^{3}-524288 y^{3} b_{3}+61440 x^{2} a_{1}-49152 x^{2} a_{2}-49152 a_{3} x^{2}-114688 x^{2} b_{1}+131072 x^{2} b_{2}+16384 x^{2} b_{3}-229376 x y a_{1}+262144 x y a_{2}+229376 a_{3} x y +393216 x y b_{1}-524288 x y b_{2}+196608 y^{2} a_{1}-262144 y^{2} a_{2}-393216 a_{3} y^{2}-786432 y^{2} b_{1}-262144 y^{2} b_{3}-32768 x a_{1}+131072 x a_{2}+131072 a_{3} x +131072 x b_{1}-65536 x b_{3}+131072 y a_{1}+65536 y a_{3}-524288 y b_{1}+65536 a_{1}-262144 a_{2}-262144 a_{3}+262144 b_{2}+262144 b_{3} = 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} -a_{3} v_{1}^{12}+12 a_{3} v_{1}^{11}-48 a_{3} v_{1}^{10} v_{2}-76 a_{3} v_{1}^{10}+480 a_{3} v_{1}^{9} v_{2}-960 a_{3} v_{1}^{8} v_{2}^{2}+320 a_{3} v_{1}^{9}-2560 a_{3} v_{1}^{8} v_{2}+7680 a_{3} v_{1}^{7} v_{2}^{2}-10240 a_{3} v_{1}^{6} v_{2}^{3}-944 a_{3} v_{1}^{8}+8960 a_{3} v_{1}^{7} v_{2}-33280 a_{3} v_{1}^{6} v_{2}^{2}+61440 a_{3} v_{1}^{5} v_{2}^{3}-61440 a_{3} v_{1}^{4} v_{2}^{4}+2240 a_{3} v_{1}^{7}-21248 a_{3} v_{1}^{6} v_{2}+92160 a_{3} v_{1}^{5} v_{2}^{2}-204800 a_{3} v_{1}^{4} v_{2}^{3}+245760 a_{3} v_{1}^{3} v_{2}^{4}-196608 a_{3} v_{1}^{2} v_{2}^{5}-3584 a_{2} v_{1}^{6}-5440 a_{3} v_{1}^{6}+37376 a_{3} v_{1}^{5} v_{2}-162816 a_{3} v_{1}^{4} v_{2}^{2}+409600 a_{3} v_{1}^{3} v_{2}^{3}-573440 a_{3} v_{1}^{2} v_{2}^{4}+393216 a_{3} v_{1} v_{2}^{5}-262144 a_{3} v_{2}^{6}+512 b_{3} v_{1}^{6}-3072 a_{1} v_{1}^{5}+18432 a_{2} v_{1}^{5}-61440 a_{2} v_{1}^{4} v_{2}+13824 a_{3} v_{1}^{5}-68608 a_{3} v_{1}^{4} v_{2}+180224 a_{3} v_{1}^{3} v_{2}^{2}-458752 a_{3} v_{1}^{2} v_{2}^{3}+655360 a_{3} v_{1} v_{2}^{4}-524288 a_{3} v_{2}^{5}-12288 b_{2} v_{1}^{5}-3072 b_{3} v_{1}^{5}+15360 a_{1} v_{1}^{4}-49152 a_{1} v_{1}^{3} v_{2}-51200 a_{2} v_{1}^{4}+196608 a_{2} v_{1}^{3} v_{2}-294912 a_{2} v_{1}^{2} v_{2}^{2}-31744 a_{3} v_{1}^{4}+131072 a_{3} v_{1}^{3} v_{2}-245760 a_{3} v_{1}^{2} v_{2}^{2}+196608 a_{3} v_{1} v_{2}^{3}-262144 a_{3} v_{2}^{4}-12288 b_{1} v_{1}^{4}+49152 b_{2} v_{1}^{4}-196608 b_{2} v_{1}^{3} v_{2}+10240 b_{3} v_{1}^{4}-98304 b_{3} v_{1}^{2} v_{2}^{2}-40960 a_{1} v_{1}^{3}+147456 a_{1} v_{1}^{2} v_{2}-196608 a_{1} v_{1} v_{2}^{2}+81920 a_{2} v_{1}^{3}-344064 a_{2} v_{1}^{2} v_{2}+393216 a_{2} v_{1} v_{2}^{2}-262144 a_{2} v_{2}^{3}+49152 a_{3} v_{1}^{3}-233472 a_{3} v_{1}^{2} v_{2}+294912 a_{3} v_{1} v_{2}^{2}-327680 a_{3} v_{2}^{3}+49152 b_{1} v_{1}^{3}-196608 b_{1} v_{1}^{2} v_{2}-114688 b_{2} v_{1}^{3}+393216 b_{2} v_{1}^{2} v_{2}-786432 b_{2} v_{1} v_{2}^{2}-20480 b_{3} v_{1}^{3}+196608 b_{3} v_{1} v_{2}^{2}-524288 b_{3} v_{2}^{3}+61440 a_{1} v_{1}^{2}-229376 a_{1} v_{1} v_{2}+196608 a_{1} v_{2}^{2}-49152 a_{2} v_{1}^{2}+262144 a_{2} v_{1} v_{2}-262144 a_{2} v_{2}^{2}-49152 