Internal problem ID [3231]
Internal file name [OUTPUT/2723_Sunday_June_05_2022_08_39_21_AM_24446730/index.tex]
Book: Differential equations for engineers by Wei-Chau XIE, Cambridge Press 2010
Section: Chapter 2. First-Order and Simple Higher-Order Differential Equations. Page
78
Problem number: 89.
ODE order: 1.
ODE degree: 3.
The type(s) of ODE detected by this program : "unknown"
Maple gives the following as the ode type
[[_1st_order, _with_linear_symmetries]]
Unable to solve or complete the solution.
\[ \boxed {y-y^{\prime } x +x^{2} {y^{\prime }}^{3}=0} \] Solving the given ode for \(y^{\prime }\) results in \(3\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} y^{\prime }&=\frac {{\left (\left (12 \sqrt {3}\, \sqrt {27 y^{2}-4 x}-108 y\right ) x \right )}^{{1}/{3}}}{6 x}+\frac {2}{{\left (\left (12 \sqrt {3}\, \sqrt {27 y^{2}-4 x}-108 y\right ) x \right )}^{{1}/{3}}} \tag {1} \\ y^{\prime }&=-\frac {{\left (\left (12 \sqrt {3}\, \sqrt {27 y^{2}-4 x}-108 y\right ) x \right )}^{{1}/{3}}}{12 x}-\frac {1}{{\left (\left (12 \sqrt {3}\, \sqrt {27 y^{2}-4 x}-108 y\right ) x \right )}^{{1}/{3}}}+\frac {i \sqrt {3}\, \left (\frac {{\left (\left (12 \sqrt {3}\, \sqrt {27 y^{2}-4 x}-108 y\right ) x \right )}^{{1}/{3}}}{6 x}-\frac {2}{{\left (\left (12 \sqrt {3}\, \sqrt {27 y^{2}-4 x}-108 y\right ) x \right )}^{{1}/{3}}}\right )}{2} \tag {2} \\ y^{\prime }&=-\frac {{\left (\left (12 \sqrt {3}\, \sqrt {27 y^{2}-4 x}-108 y\right ) x \right )}^{{1}/{3}}}{12 x}-\frac {1}{{\left (\left (12 \sqrt {3}\, \sqrt {27 y^{2}-4 x}-108 y\right ) x \right )}^{{1}/{3}}}-\frac {i \sqrt {3}\, \left (\frac {{\left (\left (12 \sqrt {3}\, \sqrt {27 y^{2}-4 x}-108 y\right ) x \right )}^{{1}/{3}}}{6 x}-\frac {2}{{\left (\left (12 \sqrt {3}\, \sqrt {27 y^{2}-4 x}-108 y\right ) x \right )}^{{1}/{3}}}\right )}{2} \tag {3} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Writing the ode as \begin {align*} y^{\prime }&=\frac {\left ({\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{2}/{3}}+12^{{1}/{3}} x \right ) 12^{{1}/{3}}}{6 x {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{1}/{3}}}\\ y^{\prime }&= \omega \left ( x,y\right ) \end {align*}
The condition of Lie symmetry is the linearized PDE given by \begin {align*} \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0\tag {A} \end {align*}
The type of this ode is not in the lookup table. To determine \(\xi ,\eta \) then (A) is solved using ansatz. Making bivariate polynomials of degree 1 to use as anstaz gives \begin{align*} \tag{1E} \xi &= x a_{2}+y a_{3}+a_{1} \\ \tag{2E} \eta &= x b_{2}+y b_{3}+b_{1} \\ \end{align*} Where the unknown coefficients are \[ \{a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}\} \] Substituting equations (1E,2E) and \(\omega \) into (A) gives \begin{equation} \tag{5E} b_{2}+\frac {\left ({\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{2}/{3}}+12^{{1}/{3}} x \right ) 12^{{1}/{3}} \left (b_{3}-a_{2}\right )}{6 x {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{1}/{3}}}-\frac {{\left ({\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{2}/{3}}+12^{{1}/{3}} x \right )}^{2} 12^{{2}/{3}} a_{3}}{36 x^{2} {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{2}/{3}}}-\left (\frac {\left (\frac {-\frac {4 \sqrt {3}\, x}{3 \sqrt {27 y^{2}-4 x}}-\frac {2 \sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right )}{3}}{{\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{1}/{3}}}+12^{{1}/{3}}\right ) 12^{{1}/{3}}}{6 x {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{1}/{3}}}-\frac {\left ({\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{2}/{3}}+12^{{1}/{3}} x \right ) 12^{{1}/{3}}}{6 x^{2} {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{1}/{3}}}-\frac {\left ({\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{2}/{3}}+12^{{1}/{3}} x \right ) 12^{{1}/{3}} \left (-\frac {2 \sqrt {3}\, x}{\sqrt {27 y^{2}-4 x}}-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right )\right )}{18 x {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{4}/{3}}}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )-\left (-\frac {\sqrt {3}\, \left (3 \sqrt {3}-\frac {27 y}{\sqrt {27 y^{2}-4 x}}\right ) 12^{{1}/{3}}}{9 {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{2}/{3}}}+\frac {\left ({\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{2}/{3}}+12^{{1}/{3}} x \right ) 12^{{1}/{3}} \sqrt {3}\, \left (3 \sqrt {3}-\frac {27 y}{\sqrt {27 y^{2}-4 x}}\right )}{18 {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{4}/{3}}}\right ) \left (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} Simplifying the above gives \begin{equation} \tag{6E} \text {Expression too large to display} \end{equation} Since the PDE has radicals, simplifying gives \[ \text {Expression too large to display} \] Looking at the above PDE shows the following are all the terms with \(\{x, y\}\) in them. \[ \left \{x, y, {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{1}/{3}}, {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{2}/{3}}, \sqrt {27 y^{2}-4 x}\right \} \] The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them \[ \left \{x = v_{1}, y = v_{2}, {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{1}/{3}} = v_{3}, {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{2}/{3}} = v_{4}, \sqrt {27 y^{2}-4 x} = v_{5}\right \} \] The above PDE (6E) now becomes \begin{equation} \tag{7E} 6 v_{1} \left (-9 \,12^{{1}/{3}} v_{4} \sqrt {3}\, v_{1}^{2} v_{2} b_{2}-9 \,12^{{1}/{3}} v_{4} \sqrt {3}\, v_{1} v_{2}^{2} a_{2}+18 \,12^{{1}/{3}} v_{4} \sqrt {3}\, v_{1} v_{2}^{2} b_{3}-2 \,12^{{1}/{3}} v_{4} \sqrt {3}\, v_{1} v_{2} a_{3}-9 \,12^{{1}/{3}} v_{4} \sqrt {3}\, v_{1} v_{2} b_{1}+3 \,12^{{1}/{3}} v_{4} v_{5} v_{1} v_{2} a_{2}-4 \,12^{{2}/{3}} \sqrt {3}\, v_{1}^{3} b_{3}-3 \,12^{{2}/{3}} v_{5} v_{1}^{3} b_{2}-2 \,12^{{2}/{3}} \sqrt {3}\, v_{1}^{2} a_{1}+2 \,12^{{2}/{3}} v_{1}^{2} v_{5} a_{3}-3 \,12^{{2}/{3}} v_{5} v_{1}^{2} b_{1}-24 v_{3} \sqrt {3}\, v_{1}^{3} b_{2}+16 v_{3} \sqrt {3}\, v_{1}^{2} a_{3}+9 \,12^{{2}/{3}} \sqrt {3}\, v_{1}^{3} v_{2} b_{2}-18 \,12^{{2}/{3}} \sqrt {3}\, v_{1}^{2} v_{2}^{2} a_{2}+36 \,12^{{2}/{3}} \sqrt {3}\, v_{1}^{2} v_{2}^{2} b_{3}+90 \,12^{{2}/{3}} \sqrt {3}\, v_{1} v_{2}^{3} a_{3}+18 \,12^{{1}/{3}} v_{4} \sqrt {3}\, v_{2}^{3} a_{3}-14 \,12^{{2}/{3}} \sqrt {3}\, v_{1}^{2} v_{2} a_{3}+9 \,12^{{2}/{3}} \sqrt {3}\, v_{1}^{2} v_{2} b_{1}+9 \,12^{{2}/{3}} \sqrt {3}\, v_{1} v_{2}^{2} a_{1}+6 \,12^{{2}/{3}} v_{5} v_{1}^{2} v_{2} a_{2}-12 \,12^{{2}/{3}} v_{5} v_{1}^{2} v_{2} b_{3}-30 \,12^{{2}/{3}} v_{5} v_{1} v_{2}^{2} a_{3}+2 \,12^{{1}/{3}} v_{4} \sqrt {3}\, v_{1}^{2} a_{2}-4 \,12^{{1}/{3}} v_{4} \sqrt {3}\, v_{1}^{2} b_{3}+18 \,12^{{1}/{3}} v_{4} \sqrt {3}\, v_{2}^{2} a_{1}+2 \,12^{{2}/{3}} \sqrt {3}\, v_{1}^{3} a_{2}+3 \,12^{{1}/{3}} v_{4} v_{5} v_{1}^{2} b_{2}-6 \,12^{{1}/{3}} v_{4} v_{5} v_{2}^{2} a_{3}+162 v_{3} \sqrt {3}\, v_{1}^{2} v_{2}^{2} b_{2}-3 \,12^{{2}/{3}} v_{5} v_{1} v_{2} a_{1}-2 \,12^{{1}/{3}} v_{4} \sqrt {3}\, v_{1} a_{1}-2 \,12^{{1}/{3}} v_{4} v_{5} v_{1} a_{3}+3 \,12^{{1}/{3}} v_{4} v_{5} v_{1} b_{1}-6 \,12^{{1}/{3}} v_{4} v_{5} v_{2} a_{1}-108 v_{3} \sqrt {3}\, v_{1} v_{2}^{2} a_{3}-54 v_{3} v_{5} v_{1}^{2} v_{2} b_{2}+36 v_{3} v_{5} v_{1} v_{2} a_{3}-6 \,12^{{1}/{3}} v_{4} v_{5} v_{1} v_{2} b_{3}\right ) = 0 \end{equation} Collecting the above on the terms \(v_i\) introduced, and these are \[ \{v_{1}, v_{2}, v_{3}, v_{4}, v_{5}\} \] Equation (7E) now becomes \begin{equation} \tag{8E} \left (36 \,12^{{2}/{3}} a_{2}-72 \,12^{{2}/{3}} b_{3}\right ) v_{2} v_{5} v_{1}^{3}+96 \sqrt {3}\, a_{3} v_{3} v_{1}^{3}-12 \,12^{{2}/{3}} \sqrt {3}\, a_{1} v_{1}^{3}+\left (-54 \,12^{{1}/{3}} \sqrt {3}\, a_{2}+108 \,12^{{1}/{3}} \sqrt {3}\, b_{3}\right ) v_{2}^{2} v_{4} v_{1}^{2}+\left (-12 \,12^{{1}/{3}} \sqrt {3}\, a_{3}-54 \,12^{{1}/{3}} \sqrt {3}\, b_{1}\right ) v_{2} v_{4} v_{1}^{2}+\left (-12 \,12^{{1}/{3}} a_{3}+18 \,12^{{1}/{3}} b_{1}\right ) v_{4} v_{5} v_{1}^{2}+\left (12 \,12^{{2}/{3}} \sqrt {3}\, a_{2}-24 \,12^{{2}/{3}} \sqrt {3}\, b_{3}\right ) v_{1}^{4}+\left (-108 \,12^{{2}/{3}} \sqrt {3}\, a_{2}+216 \,12^{{2}/{3}} \sqrt {3}\, b_{3}\right ) v_{2}^{2} v_{1}^{3}+\left (-84 \,12^{{2}/{3}} \sqrt {3}\, a_{3}+54 \,12^{{2}/{3}} \sqrt {3}\, b_{1}\right ) v_{2} v_{1}^{3}+\left (12 \,12^{{1}/{3}} \sqrt {3}\, a_{2}-24 \,12^{{1}/{3}} \sqrt {3}\, b_{3}\right ) v_{4} v_{1}^{3}+\left (12 \,12^{{2}/{3}} a_{3}-18 \,12^{{2}/{3}} b_{1}\right ) v_{5} v_{1}^{3}+108 \,12^{{1}/{3}} \sqrt {3}\, a_{3} v_{2}^{3} v_{4} v_{1}-36 \,12^{{1}/{3}} a_{3} v_{2}^{2} v_{4} v_{5} v_{1}+108 \,12^{{1}/{3}} \sqrt {3}\, a_{1} v_{2}^{2} v_{4} v_{1}-36 \,12^{{1}/{3}} a_{1} v_{2} v_{4} v_{5} v_{1}-324 b_{2} v_{2} v_{3} v_{5} v_{1}^{3}+18 \,12^{{1}/{3}} b_{2} v_{4} v_{5} v_{1}^{3}+540 \,12^{{2}/{3}} \sqrt {3}\, a_{3} v_{2}^{3} v_{1}^{2}-648 \sqrt {3}\, a_{3} v_{2}^{2} v_{3} v_{1}^{2}-180 \,12^{{2}/{3}} a_{3} v_{2}^{2} v_{5} v_{1}^{2}+54 \,12^{{2}/{3}} \sqrt {3}\, a_{1} v_{2}^{2} v_{1}^{2}+216 a_{3} v_{2} v_{3} v_{5} v_{1}^{2}+\left (18 \,12^{{1}/{3}} a_{2}-36 \,12^{{1}/{3}} b_{3}\right ) v_{2} v_{4} v_{5} v_{1}^{2}-18 \,12^{{2}/{3}} a_{1} v_{2} v_{5} v_{1}^{2}-12 \,12^{{1}/{3}} \sqrt {3}\, a_{1} v_{4} v_{1}^{2}+54 \,12^{{2}/{3}} \sqrt {3}\, b_{2} v_{2} v_{1}^{4}+972 \sqrt {3}\, b_{2} v_{2}^{2} v_{3} v_{1}^{3}-144 \sqrt {3}\, b_{2} v_{3} v_{1}^{4}-18 \,12^{{2}/{3}} b_{2} v_{5} v_{1}^{4}-54 \,12^{{1}/{3}} \sqrt {3}\, b_{2} v_{2} v_{4} v_{1}^{3} = 0 \end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve \begin {align*} 216 a_{3}&=0\\ -324 b_{2}&=0\\ -648 \sqrt {3}\, a_{3}&=0\\ 96 \sqrt {3}\, a_{3}&=0\\ -144 \sqrt {3}\, b_{2}&=0\\ 972 \sqrt {3}\, b_{2}&=0\\ -36 \,12^{{1}/{3}} a_{1}&=0\\ -36 \,12^{{1}/{3}} a_{3}&=0\\ 18 \,12^{{1}/{3}} b_{2}&=0\\ -18 \,12^{{2}/{3}} a_{1}&=0\\ -180 \,12^{{2}/{3}} a_{3}&=0\\ -18 \,12^{{2}/{3}} b_{2}&=0\\ -12 \,12^{{1}/{3}} \sqrt {3}\, a_{1}&=0\\ 108 \,12^{{1}/{3}} \sqrt {3}\, a_{1}&=0\\ 108 \,12^{{1}/{3}} \sqrt {3}\, a_{3}&=0\\ -54 \,12^{{1}/{3}} \sqrt {3}\, b_{2}&=0\\ -12 \,12^{{2}/{3}} \sqrt {3}\, a_{1}&=0\\ 54 \,12^{{2}/{3}} \sqrt {3}\, a_{1}&=0\\ 540 \,12^{{2}/{3}} \sqrt {3}\, a_{3}&=0\\ 54 \,12^{{2}/{3}} \sqrt {3}\, b_{2}&=0\\ 18 \,12^{{1}/{3}} a_{2}-36 \,12^{{1}/{3}} b_{3}&=0\\ -12 \,12^{{1}/{3}} a_{3}+18 \,12^{{1}/{3}} b_{1}&=0\\ 36 \,12^{{2}/{3}} a_{2}-72 \,12^{{2}/{3}} b_{3}&=0\\ 12 \,12^{{2}/{3}} a_{3}-18 \,12^{{2}/{3}} b_{1}&=0\\ -54 \,12^{{1}/{3}} \sqrt {3}\, a_{2}+108 \,12^{{1}/{3}} \sqrt {3}\, b_{3}&=0\\ 12 \,12^{{1}/{3}} \sqrt {3}\, a_{2}-24 \,12^{{1}/{3}} \sqrt {3}\, b_{3}&=0\\ -12 \,12^{{1}/{3}} \sqrt {3}\, a_{3}-54 \,12^{{1}/{3}} \sqrt {3}\, b_{1}&=0\\ -108 \,12^{{2}/{3}} \sqrt {3}\, a_{2}+216 \,12^{{2}/{3}} \sqrt {3}\, b_{3}&=0\\ 12 \,12^{{2}/{3}} \sqrt {3}\, a_{2}-24 \,12^{{2}/{3}} \sqrt {3}\, b_{3}&=0\\ -84 \,12^{{2}/{3}} \sqrt {3}\, a_{3}+54 \,12^{{2}/{3}} \sqrt {3}\, b_{1}&=0 \end {align*}
Solving the above equations for the unknowns gives \begin {align*} a_{1}&=0\\ a_{2}&=2 b_{3}\\ a_{3}&=0\\ b_{1}&=0\\ b_{2}&=0\\ b_{3}&=b_{3} \end {align*}
Substituting the above solution in the anstaz (1E,2E) (using \(1\) as arbitrary value for any unknown in the RHS) gives \begin{align*} \xi &= 2 x \\ \eta &= y \\ \end{align*} The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( x,y\right ) \to \left ( R,S \right )\) where \(\left ( R,S \right )\) are the canonical coordinates which make the original ode become a quadrature and hence solved by integration.
