problem 86

Internal problem ID [3228]
Internal ﬁle name [OUTPUT/2720_Sunday_June_05_2022_08_39_16_AM_49114467/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: 86.
ODE order: 1.
ODE degree: 3.

The type(s) of ODE detected by this program : "ﬁrst_order_ode_lie_symmetry_calculated"

\[ \boxed {y-2 x y^{\prime }-y^{2} {y^{\prime }}^{3}=0} \] 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 (108 y^{2}+12 \sqrt {3}\, \sqrt {27 y^{4}+32 x^{3}}\right )^{\frac {1}{3}}}{6 y}-\frac {4 x}{y \left (108 y^{2}+12 \sqrt {3}\, \sqrt {27 y^{4}+32 x^{3}}\right )^{\frac {1}{3}}} \tag {1} \\ y^{\prime }&=-\frac {\left (108 y^{2}+12 \sqrt {3}\, \sqrt {27 y^{4}+32 x^{3}}\right )^{\frac {1}{3}}}{12 y}+\frac {2 x}{y \left (108 y^{2}+12 \sqrt {3}\, \sqrt {27 y^{4}+32 x^{3}}\right )^{\frac {1}{3}}}+\frac {i \sqrt {3}\, \left (\frac {\left (108 y^{2}+12 \sqrt {3}\, \sqrt {27 y^{4}+32 x^{3}}\right )^{\frac {1}{3}}}{6 y}+\frac {4 x}{y \left (108 y^{2}+12 \sqrt {3}\, \sqrt {27 y^{4}+32 x^{3}}\right )^{\frac {1}{3}}}\right )}{2} \tag {2} \\ y^{\prime }&=-\frac {\left (108 y^{2}+12 \sqrt {3}\, \sqrt {27 y^{4}+32 x^{3}}\right )^{\frac {1}{3}}}{12 y}+\frac {2 x}{y \left (108 y^{2}+12 \sqrt {3}\, \sqrt {27 y^{4}+32 x^{3}}\right )^{\frac {1}{3}}}-\frac {i \sqrt {3}\, \left (\frac {\left (108 y^{2}+12 \sqrt {3}\, \sqrt {27 y^{4}+32 x^{3}}\right )^{\frac {1}{3}}}{6 y}+\frac {4 x}{y \left (108 y^{2}+12 \sqrt {3}\, \sqrt {27 y^{4}+32 x^{3}}\right )^{\frac {1}{3}}}\right )}{2} \tag {3} \end {align*}