a_{3} v_{1}^{2}+229376 a_{3} v_{1} v_{2}-393216 a_{3} v_{2}^{2}-114688 b_{1} v_{1}^{2}+393216 b_{1} v_{1} v_{2}-786432 b_{1} v_{2}^{2}+131072 b_{2} v_{1}^{2}-524288 b_{2} v_{1} v_{2}+16384 b_{3} v_{1}^{2}-262144 b_{3} v_{2}^{2}-32768 a_{1} v_{1}+131072 a_{1} v_{2}+131072 a_{2} v_{1}+131072 a_{3} v_{1}+65536 a_{3} v_{2}+131072 b_{1} v_{1}-524288 b_{1} v_{2}-65536 b_{3} v_{1}+65536 a_{1}-262144 a_{2}-262144 a_{3}+262144 b_{2}+262144 b_{3} = 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} 65536 a_{1}-262144 a_{2}-262144 a_{3}+262144 b_{2}+262144 b_{3}-48 a_{3} v_{1}^{10} v_{2}+480 a_{3} v_{1}^{9} v_{2}-960 a_{3} v_{1}^{8} v_{2}^{2}-2560 a_{3} v_{1}^{8} v_{2}+7680 a_{3} v_{1}^{7} v_{2}^{2}-10240 a_{3} v_{1}^{6} v_{2}^{3}+8960 a_{3} v_{1}^{7} v_{2}-33280 a_{3} v_{1}^{6} v_{2}^{2}+61440 a_{3} v_{1}^{5} v_{2}^{3}-61440 a_{3} v_{1}^{4} v_{2}^{4}-21248 a_{3} v_{1}^{6} v_{2}+92160 a_{3} v_{1}^{5} v_{2}^{2}-204800 a_{3} v_{1}^{4} v_{2}^{3}+245760 a_{3} v_{1}^{3} v_{2}^{4}-196608 a_{3} v_{1}^{2} v_{2}^{5}+37376 a_{3} v_{1}^{5} v_{2}-162816 a_{3} v_{1}^{4} v_{2}^{2}+409600 a_{3} v_{1}^{3} v_{2}^{3}-573440 a_{3} v_{1}^{2} v_{2}^{4}+393216 a_{3} v_{1} v_{2}^{5}+180224 a_{3} v_{1}^{3} v_{2}^{2}-458752 a_{3} v_{1}^{2} v_{2}^{3}+655360 a_{3} v_{1} v_{2}^{4}+196608 a_{3} v_{1} v_{2}^{3}+12 a_{3} v_{1}^{11}-76 a_{3} v_{1}^{10}+320 a_{3} v_{1}^{9}-944 a_{3} v_{1}^{8}+2240 a_{3} v_{1}^{7}-262144 a_{3} v_{2}^{6}-524288 a_{3} v_{2}^{5}-262144 a_{3} v_{2}^{4}-a_{3} v_{1}^{12}+\left (-196608 a_{1}+393216 a_{2}+294912 a_{3}-786432 b_{2}+196608 b_{3}\right ) v_{1} v_{2}^{2}+\left (-229376 a_{1}+262144 a_{2}+229376 a_{3}+393216 b_{1}-524288 b_{2}\right ) v_{1} v_{2}+\left (-61440 a_{2}-68608 a_{3}\right ) v_{1}^{4} v_{2}+\left (-49152 a_{1}+196608 a_{2}+131072 a_{3}-196608 b_{2}\right ) v_{1}^{3} v_{2}+\left (-294912 a_{2}-245760 a_{3}-98304 b_{3}\right ) v_{1}^{2} v_{2}^{2}+\left (147456 a_{1}-344064 a_{2}-233472 a_{3}-196608 b_{1}+393216 b_{2}\right ) v_{1}^{2} v_{2}+\left (61440 a_{1}-49152 a_{2}-49152 a_{3}-114688 b_{1}+131072 b_{2}+16384 b_{3}\right ) v_{1}^{2}+\left (-32768 a_{1}+131072 a_{2}+131072 a_{3}+131072 b_{1}-65536 b_{3}\right ) v_{1}+\left (-262144 a_{2}-327680 a_{3}-524288 b_{3}\right ) v_{2}^{3}+\left (196608 a_{1}-262144 a_{2}-393216 a_{3}-786432 b_{1}-262144 b_{3}\right ) v_{2}^{2}+\left (131072 a_{1}+65536 a_{3}-524288 b_{1}\right ) v_{2}+\left (-3584 a_{2}-5440 a_{3}+512 b_{3}\right ) v_{1}^{6}+\left (-3072 a_{1}+18432 a_{2}+13824 a_{3}-12288 b_{2}-3072 b_{3}\right ) v_{1}^{5}+\left (15360 a_{1}-51200 a_{2}-31744 a_{3}-12288 b_{1}+49152 b_{2}+10240 b_{3}\right ) v_{1}^{4}+\left (-40960 a_{1}+81920 a_{2}+49152 a_{3}+49152 