Internal problem ID [9268]
Internal file name [OUTPUT/8204_Monday_June_06_2022_02_16_17_AM_40150835/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 935.
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]
\[ \boxed {y^{\prime }-y^{2}-\frac {7 x^{2} y}{2}+2 x y-y^{3}-\frac {3 y^{2} x^{2}}{4}+3 y^{2} x -\frac {3 y x^{4}}{16}+\frac {3 x^{3} y}{2}=-\frac {1}{2} x +1+\frac {13}{16} x^{4}-\frac {3}{2} x^{3}+x^{2}+\frac {1}{64} x^{6}-\frac {3}{16} x^{5}} \]
Writing the ode as \begin {align*} y^{\prime }&=-\frac {1}{2} x +1+y^{2}+\frac {7}{2} x^{2} y -2 x y +\frac {13}{16} x^{4}-\frac {3}{2} x^{3}+x^{2}+y^{3}+\frac {3}{4} x^{2} y^{2}-3 x \,y^{2}+\frac {3}{16} y \,x^{4}-\frac {3}{2} x^{3} y +\frac {1}{64} x^{6}-\frac {3}{16} 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}{2} x +1+y^{2}+\frac {7}{2} x^{2} y -2 x y +\frac {13}{16} x^{4}-\frac {3}{2} x^{3}+x^{2}+y^{3}+\frac {3}{4} x^{2} y^{2}-3 x \,y^{2}+\frac {3}{16} y \,x^{4}-\frac {3}{2} x^{3} y +\frac {1}{64} x^{6}-\frac {3}{16} x^{5}\right ) \left (b_{3}-a_{2}\right )-\left (-\frac {1}{2} x +1+y^{2}+\frac {7}{2} x^{2} y -2 x y +\frac {13}{16} x^{4}-\frac {3}{2} x^{3}+x^{2}+y^{3}+\frac {3}{4} x^{2} y^{2}-3 x \,y^{2}+\frac {3}{16} y \,x^{4}-\frac {3}{2} x^{3} y +\frac {1}{64} x^{6}-\frac {3}{16} x^{5}\right )^{2} a_{3}-\left (-\frac {1}{2}+7 x y -2 y +\frac {13}{4} x^{3}-\frac {9}{2} x^{2}+2 x +\frac {3}{2} x \,y^{2}-3 y^{2}+\frac {3}{4} x^{3} y -\frac {9}{2} x^{2} y +\frac {3}{32} x^{5}-\frac {15}{16} x^{4}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )-\left (2 y +\frac {7}{2} x^{2}-2 x +3 y^{2}+\frac {3}{2} x^{2} y -6 x y +\frac {3}{16} x^{4}-\frac {3}{2} x^{3}\right ) \left (x b_{2}+y b_{3}+b_{1}\right ) = 0 \end{equation} Putting the above in normal form gives \[ -\frac {3}{2} x^{3} y b_{2}-\frac {3}{2} x^{2} y b_{1}+6 x^{2} y b_{2}+6 x y b_{1}+2 y a_{1}-\frac {13}{4} x^{3} a_{1}+\frac {9}{2} x^{2} a_{1}-2 x a_{1}+3 y^{2} a_{1}-\frac {3}{32} x^{5} a_{1}+\frac {15}{16} x^{4} a_{1}-2 y b_{1}-\frac {7}{2} x^{3} b_{2}-\frac {7}{2} x^{2} b_{1}+2 x^{2} b_{2}+2 x b_{1}-3 y^{2} b_{1}-\frac {3}{16} x^{5} b_{2}-\frac {3}{16} x^{4} b_{1}+\frac {3}{2} x^{4} b_{2}+\frac {3}{2} x^{3} b_{1}+a_{3} x -\frac {131}{32} a_{3} x^{6}-a_{3} y^{6}-\frac {1}{4096} a_{3} x^{12}+\frac {45}{128} a_{3} x^{9}-\frac {321}{256} a_{3} x^{8}-\frac {33}{8} a_{3} x^{4}-\frac {31}{512} a_{3} x^{10}+\frac {181}{64} a_{3} x^{7}-a_{3} y^{4}-\frac {9}{4} a_{3} x^{2}+4 a_{3} x^{3}+\frac {67}{16} a_{3} x^{5}-2 a_{3} y^{5}+a_{3} y^{3}+\frac {3}{512} a_{3} x^{11}-y^{3} a_{2}+\frac {1}{64} x^{6} b_{3}-\frac {7}{64} x^{6} a_{2}-\frac {3}{16} x^{5} b_{3}+\frac {9}{8} x^{5} a_{2}-\frac {1}{2} x b_{3}-y^{2} b_{3}-y^{2} a_{2}+\frac {13}{16} x^{4} b_{3}-\frac {65}{16} x^{4} a_{2}-\frac {3}{2} x^{3} b_{3}+6 x^{3} a_{2}+x^{2} b_{3}-3 x^{2} a_{2}-2 y^{3} b_{3}+\frac {1}{2} a_{1}-a_{2}-a_{3}+b_{2}+b_{3}+x a_{2}+\frac {1}{2} y a_{3}-\frac {21}{2} x^{2} y a_{2}+4 x y a_{2}-\frac {3}{4} x^{2} y^{2} b_{3}-\frac {9}{4} x^{2} y^{2} a_{2}+3 x \,y^{2} b_{3}+6 x \,y^{2} a_{2}-\frac {15}{16} y \,x^{4} a_{2}+6 x^{3} y a_{2}+2 a_{3} x y -\frac {243}{8} a_{3} y^{2} x^{4}-21 a_{3} x^{2} y^{3}+\frac {7}{2} a_{3} x \,y^{3}+23 a_{3} x^{3} y^{2}-\frac {35}{2} a_{3} y^{4} x^{2}-\frac {9}{2} a_{3} x^{2} y -\frac {125}{128} a_{3} x^{8} y -6 a_{3} x^{2} y^{2}-\frac {3}{2} a_{3} x^{2} y^{5}+\frac {29}{4} a_{3} x^{3} y -\frac {223}{16} a_{3} y \,x^{4}+6 a_{3} x \,y^{5}-\frac {95}{16} a_{3} x^{6} y^{2}+30 a_{3} x^{3} y^{3}+\frac {539}{32} a_{3} x^{5} y -\frac {65}{4} a_{3} x^{4} y^{3}+\frac {75}{4} a_{3} x^{5} y^{2}-\frac {5}{16} a_{3} x^{6} y^{3}+\frac {15}{16} a_{3} x^{7} y^{2}+10 a_{3} y^{4} x -\frac {15}{16} a_{3} y^{4} x^{4}-\frac {15}{256} a_{3} y^{2} x^{8}+\frac {15}{128} a_{3} y \,x^{9}+\frac {35}{8} a_{3} y \,x^{7}-\frac {181}{16} a_{3} x^{6} y +\frac {15}{2} a_{3} y^{4} x^{3}+\frac {15}{4} a_{3} y^{3} x^{5}-\frac {3}{512} a_{3} y \,x^{10}-7 x y a_{1}-\frac {3}{2} x \,y^{2} a_{1}-\frac {3}{4} x^{3} y a_{1}+\frac {9}{2} x^{2} y a_{1}-2 x y b_{2}-3 x \,y^{2} b_{2} = 0 \] Setting the numerator to zero gives \begin{equation} \tag{6E} -a_{3} x^{12}+24 a_{3} x^{11}-24 a_{3} y \,x^{10}-248 a_{3} x^{10}+480 a_{3} y \,x^{9}-240 a_{3} y^{2} x^{8}+1440 a_{3} x^{9}-4000 a_{3} x^{8} y +3840 a_{3} x^{7} y^{2}-1280 a_{3} x^{6} y^{3}-5136 a_{3} x^{8}+17920 a_{3} y \,x^{7}-24320 a_{3} x^{6} y^{2}+15360 a_{3} y^{3} x^{5}-3840 a_{3} y^{4} x^{4}+11584 a_{3} x^{7}-46336 a_{3} x^{6} y +76800 a_{3} x^{5} y^{2}-66560 a_{3} x^{4} y^{3}+30720 a_{3} y^{4} x^{3}-6144 a_{3} x^{2} y^{5}-448 x^{6} a_{2}-16768 a_{3} x^{6}+64 x^{6} b_{3}+68992 a_{3} x^{5} y -124416 a_{3} y^{2} x^{4}+122880 a_{3} x^{3} y^{3}-71680 a_{3} y^{4} x^{2}+24576 a_{3} x \,y^{5}-4096 a_{3} y^{6}-384 x^{5} a_{1}+4608 x^{5} a_{2}+17152 a_{3} x^{5}-768 