Writing the ode as \begin {align*} y^{\prime }&=\frac {\left ({\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {2}{3}}-2 \,12^{\frac {1}{3}} x \right ) 12^{\frac {1}{3}}}{6 y {\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {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 (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {2}{3}}-2 \,12^{\frac {1}{3}} x \right ) 12^{\frac {1}{3}} \left (b_{3}-a_{2}\right )}{6 y {\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {1}{3}}}-\frac {{\left ({\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {2}{3}}-2 \,12^{\frac {1}{3}} x \right )}^{2} 12^{\frac {2}{3}} a_{3}}{36 y^{2} {\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {2}{3}}}-\left (\frac {\left (\frac {32 \sqrt {3}\, x^{2}}{{\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {1}{3}} \sqrt {27 y^{4}+32 x^{3}}}-2 \,12^{\frac {1}{3}}\right ) 12^{\frac {1}{3}}}{6 y {\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {1}{3}}}-\frac {8 \left ({\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {2}{3}}-2 \,12^{\frac {1}{3}} x \right ) 12^{\frac {1}{3}} \sqrt {3}\, x^{2}}{3 y {\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {4}{3}} \sqrt {27 y^{4}+32 x^{3}}}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )-\left (\frac {\sqrt {3}\, \left (\frac {54 y^{3}}{\sqrt {27 y^{4}+32 x^{3}}}+6 \sqrt {3}\, y \right ) 12^{\frac {1}{3}}}{9 {\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {2}{3}} y}-\frac {\left ({\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {2}{3}}-2 \,12^{\frac {1}{3}} x \right ) 12^{\frac {1}{3}}}{6 y^{2} {\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {1}{3}}}-\frac {\left ({\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {2}{3}}-2 \,12^{\frac {1}{3}} x \right ) 12^{\frac {1}{3}} \sqrt {3}\, \left (\frac {54 y^{3}}{\sqrt {27 y^{4}+32 x^{3}}}+6 \sqrt {3}\, y \right )}{18 y {\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {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. \[ \left \{x, y, {\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {1}{3}}, {\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {2}{3}}, \sqrt {27 y^{4}+32 x^{3}}\right \} \] The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them \[ \left \{x = v_{1}, y = v_{2}, {\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {1}{3}} = v_{3}, {\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {2}{3}} = v_{4}, \sqrt {27 y^{4}+32 x^{3}} = v_{5}\right \} \] The above PDE (6E) now becomes \begin{equation} \tag{7E} -384 \sqrt {3}\, 12^{\frac {2}{3}} v_{1}^{5} b_{2}+324 \sqrt {3}\, 12^{\frac {2}{3}} v_{2}^{5} a_{1}-384 \sqrt {3}\, 12^{\frac {2}{3}} v_{1}^{4} b_{1}+972 \sqrt {3}\, v_{3} v_{2}^{6} b_{2}+1536 \sqrt {3}\, v_{1}^{4} v_{3} a_{3}-54 v_{5} 12^{\frac {2}{3}} v_{2}^{4} a_{3}+108 v_{5} 12^{\frac {2}{3}} v_{2}^{3} a_{1}+324 v_{3} v_{5} v_{2}^{4} b_{2}-96 \,12^{\frac {2}{3}} v_{1}^{3} v_{5} a_{3}-180 v_{5} 12^{\frac {2}{3}} v_{1} v_{2}^{2} b_{1}+18 v_{5} 12^{\frac {1}{3}} v_{4} v_{2}^{2} b_{1}+576 \sqrt {3}\, 12^{\frac {2}{3}} v_{1}^{4} v_{2} a_{2}-384 \sqrt {3}\, 12^{\frac {2}{3}} v_{1}^{3} v_{2}^{2} a_{3}-540 \sqrt {3}\, 12^{\frac {2}{3}} v_{1} v_{2}^{4} b_{1}+54 \sqrt {3}\, 12^{\frac {1}{3}} v_{4} v_{2}^{4} b_{1}+192 \sqrt {3}\, 12^{\frac {2}{3}} v_{1}^{3} v_{2} a_{1}-162 \,12^{\frac {1}{3}} \sqrt {3}\, v_{4} v_{2}^{5} a_{2}+192 \,12^{\frac {1}{3}} \sqrt {3}\, v_{1}^{4} v_{4} b_{2}+192 \,12^{\frac {1}{3}} \sqrt {3}\, v_{1}^{3} v_{4} b_{1}+1152 \sqrt {3}\, v_{1}^{3} v_{3} v_{2}^{2} b_{2}+1296 \sqrt {3}\, v_{1} v_{3} v_{2}^{4} a_{3}+432 v_{1} v_{3} v_{5} v_{2}^{2} a_{3}-54 \,12^{\frac {1}{3}} v_{4} v_{5} v_{2}^{3} a_{2}+216 v_{5} 12^{\frac {2}{3}} v_{1} v_{2}^{3} a_{2}+72 v_{5} 12^{\frac {1}{3}} v_{4} v_{2}^{3} b_{3}-540 \sqrt {3}\, 12^{\frac {2}{3}} v_{1}^{2} v_{2}^{4} b_{2}-864 \sqrt {3}\, 12^{\frac {2}{3}} v_{1} v_{2}^{5} b_{3}+216 \sqrt {3}\, 12^{\frac {1}{3}} v_{4} v_{2}^{5} b_{3}-48 v_{5} 12^{\frac {1}{3}} v_{4} v_{1}^{2} a_{3}+648 \sqrt {3}\, 12^{\frac {2}{3}} v_{1} v_{2}^{5} a_{2}-768 \sqrt {3}\, 12^{\frac {2}{3}} v_{1}^{4} v_{2} b_{3}-180 v_{5} 12^{\frac {2}{3}} v_{1}^{2} v_{2}^{2} b_{2}-288 v_{5} 12^{\frac {2}{3}} v_{1} v_{2}^{3} b_{3}+384 \,12^{\frac {1}{3}} \sqrt {3}\, v_{1}^{3} v_{4} v_{2} b_{3}+18 v_{5} 12^{\frac {1}{3}} v_{4} v_{1} v_{2}^{2} b_{2}+54 \sqrt {3}\, 12^{\frac {1}{3}} v_{4} v_{1} v_{2}^{4} b_{2}-288 \sqrt {3}\, 12^{\frac {1}{3}} v_{4} v_{1}^{3} v_{2} a_{2}-96 \sqrt {3}\, 12^{\frac {1}{3}} v_{4} v_{1}^{2} v_{2}^{2} a_{3}-96 \sqrt {3}\, 12^{\frac {1}{3}} v_{4} v_{1}^{2} v_{2} a_{1}-162 \sqrt {3}\, 12^{\frac {2}{3}} v_{2}^{6} a_{3} = 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} -384 \sqrt {3}\, 12^{\frac {2}{3}} v_{1}^{5} b_{2}+324 \sqrt {3}\, 12^{\frac {2}{3}} v_{2}^{5} a_{1}-384 \sqrt {3}\, 12^{\frac {2}{3}} v_{1}^{4} b_{1}+972 \sqrt {3}\, v_{3} v_{2}^{6} b_{2}+1536 \sqrt {3}\, v_{1}^{4} v_{3} a_{3}-54 v_{5} 12^{\frac {2}{3}} v_{2}^{4} a_{3}+108 v_{5} 12^{\frac {2}{3}} v_{2}^{3} a_{1}+324 v_{3} v_{5} v_{2}^{4} b_{2}-96 \,12^{\frac {2}{3}} v_{1}^{3} v_{5} a_{3}+\left (576 \,12^{\frac {2}{3}} \sqrt {3}\, a_{2}-768 \,12^{\frac {2}{3}} \sqrt {3}\, b_{3}\right ) v_{1}^{4} v_{2}+\left (648 \,12^{\frac {2}{3}} \sqrt {3}\, a_{2}-864 \,12^{\frac {2}{3}} \sqrt {3}\, b_{3}\right ) v_{1} v_{2}^{5}+\left (-162 \,12^{\frac {1}{3}} \sqrt {3}\, a_{2}+216 \,12^{\frac {1}{3}} \sqrt {3}\, b_{3}\right ) v_{2}^{5} v_{4}+\left (216 \,12^{\frac {2}{3}} a_{2}-288 \,12^{\frac {2}{3}} b_{3}\right ) v_{1} v_{2}^{3} v_{5}+\left (-54 \,12^{\frac {1}{3}} a_{2}+72 \,12^{\frac {1}{3}} b_{3}\right ) v_{2}^{3} v_{4} v_{5}+\left (-288 \,12^{\frac {1}{3}} \sqrt {3}\, a_{2}+384 \,12^{\frac {1}{3}} \sqrt {3}\, b_{3}\right ) v_{1}^{3} v_{2} v_{4}-180 v_{5} 12^{\frac {2}{3}} v_{1} v_{2}^{2} b_{1}+18 v_{5} 12^{\frac {1}{3}} v_{4} v_{2}^{2} b_{1}-384 \sqrt {3}\, 12^{\frac {2}{3}} v_{1}^{3} v_{2}^{2} a_{3}-540 \sqrt {3}\, 12^{\frac {2}{3}} v_{1} v_{2}^{4} b_{1}+54 \sqrt {3}\, 12^{\frac {1}{3}} v_{4} v_{2}^{4} b_{1}+192 \sqrt {3}\, 12^{\frac {2}{3}} v_{1}^{3} v_{2} a_{1}+192 \,12^{\frac {1}{3}} \sqrt {3}\, v_{1}^{4} v_{4} b_{2}+192 \,12^{\frac {1}{3}} \sqrt {3}\, v_{1}^{3} v_{4} b_{1}+1152 \sqrt {3}\, v_{1}^{3} v_{3} v_{2}^{2} b_{2}+1296 \sqrt {3}\, v_{1} v_{3} v_{2}^{4} a_{3}+432 v_{1} v_{3} v_{5} v_{2}^{2} a_{3}-540 \sqrt {3}\, 12^{\frac {2}{3}} v_{1}^{2} v_{2}^{4} b_{2}-48 v_{5} 12^{\frac {1}{3}} v_{4} v_{1}^{2} a_{3}-180 v_{5} 12^{\frac {2}{3}} v_{1}^{2} v_{2}^{2} b_{2}+18 v_{5} 12^{\frac {1}{3}} v_{4} v_{1} v_{2}^{2} b_{2}+54 \sqrt {3}\, 12^{\frac {1}{3}} v_{4} v_{1} v_{2}^{4} b_{2}-96 \sqrt {3}\, 12^{\frac {1}{3}} v_{4} v_{1}^{2} v_{2}^{2} a_{3}-96 \sqrt {3}\, 12^{\frac {1}{3}} v_{4} v_{1}^{2} v_{2} a_{1}-162 \sqrt {3}\, 12^{\frac {2}{3}} v_{2}^{6} a_{3} = 0 \end{equation} Setting each coeﬃcients in (8E) to zero gives the following equations to solve \begin {align*} 432 a_{3}&=0\\ 324 b_{2}&=0\\ 1296 \sqrt {3}\, a_{3}&=0\\ 1536 \sqrt {3}\, a_{3}&=0\\ 972 \sqrt {3}\, b_{2}&=0\\ 1152 \sqrt {3}\, b_{2}&=0\\ -48 \,12^{\frac {1}{3}} a_{3}&=0\\ 18 \,12^{\frac {1}{3}} b_{1}&=0\\ 18 \,12^{\frac {1}{3}} b_{2}&=0\\ 108 \,12^{\frac {2}{3}} a_{1}&=0\\ -96 \,12^{\frac {2}{3}} a_{3}&=0\\ -54 \,12^{\frac {2}{3}} a_{3}&=0\\ -180 \,12^{\frac {2}{3}} b_{1}&=0\\ -180 \,12^{\frac {2}{3}} b_{2}&=0\\ -96 \,12^{\frac {1}{3}} \sqrt {3}\, a_{1}&=0\\ -96 \,12^{\frac {1}{3}} \sqrt {3}\, a_{3}&=0\\ 54 \,12^{\frac {1}{3}} \sqrt {3}\, b_{1}&=0\\ 192 \,12^{\frac {1}{3}} \sqrt {3}\, b_{1}&=0\\ 54 \,12^{\frac {1}{3}} \sqrt {3}\, b_{2}&=0\\ 192 \,12^{\frac {1}{3}} \sqrt {3}\, b_{2}&=0\\ 192 \,12^{\frac {2}{3}} \sqrt {3}\, a_{1}&=0\\ 324 \,12^{\frac {2}{3}} \sqrt {3}\, a_{1}&=0\\ -384 \,12^{\frac {2}{3}} \sqrt {3}\, a_{3}&=0\\ -162 \,12^{\frac {2}{3}} \sqrt {3}\, a_{3}&=0\\ -540 \,12^{\frac {2}{3}} \sqrt {3}\, b_{1}&=0\\ -384 \,12^{\frac {2}{3}} \sqrt {3}\, b_{1}&=0\\ -540 \,12^{\frac {2}{3}} \sqrt {3}\, b_{2}&=0\\ -384 \,12^{\frac {2}{3}} \sqrt {3}\, b_{2}&=0\\ -54 \,12^{\frac {1}{3}} a_{2}+72 \,12^{\frac {1}{3}} b_{3}&=0\\ 216 \,12^{\frac {2}{3}} a_{2}-288 \,12^{\frac {2}{3}} b_{3}&=0\\ -288 \,12^{\frac {1}{3}} \sqrt {3}\, a_{2}+384 \,12^{\frac {1}{3}} \sqrt {3}\, b_{3}&=0\\ -162 \,12^{\frac {1}{3}} \sqrt {3}\, a_{2}+216 \,12^{\frac {1}{3}} \sqrt {3}\, b_{3}&=0\\ 576 \,12^{\frac {2}{3}} \sqrt {3}\, a_{2}-768 \,12^{\frac {2}{3}} \sqrt {3}\, b_{3}&=0\\ 648 \,12^{\frac {2}{3}} \sqrt {3}\, a_{2}-864 \,12^{\frac {2}{3}} \sqrt {3}\, b_{3}&=0 \end {align*}