b_{1}-114688 b_{2}-20480 b_{3}\right ) v_{1}^{3} = 0 \end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve \begin {align*} -573440 a_{3}&=0\\ -524288 a_{3}&=0\\ -458752 a_{3}&=0\\ -262144 a_{3}&=0\\ -204800 a_{3}&=0\\ -196608 a_{3}&=0\\ -162816 a_{3}&=0\\ -61440 a_{3}&=0\\ -33280 a_{3}&=0\\ -21248 a_{3}&=0\\ -10240 a_{3}&=0\\ -2560 a_{3}&=0\\ -960 a_{3}&=0\\ -944 a_{3}&=0\\ -76 a_{3}&=0\\ -48 a_{3}&=0\\ -a_{3}&=0\\ 12 a_{3}&=0\\ 320 a_{3}&=0\\ 480 a_{3}&=0\\ 2240 a_{3}&=0\\ 7680 a_{3}&=0\\ 8960 a_{3}&=0\\ 37376 a_{3}&=0\\ 61440 a_{3}&=0\\ 92160 a_{3}&=0\\ 180224 a_{3}&=0\\ 196608 a_{3}&=0\\ 245760 a_{3}&=0\\ 393216 a_{3}&=0\\ 409600 a_{3}&=0\\ 655360 a_{3}&=0\\ -61440 a_{2}-68608 a_{3}&=0\\ 131072 a_{1}+65536 a_{3}-524288 b_{1}&=0\\ -294912 a_{2}-245760 a_{3}-98304 b_{3}&=0\\ -262144 a_{2}-327680 a_{3}-524288 b_{3}&=0\\ -3584 a_{2}-5440 a_{3}+512 b_{3}&=0\\ -49152 a_{1}+196608 a_{2}+131072 a_{3}-196608 b_{2}&=0\\ -229376 a_{1}+262144 a_{2}+229376 a_{3}+393216 b_{1}-524288 b_{2}&=0\\ -196608 a_{1}+393216 a_{2}+294912 a_{3}-786432 b_{2}+196608 b_{3}&=0\\ -32768 a_{1}+131072 a_{2}+131072 a_{3}+131072 b_{1}-65536 b_{3}&=0\\ -3072 a_{1}+18432 a_{2}+13824 a_{3}-12288 b_{2}-3072 b_{3}&=0\\ 65536 a_{1}-262144 a_{2}-262144 a_{3}+262144 b_{2}+262144 b_{3}&=0\\ 147456 a_{1}-344064 a_{2}-233472 a_{3}-196608 b_{1}+393216 b_{2}&=0\\ 196608 a_{1}-262144 a_{2}-393216 a_{3}-786432 b_{1}-262144 b_{3}&=0\\ -40960 a_{1}+81920 a_{2}+49152 a_{3}+49152 b_{1}-114688 b_{2}-20480 b_{3}&=0\\ 15360 a_{1}-51200 a_{2}-31744 a_{3}-12288 b_{1}+49152 b_{2}+10240 b_{3}&=0\\ 61440 a_{1}-49152 a_{2}-49152 a_{3}-114688 b_{1}+131072 b_{2}+16384 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}&=\frac {a_{1}}{4}\\ b_{2}&=-\frac {a_{1}}{4}\\ 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}{4}+\frac {1}{4} \\ \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}{4}+\frac {1}{4}}{1}\\ &= -\frac {x}{4}+\frac {1}{4} \end {align*}
This is easily solved to give \begin {align*} y = -\frac {1}{8} x^{2}+\frac {1}{4} x +c_{1} \end {align*}
Where now the coordinate \(R\) is taken as the constant of integration. Hence \begin {align*} R &= \frac {1}{8} x^{2}-\frac {1}{4} 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 {1}{4} x +1+y^{2}+\frac {7}{16} x^{2} y -\frac {1}{2} x y +\frac {5}{128} x^{4}-\frac {5}{64} x^{3}+\frac {1}{16} x^{2}+y^{3}+\frac {3}{8} x^{2} y^{2}-\frac {3}{4} x \,y^{2}+\frac {3}{64} y \,x^{4}-\frac {3}{16} x^{3} y +\frac {1}{512} x^{6}-\frac {3}{256} x^{5} \end {align*}