x^{5} b_{2}-768 x^{5} b_{3}-3840 y \,x^{4} a_{2}-57088 a_{3} y \,x^{4}+94208 a_{3} x^{3} y^{2}-86016 a_{3} x^{2} y^{3}+40960 a_{3} y^{4} x -8192 a_{3} y^{5}+3840 x^{4} a_{1}-16640 x^{4} a_{2}-16896 a_{3} x^{4}-768 x^{4} b_{1}+6144 x^{4} b_{2}+3328 x^{4} b_{3}-3072 x^{3} y a_{1}+24576 x^{3} y a_{2}+29696 a_{3} x^{3} y -6144 x^{3} y b_{2}-9216 x^{2} y^{2} a_{2}-24576 a_{3} x^{2} y^{2}-3072 x^{2} y^{2} b_{3}+14336 a_{3} x \,y^{3}-4096 a_{3} y^{4}-13312 x^{3} a_{1}+24576 x^{3} a_{2}+16384 a_{3} x^{3}+6144 x^{3} b_{1}-14336 x^{3} b_{2}-6144 x^{3} b_{3}+18432 x^{2} y a_{1}-43008 x^{2} y a_{2}-18432 a_{3} x^{2} y -6144 x^{2} y b_{1}+24576 x^{2} y b_{2}-6144 x \,y^{2} a_{1}+24576 x \,y^{2} a_{2}-12288 x \,y^{2} b_{2}+12288 x \,y^{2} b_{3}-4096 y^{3} a_{2}+4096 a_{3} y^{3}-8192 y^{3} b_{3}+18432 x^{2} a_{1}-12288 x^{2} a_{2}-9216 a_{3} x^{2}-14336 x^{2} b_{1}+8192 x^{2} b_{2}+4096 x^{2} b_{3}-28672 x y a_{1}+16384 x y a_{2}+8192 a_{3} x y +24576 x y b_{1}-8192 x y b_{2}+12288 y^{2} a_{1}-4096 y^{2} a_{2}-12288 y^{2} b_{1}-4096 y^{2} b_{3}-8192 x a_{1}+4096 x a_{2}+4096 a_{3} x +8192 x b_{1}-2048 x b_{3}+8192 y a_{1}+2048 y a_{3}-8192 y b_{1}+2048 a_{1}-4096 a_{2}-4096 a_{3}+4096 b_{2}+4096 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}+24 a_{3} v_{1}^{11}-24 a_{3} v_{1}^{10} v_{2}-248 a_{3} v_{1}^{10}+480 a_{3} v_{1}^{9} v_{2}-240 a_{3} v_{1}^{8} v_{2}^{2}+1440 a_{3} v_{1}^{9}-4000 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}-5136 a_{3} v_{1}^{8}+17920 a_{3} v_{1}^{7} v_{2}-24320 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}+11584 a_{3} v_{1}^{7}-46336 a_{3} v_{1}^{6} v_{2}+76800 a_{3} v_{1}^{5} v_{2}^{2}-66560 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}-448 a_{2} v_{1}^{6}-16768 a_{3} v_{1}^{6}+68992 a_{3} v_{1}^{5} v_{2}-124416 a_{3} v_{1}^{4} v_{2}^{2}+122880 a_{3} v_{1}^{3} v_{2}^{3}-71680 a_{3} v_{1}^{2} v_{2}^{4}+24576 a_{3} v_{1} v_{2}^{5}-4096 a_{3} v_{2}^{6}+64 b_{3} v_{1}^{6}-384 a_{1} v_{1}^{5}+4608 a_{2} v_{1}^{5}-3840 a_{2} v_{1}^{4} v_{2}+17152 a_{3} v_{1}^{5}-57088 a_{3} v_{1}^{4} v_{2}+94208 a_{3} v_{1}^{3} v_{2}^{2}-86016 a_{3} v_{1}^{2} v_{2}^{3}+40960 a_{3} v_{1} v_{2}^{4}-8192 a_{3} v_{2}^{5}-768 b_{2} v_{1}^{5}-768 b_{3} v_{1}^{5}+3840 a_{1} v_{1}^{4}-3072 a_{1} v_{1}^{3} v_{2}-16640 a_{2} v_{1}^{4}+24576 a_{2} v_{1}^{3} v_{2}-9216 a_{2} v_{1}^{2} v_{2}^{2}-16896 a_{3} v_{1}^{4}+29696 