Solving the above equations for the unknowns gives \begin {align*} a_{1}&=0\\ a_{2}&=\frac {4 b_{3}}{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 &= \frac {4 x}{3} \\ \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 \(S(R)\). Therefore \begin {align*} \frac {dy}{dx} &= \frac {\eta }{\xi }\\ &= \frac {y}{\frac {4 x}{3}}\\ &= \frac {3 y}{4 x} \end {align*}

This is easily solved to give \begin {align*} y = c_{1} x^{\frac {3}{4}} \end {align*}

Where now the coordinate \(R\) is taken as the constant of integration. Hence \begin {align*} R &= \frac {y}{x^{\frac {3}{4}}} \end {align*}

And \(S\) is found from \begin {align*} dS &= \frac {dx}{\xi } \\ &= \frac {dx}{\frac {4 x}{3}} \end {align*}

Integrating gives \begin {align*} S &= \int { \frac {dx}{T}}\\ &= \frac {3 \ln \left (x \right )}{4} \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 {\left ({\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {2}{3}}-2 \,12^{\frac {1}{3}} x \right ) 12^{\frac {1}{3}}}{6 y {\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {1}{3}}} \end {align*}

Evaluating all the partial derivatives gives \begin {align*} R_{x} &= -\frac {3 y}{4 x^{\frac {7}{4}}}\\ R_{y} &= \frac {1}{x^{\frac {3}{4}}}\\ S_{x} &= \frac {3}{4 x}\\ 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 {9 x^{\frac {3}{4}} y 3^{\frac {1}{6}} \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )^{\frac {1}{3}}}{2 \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )^{\frac {2}{3}} x 6^{\frac {2}{3}}-8 \,3^{\frac {2}{3}} 2^{\frac {1}{3}} x^{2}-9 \,3^{\frac {1}{6}} \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )^{\frac {1}{3}} y^{2}}\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 {9 R 3^{\frac {1}{6}} \left (3 \sqrt {3}\, R^{2}+\sqrt {27 R^{4}+32}\right )^{\frac {1}{3}}}{2 \left (3 \sqrt {3}\, R^{2}+\sqrt {27 R^{4}+32}\right )^{\frac {2}{3}} 6^{\frac {2}{3}}-9 \,3^{\frac {1}{6}} \left (3 \sqrt {3}\, R^{2}+\sqrt {27 R^{4}+32}\right )^{\frac {1}{3}} R^{2}-8 \,3^{\frac {2}{3}} 2^{\frac {1}{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 {9 R \left (3 \sqrt {3}\, R^{2}+\sqrt {27 R^{4}+32}\right )^{\frac {1}{3}} 3^{\frac {1}{6}}}{2 \,36^{\frac {1}{3}} {\left (\left (3 \sqrt {3}\, R^{2}+\sqrt {27 R^{4}+32}\right )^{2}\right )}^{\frac {1}{3}}-9 \,3^{\frac {1}{6}} \left (3 \sqrt {3}\, R^{2}+\sqrt {27 R^{4}+32}\right )^{\frac {1}{3}} R^{2}-8 \,18^{\frac {1}{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*} \frac {3 \ln \left (x \right )}{4} = \int _{}^{\frac {y}{x^{\frac {3}{4}}}}\frac {9 \textit {\_a} \left (3 \sqrt {3}\, \textit {\_a}^{2}+\sqrt {27 \textit {\_a}^{4}+32}\right )^{\frac {1}{3}} 3^{\frac {1}{6}}}{2 \,36^{\frac {1}{3}} {\left (\left (3 \sqrt {3}\, \textit {\_a}^{2}+\sqrt {27 \textit {\_a}^{4}+32}\right )^{2}\right )}^{\frac {1}{3}}-9 \,3^{\frac {1}{6}} \left (3 \sqrt {3}\, \textit {\_a}^{2}+\sqrt {27 \textit {\_a}^{4}+32}\right )^{\frac {1}{3}} \textit {\_a}^{2}-8 \,18^{\frac {1}{3}}}d \textit {\_a} +c_{1} \end {align*}