Evaluating all the partial derivatives gives \begin {align*} R_{x} &= \frac {x}{4}-\frac {1}{4}\\ 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 {512}{x^{6}-6 x^{5}+\left (24 y +20\right ) x^{4}+\left (-96 y -40\right ) x^{3}+\left (192 y^{2}+224 y +32\right ) x^{2}+\left (-384 y^{2}-256 y \right ) x +512 y^{3}+512 y^{2}+384}\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 {4}{4 R^{3}+4 R^{2}+3} \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 {4}{4 R^{3}+4 R^{2}+3}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 {4}{4 \left (\frac {1}{8} x^{2}-\frac {1}{4} x +\textit {\_a} \right )^{3}+4 \left (\frac {1}{8} x^{2}-\frac {1}{4} x +\textit {\_a} \right )^{2}+3}d \textit {\_a} +c_{1} \end {align*}
Which simplifies to \begin {align*} x = \int _{}^{y}\frac {4}{4 \left (\frac {1}{8} x^{2}-\frac {1}{4} x +\textit {\_a} \right )^{3}+4 \left (\frac {1}{8} x^{2}-\frac {1}{4} x +\textit {\_a} \right )^{2}+3}d \textit {\_a} +c_{1} \end {align*}
This results in \begin {align*} x = \int _{}^{y}\frac {4}{4 \left (\frac {1}{8} x^{2}-\frac {1}{4} x +\textit {\_a} \right )^{3}+4 \left (\frac {1}{8} x^{2}-\frac {1}{4} x +\textit {\_a} \right )^{2}+3}d \textit {\_a} +c_{1} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} x &= \int _{}^{y}\frac {4}{4 \left (\frac {1}{8} x^{2}-\frac {1}{4} x +\textit {\_a} \right )^{3}+4 \left (\frac {1}{8} x^{2}-\frac {1}{4} x +\textit {\_a} \right )^{2}+3}d \textit {\_a} +c_{1} \\ \end{align*}
Verification of solutions
\[ x = \int _{}^{y}\frac {4}{4 \left (\frac {1}{8} x^{2}-\frac {1}{4} x +\textit {\_a} \right )^{3}+4 \left (\frac {1}{8} x^{2}-\frac {1}{4} x +\textit {\_a} \right )^{2}+3}d \textit {\_a} +c_{1} \] 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 }&=y^{3}+\left (1+\frac {3}{8} x^{2}-\frac {3}{4} x \right ) y^{2}+\left (\frac {7}{16} x^{2}-\frac {1}{2} x +\frac {3}{64} x^{4}-\frac {3}{16} x^{3}\right ) y-\frac {x}{4}+1+\frac {5 x^{4}}{128}-\frac {5 x^{3}}{64}+\frac {x^{2}}{16}+\frac {x^{6}}{512}-\frac {3 x^{5}}{256}\tag {1} \end {align*}
Therefore \begin {align*} f_0(x) &= -\frac {1}{4} x +1+\frac {5}{128} x^{4}-\frac {5}{64} x^{3}+\frac {1}{16} x^{2}+\frac {1}{512} x^{6}-\frac {3}{256} x^{5}\\ f_1(x) &= \frac {7}{16} x^{2}-\frac {1}{2} x +\frac {3}{64} x^{4}-\frac {3}{16} x^{3}\\ f_2(x) &= 1+\frac {3}{8} x^{2}-\frac {3}{4} x\\ f_3(x) &= 1 \end {align*}
Since \(f_2(x)=1+\frac {3}{8} x^{2}-\frac {3}{4} x\) 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 {1+\frac {3}{8} x^{2}-\frac {3}{4} x}{3} \right ) \\ &= u \left (x \right )-\frac {1}{3}-\frac {x^{2}}{8}+\frac {x}{4} \end {align*}
The above transformation applied to (1) gives a new ODE as \begin {align*} u^{\prime }\left (x \right ) = \frac {89}{108}+u \left (x \right )^{3}-\frac {u \left (x \right )}{3}\tag {2} \end {align*}
The above ODE (2) can now be solved as separable.