a_{3} v_{1}^{3} v_{2}-24576 a_{3} v_{1}^{2} v_{2}^{2}+14336 a_{3} v_{1} v_{2}^{3}-4096 a_{3} v_{2}^{4}-768 b_{1} v_{1}^{4}+6144 b_{2} v_{1}^{4}-6144 b_{2} v_{1}^{3} v_{2}+3328 b_{3} v_{1}^{4}-3072 b_{3} v_{1}^{2} v_{2}^{2}-13312 a_{1} v_{1}^{3}+18432 a_{1} v_{1}^{2} v_{2}-6144 a_{1} v_{1} v_{2}^{2}+24576 a_{2} v_{1}^{3}-43008 a_{2} v_{1}^{2} v_{2}+24576 a_{2} v_{1} v_{2}^{2}-4096 a_{2} v_{2}^{3}+16384 a_{3} v_{1}^{3}-18432 a_{3} v_{1}^{2} v_{2}+4096 a_{3} v_{2}^{3}+6144 b_{1} v_{1}^{3}-6144 b_{1} v_{1}^{2} v_{2}-14336 b_{2} v_{1}^{3}+24576 b_{2} v_{1}^{2} v_{2}-12288 b_{2} v_{1} v_{2}^{2}-6144 b_{3} v_{1}^{3}+12288 b_{3} v_{1} v_{2}^{2}-8192 b_{3} v_{2}^{3}+18432 a_{1} v_{1}^{2}-28672 a_{1} v_{1} v_{2}+12288 a_{1} v_{2}^{2}-12288 a_{2} v_{1}^{2}+16384 a_{2} v_{1} v_{2}-4096 a_{2} v_{2}^{2}-9216 a_{3} v_{1}^{2}+8192 a_{3} v_{1} v_{2}-14336 b_{1} v_{1}^{2}+24576 b_{1} v_{1} v_{2}-12288 b_{1} v_{2}^{2}+8192 b_{2} v_{1}^{2}-8192 b_{2} v_{1} v_{2}+4096 b_{3} v_{1}^{2}-4096 b_{3} v_{2}^{2}-8192 a_{1} v_{1}+8192 a_{1} v_{2}+4096 a_{2} v_{1}+4096 a_{3} v_{1}+2048 a_{3} v_{2}+8192 b_{1} v_{1}-8192 b_{1} v_{2}-2048 b_{3} v_{1}+2048 a_{1}-4096 a_{2}-4096 a_{3}+4096 b_{2}+4096 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} \left (-3840 a_{2}-57088 a_{3}\right ) v_{1}^{4} v_{2}+\left (-3072 a_{1}+24576 a_{2}+29696 a_{3}-6144 b_{2}\right ) v_{1}^{3} v_{2}+\left (-9216 a_{2}-24576 a_{3}-3072 b_{3}\right ) v_{1}^{2} v_{2}^{2}+\left (18432 a_{1}-43008 a_{2}-18432 a_{3}-6144 b_{1}+24576 b_{2}\right ) v_{1}^{2} v_{2}+\left (-6144 a_{1}+24576 a_{2}-12288 b_{2}+12288 b_{3}\right ) v_{1} v_{2}^{2}+\left (-28672 a_{1}+16384 a_{2}+8192 a_{3}+24576 b_{1}-8192 b_{2}\right ) v_{1} v_{2}+2048 a_{1}-4096 a_{2}-4096 a_{3}+4096 b_{2}+4096 b_{3}+\left (18432 a_{1}-12288 a_{2}-9216 a_{3}-14336 b_{1}+8192 b_{2}+4096 b_{3}\right ) v_{1}^{2}+\left (-8192 a_{1}+4096 a_{2}+4096 a_{3}+8192 b_{1}-2048 b_{3}\right ) v_{1}+\left (-4096 a_{2}+4096 a_{3}-8192 b_{3}\right ) v_{2}^{3}+\left (12288 a_{1}-4096 a_{2}-12288 b_{1}-4096 b_{3}\right ) v_{2}^{2}+\left (8192 a_{1}+2048 a_{3}-8192 b_{1}\right ) v_{2}+\left (-448 a_{2}-16768 a_{3}+64 b_{3}\right ) v_{1}^{6}+\left (-384 a_{1}+4608 a_{2}+17152 a_{3}-768 b_{2}-768 b_{3}\right ) v_{1}^{5}+\left (3840 a_{1}-16640 a_{2}-16896 a_{3}-768 b_{1}+6144 b_{2}+3328 b_{3}\right ) v_{1}^{4}+\left (-13312 a_{1}+24576 a_{2}+16384 a_{3}+6144 b_{1}-14336 