Which simpliﬁes to \begin {align*} \frac {3 \ln \left (x \right )}{4} = \int _{}^{\frac {y}{x^{\frac {3}{4}}}}\frac {9 \textit {\_a} \left (3 \sqrt {3}\, \textit {\_a}^{2}+\sqrt {27 \textit {\_a}^{4}+32}\right )^{\frac {1}{3}} 3^{\frac {1}{6}}}{2 \,36^{\frac {1}{3}} {\left (\left (3 \sqrt {3}\, \textit {\_a}^{2}+\sqrt {27 \textit {\_a}^{4}+32}\right )^{2}\right )}^{\frac {1}{3}}-9 \,3^{\frac {1}{6}} \left (3 \sqrt {3}\, \textit {\_a}^{2}+\sqrt {27 \textit {\_a}^{4}+32}\right )^{\frac {1}{3}} \textit {\_a}^{2}-8 \,18^{\frac {1}{3}}}d \textit {\_a} +c_{1} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} \frac {3 \ln \left (x \right )}{4} &= \int _{}^{\frac {y}{x^{\frac {3}{4}}}}\frac {9 \textit {\_a} \left (3 \sqrt {3}\, \textit {\_a}^{2}+\sqrt {27 \textit {\_a}^{4}+32}\right )^{\frac {1}{3}} 3^{\frac {1}{6}}}{2 \,36^{\frac {1}{3}} {\left (\left (3 \sqrt {3}\, \textit {\_a}^{2}+\sqrt {27 \textit {\_a}^{4}+32}\right )^{2}\right )}^{\frac {1}{3}}-9 \,3^{\frac {1}{6}} \left (3 \sqrt {3}\, \textit {\_a}^{2}+\sqrt {27 \textit {\_a}^{4}+32}\right )^{\frac {1}{3}} \textit {\_a}^{2}-8 \,18^{\frac {1}{3}}}d \textit {\_a} +c_{1} \\ \end{align*}

Veriﬁcation of solutions

\[ \frac {3 \ln \left (x \right )}{4} = \int _{}^{\frac {y}{x^{\frac {3}{4}}}}\frac {9 \textit {\_a} \left (3 \sqrt {3}\, \textit {\_a}^{2}+\sqrt {27 \textit {\_a}^{4}+32}\right )^{\frac {1}{3}} 3^{\frac {1}{6}}}{2 \,36^{\frac {1}{3}} {\left (\left (3 \sqrt {3}\, \textit {\_a}^{2}+\sqrt {27 \textit {\_a}^{4}+32}\right )^{2}\right )}^{\frac {1}{3}}-9 \,3^{\frac {1}{6}} \left (3 \sqrt {3}\, \textit {\_a}^{2}+\sqrt {27 \textit {\_a}^{4}+32}\right )^{\frac {1}{3}} \textit {\_a}^{2}-8 \,18^{\frac {1}{3}}}d \textit {\_a} +c_{1} \] Veriﬁed OK.