Integrating both sides gives \begin {align*} \int _{}^{u \left (x \right )}\frac {1}{\frac {89}{108}+\textit {\_a}^{3}-\frac {1}{3} \textit {\_a}}d \textit {\_a} = x +c_{2} \end {align*}
Substituting \(u=y-\frac {1}{3}-\frac {x^{2}}{8}+\frac {x}{4}\) in the above solution gives \begin {align*} \int _{}^{y-\frac {1}{3}-\frac {x^{2}}{8}+\frac {x}{4}}\frac {1}{\frac {89}{108}+\textit {\_a}^{3}-\frac {1}{3} \textit {\_a}}d \textit {\_a} = x +c_{2} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{y-\frac {1}{3}-\frac {x^{2}}{8}+\frac {x}{4}}\frac {1}{\frac {89}{108}+\textit {\_a}^{3}-\frac {1}{3} \textit {\_a}}d \textit {\_a} &= x +c_{2} \\ \end{align*}
Verification of solutions
\[ \int _{}^{y-\frac {1}{3}-\frac {x^{2}}{8}+\frac {x}{4}}\frac {1}{\frac {89}{108}+\textit {\_a}^{3}-\frac {1}{3} \textit {\_a}}d \textit {\_a} = x +c_{2} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-y^{2}-\frac {7 x^{2} y}{16}+\frac {x y}{2}-y^{3}-\frac {3 y^{2} x^{2}}{8}+\frac {3 y^{2} x}{4}-\frac {3 y x^{4}}{64}+\frac {3 x^{3} y}{16}=-\frac {1}{4} x +1+\frac {5}{128} x^{4}-\frac {5}{64} x^{3}+\frac {1}{16} x^{2}+\frac {1}{512} x^{6}-\frac {3}{256} x^{5} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {x}{4}+1+y^{2}+\frac {7 x^{2} y}{16}-\frac {x y}{2}+\frac {5 x^{4}}{128}-\frac {5 x^{3}}{64}+\frac {x^{2}}{16}+y^{3}+\frac {3 y^{2} x^{2}}{8}-\frac {3 y^{2} x}{4}+\frac {3 y x^{4}}{64}-\frac {3 x^{3} y}{16}+\frac {x^{6}}{512}-\frac {3 x^{5}}{256} \end {array} \]
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: 0.031 (sec). Leaf size: 39
dsolve(diff(y(x),x) = -1/4*x+1+y(x)^2+7/16*x^2*y(x)-1/2*x*y(x)+5/128*x^4-5/64*x^3+1/16*x^2+y(x)^3+3/8*x^2*y(x)^2-3/4*x*y(x)^2+3/64*y(x)*x^4-3/16*x^3*y(x)+1/512*x^6-3/256*x^5,y(x), singsol=all)
\[ y \left (x \right ) = -\frac {x^{2}}{8}+\frac {x}{4}+\operatorname {RootOf}\left (-x +4 \left (\int _{}^{\textit {\_Z}}\frac {1}{4 \textit {\_a}^{3}+4 \textit {\_a}^{2}+3}d \textit {\_a} \right )+c_{1} \right ) \]
✓ Solution by Mathematica
Time used: 0.194 (sec). Leaf size: 99
DSolve[y'[x] == 1 - x/4 + x^2/16 - (5*x^3)/64 + (5*x^4)/128 - (3*x^5)/256 + x^6/512 - (x*y[x])/2 + (7*x^2*y[x])/16 - (3*x^3*y[x])/16 + (3*x^4*y[x])/64 + y[x]^2 - (3*x*y[x]^2)/4 + (3*x^2*y[x]^2)/8 + y[x]^3,y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [-\frac {89}{3} \text {RootSum}\left [-89 \text {$\#$1}^3+6 \sqrt [3]{178} \text {$\#$1}-89\&,\frac {\log \left (\frac {2^{2/3} \left (\frac {1}{8} \left (3 x^2-6 x+8\right )+3 y(x)\right )}{\sqrt [3]{89}}-\text {$\#$1}\right )}{2 \sqrt [3]{178}-89 \text {$\#$1}^2}\&\right ]=\frac {89^{2/3} x}{18 \sqrt [3]{2}}+c_1,y(x)\right ] \]