The characteristic pde which is used to find the canonical coordinates is \begin {align*} \frac {d x}{\xi } &= \frac {d y}{\eta } = dS \tag {1} \end {align*}
The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial y}\right ) S(x,y) = 1\). Starting with the first pair of ode’s in (1) gives an ode to solve for the independent variable \(R\) in the canonical coordinates, where \(S(R)\). Unable to determine \(R\). Terminating
Unable to determine ODE type.
Solving equation (2)
Writing the ode as \begin {align*} y^{\prime }&=-\frac {\left (i x \sqrt {3}\, 12^{{1}/{3}}+2 {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{2}/{3}}-12^{{1}/{3}} x \right ) 12^{{1}/{3}}}{6 {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{1}/{3}} x \left (1+i \sqrt {3}\right )}\\ 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 (i x \sqrt {3}\, 12^{{1}/{3}}+2 {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{2}/{3}}-12^{{1}/{3}} x \right ) 12^{{1}/{3}} \left (b_{3}-a_{2}\right )}{6 {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{1}/{3}} x \left (1+i \sqrt {3}\right )}-\frac {\left (i x \sqrt {3}\, 12^{{1}/{3}}+2 {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{2}/{3}}-12^{{1}/{3}} x \right )^{2} 12^{{2}/{3}} a_{3}}{36 {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{2}/{3}} x^{2} \left (1+i \sqrt {3}\right )^{2}}-\left (-\frac {\left (i \sqrt {3}\, 12^{{1}/{3}}+\frac {-\frac {8 \sqrt {3}\, x}{3 \sqrt {27 y^{2}-4 x}}-\frac {4 \sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right )}{3}}{{\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{1}/{3}}}-12^{{1}/{3}}\right ) 12^{{1}/{3}}}{6 {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{1}/{3}} x \left (1+i \sqrt {3}\right )}+\frac {\left (i x \sqrt {3}\, 12^{{1}/{3}}+2 {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{2}/{3}}-12^{{1}/{3}} x \right ) 12^{{1}/{3}} \left (-\frac {2 \sqrt {3}\, x}{\sqrt {27 y^{2}-4 x}}-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right )\right )}{18 {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{4}/{3}} x \left (1+i \sqrt {3}\right )}+\frac {\left (i x \sqrt {3}\, 12^{{1}/{3}}+2 {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{2}/{3}}-12^{{1}/{3}} x \right ) 12^{{1}/{3}}}{6 {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{1}/{3}} x^{2} \left (1+i \sqrt {3}\right )}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )-\left (\frac {2 \sqrt {3}\, \left (3 \sqrt {3}-\frac {27 y}{\sqrt {27 y^{2}-4 x}}\right ) 12^{{1}/{3}}}{9 {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{2}/{3}} \left (1+i \sqrt {3}\right )}-\frac {\left (i x \sqrt {3}\, 12^{{1}/{3}}+2 {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{2}/{3}}-12^{{1}/{3}} x \right ) 12^{{1}/{3}} \sqrt {3}\, \left (3 \sqrt {3}-\frac {27 y}{\sqrt {27 y^{2}-4 x}}\right )}{18 {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{4}/{3}} \left (1+i \sqrt {3}\right )}\right ) \left (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} Simplifying the above gives \begin{equation} \tag{6E} \text {Expression too large to display} \end{equation} Since the PDE has radicals, simplifying gives \[ \text {Expression too large to display} \] Looking at the above PDE shows the following are all the terms with \(\{x, y\}\) in them. \[ \left \{x, y, {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{1}/{3}}, {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{2}/{3}}, \sqrt {27 y^{2}-4 x}\right \} \] The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them \[ \left \{x = v_{1}, y = v_{2}, {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{1}/{3}} = v_{3}, {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{2}/{3}} = v_{4}, \sqrt {27 y^{2}-4 x} = v_{5}\right \} \] The above PDE (6E) now becomes \begin{equation} \tag{7E} 12 v_{1} \left (-3 i 12^{{1}/{3}} v_{4} \sqrt {3}\, v_{5} v_{1}^{2} b_{2}+6 i 12^{{1}/{3}} v_{4} \sqrt {3}\, v_{5} v_{2}^{2} a_{3}+2 i 12^{{1}/{3}} v_{4} \sqrt {3}\, v_{5} v_{1} a_{3}-3 i 12^{{1}/{3}} v_{4} \sqrt {3}\, v_{5} v_{1} b_{1}+6 i 12^{{1}/{3}} v_{4} \sqrt {3}\, v_{5} v_{2} a_{1}-54 i v_{3} \sqrt {3}\, v_{5} v_{1}^{2} v_{2} b_{2}+36 i v_{3} \sqrt {3}\, v_{5} v_{1} v_{2} a_{3}+9 \,12^{{1}/{3}} v_{4} \sqrt {3}\, v_{1}^{2} v_{2} b_{2}+9 \,12^{{1}/{3}} v_{4} \sqrt {3}\, v_{1} v_{2}^{2} a_{2}-18 \,12^{{1}/{3}} v_{4} \sqrt {3}\, v_{1} v_{2}^{2} b_{3}+2 \,12^{{1}/{3}} v_{4} \sqrt {3}\, v_{1} v_{2} a_{3}+9 \,12^{{1}/{3}} v_{4} \sqrt {3}\, v_{1} v_{2} b_{1}-3 \,12^{{1}/{3}} v_{4} v_{5} v_{1} v_{2} a_{2}+6 \,12^{{1}/{3}} v_{4} v_{5} v_{1} v_{2} b_{3}+27 i 12^{{1}/{3}} v_{4} v_{1}^{2} v_{2} b_{2}+27 i 12^{{1}/{3}} v_{4} v_{1} v_{2}^{2} a_{2}-54 i 12^{{1}/{3}} v_{4} v_{1} v_{2}^{2} b_{3}+6 i 12^{{1}/{3}} v_{4} v_{1} v_{2} a_{3}+27 i 12^{{1}/{3}} v_{4} v_{1} v_{2} b_{1}+18 \,12^{{2}/{3}} \sqrt {3}\, v_{1}^{3} v_{2} b_{2}-36 \,12^{{2}/{3}} \sqrt {3}\, v_{1}^{2} v_{2}^{2} a_{2}+72 \,12^{{2}/{3}} \sqrt {3}\, v_{1}^{2} v_{2}^{2} b_{3}+180 \,12^{{2}/{3}} \sqrt {3}\, v_{1} v_{2}^{3} a_{3}-18 \,12^{{1}/{3}} v_{4} \sqrt {3}\, v_{2}^{3} a_{3}-28 \,12^{{2}/{3}} \sqrt {3}\, v_{1}^{2} v_{2} a_{3}+18 \,12^{{2}/{3}} \sqrt {3}\, v_{1}^{2} v_{2} b_{1}+18 \,12^{{2}/{3}} \sqrt {3}\, v_{1} v_{2}^{2} a_{1}+12 \,12^{{2}/{3}} v_{5} v_{1}^{2} v_{2} a_{2}-24 \,12^{{2}/{3}} v_{5} v_{1}^{2} v_{2} b_{3}-60 \,12^{{2}/{3}} v_{5} v_{1} v_{2}^{2} a_{3}-2 \,12^{{1}/{3}} v_{4} \sqrt {3}\, v_{1}^{2} a_{2}+4 \,12^{{1}/{3}} v_{4} \sqrt {3}\, v_{1}^{2} b_{3}-18 \,12^{{1}/{3}} v_{4} \sqrt {3}\, v_{2}^{2} a_{1}-3 \,12^{{1}/{3}} v_{4} v_{5} v_{1}^{2} b_{2}+6 \,12^{{1}/{3}} v_{4} v_{5} v_{2}^{2} a_{3}-162 v_{3} \sqrt {3}\, v_{1}^{2} v_{2}^{2} b_{2}-6 \,12^{{2}/{3}} v_{5} v_{1} v_{2} a_{1}+2 \,12^{{1}/{3}} v_{4} \sqrt {3}\, v_{1} a_{1}+2 \,12^{{1}/{3}} v_{4} v_{5} v_{1} a_{3}-3 \,12^{{1}/{3}} v_{4} v_{5} v_{1} b_{1}+6 \,12^{{1}/{3}} v_{4} v_{5} v_{2} a_{1}+108 v_{3} \sqrt {3}\, v_{1} v_{2}^{2} a_{3}+54 v_{3} v_{5} v_{1}^{2} v_{2} b_{2}-36 v_{3} v_{5} v_{1} v_{2} a_{3}-54 i 12^{{1}/{3}} v_{4} v_{2}^{3} a_{3}-6 i 12^{{1}/{3}} v_{4} v_{1}^{2} a_{2}+12 i 12^{{1}/{3}} v_{4} v_{1}^{2} b_{3}-54 i 12^{{1}/{3}} v_{4} v_{2}^{2} a_{1}+486 i v_{3} v_{1}^{2} v_{2}^{2} b_{2}+6 i 12^{{1}/{3}} v_{4} v_{1} a_{1}-324 i v_{3} v_{1} v_{2}^{2} a_{3}-3 i 12^{{1}/{3}} v_{4} \sqrt {3}\, v_{5} v_{1} v_{2} a_{2}+6 i 12^{{1}/{3}} v_{4} \sqrt {3}\, v_{5} v_{1} v_{2} b_{3}-72 i v_{3} v_{1}^{3} b_{2}+48 i v_{3} v_{1}^{2} a_{3}+4 \,12^{{2}/{3}} \sqrt {3}\, v_{1}^{3} a_{2}-8 \,12^{{2}/{3}} \sqrt {3}\, v_{1}^{3} b_{3}-6 \,12^{{2}/{3}} v_{5} v_{1}^{3} b_{2}-4 \,12^{{2}/{3}} \sqrt {3}\, v_{1}^{2} a_{1}+4 \,12^{{2}/{3}} v_{1}^{2} v_{5} a_{3}-6 \,12^{{2}/{3}} v_{5} v_{1}^{2} b_{1}+24 v_{3} \sqrt {3}\, v_{1}^{3} b_{2}-16 v_{3} \sqrt {3}\, v_{1}^{2} a_{3}\right ) = 0 \end{equation} Collecting the above on the terms \(v_i\) introduced, and these are \[ \{v_{1}, v_{2}, v_{3}, v_{4}, v_{5}\} \] Equation (7E) now becomes \begin{equation} \tag{8E} \left (72 i 12^{{1}/{3}} a_{1}+24 \,12^{{1}/{3}} \sqrt {3}\, a_{1}\right ) v_{4} v_{1}^{2}+\left (-864 i b_{2}+288 \sqrt {3}\, b_{2}\right ) v_{3} v_{1}^{4}+\left (576 i a_{3}-192 \sqrt {3}\, a_{3}\right ) v_{3} v_{1}^{3}+\left (-72 i 12^{{1}/{3}} a_{2}+144 i 12^{{1}/{3}} b_{3}-24 \,12^{{1}/{3}} \sqrt {3}\, a_{2}+48 \,12^{{1}/{3}} \sqrt {3}\, b_{3}\right ) v_{4} v_{1}^{3}+\left (48 \,12^{{2}/{3}} \sqrt {3}\, a_{2}-96 \,12^{{2}/{3}} \sqrt {3}\, b_{3}\right ) v_{1}^{4}+\left (-432 \,12^{{2}/{3}} \sqrt {3}\, a_{2}+864 \,12^{{2}/{3}} \sqrt {3}\, b_{3}\right ) v_{2}^{2} v_{1}^{3}+\left (-336 \,12^{{2}/{3}} \sqrt {3}\, a_{3}+216 \,12^{{2}/{3}} \sqrt {3}\, b_{1}\right ) v_{2} v_{1}^{3}+\left (48 \,12^{{2}/{3}} a_{3}-72 \,12^{{2}/{3}} b_{1}\right ) v_{5} v_{1}^{3}-72 \,12^{{2}/{3}} b_{2} v_{5} v_{1}^{4}+\left (144 \,12^{{2}/{3}} a_{2}-288 \,12^{{2}/{3}} b_{3}\right ) v_{2} v_{5} v_{1}^{3}-48 \,12^{{2}/{3}} \sqrt {3}\, a_{1} v_{1}^{3}+216 \,12^{{2}/{3}} \sqrt {3}\, b_{2} v_{2} v_{1}^{4}+2160 \,12^{{2}/{3}} \sqrt {3}\, a_{3} v_{2}^{3} v_{1}^{2}-720 \,12^{{2}/{3}} a_{3} v_{2}^{2} v_{5} v_{1}^{2}+216 \,12^{{2}/{3}} \sqrt {3}\, a_{1} v_{2}^{2} v_{1}^{2}-72 \,12^{{2}/{3}} a_{1} v_{2} v_{5} v_{1}^{2}+\left (-648 i \sqrt {3}\, b_{2}+648 b_{2}\right ) v_{2} v_{3} v_{5} v_{1}^{3}+\left (432 i \sqrt {3}\, a_{3}-432 a_{3}\right ) v_{2} v_{3} v_{5} v_{1}^{2}+\left (-36 i 12^{{1}/{3}} \sqrt {3}\, a_{2}+72 i 12^{{1}/{3}} \sqrt {3}\, b_{3}-36 \,12^{{1}/{3}} a_{2}+72 \,12^{{1}/{3}} b_{3}\right ) v_{2} v_{4} v_{5} v_{1}^{2}+\left (72 i 12^{{1}/{3}} \sqrt {3}\, a_{3}+72 \,12^{{1}/{3}} a_{3}\right ) v_{2}^{2} v_{4} v_{5} v_{1}+\left (72 i 12^{{1}/{3}} \sqrt {3}\, a_{1}+72 \,12^{{1}/{3}} a_{1}\right ) v_{2} v_{4} v_{5} v_{1}+\left (5832 i b_{2}-1944 \sqrt {3}\, b_{2}\right ) v_{2}^{2} v_{3} v_{1}^{3}+\left (324 i 12^{{1}/{3}} b_{2}+108 \,12^{{1}/{3}} \sqrt {3}\, b_{2}\right ) v_{2} v_{4} v_{1}^{3}+\left (-36 i 12^{{1}/{3}} \sqrt {3}\, b_{2}-36 \,12^{{1}/{3}} b_{2}\right ) v_{4} v_{5} v_{1}^{3}+\left (-3888 i a_{3}+1296 \sqrt {3}\, a_{3}\right ) v_{2}^{2} v_{3} v_{1}^{2}+\left (324 i 12^{{1}/{3}} a_{2}-648 i 12^{{1}/{3}} b_{3}+108 \,12^{{1}/{3}} \sqrt {3}\, a_{2}-216 \,12^{{1}/{3}} \sqrt {3}\, b_{3}\right ) v_{2}^{2} v_{4} v_{1}^{2}+\left (72 i 12^{{1}/{3}} a_{3}+324 i 12^{{1}/{3}} b_{1}+24 \,12^{{1}/{3}} \sqrt {3}\, a_{3}+108 \,12^{{1}/{3}} \sqrt {3}\, b_{1}\right ) v_{2} v_{4} v_{1}^{2}+\left (24 i 12^{{1}/{3}} \sqrt {3}\, a_{3}-36 i 12^{{1}/{3}} \sqrt {3}\, b_{1}+24 \,12^{{1}/{3}} a_{3}-36 \,12^{{1}/{3}} b_{1}\right ) v_{4} v_{5} v_{1}^{2}+\left (-648 i 12^{{1}/{3}} a_{3}-216 \,12^{{1}/{3}} \sqrt {3}\, a_{3}\right ) v_{2}^{3} v_{4} v_{1}+\left (-648 i 12^{{1}/{3}} a_{1}-216 \,12^{{1}/{3}} \sqrt {3}\, a_{1}\right ) v_{2}^{2} v_{4} v_{1} = 0 \end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve \begin {align*} -72 \,12^{{2}/{3}} a_{1}&=0\\ -720 \,12^{{2}/{3}} a_{3}&=0\\ -72 \,12^{{2}/{3}} b_{2}&=0\\ -48 \,12^{{2}/{3}} \sqrt {3}\, a_{1}&=0\\ 216 \,12^{{2}/{3}} \sqrt {3}\, a_{1}&=0\\ 2160 \,12^{{2}/{3}} \sqrt {3}\, a_{3}&=0\\ 216 \,12^{{2}/{3}} \sqrt {3}\, b_{2}&=0\\ 144 \,12^{{2}/{3}} a_{2}-288 \,12^{{2}/{3}} b_{3}&=0\\ 48 \,12^{{2}/{3}} a_{3}-72 \,12^{{2}/{3}} b_{1}&=0\\ -3888 i a_{3}+1296 \sqrt {3}\, a_{3}&=0\\ -864 i b_{2}+288 \sqrt {3}\, b_{2}&=0\\ 576 i a_{3}-192 \sqrt {3}\, a_{3}&=0\\ 5832 i b_{2}-1944 \sqrt {3}\, b_{2}&=0\\ -432 \,12^{{2}/{3}} \sqrt {3}\, a_{2}+864 \,12^{{2}/{3}} \sqrt {3}\, b_{3}&=0\\ 48 \,12^{{2}/{3}} \sqrt {3}\, a_{2}-96 \,12^{{2}/{3}} \sqrt {3}\, b_{3}&=0\\ -336 \,12^{{2}/{3}} \sqrt {3}\, a_{3}+216 \,12^{{2}/{3}} \sqrt {3}\, b_{1}&=0\\ -648 i \sqrt {3}\, b_{2}+648 b_{2}&=0\\ -648 i 12^{{1}/{3}} a_{1}-216 \,12^{{1}/{3}} \sqrt {3}\, a_{1}&=0\\ -648 i 12^{{1}/{3}} a_{3}-216 \,12^{{1}/{3}} \sqrt {3}\, a_{3}&=0\\ 72 i 12^{{1}/{3}} a_{1}+24 \,12^{{1}/{3}} \sqrt {3}\, a_{1}&=0\\ 324 i 12^{{1}/{3}} b_{2}+108 \,12^{{1}/{3}} \sqrt {3}\, b_{2}&=0\\ 432 i \sqrt {3}\, a_{3}-432 a_{3}&=0\\ -36 i 12^{{1}/{3}} \sqrt {3}\, b_{2}-36 \,12^{{1}/{3}} b_{2}&=0\\ 72 i 12^{{1}/{3}} \sqrt {3}\, a_{1}+72 \,12^{{1}/{3}} a_{1}&=0\\ 72 i 12^{{1}/{3}} \sqrt {3}\, a_{3}+72 \,12^{{1}/{3}} a_{3}&=0\\ -72 i 12^{{1}/{3}} a_{2}+144 i 12^{{1}/{3}} b_{3}-24 \,12^{{1}/{3}} \sqrt {3}\, a_{2}+48 \,12^{{1}/{3}} \sqrt {3}\, b_{3}&=0\\ 72 i 12^{{1}/{3}} a_{3}+324 i 12^{{1}/{3}} b_{1}+24 \,12^{{1}/{3}} \sqrt {3}\, a_{3}+108 \,12^{{1}/{3}} \sqrt {3}\, b_{1}&=0\\ 324 i 12^{{1}/{3}} a_{2}-648 i 12^{{1}/{3}} b_{3}+108 \,12^{{1}/{3}} \sqrt {3}\, a_{2}-216 \,12^{{1}/{3}} \sqrt {3}\, b_{3}&=0\\ -36 i 12^{{1}/{3}} \sqrt {3}\, a_{2}+72 i 12^{{1}/{3}} \sqrt {3}\, b_{3}-36 \,12^{{1}/{3}} a_{2}+72 \,12^{{1}/{3}} b_{3}&=0\\ 24 i 12^{{1}/{3}} \sqrt {3}\, a_{3}-36 i 12^{{1}/{3}} \sqrt {3}\, b_{1}+24 \,12^{{1}/{3}} a_{3}-36 \,12^{{1}/{3}} b_{1}&=0 \end {align*}
Solving the above equations for the unknowns gives \begin {align*} a_{1}&=0\\ a_{2}&=2 b_{3}\\ a_{3}&=0\\ b_{1}&=0\\ b_{2}&=0\\ b_{3}&=b_{3} \end {align*}
Substituting the above solution in the anstaz (1E,2E) (using \(1\) as arbitrary value for any unknown in the RHS) gives \begin{align*} \xi &= 2 x \\ \eta &= y \\ \end{align*} The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( x,y\right ) \to \left ( R,S \right )\) where \(\left ( R,S \right )\) are the canonical coordinates which make the original ode become a quadrature and hence solved by integration.
The characteristic pde which is used to find the canonical coordinates is \begin {align*} \frac {d x}{\xi } &= \frac {d y}{\eta } = dS \tag {1} \end {align*}
The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial y}\right ) S(x,y) = 1\). Starting with the first pair of ode’s in (1) gives an ode to solve for the independent variable \(R\) in the canonical coordinates, where \(S(R)\). Unable to determine \(R\). Terminating
Unable to determine ODE type.
Solving equation (3)
Writing the ode as \begin {align*} y^{\prime }&=\frac {\left (-i x \sqrt {3}\, 12^{{1}/{3}}+2 {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{2}/{3}}-12^{{1}/{3}} x \right ) 12^{{1}/{3}}}{6 {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{1}/{3}} x \left (i \sqrt {3}-1\right )}\\ 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 (-i x \sqrt {3}\, 12^{{1}/{3}}+2 {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{2}/{3}}-12^{{1}/{3}} x \right ) 12^{{1}/{3}} \left (b_{3}-a_{2}\right )}{6 {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{1}/{3}} x \left (i \sqrt {3}-1\right )}-\frac {\left (-i x \sqrt {3}\, 12^{{1}/{3}}+2 {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{2}/{3}}-12^{{1}/{3}} x \right )^{2} 12^{{2}/{3}} a_{3}}{36 {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{2}/{3}} x^{2} \left (i \sqrt {3}-1\right )^{2}}-\left (\frac {\left (-i \sqrt {3}\, 12^{{1}/{3}}+\frac {-\frac {8 \sqrt {3}\, x}{3 \sqrt {27 y^{2}-4 x}}-\frac {4 \sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right )}{3}}{{\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{1}/{3}}}-12^{{1}/{3}}\right ) 12^{{1}/{3}}}{6 {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{1}/{3}} x \left (i \sqrt {3}-1\right )}-\frac {\left (-i x \sqrt {3}\, 12^{{1}/{3}}+2 {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{2}/{3}}-12^{{1}/{3}} x \right ) 12^{{1}/{3}} \left (-\frac {2 \sqrt {3}\, x}{\sqrt {27 y^{2}-4 x}}-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right )\right )}{18 {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{4}/{3}} x \left (i \sqrt {3}-1\right )}-\frac {\left (-i x \sqrt {3}\, 12^{{1}/{3}}+2 {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{2}/{3}}-12^{{1}/{3}} x \right ) 12^{{1}/{3}}}{6 {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{1}/{3}} x^{2} \left (i \sqrt {3}-1\right )}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )-\left (-\frac {2 \sqrt {3}\, \left (3 \sqrt {3}-\frac {27 y}{\sqrt {27 y^{2}-4 