b_{2}-6144 b_{3}\right ) v_{1}^{3}+122880 a_{3} v_{1}^{3} v_{2}^{3}-71680 a_{3} v_{1}^{2} v_{2}^{4}+24576 a_{3} v_{1} v_{2}^{5}+94208 a_{3} v_{1}^{3} v_{2}^{2}-86016 a_{3} v_{1}^{2} v_{2}^{3}+40960 a_{3} v_{1} v_{2}^{4}+14336 a_{3} v_{1} v_{2}^{3}-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}-4000 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}+17920 a_{3} v_{1}^{7} v_{2}-24320 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}-46336 a_{3} v_{1}^{6} v_{2}+76800 a_{3} v_{1}^{5} v_{2}^{2}-66560 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}+68992 a_{3} v_{1}^{5} v_{2}-124416 a_{3} v_{1}^{4} v_{2}^{2}-a_{3} v_{1}^{12}+24 a_{3} v_{1}^{11}-248 a_{3} v_{1}^{10}+1440 a_{3} v_{1}^{9}-5136 a_{3} v_{1}^{8}+11584 a_{3} v_{1}^{7}-4096 a_{3} v_{2}^{6}-8192 a_{3} v_{2}^{5}-4096 a_{3} v_{2}^{4} = 0 \end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve \begin {align*} -124416 a_{3}&=0\\ -86016 a_{3}&=0\\ -71680 a_{3}&=0\\ -66560 a_{3}&=0\\ -46336 a_{3}&=0\\ -24320 a_{3}&=0\\ -8192 a_{3}&=0\\ -6144 a_{3}&=0\\ -5136 a_{3}&=0\\ -4096 a_{3}&=0\\ -4000 a_{3}&=0\\ -3840 a_{3}&=0\\ -1280 a_{3}&=0\\ -248 a_{3}&=0\\ -240 a_{3}&=0\\ -24 a_{3}&=0\\ -a_{3}&=0\\ 24 a_{3}&=0\\ 480 a_{3}&=0\\ 1440 a_{3}&=0\\ 3840 a_{3}&=0\\ 11584 a_{3}&=0\\ 14336 a_{3}&=0\\ 15360 a_{3}&=0\\ 17920 a_{3}&=0\\ 24576 a_{3}&=0\\ 30720 a_{3}&=0\\ 40960 a_{3}&=0\\ 68992 a_{3}&=0\\ 76800 a_{3}&=0\\ 94208 a_{3}&=0\\ 122880 a_{3}&=0\\ -3840 a_{2}-57088 a_{3}&=0\\ 8192 a_{1}+2048 a_{3}-8192 b_{1}&=0\\ -9216 a_{2}-24576 a_{3}-3072 b_{3}&=0\\ -4096 a_{2}+4096 a_{3}-8192 b_{3}&=0\\ -448 a_{2}-16768 a_{3}+64 b_{3}&=0\\ -6144 a_{1}+24576 a_{2}-12288 b_{2}+12288 b_{3}&=0\\ -3072 a_{1}+24576 a_{2}+29696 a_{3}-6144 b_{2}&=0\\ 12288 a_{1}-4096 a_{2}-12288 b_{1}-4096 b_{3}&=0\\ -28672 a_{1}+16384 a_{2}+8192 a_{3}+24576 b_{1}-8192 b_{2}&=0\\ -8192 a_{1}+4096 a_{2}+4096 a_{3}+8192 b_{1}-2048 b_{3}&=0\\ -384 a_{1}+4608 a_{2}+17152 a_{3}-768 b_{2}-768 b_{3}&=0\\ 2048 a_{1}-4096 a_{2}-4096 a_{3}+4096 b_{2}+4096 b_{3}&=0\\ 18432 a_{1}-43008 a_{2}-18432 a_{3}-6144 b_{1}+24576 b_{2}&=0\\ -13312 a_{1}+24576 a_{2}+16384 a_{3}+6144 b_{1}-14336 b_{2}-6144 b_{3}&=0\\ 3840 a_{1}-16640 a_{2}-16896 a_{3}-768 b_{1}+6144 b_{2}+3328 b_{3}&=0\\ 18432 a_{1}-12288 a_{2}-9216 a_{3}-14336 b_{1}+8192 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*} 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 {x}{2}+1 - \left (-\frac {1}{2} x +1+y^{2}+\frac {7}{2} x^{2} y -2 x y +\frac {13}{16} x^{4}-\frac {3}{2} x^{3}+x^{2}+y^{3}+\frac {3}{4} x^{2} y^{2}-3 x \,y^{2}+\frac {3}{16} y \,x^{4}-\frac {3}{2} x^{3} y +\frac {1}{64} x^{6}-\frac {3}{16} x^{5}\right ) \left (1\right ) \\ &= -y^{2}-\frac {7}{2} x^{2} y +2 x y -\frac {13}{16} x^{4}+\frac {3}{2} x^{3}-x^{2}-y^{3}-\frac {3}{4} x^{2} y^{2}+3 x \,y^{2}-\frac {3}{16} y \,x^{4}+\frac {3}{2} x^{3} y -\frac {1}{64} x^{6}+\frac {3}{16} x^{5}\\ \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}{-y^{2}-\frac {7}{2} x^{2} y +2 x y -\frac {13}{16} x^{4}+\frac {3}{2} x^{3}-x^{2}-y^{3}-\frac {3}{4} x^{2} y^{2}+3 x \,y^{2}-\frac {3}{16} y \,x^{4}+\frac {3}{2} x^{3} y -\frac {1}{64} x^{6}+\frac {3}{16} x^{5}}} dy \end {align*}
Which results in \begin {align*} S&= -\ln \left (x^{2}-4 x +4 y +4\right )+\frac {4}{x^{2}-4 x +4 y}+\ln \left (x^{2}-4 x +4 y \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,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}{2} x +1+y^{2}+\frac {7}{2} x^{2} y -2 x y +\frac {13}{16} x^{4}-\frac {3}{2} x^{3}+x^{2}+y^{3}+\frac {3}{4} x^{2} y^{2}-3 x \,y^{2}+\frac {3}{16} y \,x^{4}-\frac {3}{2} x^{3} y +\frac {1}{64} x^{6}-\frac {3}{16} x^{5} \end {align*}
Evaluating all the partial derivatives gives \begin {align*} R_{x} &= 1\\ R_{y} &= 0\\ S_{x} &= \frac {-32 x +64}{\left (x^{2}-4 x +4 y +4\right ) \left (x^{2}-4 x +4 y \right )^{2}}\\ S_{y} &= -\frac {64}{\left (x^{2}-4 x +4 y +4\right ) \left (x^{2}-4 x +4 y \right )^{2}} \end {align*}
Substituting all the above in (2) and simplifying gives the ode in canonical coordinates. \begin {align*} \frac {dS}{dR} &= -1\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} &= -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 ) = -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*} \frac {\left (-x^{2}+4 x -4 y\right ) \ln \left (x^{2}-4 x +4 y+4\right )+4+\left (x^{2}-4 x +4 y\right ) \ln \left (x^{2}-4 x +4 y\right )}{x^{2}-4 x +4 y} = -x +c_{1} \end {align*}
Which simplifies to \begin {align*} \frac {\left (-x^{2}+4 x -4 y\right ) \ln \left (x^{2}-4 x +4 y+4\right )+4+\left (x^{2}-4 x +4 y\right ) \ln \left (x^{2}-4 x +4 y\right )}{x^{2}-4 x +4 y} = -x +c_{1} \end {align*}
The following diagram shows solution curves of the original ode and how they transform in the canonical coordinates space using the mapping shown.