Writing the ode as \begin {align*} y^{\prime }&=\frac {\left (i \sqrt {3}\, x 12^{\frac {1}{3}}-12^{\frac {1}{3}} x -{\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {2}{3}}\right ) 12^{\frac {1}{3}}}{3 y {\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {1}{3}} \left (1+i \sqrt {3}\right )}\\ y^{\prime }&= \omega \left ( x,y\right ) \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 \sqrt {3}\, x 12^{\frac {1}{3}}-12^{\frac {1}{3}} x -{\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {2}{3}}\right ) 12^{\frac {1}{3}} \left (b_{3}-a_{2}\right )}{3 y {\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {1}{3}} \left (1+i \sqrt {3}\right )}-\frac {\left (i \sqrt {3}\, x 12^{\frac {1}{3}}-12^{\frac {1}{3}} x -{\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {2}{3}}\right )^{2} 12^{\frac {2}{3}} a_{3}}{9 y^{2} {\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {2}{3}} \left (1+i \sqrt {3}\right )^{2}}-\left (\frac {\left (i \sqrt {3}\, 12^{\frac {1}{3}}-12^{\frac {1}{3}}-\frac {32 \sqrt {3}\, x^{2}}{{\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {1}{3}} \sqrt {27 y^{4}+32 x^{3}}}\right ) 12^{\frac {1}{3}}}{3 y {\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {1}{3}} \left (1+i \sqrt {3}\right )}-\frac {16 \left (i \sqrt {3}\, x 12^{\frac {1}{3}}-12^{\frac {1}{3}} x -{\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {2}{3}}\right ) 12^{\frac {1}{3}} \sqrt {3}\, x^{2}}{3 y {\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {4}{3}} \left (1+i \sqrt {3}\right ) \sqrt {27 y^{4}+32 x^{3}}}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )-\left (-\frac {2 \sqrt {3}\, \left (\frac {54 y^{3}}{\sqrt {27 y^{4}+32 x^{3}}}+6 \sqrt {3}\, y \right ) 12^{\frac {1}{3}}}{9 {\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {2}{3}} y \left (1+i \sqrt {3}\right )}-\frac {\left (i \sqrt {3}\, x 12^{\frac {1}{3}}-12^{\frac {1}{3}} x -{\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {2}{3}}\right ) 12^{\frac {1}{3}}}{3 y^{2} {\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {1}{3}} \left (1+i \sqrt {3}\right )}-\frac {\left (i \sqrt {3}\, x 12^{\frac {1}{3}}-12^{\frac {1}{3}} x -{\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {2}{3}}\right ) 12^{\frac {1}{3}} \sqrt {3}\, \left (\frac {54 y^{3}}{\sqrt {27 y^{4}+32 x^{3}}}+6 \sqrt {3}\, y \right )}{9 y {\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {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. \[ \left \{x, y, {\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {1}{3}}, {\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {2}{3}}, \sqrt {27 y^{4}+32 x^{3}}\right \} \] The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them \[ \left \{x = v_{1}, y = v_{2}, {\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {1}{3}} = v_{3}, {\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {2}{3}} = v_{4}, \sqrt {27 y^{4}+32 x^{3}} = v_{5}\right \} \] The above PDE (6E) now becomes \begin{equation} \tag{7E} \text {Expression too large to display} \end{equation} Collecting the above on the terms \(v_i\) introduced, and these are \[ \{v_{1}, v_{2}, v_{3}, v_{4}, v_{5}\} \] Equation (7E) now becomes \begin{equation} \tag{8E} -96 \,12^{\frac {2}{3}} v_{1}^{3} v_{5} a_{3}-384 \,12^{\frac {2}{3}} \sqrt {3}\, v_{1}^{5} b_{2}+324 \,12^{\frac {2}{3}} \sqrt {3}\, v_{2}^{5} a_{1}-384 \,12^{\frac {2}{3}} \sqrt {3}\, v_{1}^{4} b_{1}-54 \,12^{\frac {2}{3}} v_{5} v_{2}^{4} a_{3}+108 \,12^{\frac {2}{3}} v_{5} v_{2}^{3} a_{1}-162 \,12^{\frac {2}{3}} \sqrt {3}\, v_{2}^{6} a_{3}+\left (648 \,12^{\frac {2}{3}} \sqrt {3}\, a_{2}-864 \,12^{\frac {2}{3}} \sqrt {3}\, b_{3}\right ) v_{1} v_{2}^{5}+\left (1728 i b_{2}-576 \sqrt {3}\, b_{2}\right ) v_{1}^{3} v_{2}^{2} v_{3}-180 \,12^{\frac {2}{3}} v_{5} v_{1}^{2} v_{2}^{2} b_{2}-540 \,12^{\frac {2}{3}} \sqrt {3}\, v_{1}^{2} v_{2}^{4} b_{2}-180 \,12^{\frac {2}{3}} v_{5} v_{1} v_{2}^{2} b_{1}-384 \,12^{\frac {2}{3}} \sqrt {3}\, v_{1}^{3} v_{2}^{2} a_{3}-540 \,12^{\frac {2}{3}} \sqrt {3}\, v_{1} v_{2}^{4} b_{1}+192 \,12^{\frac {2}{3}} \sqrt {3}\, v_{1}^{3} v_{2} a_{1}+\left (2304 i a_{3}-768 \sqrt {3}\, a_{3}\right ) v_{1}^{4} v_{3}+\left (-288 i 12^{\frac {1}{3}} b_{2}-96 \,12^{\frac {1}{3}} \sqrt {3}\, b_{2}\right ) v_{1}^{4} v_{4}+\left (-288 i 12^{\frac {1}{3}} b_{1}-96 \,12^{\frac {1}{3}} \sqrt {3}\, b_{1}\right ) v_{1}^{3} v_{4}+\left (1458 i b_{2}-486 \sqrt {3}\, b_{2}\right ) v_{2}^{6} v_{3}+\left (243 i 12^{\frac {1}{3}} a_{2}-324 i 12^{\frac {1}{3}} b_{3}+81 \,12^{\frac {1}{3}} \sqrt {3}\, a_{2}-108 \,12^{\frac {1}{3}} \sqrt {3}\, b_{3}\right ) v_{2}^{5} v_{4}+\left (-81 i 12^{\frac {1}{3}} b_{1}-27 \,12^{\frac {1}{3}} \sqrt {3}\, b_{1}\right ) v_{2}^{4} v_{4}+\left (576 \,12^{\frac {2}{3}} \sqrt {3}\, a_{2}-768 \,12^{\frac {2}{3}} \sqrt {3}\, b_{3}\right ) v_{1}^{4} v_{2}+\left (216 i \sqrt {3}\, a_{3}-216 a_{3}\right ) v_{1} v_{2}^{2} v_{3} v_{5}+\left (-9 i 12^{\frac {1}{3}} \sqrt {3}\, b_{2}-9 \,12^{\frac {1}{3}} b_{2}\right ) v_{1} v_{2}^{2} v_{4} v_{5}+\left (432 i 12^{\frac {1}{3}} a_{2}-576 i 12^{\frac {1}{3}} b_{3}+144 \,12^{\frac {1}{3}} \sqrt {3}\, a_{2}-192 \,12^{\frac {1}{3}} \sqrt {3}\, b_{3}\right ) v_{1}^{3} v_{2} v_{4}+\left (144 i 12^{\frac {1}{3}} a_{3}+48 \,12^{\frac {1}{3}} \sqrt {3}\, a_{3}\right ) v_{1}^{2} v_{2}^{2} v_{4}+\left (144 i 12^{\frac {1}{3}} a_{1}+48 \,12^{\frac {1}{3}} \sqrt {3}\, a_{1}\right ) v_{1}^{2} v_{2} v_{4}+\left (24 i 12^{\frac {1}{3}} \sqrt {3}\, a_{3}+24 \,12^{\frac {1}{3}} a_{3}\right ) v_{1}^{2} v_{4} v_{5}+\left (1944 i a_{3}-648 \sqrt {3}\, a_{3}\right ) v_{1} v_{2}^{4} v_{3}+\left (-81 i 12^{\frac {1}{3}} b_{2}-27 \,12^{\frac {1}{3}} \sqrt {3}\, b_{2}\right ) v_{1} v_{2}^{4} v_{4}+\left (162 i \sqrt {3}\, b_{2}-162 b_{2}\right ) v_{2}^{4} v_{3} v_{5}+\left (27 i 12^{\frac {1}{3}} \sqrt {3}\, a_{2}-36 i 12^{\frac {1}{3}} \sqrt {3}\, b_{3}+27 \,12^{\frac {1}{3}} a_{2}-36 \,12^{\frac {1}{3}} b_{3}\right ) v_{2}^{3} v_{4} v_{5}+\left (-9 i 12^{\frac {1}{3}} \sqrt {3}\, b_{1}-9 \,12^{\frac {1}{3}} b_{1}\right ) v_{2}^{2} v_{4} v_{5}+\left (216 \,12^{\frac {2}{3}} a_{2}-288 \,12^{\frac {2}{3}} b_{3}\right ) v_{1} v_{2}^{3} v_{5} = 0 \end{equation} Setting each coeﬃcients in (8E) to zero gives the following equations to solve \begin {align*} 108 \,12^{\frac {2}{3}} a_{1}&=0\\ -96 \,12^{\frac {2}{3}} a_{3}&=0\\ -54 \,12^{\frac {2}{3}} a_{3}&=0\\ -180 \,12^{\frac {2}{3}} b_{1}&=0\\ -180 \,12^{\frac {2}{3}} b_{2}&=0\\ 192 \,12^{\frac {2}{3}} \sqrt {3}\, a_{1}&=0\\ 324 \,12^{\frac {2}{3}} \sqrt {3}\, a_{1}&=0\\ -384 \,12^{\frac {2}{3}} \sqrt {3}\, a_{3}&=0\\ -162 \,12^{\frac {2}{3}} \sqrt {3}\, a_{3}&=0\\ -540 \,12^{\frac {2}{3}} \sqrt {3}\, b_{1}&=0\\ -384 \,12^{\frac {2}{3}} \sqrt {3}\, b_{1}&=0\\ -540 \,12^{\frac {2}{3}} \sqrt {3}\, b_{2}&=0\\ -384 \,12^{\frac {2}{3}} \sqrt {3}\, b_{2}&=0\\ 216 \,12^{\frac {2}{3}} a_{2}-288 \,12^{\frac {2}{3}} b_{3}&=0\\ 1458 i b_{2}-486 \sqrt {3}\, b_{2}&=0\\ 1728 i b_{2}-576 \sqrt {3}\, b_{2}&=0\\ 1944 i a_{3}-648 \sqrt {3}\, a_{3}&=0\\ 2304 i a_{3}-768 \sqrt {3}\, a_{3}&=0\\ 576 \,12^{\frac {2}{3}} \sqrt {3}\, a_{2}-768 \,12^{\frac {2}{3}} \sqrt {3}\, b_{3}&=0\\ 648 \,12^{\frac {2}{3}} \sqrt {3}\, a_{2}-864 \,12^{\frac {2}{3}} \sqrt {3}\, b_{3}&=0\\ -288 i 12^{\frac {1}{3}} b_{1}-96 \,12^{\frac {1}{3}} \sqrt {3}\, b_{1}&=0\\ -288 i 12^{\frac {1}{3}} b_{2}-96 \,12^{\frac {1}{3}} \sqrt {3}\, b_{2}&=0\\ -81 i 12^{\frac {1}{3}} b_{1}-27 \,12^{\frac {1}{3}} \sqrt {3}\, b_{1}&=0\\ -81 i 12^{\frac {1}{3}} b_{2}-27 \,12^{\frac {1}{3}} \sqrt {3}\, b_{2}&=0\\ 144 i 12^{\frac {1}{3}} a_{1}+48 \,12^{\frac {1}{3}} \sqrt {3}\, a_{1}&=0\\ 144 i 12^{\frac {1}{3}} a_{3}+48 \,12^{\frac {1}{3}} \sqrt {3}\, a_{3}&=0\\ 162 i \sqrt {3}\, b_{2}-162 b_{2}&=0\\ 216 i \sqrt {3}\, a_{3}-216 a_{3}&=0\\ -9 i 12^{\frac {1}{3}} \sqrt {3}\, b_{1}-9 \,12^{\frac {1}{3}} b_{1}&=0\\ -9 i 12^{\frac {1}{3}} \sqrt {3}\, b_{2}-9 \,12^{\frac {1}{3}} b_{2}&=0\\ 24 i 12^{\frac {1}{3}} \sqrt {3}\, a_{3}+24 \,12^{\frac {1}{3}} a_{3}&=0\\ 243 i 12^{\frac {1}{3}} a_{2}-324 i 12^{\frac {1}{3}} b_{3}+81 \,12^{\frac {1}{3}} \sqrt {3}\, a_{2}-108 \,12^{\frac {1}{3}} \sqrt {3}\, b_{3}&=0\\ 432 i 12^{\frac {1}{3}} a_{2}-576 i 12^{\frac {1}{3}} b_{3}+144 \,12^{\frac {1}{3}} \sqrt {3}\, a_{2}-192 \,12^{\frac {1}{3}} \sqrt {3}\, b_{3}&=0\\ 27 i 12^{\frac {1}{3}} \sqrt {3}\, a_{2}-36 i 12^{\frac {1}{3}} \sqrt {3}\, b_{3}+27 \,12^{\frac {1}{3}} a_{2}-36 \,12^{\frac {1}{3}} b_{3}&=0 \end {align*}