x}}\right ) 12^{{1}/{3}}}{9 {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{2}/{3}} \left (i \sqrt {3}-1\right )}+\frac {\left (-i x \sqrt {3}\, 12^{{1}/{3}}+2 {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{2}/{3}}-12^{{1}/{3}} x \right ) 12^{{1}/{3}} \sqrt {3}\, \left (3 \sqrt {3}-\frac {27 y}{\sqrt {27 y^{2}-4 x}}\right )}{18 {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{4}/{3}} \left (i \sqrt {3}-1\right )}\right ) \left (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} Simplifying the above gives \begin{equation} \tag{6E} \text {Expression too large to display} \end{equation} Since the PDE has radicals, simplifying gives \[ \text {Expression too large to display} \] Looking at the above PDE shows the following are all the terms with \(\{x, y\}\) in them. \[ \left \{x, y, {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{1}/{3}}, {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{2}/{3}}, \sqrt {27 y^{2}-4 x}\right \} \] The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them \[ \left \{x = v_{1}, y = v_{2}, {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{1}/{3}} = v_{3}, {\left (-\sqrt {3}\, \left (3 \sqrt {3}\, y -\sqrt {27 y^{2}-4 x}\right ) x \right )}^{{2}/{3}} = v_{4}, \sqrt {27 y^{2}-4 x} = v_{5}\right \} \] The above PDE (6E) now becomes \begin{equation} \tag{7E} 12 v_{1} \left (18 \,12^{{2}/{3}} \sqrt {3}\, v_{1}^{3} v_{2} b_{2}-36 \,12^{{2}/{3}} \sqrt {3}\, v_{1}^{2} v_{2}^{2} a_{2}+72 \,12^{{2}/{3}} \sqrt {3}\, v_{1}^{2} v_{2}^{2} b_{3}+180 \,12^{{2}/{3}} \sqrt {3}\, v_{1} v_{2}^{3} a_{3}-18 \,12^{{1}/{3}} v_{4} \sqrt {3}\, v_{2}^{3} a_{3}-28 \,12^{{2}/{3}} \sqrt {3}\, v_{1}^{2} v_{2} a_{3}+18 \,12^{{2}/{3}} \sqrt {3}\, v_{1}^{2} v_{2} b_{1}+18 \,12^{{2}/{3}} \sqrt {3}\, v_{1} v_{2}^{2} a_{1}+12 \,12^{{2}/{3}} v_{5} v_{1}^{2} v_{2} a_{2}-24 \,12^{{2}/{3}} v_{5} v_{1}^{2} v_{2} b_{3}-60 \,12^{{2}/{3}} v_{5} v_{1} v_{2}^{2} a_{3}-2 \,12^{{1}/{3}} v_{4} \sqrt {3}\, v_{1}^{2} a_{2}+4 \,12^{{1}/{3}} v_{4} \sqrt {3}\, v_{1}^{2} b_{3}-18 \,12^{{1}/{3}} v_{4} \sqrt {3}\, v_{2}^{2} a_{1}-3 \,12^{{1}/{3}} v_{4} v_{5} v_{1}^{2} b_{2}+6 \,12^{{1}/{3}} v_{4} v_{5} v_{2}^{2} a_{3}-162 v_{3} \sqrt {3}\, v_{1}^{2} v_{2}^{2} b_{2}-6 \,12^{{2}/{3}} v_{5} v_{1} v_{2} a_{1}+2 \,12^{{1}/{3}} v_{4} \sqrt {3}\, v_{1} a_{1}+2 \,12^{{1}/{3}} v_{4} v_{5} v_{1} a_{3}-3 \,12^{{1}/{3}} v_{4} v_{5} v_{1} b_{1}+6 \,12^{{1}/{3}} v_{4} v_{5} v_{2} a_{1}+108 v_{3} \sqrt {3}\, v_{1} v_{2}^{2} a_{3}+54 v_{3} v_{5} v_{1}^{2} v_{2} b_{2}-36 v_{3} v_{5} v_{1} v_{2} a_{3}+54 i 12^{{1}/{3}} v_{4} v_{2}^{3} a_{3}+6 i 12^{{1}/{3}} v_{4} v_{1}^{2} a_{2}-12 i 12^{{1}/{3}} v_{4} v_{1}^{2} b_{3}+54 i 12^{{1}/{3}} v_{4} v_{2}^{2} a_{1}-486 i v_{3} v_{1}^{2} v_{2}^{2} b_{2}-6 i 12^{{1}/{3}} v_{4} v_{1} a_{1}+324 i v_{3} v_{1} v_{2}^{2} a_{3}-6 i 12^{{1}/{3}} v_{4} \sqrt {3}\, v_{5} v_{2}^{2} a_{3}-2 i 12^{{1}/{3}} v_{4} \sqrt {3}\, v_{5} v_{1} a_{3}+3 i 12^{{1}/{3}} v_{4} \sqrt {3}\, v_{5} v_{1} b_{1}-6 i 12^{{1}/{3}} v_{4} \sqrt {3}\, v_{5} v_{2} a_{1}+54 i v_{3} \sqrt {3}\, v_{5} v_{1}^{2} v_{2} b_{2}-36 i v_{3} \sqrt {3}\, v_{5} v_{1} v_{2} a_{3}+3 i 12^{{1}/{3}} v_{4} \sqrt {3}\, v_{5} v_{1}^{2} b_{2}+4 \,12^{{2}/{3}} \sqrt {3}\, v_{1}^{3} a_{2}-8 \,12^{{2}/{3}} \sqrt {3}\, v_{1}^{3} b_{3}-6 \,12^{{2}/{3}} v_{5} v_{1}^{3} b_{2}-4 \,12^{{2}/{3}} \sqrt {3}\, v_{1}^{2} a_{1}+4 \,12^{{2}/{3}} v_{1}^{2} v_{5} a_{3}-6 \,12^{{2}/{3}} v_{5} v_{1}^{2} b_{1}+24 v_{3} \sqrt {3}\, v_{1}^{3} b_{2}-16 v_{3} \sqrt {3}\, v_{1}^{2} a_{3}+72 i v_{3} v_{1}^{3} b_{2}-48 i v_{3} v_{1}^{2} a_{3}+9 \,12^{{1}/{3}} v_{4} \sqrt {3}\, v_{1}^{2} v_{2} b_{2}+9 \,12^{{1}/{3}} v_{4} \sqrt {3}\, v_{1} v_{2}^{2} a_{2}-18 \,12^{{1}/{3}} v_{4} \sqrt {3}\, v_{1} v_{2}^{2} b_{3}+2 \,12^{{1}/{3}} v_{4} \sqrt {3}\, v_{1} v_{2} a_{3}+9 \,12^{{1}/{3}} v_{4} \sqrt {3}\, v_{1} v_{2} b_{1}-3 \,12^{{1}/{3}} v_{4} v_{5} v_{1} v_{2} a_{2}+6 \,12^{{1}/{3}} v_{4} v_{5} v_{1} v_{2} b_{3}-27 i 12^{{1}/{3}} v_{4} v_{1}^{2} v_{2} b_{2}-27 i 12^{{1}/{3}} v_{4} v_{1} v_{2}^{2} a_{2}+54 i 12^{{1}/{3}} v_{4} v_{1} v_{2}^{2} b_{3}-6 i 12^{{1}/{3}} v_{4} v_{1} v_{2} a_{3}-27 i 12^{{1}/{3}} v_{4} v_{1} v_{2} b_{1}+3 i 12^{{1}/{3}} v_{4} \sqrt {3}\, v_{5} v_{1} v_{2} a_{2}-6 i 12^{{1}/{3}} v_{4} \sqrt {3}\, v_{5} v_{1} v_{2} b_{3}\right ) = 0 \end{equation} Collecting the above on the terms \(v_i\) introduced, and these are \[ \{v_{1}, v_{2}, v_{3}, v_{4}, v_{5}\} \] Equation (7E) now becomes \begin{equation} \tag{8E} \left (144 \,12^{{2}/{3}} a_{2}-288 \,12^{{2}/{3}} b_{3}\right ) v_{2} v_{5} v_{1}^{3}-48 \,12^{{2}/{3}} \sqrt {3}\, a_{1} v_{1}^{3}-72 \,12^{{2}/{3}} b_{2} v_{5} v_{1}^{4}+\left (-432 i \sqrt {3}\, a_{3}-432 a_{3}\right ) v_{2} v_{3} v_{5} v_{1}^{2}+\left (36 i 12^{{1}/{3}} \sqrt {3}\, a_{2}-72 i 12^{{1}/{3}} \sqrt {3}\, b_{3}-36 \,12^{{1}/{3}} a_{2}+72 \,12^{{1}/{3}} b_{3}\right ) v_{2} v_{4} v_{5} v_{1}^{2}+\left (-72 i 12^{{1}/{3}} \sqrt {3}\, a_{3}+72 \,12^{{1}/{3}} a_{3}\right ) v_{2}^{2} v_{4} v_{5} v_{1}+\left (-72 i 12^{{1}/{3}} \sqrt {3}\, a_{1}+72 \,12^{{1}/{3}} a_{1}\right ) v_{2} v_{4} v_{5} v_{1}+\left (648 i \sqrt {3}\, b_{2}+648 b_{2}\right ) v_{2} v_{3} v_{5} v_{1}^{3}+\left (-432 \,12^{{2}/{3}} \sqrt {3}\, a_{2}+864 \,12^{{2}/{3}} \sqrt {3}\, b_{3}\right ) v_{2}^{2} v_{1}^{3}+\left (-336 \,12^{{2}/{3}} \sqrt {3}\, a_{3}+216 \,12^{{2}/{3}} \sqrt {3}\, b_{1}\right ) v_{2} v_{1}^{3}+\left (48 \,12^{{2}/{3}} a_{3}-72 \,12^{{2}/{3}} b_{1}\right ) v_{5} v_{1}^{3}+\left (-576 i a_{3}-192 \sqrt {3}\, a_{3}\right ) v_{3} v_{1}^{3}+\left (72 i 12^{{1}/{3}} a_{2}-144 i 12^{{1}/{3}} b_{3}-24 \,12^{{1}/{3}} \sqrt {3}\, a_{2}+48 \,12^{{1}/{3}} \sqrt {3}\, b_{3}\right ) v_{4} v_{1}^{3}+\left (864 i b_{2}+288 \sqrt {3}\, b_{2}\right ) v_{3} v_{1}^{4}+\left (-72 i 12^{{1}/{3}} a_{1}+24 \,12^{{1}/{3}} \sqrt {3}\, a_{1}\right ) v_{4} v_{1}^{2}+\left (-5832 i b_{2}-1944 \sqrt {3}\, b_{2}\right ) v_{2}^{2} v_{3} v_{1}^{3}+\left (-324 i 12^{{1}/{3}} b_{2}+108 \,12^{{1}/{3}} \sqrt {3}\, b_{2}\right ) v_{2} v_{4} v_{1}^{3}+\left (36 i 12^{{1}/{3}} \sqrt {3}\, b_{2}-36 \,12^{{1}/{3}} b_{2}\right ) v_{4} v_{5} v_{1}^{3}+\left (3888 i a_{3}+1296 \sqrt {3}\, a_{3}\right ) v_{2}^{2} v_{3} v_{1}^{2}+\left (-324 i 12^{{1}/{3}} a_{2}+648 i 12^{{1}/{3}} b_{3}+108 \,12^{{1}/{3}} \sqrt {3}\, a_{2}-216 \,12^{{1}/{3}} \sqrt {3}\, b_{3}\right ) v_{2}^{2} v_{4} v_{1}^{2}+\left (-72 i 12^{{1}/{3}} a_{3}-324 i 12^{{1}/{3}} b_{1}+24 \,12^{{1}/{3}} \sqrt {3}\, a_{3}+108 \,12^{{1}/{3}} \sqrt {3}\, b_{1}\right ) v_{2} v_{4} v_{1}^{2}+\left (-24 i 12^{{1}/{3}} \sqrt {3}\, a_{3}+36 i 12^{{1}/{3}} \sqrt {3}\, b_{1}+24 \,12^{{1}/{3}} a_{3}-36 \,12^{{1}/{3}} b_{1}\right ) v_{4} v_{5} v_{1}^{2}+\left (648 i 12^{{1}/{3}} a_{3}-216 \,12^{{1}/{3}} \sqrt {3}\, a_{3}\right ) v_{2}^{3} v_{4} v_{1}+\left (648 i 12^{{1}/{3}} a_{1}-216 \,12^{{1}/{3}} \sqrt {3}\, a_{1}\right ) v_{2}^{2} v_{4} v_{1}+2160 \,12^{{2}/{3}} \sqrt {3}\, a_{3} v_{2}^{3} v_{1}^{2}-720 \,12^{{2}/{3}} a_{3} v_{2}^{2} v_{5} v_{1}^{2}+216 \,12^{{2}/{3}} \sqrt {3}\, a_{1} v_{2}^{2} v_{1}^{2}-72 \,12^{{2}/{3}} a_{1} v_{2} v_{5} v_{1}^{2}+216 \,12^{{2}/{3}} \sqrt {3}\, b_{2} v_{2} v_{1}^{4}+\left (48 \,12^{{2}/{3}} \sqrt {3}\, a_{2}-96 \,12^{{2}/{3}} \sqrt {3}\, b_{3}\right ) v_{1}^{4} = 0 \end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve \begin {align*} -72 \,12^{{2}/{3}} a_{1}&=0\\ -720 \,12^{{2}/{3}} a_{3}&=0\\ -72 \,12^{{2}/{3}} b_{2}&=0\\ -48 \,12^{{2}/{3}} \sqrt {3}\, a_{1}&=0\\ 216 \,12^{{2}/{3}} \sqrt {3}\, a_{1}&=0\\ 2160 \,12^{{2}/{3}} \sqrt {3}\, a_{3}&=0\\ 216 \,12^{{2}/{3}} \sqrt {3}\, b_{2}&=0\\ 144 \,12^{{2}/{3}} a_{2}-288 \,12^{{2}/{3}} b_{3}&=0\\ 48 \,12^{{2}/{3}} a_{3}-72 \,12^{{2}/{3}} b_{1}&=0\\ -5832 i b_{2}-1944 \sqrt {3}\, b_{2}&=0\\ -576 i a_{3}-192 \sqrt {3}\, a_{3}&=0\\ 864 i b_{2}+288 \sqrt {3}\, b_{2}&=0\\ 3888 i a_{3}+1296 \sqrt {3}\, a_{3}&=0\\ -432 \,12^{{2}/{3}} \sqrt {3}\, a_{2}+864 \,12^{{2}/{3}} \sqrt {3}\, b_{3}&=0\\ 48 \,12^{{2}/{3}} \sqrt {3}\, a_{2}-96 \,12^{{2}/{3}} \sqrt {3}\, b_{3}&=0\\ -336 \,12^{{2}/{3}} \sqrt {3}\, a_{3}+216 \,12^{{2}/{3}} \sqrt {3}\, b_{1}&=0\\ -432 i \sqrt {3}\, a_{3}-432 a_{3}&=0\\ -324 i 12^{{1}/{3}} b_{2}+108 \,12^{{1}/{3}} \sqrt {3}\, b_{2}&=0\\ -72 i 12^{{1}/{3}} a_{1}+24 \,12^{{1}/{3}} \sqrt {3}\, a_{1}&=0\\ 648 i \sqrt {3}\, b_{2}+648 b_{2}&=0\\ 648 i 12^{{1}/{3}} a_{1}-216 \,12^{{1}/{3}} \sqrt {3}\, a_{1}&=0\\ 648 i 12^{{1}/{3}} a_{3}-216 \,12^{{1}/{3}} \sqrt {3}\, a_{3}&=0\\ -72 i 12^{{1}/{3}} \sqrt {3}\, a_{1}+72 \,12^{{1}/{3}} a_{1}&=0\\ -72 i 12^{{1}/{3}} \sqrt {3}\, a_{3}+72 \,12^{{1}/{3}} a_{3}&=0\\ 36 i 12^{{1}/{3}} \sqrt {3}\, b_{2}-36 \,12^{{1}/{3}} b_{2}&=0\\ -324 i 12^{{1}/{3}} a_{2}+648 i 12^{{1}/{3}} b_{3}+108 \,12^{{1}/{3}} \sqrt {3}\, a_{2}-216 \,12^{{1}/{3}} \sqrt {3}\, b_{3}&=0\\ -72 i 12^{{1}/{3}} a_{3}-324 i 12^{{1}/{3}} b_{1}+24 \,12^{{1}/{3}} \sqrt {3}\, a_{3}+108 \,12^{{1}/{3}} \sqrt {3}\, b_{1}&=0\\ 72 i 12^{{1}/{3}} a_{2}-144 i 12^{{1}/{3}} b_{3}-24 \,12^{{1}/{3}} \sqrt {3}\, a_{2}+48 \,12^{{1}/{3}} \sqrt {3}\, b_{3}&=0\\ -24 i 12^{{1}/{3}} \sqrt {3}\, a_{3}+36 i 12^{{1}/{3}} \sqrt {3}\, b_{1}+24 \,12^{{1}/{3}} a_{3}-36 \,12^{{1}/{3}} b_{1}&=0\\ 36 i 12^{{1}/{3}} \sqrt {3}\, a_{2}-72 i 12^{{1}/{3}} \sqrt {3}\, b_{3}-36 \,12^{{1}/{3}} a_{2}+72 \,12^{{1}/{3}} b_{3}&=0 \end {align*}
Solving the above equations for the unknowns gives \begin {align*} a_{1}&=0\\ a_{2}&=2 b_{3}\\ a_{3}&=0\\ b_{1}&=0\\ b_{2}&=0\\ b_{3}&=b_{3} \end {align*}
Substituting the above solution in the anstaz (1E,2E) (using \(1\) as arbitrary value for any unknown in the RHS) gives \begin{align*} \xi &= 2 x \\ \eta &= y \\ \end{align*} The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( x,y\right ) \to \left ( R,S \right )\) where \(\left ( R,S \right )\) are the canonical coordinates which make the original ode become a quadrature and hence solved by integration.