|
Original ode in \(x,y\) coordinates |
Canonical coordinates transformation |
ODE in canonical coordinates \((R,S)\) |
| \( \frac {dy}{dx} = -\frac {1}{2} x +1+y^{2}+\frac {7}{2} x^{2} y -2 x y +\frac {13}{16} x^{4}-\frac {3}{2} x^{3}+x^{2}+y^{3}+\frac {3}{4} x^{2} y^{2}-3 x \,y^{2}+\frac {3}{16} y \,x^{4}-\frac {3}{2} x^{3} y +\frac {1}{64} x^{6}-\frac {3}{16} x^{5}\) |
|
\( \frac {d S}{d R} = -1\) |
| |
\(\!\begin {aligned} R&= x\\ S&= \frac {\left (-x^{2}+4 x -4 y \right ) \ln \left (x^{2}-4 x +4 y +4\right )+4+\left (x^{2}-4 x +4 y \right ) \ln \left (x^{2}-4 x +4 y \right )}{x^{2}-4 x +4 y} \end {aligned} \) |
|
Summary
The solution(s) found are the following \begin{align*} \tag{1} \frac {\left (-x^{2}+4 x -4 y\right ) \ln \left (x^{2}-4 x +4 y+4\right )+4+\left (x^{2}-4 x +4 y\right ) \ln \left (x^{2}-4 x +4 y\right )}{x^{2}-4 x +4 y} &= -x +c_{1} \\ \end{align*}
Verification of solutions
\[ \frac {\left (-x^{2}+4 x -4 y\right ) \ln \left (x^{2}-4 x +4 y+4\right )+4+\left (x^{2}-4 x +4 y\right ) \ln \left (x^{2}-4 x +4 y\right )}{x^{2}-4 x +4 y} = -x +c_{1} \] 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}{4} x^{2}-3 x \right ) y^{2}+\left (\frac {7}{2} x^{2}-2 x +\frac {3}{16} x^{4}-\frac {3}{2} x^{3}\right ) y-\frac {x}{2}+1+\frac {13 x^{4}}{16}-\frac {3 x^{3}}{2}+x^{2}+\frac {x^{6}}{64}-\frac {3 x^{5}}{16}\tag {1} \end {align*}
Therefore \begin {align*} f_0(x) &= -\frac {1}{2} x +1+\frac {13}{16} x^{4}-\frac {3}{2} x^{3}+x^{2}+\frac {1}{64} x^{6}-\frac {3}{16} x^{5}\\ f_1(x) &= \frac {7}{2} x^{2}-2 x +\frac {3}{16} x^{4}-\frac {3}{2} x^{3}\\ f_2(x) &= 1+\frac {3}{4} x^{2}-3 x\\ f_3(x) &= 1 \end {align*}
Since \(f_2(x)=1+\frac {3}{4} x^{2}-3 x\) 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}{4} x^{2}-3 x}{3} \right ) \\ &= u \left (x \right )-\frac {1}{3}-\frac {x^{2}}{4}+x \end {align*}
The above transformation applied to (1) gives a new ODE as \begin {align*} u^{\prime }\left (x \right ) = \frac {2}{27}+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 \frac {1}{\frac {2}{27}+u^{3}-\frac {1}{3} u}d u &= \int {dx}\\ \int _{}^{u \left (x \right )}\frac {1}{\frac {2}{27}+\textit {\_a}^{3}-\frac {1}{3} \textit {\_a}}d \textit {\_a}&= x +c_{2} \end {align*}
Substituting \(u=y-\frac {1}{3}-\frac {x^{2}}{4}+x\) in the above solution gives \begin {align*} \int _{}^{y-\frac {1}{3}-\frac {x^{2}}{4}+x}\frac {1}{\frac {2}{27}+\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}}{4}+x}\frac {1}{\frac {2}{27}+\textit {\_a}^{3}-\frac {1}{3} \textit {\_a}}d \textit {\_a} &= x +c_{2} \\ \end{align*}