Solving the above equations for the unknowns gives \begin {align*} a_{1}&=0\\ a_{2}&=\frac {4 b_{3}}{3}\\ a_{3}&=0\\ b_{1}&=0\\ b_{2}&=0\\ b_{3}&=b_{3} \end {align*}

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*}

Writing the ode as \begin {align*} y^{\prime }&=\frac {\left (i \sqrt {3}\, x 12^{\frac {1}{3}}+12^{\frac {1}{3}} x +{\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {2}{3}}\right ) 12^{\frac {1}{3}}}{3 y {\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {1}{3}} \left (i \sqrt {3}-1\right )}\\ y^{\prime }&= \omega \left ( x,y\right ) \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 \sqrt {3}\, x 12^{\frac {1}{3}}+12^{\frac {1}{3}} x +{\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {2}{3}}\right ) 12^{\frac {1}{3}} \left (b_{3}-a_{2}\right )}{3 y {\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {1}{3}} \left (i \sqrt {3}-1\right )}-\frac {\left (i \sqrt {3}\, x 12^{\frac {1}{3}}+12^{\frac {1}{3}} x +{\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {2}{3}}\right )^{2} 12^{\frac {2}{3}} a_{3}}{9 y^{2} {\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {2}{3}} \left (i \sqrt {3}-1\right )^{2}}-\left (\frac {\left (i \sqrt {3}\, 12^{\frac {1}{3}}+12^{\frac {1}{3}}+\frac {32 \sqrt {3}\, x^{2}}{{\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {1}{3}} \sqrt {27 y^{4}+32 x^{3}}}\right ) 12^{\frac {1}{3}}}{3 y {\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {1}{3}} \left (i \sqrt {3}-1\right )}-\frac {16 \left (i \sqrt {3}\, x 12^{\frac {1}{3}}+12^{\frac {1}{3}} x +{\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {2}{3}}\right ) 12^{\frac {1}{3}} \sqrt {3}\, x^{2}}{3 y {\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {4}{3}} \left (i \sqrt {3}-1\right ) \sqrt {27 y^{4}+32 x^{3}}}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )-\left (\frac {2 \sqrt {3}\, \left (\frac {54 y^{3}}{\sqrt {27 y^{4}+32 x^{3}}}+6 \sqrt {3}\, y \right ) 12^{\frac {1}{3}}}{9 {\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {2}{3}} y \left (i \sqrt {3}-1\right )}-\frac {\left (i \sqrt {3}\, x 12^{\frac {1}{3}}+12^{\frac {1}{3}} x +{\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {2}{3}}\right ) 12^{\frac {1}{3}}}{3 y^{2} {\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {1}{3}} \left (i \sqrt {3}-1\right )}-\frac {\left (i \sqrt {3}\, x 12^{\frac {1}{3}}+12^{\frac {1}{3}} x +{\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {2}{3}}\right ) 12^{\frac {1}{3}} \sqrt {3}\, \left (\frac {54 y^{3}}{\sqrt {27 y^{4}+32 x^{3}}}+6 \sqrt {3}\, y \right )}{9 y {\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {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. \[ \left \{x, y, {\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {1}{3}}, {\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {2}{3}}, \sqrt {27 y^{4}+32 x^{3}}\right \} \] The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them \[ \left \{x = v_{1}, y = v_{2}, {\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {1}{3}} = v_{3}, {\left (\sqrt {3}\, \left (\sqrt {27 y^{4}+32 x^{3}}+3 \sqrt {3}\, y^{2}\right )\right )}^{\frac {2}{3}} = v_{4}, \sqrt {27 y^{4}+32 x^{3}} = v_{5}\right \} \] The above PDE (6E) now becomes \begin{equation} \tag{7E} \text {Expression too large to display} \end{equation} Collecting the above on the terms \(v_i\) introduced, and these are \[ \{v_{1}, v_{2}, v_{3}, v_{4}, v_{5}\} \] Equation (7E) now becomes \begin{equation} \tag{8E} \left (-216 i \sqrt {3}\, a_{3}-216 a_{3}\right ) v_{1} v_{2}^{2} v_{3} v_{5}+\left (9 i \sqrt {3}\, 12^{\frac {1}{3}} b_{2}-9 \,12^{\frac {1}{3}} b_{2}\right ) v_{1} v_{2}^{2} v_{4} v_{5}+\left (216 \,12^{\frac {2}{3}} a_{2}-288 \,12^{\frac {2}{3}} b_{3}\right ) v_{1} v_{2}^{3} v_{5}-540 \sqrt {3}\, 12^{\frac {2}{3}} v_{1}^{2} v_{2}^{4} b_{2}-180 \,12^{\frac {2}{3}} v_{5} v_{1} v_{2}^{2} b_{1}-384 \sqrt {3}\, 12^{\frac {2}{3}} v_{1}^{3} v_{2}^{2} a_{3}-540 \sqrt {3}\, 12^{\frac {2}{3}} v_{1} v_{2}^{4} b_{1}+192 \sqrt {3}\, 12^{\frac {2}{3}} v_{1}^{3} v_{2} a_{1}-180 \,12^{\frac {2}{3}} v_{5} v_{1}^{2} v_{2}^{2} b_{2}+\left (81 i 12^{\frac {1}{3}} b_{1}-27 \sqrt {3}\, 12^{\frac {1}{3}} b_{1}\right ) v_{2}^{4} v_{4}+\left (-2304 i a_{3}-768 \sqrt {3}\, a_{3}\right ) v_{1}^{4} v_{3}+\left (288 i 12^{\frac {1}{3}} b_{2}-96 \sqrt {3}\, 12^{\frac {1}{3}} b_{2}\right ) v_{1}^{4} v_{4}+\left (288 i 12^{\frac {1}{3}} b_{1}-96 \sqrt {3}\, 12^{\frac {1}{3}} b_{1}\right ) v_{1}^{3} v_{4}+\left (-1458 i b_{2}-486 \sqrt {3}\, b_{2}\right ) v_{2}^{6} v_{3}+\left (-243 i 12^{\frac {1}{3}} a_{2}+324 i 12^{\frac {1}{3}} b_{3}+81 \sqrt {3}\, 12^{\frac {1}{3}} a_{2}-108 \sqrt {3}\, 12^{\frac {1}{3}} b_{3}\right ) v_{2}^{5} v_{4}+\left (576 \sqrt {3}\, 12^{\frac {2}{3}} a_{2}-768 \sqrt {3}\, 12^{\frac {2}{3}} b_{3}\right ) v_{1}^{4} v_{2}+\left (-162 i \sqrt {3}\, b_{2}-162 b_{2}\right ) v_{2}^{4} v_{3} v_{5}+\left (-27 i \sqrt {3}\, 12^{\frac {1}{3}} a_{2}+36 i \sqrt {3}\, 12^{\frac {1}{3}} b_{3}+27 \,12^{\frac {1}{3}} a_{2}-36 \,12^{\frac {1}{3}} b_{3}\right ) v_{2}^{3} v_{4} v_{5}+\left (9 i \sqrt {3}\, 12^{\frac {1}{3}} b_{1}-9 \,12^{\frac {1}{3}} b_{1}\right ) v_{2}^{2} v_{4} v_{5}+\left (648 \sqrt {3}\, 12^{\frac {2}{3}} a_{2}-864 \sqrt {3}\, 12^{\frac {2}{3}} b_{3}\right ) v_{1} v_{2}^{5}+\left (-1728 i b_{2}-576 \sqrt {3}\, b_{2}\right ) v_{1}^{3} v_{2}^{2} v_{3}+\left (-432 i 12^{\frac {1}{3}} a_{2}+576 i 12^{\frac {1}{3}} b_{3}+144 \sqrt {3}\, 12^{\frac {1}{3}} a_{2}-192 \sqrt {3}\, 12^{\frac {1}{3}} b_{3}\right ) v_{1}^{3} v_{2} v_{4}+\left (-144 i 12^{\frac {1}{3}} a_{3}+48 \sqrt {3}\, 12^{\frac {1}{3}} a_{3}\right ) v_{1}^{2} v_{2}^{2} v_{4}+\left (-144 i 12^{\frac {1}{3}} a_{1}+48 \sqrt {3}\, 12^{\frac {1}{3}} a_{1}\right ) v_{1}^{2} v_{2} v_{4}+\left (-24 i \sqrt {3}\, 12^{\frac {1}{3}} a_{3}+24 \,12^{\frac {1}{3}} a_{3}\right ) v_{1}^{2} v_{4} v_{5}+\left (-1944 i a_{3}-648 \sqrt {3}\, a_{3}\right ) v_{1} v_{2}^{4} v_{3}+\left (81 i 12^{\frac {1}{3}} b_{2}-27 \sqrt {3}\, 12^{\frac {1}{3}} b_{2}\right ) v_{1} v_{2}^{4} v_{4}-96 \,12^{\frac {2}{3}} v_{1}^{3} v_{5} a_{3}-54 \,12^{\frac {2}{3}} v_{5} v_{2}^{4} a_{3}+108 \,12^{\frac {2}{3}} v_{5} v_{2}^{3} a_{1}-162 \sqrt {3}\, 12^{\frac {2}{3}} v_{2}^{6} a_{3}-384 \sqrt {3}\, 12^{\frac {2}{3}} v_{1}^{5} b_{2}+324 \sqrt {3}\, 12^{\frac {2}{3}} v_{2}^{5} a_{1}-384 \sqrt {3}\, 12^{\frac {2}{3}} v_{1}^{4} b_{1} = 0 \end{equation} Setting each coeﬃcients in (8E) to zero gives the following equations to solve \begin {align*} 108 \,12^{\frac {2}{3}} a_{1}&=0\\ -96 \,12^{\frac {2}{3}} a_{3}&=0\\ -54 \,12^{\frac {2}{3}} a_{3}&=0\\ -180 \,12^{\frac {2}{3}} b_{1}&=0\\ -180 \,12^{\frac {2}{3}} b_{2}&=0\\ 192 \sqrt {3}\, 12^{\frac {2}{3}} a_{1}&=0\\ 324 \sqrt {3}\, 12^{\frac {2}{3}} a_{1}&=0\\ -384 \sqrt {3}\, 12^{\frac {2}{3}} a_{3}&=0\\ -162 \sqrt {3}\, 12^{\frac {2}{3}} a_{3}&=0\\ -540 \sqrt {3}\, 12^{\frac {2}{3}} b_{1}&=0\\ -384 \sqrt {3}\, 12^{\frac {2}{3}} b_{1}&=0\\ -540 \sqrt {3}\, 12^{\frac {2}{3}} b_{2}&=0\\ -384 \sqrt {3}\, 12^{\frac {2}{3}} b_{2}&=0\\ 216 \,12^{\frac {2}{3}} a_{2}-288 \,12^{\frac {2}{3}} b_{3}&=0\\ -2304 i a_{3}-768 \sqrt {3}\, a_{3}&=0\\ -1944 i a_{3}-648 \sqrt {3}\, a_{3}&=0\\ -1728 i b_{2}-576 \sqrt {3}\, b_{2}&=0\\ -1458 i b_{2}-486 \sqrt {3}\, b_{2}&=0\\ 576 \sqrt {3}\, 12^{\frac {2}{3}} a_{2}-768 \sqrt {3}\, 12^{\frac {2}{3}} b_{3}&=0\\ 648 \sqrt {3}\, 12^{\frac {2}{3}} a_{2}-864 \sqrt {3}\, 12^{\frac {2}{3}} b_{3}&=0\\ -216 i \sqrt {3}\, a_{3}-216 a_{3}&=0\\ -162 i \sqrt {3}\, b_{2}-162 b_{2}&=0\\ -144 i 12^{\frac {1}{3}} a_{1}+48 \sqrt {3}\, 12^{\frac {1}{3}} a_{1}&=0\\ -144 i 12^{\frac {1}{3}} a_{3}+48 \sqrt {3}\, 12^{\frac {1}{3}} a_{3}&=0\\ 81 i 12^{\frac {1}{3}} b_{1}-27 \sqrt {3}\, 12^{\frac {1}{3}} b_{1}&=0\\ 81 i 12^{\frac {1}{3}} b_{2}-27 \sqrt {3}\, 12^{\frac {1}{3}} b_{2}&=0\\ 288 i 12^{\frac {1}{3}} b_{1}-96 \sqrt {3}\, 12^{\frac {1}{3}} b_{1}&=0\\ 288 i 12^{\frac {1}{3}} b_{2}-96 \sqrt {3}\, 12^{\frac {1}{3}} b_{2}&=0\\ -24 i \sqrt {3}\, 12^{\frac {1}{3}} a_{3}+24 \,12^{\frac {1}{3}} a_{3}&=0\\ 9 i \sqrt {3}\, 12^{\frac {1}{3}} b_{1}-9 \,12^{\frac {1}{3}} b_{1}&=0\\ 9 i \sqrt {3}\, 12^{\frac {1}{3}} b_{2}-9 \,12^{\frac {1}{3}} b_{2}&=0\\ -432 i 12^{\frac {1}{3}} a_{2}+576 i 12^{\frac {1}{3}} b_{3}+144 \sqrt {3}\, 12^{\frac {1}{3}} a_{2}-192 \sqrt {3}\, 12^{\frac {1}{3}} b_{3}&=0\\ -243 i 12^{\frac {1}{3}} a_{2}+324 i 12^{\frac {1}{3}} b_{3}+81 \sqrt {3}\, 12^{\frac {1}{3}} a_{2}-108 \sqrt {3}\, 12^{\frac {1}{3}} b_{3}&=0\\ -27 i \sqrt {3}\, 12^{\frac {1}{3}} a_{2}+36 i \sqrt {3}\, 12^{\frac {1}{3}} b_{3}+27 \,12^{\frac {1}{3}} a_{2}-36 \,12^{\frac {1}{3}} b_{3}&=0 \end {align*}