The characteristic pde which is used to find the canonical coordinates is \begin {align*} \frac {d x}{\xi } &= \frac {d y}{\eta } = dS \tag {1} \end {align*}
The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial y}\right ) S(x,y) = 1\). Starting with the first pair of ode’s in (1) gives an ode to solve for the independent variable \(R\) in the canonical coordinates, where \(S(R)\). Unable to determine \(R\). Terminating
Unable to determine ODE type.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y-y^{\prime } x +x^{2} {y^{\prime }}^{3}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=\frac {{\left (\left (12 \sqrt {3}\, \sqrt {27 y^{2}-4 x}-108 y\right ) x \right )}^{{1}/{3}}}{6 x}+\frac {2}{{\left (\left (12 \sqrt {3}\, \sqrt {27 y^{2}-4 x}-108 y\right ) x \right )}^{{1}/{3}}}, y^{\prime }=-\frac {{\left (\left (12 \sqrt {3}\, \sqrt {27 y^{2}-4 x}-108 y\right ) x \right )}^{{1}/{3}}}{12 x}-\frac {1}{{\left (\left (12 \sqrt {3}\, \sqrt {27 y^{2}-4 x}-108 y\right ) x \right )}^{{1}/{3}}}-\frac {\mathrm {I} \sqrt {3}\, \left (\frac {{\left (\left (12 \sqrt {3}\, \sqrt {27 y^{2}-4 x}-108 y\right ) x \right )}^{{1}/{3}}}{6 x}-\frac {2}{{\left (\left (12 \sqrt {3}\, \sqrt {27 y^{2}-4 x}-108 y\right ) x \right )}^{{1}/{3}}}\right )}{2}, y^{\prime }=-\frac {{\left (\left (12 \sqrt {3}\, \sqrt {27 y^{2}-4 x}-108 y\right ) x \right )}^{{1}/{3}}}{12 x}-\frac {1}{{\left (\left (12 \sqrt {3}\, \sqrt {27 y^{2}-4 x}-108 y\right ) x \right )}^{{1}/{3}}}+\frac {\mathrm {I} \sqrt {3}\, \left (\frac {{\left (\left (12 \sqrt {3}\, \sqrt {27 y^{2}-4 x}-108 y\right ) x \right )}^{{1}/{3}}}{6 x}-\frac {2}{{\left (\left (12 \sqrt {3}\, \sqrt {27 y^{2}-4 x}-108 y\right ) x \right )}^{{1}/{3}}}\right )}{2}\right ] \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {{\left (\left (12 \sqrt {3}\, \sqrt {27 y^{2}-4 x}-108 y\right ) x \right )}^{{1}/{3}}}{6 x}+\frac {2}{{\left (\left (12 \sqrt {3}\, \sqrt {27 y^{2}-4 x}-108 y\right ) x \right )}^{{1}/{3}}} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {{\left (\left (12 \sqrt {3}\, \sqrt {27 y^{2}-4 x}-108 y\right ) x \right )}^{{1}/{3}}}{12 x}-\frac {1}{{\left (\left (12 \sqrt {3}\, \sqrt {27 y^{2}-4 x}-108 y\right ) x \right )}^{{1}/{3}}}-\frac {\mathrm {I} \sqrt {3}\, \left (\frac {{\left (\left (12 \sqrt {3}\, \sqrt {27 y^{2}-4 x}-108 y\right ) x \right )}^{{1}/{3}}}{6 x}-\frac {2}{{\left (\left (12 \sqrt {3}\, \sqrt {27 y^{2}-4 x}-108 y\right ) x \right )}^{{1}/{3}}}\right )}{2} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {{\left (\left (12 \sqrt {3}\, \sqrt {27 y^{2}-4 x}-108 y\right ) x \right )}^{{1}/{3}}}{12 x}-\frac {1}{{\left (\left (12 \sqrt {3}\, \sqrt {27 y^{2}-4 x}-108 y\right ) x \right )}^{{1}/{3}}}+\frac {\mathrm {I} \sqrt {3}\, \left (\frac {{\left (\left (12 \sqrt {3}\, \sqrt {27 y^{2}-4 x}-108 y\right ) x \right )}^{{1}/{3}}}{6 x}-\frac {2}{{\left (\left (12 \sqrt {3}\, \sqrt {27 y^{2}-4 x}-108 y\right ) x \right )}^{{1}/{3}}}\right )}{2} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\mathit {workingODE} , \mathit {workingODE} , \mathit {workingODE}\right \} \end {array} \]
Maple trace
`Methods for first order ODEs: -> Solving 1st order ODE of high degree, 1st attempt trying 1st order WeierstrassP solution for high degree ODE trying 1st order WeierstrassPPrime solution for high degree ODE trying 1st order JacobiSN solution for high degree ODE trying 1st order ODE linearizable_by_differentiation trying differential order: 1; missing variables trying dAlembert trying simple symmetries for implicit equations Successful isolation of dy/dx: 3 solutions were found. Trying to solve each resulting ODE. *** Sublevel 2 *** Methods for first order ODEs: --- Trying classification methods --- trying homogeneous types: trying exact Looking for potential symmetries trying an equivalence to an Abel ODE trying 1st order ODE linearizable_by_differentiation -> Solving 1st order ODE of high degree, Lie methods, 1st trial `, `-> Computing symmetries using: way = 2 `, `-> Computing symmetries using: way = 2 -> Solving 1st order ODE of high degree, 2nd attempt. Trying parametric methods -> Calling odsolve with the ODE`, diff(y(x), x) = (3*y(x)*x-(-4*y(x)*x+1)^(1/2)-1)/(x^2*(1+(-4*y(x)*x+1)^(1/2))), y(x)` *** Sublev Methods for first order ODEs: --- Trying classification methods --- trying homogeneous types: trying homogeneous G <- homogeneous successful -> Calling odsolve with the ODE`, diff(y(x), x) = (-3*y(x)*x-(-4*y(x)*x+1)^(1/2)+1)/(x^2*(-1+(-4*y(x)*x+1)^(1/2))), y(x)` *** Subl Methods for first order ODEs: --- Trying classification methods --- trying homogeneous types: trying homogeneous G <- homogeneous successful <- 1st order, parametric methods successful`
✓ Solution by Maple
Time used: 0.14 (sec). Leaf size: 123
dsolve(y(x)=x*diff(y(x),x)-x^2* (diff(y(x),x))^3,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= -x^{2} \operatorname {RootOf}\left (4 \textit {\_Z}^{4} c_{1} x^{2}+8 \textit {\_Z}^{2} c_{1} x -\textit {\_Z} +4 c_{1} \right )^{3}+x \operatorname {RootOf}\left (4 \textit {\_Z}^{4} c_{1} x^{2}+8 \textit {\_Z}^{2} c_{1} x -\textit {\_Z} +4 c_{1} \right ) \\ y \left (x \right ) &= -x^{2} \operatorname {RootOf}\left (4 \textit {\_Z}^{4} c_{1} x^{2}-16 \textit {\_Z}^{2} c_{1} x -\textit {\_Z} +16 c_{1} \right )^{3}+x \operatorname {RootOf}\left (4 \textit {\_Z}^{4} c_{1} x^{2}-16 \textit {\_Z}^{2} c_{1} x -\textit {\_Z} +16 c_{1} \right ) \\ \end{align*}
✗ Solution by Mathematica
Time used: 0.0 (sec). Leaf size: 0
DSolve[y[x]==x*y'[x]-x^2*(y'[x])^3,y[x],x,IncludeSingularSolutions -> True]
Timed out