Verification of solutions
\[ \int _{}^{y-\frac {1}{3}-\frac {x^{2}}{4}+x}\frac {1}{\frac {2}{27}+\textit {\_a}^{3}-\frac {1}{3} \textit {\_a}}d \textit {\_a} = x +c_{2} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-y^{2}-\frac {7 x^{2} y}{2}+2 x y-y^{3}-\frac {3 y^{2} x^{2}}{4}+3 y^{2} x -\frac {3 y x^{4}}{16}+\frac {3 x^{3} y}{2}=-\frac {1}{2} x +1+\frac {13}{16} x^{4}-\frac {3}{2} x^{3}+x^{2}+\frac {1}{64} x^{6}-\frac {3}{16} x^{5} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {x}{2}+1+y^{2}+\frac {7 x^{2} y}{2}-2 x y+\frac {13 x^{4}}{16}-\frac {3 x^{3}}{2}+x^{2}+y^{3}+\frac {3 y^{2} x^{2}}{4}-3 y^{2} x +\frac {3 y x^{4}}{16}-\frac {3 x^{3} y}{2}+\frac {x^{6}}{64}-\frac {3 x^{5}}{16} \end {array} \]
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: 0.063 (sec). Leaf size: 55
dsolve(diff(y(x),x) = -1/2*x+1+y(x)^2+7/2*x^2*y(x)-2*x*y(x)+13/16*x^4-3/2*x^3+x^2+y(x)^3+3/4*x^2*y(x)^2-3*x*y(x)^2+3/16*y(x)*x^4-3/2*x^3*y(x)+1/64*x^6-3/16*x^5,y(x), singsol=all)
\[ y \left (x \right ) = \frac {{\mathrm e}^{\operatorname {RootOf}\left (\ln \left ({\mathrm e}^{\textit {\_Z}}-4\right ) {\mathrm e}^{\textit {\_Z}}+c_{1} {\mathrm e}^{\textit {\_Z}}-{\mathrm e}^{\textit {\_Z}} \textit {\_Z} +x \,{\mathrm e}^{\textit {\_Z}}-4 \ln \left ({\mathrm e}^{\textit {\_Z}}-4\right )-4 c_{1} +4 \textit {\_Z} -4 x +4\right )}}{4}-1-\frac {x^{2}}{4}+x \]
✓ Solution by Mathematica
Time used: 37.054 (sec). Leaf size: 248
DSolve[y'[x] == 1 - x/2 + x^2 - (3*x^3)/2 + (13*x^4)/16 - (3*x^5)/16 + x^6/64 - 2*x*y[x] + (7*x^2*y[x])/2 - (3*x^3*y[x])/2 + (3*x^4*y[x])/16 + y[x]^2 - 3*x*y[x]^2 + (3*x^2*y[x]^2)/4 + y[x]^3,y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [\frac {\sqrt [3]{2} \left (\frac {\frac {1}{4} \left (3 x^2-12 x+4\right )+3 y(x)}{\sqrt [3]{2}}+2^{2/3}\right ) \left (2^{2/3}-2^{2/3} \left (\frac {1}{4} \left (3 x^2-12 x+4\right )+3 y(x)\right )\right ) \left (\left (\frac {1}{4} \left (-3 x^2+12 x-4\right )-3 y(x)+1\right ) \log \left (2^{2/3}-2^{2/3} \left (\frac {1}{4} \left (3 x^2-12 x+4\right )+3 y(x)\right )\right )+\left (\frac {1}{4} \left (3 x^2-12 x+4\right )+3 y(x)-1\right ) \log \left (2 \left (\frac {\frac {1}{4} \left (3 x^2-12 x+4\right )+3 y(x)}{\sqrt [3]{2}}+2^{2/3}\right )\right )-3\right )}{9 \left (-\left (\frac {1}{4} \left (3 x^2-12 x+4\right )+3 y(x)\right )^3+3 \left (\frac {1}{4} \left (3 x^2-12 x+4\right )+3 y(x)\right )-2\right )}=\frac {1}{9} 2^{2/3} x+c_1,y(x)\right ] \]