Solving the above equations for the unknowns gives \begin {align*} a_{1}&=0\\ a_{2}&=\frac {4 b_{3}}{3}\\ a_{3}&=0\\ b_{1}&=0\\ b_{2}&=0\\ b_{3}&=b_{3} \end {align*}

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*}

1.83.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y-2 x y^{\prime }-y^{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 (108 y^{2}+12 \sqrt {3}\, \sqrt {27 y^{4}+32 x^{3}}\right )^{\frac {1}{3}}}{6 y}-\frac {4 x}{y \left (108 y^{2}+12 \sqrt {3}\, \sqrt {27 y^{4}+32 x^{3}}\right )^{\frac {1}{3}}}, y^{\prime }=-\frac {\left (108 y^{2}+12 \sqrt {3}\, \sqrt {27 y^{4}+32 x^{3}}\right )^{\frac {1}{3}}}{12 y}+\frac {2 x}{y \left (108 y^{2}+12 \sqrt {3}\, \sqrt {27 y^{4}+32 x^{3}}\right )^{\frac {1}{3}}}-\frac {\mathrm {I} \sqrt {3}\, \left (\frac {\left (108 y^{2}+12 \sqrt {3}\, \sqrt {27 y^{4}+32 x^{3}}\right )^{\frac {1}{3}}}{6 y}+\frac {4 x}{y \left (108 y^{2}+12 \sqrt {3}\, \sqrt {27 y^{4}+32 x^{3}}\right )^{\frac {1}{3}}}\right )}{2}, y^{\prime }=-\frac {\left (108 y^{2}+12 \sqrt {3}\, \sqrt {27 y^{4}+32 x^{3}}\right )^{\frac {1}{3}}}{12 y}+\frac {2 x}{y \left (108 y^{2}+12 \sqrt {3}\, \sqrt {27 y^{4}+32 x^{3}}\right )^{\frac {1}{3}}}+\frac {\mathrm {I} \sqrt {3}\, \left (\frac {\left (108 y^{2}+12 \sqrt {3}\, \sqrt {27 y^{4}+32 x^{3}}\right )^{\frac {1}{3}}}{6 y}+\frac {4 x}{y \left (108 y^{2}+12 \sqrt {3}\, \sqrt {27 y^{4}+32 x^{3}}\right )^{\frac {1}{3}}}\right )}{2}\right ] \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {\left (108 y^{2}+12 \sqrt {3}\, \sqrt {27 y^{4}+32 x^{3}}\right )^{\frac {1}{3}}}{6 y}-\frac {4 x}{y \left (108 y^{2}+12 \sqrt {3}\, \sqrt {27 y^{4}+32 x^{3}}\right )^{\frac {1}{3}}} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {\left (108 y^{2}+12 \sqrt {3}\, \sqrt {27 y^{4}+32 x^{3}}\right )^{\frac {1}{3}}}{12 y}+\frac {2 x}{y \left (108 y^{2}+12 \sqrt {3}\, \sqrt {27 y^{4}+32 x^{3}}\right )^{\frac {1}{3}}}-\frac {\mathrm {I} \sqrt {3}\, \left (\frac {\left (108 y^{2}+12 \sqrt {3}\, \sqrt {27 y^{4}+32 x^{3}}\right )^{\frac {1}{3}}}{6 y}+\frac {4 x}{y \left (108 y^{2}+12 \sqrt {3}\, \sqrt {27 y^{4}+32 x^{3}}\right )^{\frac {1}{3}}}\right )}{2} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {\left (108 y^{2}+12 \sqrt {3}\, \sqrt {27 y^{4}+32 x^{3}}\right )^{\frac {1}{3}}}{12 y}+\frac {2 x}{y \left (108 y^{2}+12 \sqrt {3}\, \sqrt {27 y^{4}+32 x^{3}}\right )^{\frac {1}{3}}}+\frac {\mathrm {I} \sqrt {3}\, \left (\frac {\left (108 y^{2}+12 \sqrt {3}\, \sqrt {27 y^{4}+32 x^{3}}\right )^{\frac {1}{3}}}{6 y}+\frac {4 x}{y \left (108 y^{2}+12 \sqrt {3}\, \sqrt {27 y^{4}+32 x^{3}}\right )^{\frac {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: 
   *** Sublevel 2 *** 
   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 simple symmetries for implicit equations 
   Successful isolation of dy/dx: 3 solutions were found. Trying to solve each resulting ODE. 
      *** Sublevel 3 *** 
      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 
   trying dAlembert 
   -> Calling odsolve with the ODE`, diff(y(x), x) = (-2*y(x)^2*x^3-y(x))/(2*x^4*y(x)+x), y(x)`      *** Sublevel 3 *** 
      Methods for first order ODEs: 
      --- Trying classification methods --- 
      trying a quadrature 
      trying 1st order linear 
      <- 1st order linear successful 
   <- 1st order, parametric methods successful`

\begin{align*} y \left (x \right ) &= -\frac {2 \left (-x^{3}\right )^{\frac {1}{4}} 6^{\frac {1}{4}}}{3} \\ y \left (x \right ) &= \frac {2 \left (-x^{3}\right )^{\frac {1}{4}} 6^{\frac {1}{4}}}{3} \\ y \left (x \right ) &= -\frac {2 i \left (-x^{3}\right )^{\frac {1}{4}} 6^{\frac {1}{4}}}{3} \\ y \left (x \right ) &= \frac {2 i \left (-x^{3}\right )^{\frac {1}{4}} 6^{\frac {1}{4}}}{3} \\ y \left (x \right ) &= 0 \\ y \left (x \right ) &= \sqrt {c_{1} \left (c_{1}^{2}+2 x \right )} \\ y \left (x \right ) &= -\sqrt {c_{1} \left (c_{1}^{2}+2 x \right )} \\ \end{align*}

\begin{align*} y(x)\to -\sqrt {2 c_1 x+c_1{}^3} \\ y(x)\to \sqrt {2 c_1 x+c_1{}^3} \\ y(x)\to (-1-i) \left (\frac {2}{3}\right )^{3/4} x^{3/4} \\ y(x)\to (1-i) \left (\frac {2}{3}\right )^{3/4} x^{3/4} \\ y(x)\to (-1+i) \left (\frac {2}{3}\right )^{3/4} x^{3/4} \\ y(x)\to (1+i) \left (\frac {2}{3}\right )^{3/4} x^{3/4} \\ \end{align*}

1.83 problem 86

1.83.1 Maple step by step solution