problem 90

Internal problem ID [3232]
Internal ﬁle name [OUTPUT/2724_Sunday_June_05_2022_08_39_25_AM_11199866/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: 90.
ODE order: 1.
ODE degree: 3.

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

\[ \boxed {y \left (y-2 x y^{\prime }\right )^{3}-{y^{\prime }}^{2}=0} \] 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 (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 y^{2} x^{2}-1}\, y^{3} x^{3}+36 y^{2} x^{2}-1\right )^{\frac {1}{3}}}{24 y x^{3}}-\frac {24 y^{2} x^{2}-1}{24 y x^{3} \left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 y^{2} x^{2}-1}\, y^{3} x^{3}+36 y^{2} x^{2}-1\right )^{\frac {1}{3}}}+\frac {12 y^{2} x^{2}-1}{24 y x^{3}} \tag {1} \\ y^{\prime }&=-\frac {\left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 y^{2} x^{2}-1}\, y^{3} x^{3}+36 y^{2} x^{2}-1\right )^{\frac {1}{3}}}{48 y x^{3}}+\frac {24 y^{2} x^{2}-1}{48 y x^{3} \left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 y^{2} x^{2}-1}\, y^{3} x^{3}+36 y^{2} x^{2}-1\right )^{\frac {1}{3}}}+\frac {12 y^{2} x^{2}-1}{24 y x^{3}}+\frac {i \sqrt {3}\, \left (\frac {\left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 y^{2} x^{2}-1}\, y^{3} x^{3}+36 y^{2} x^{2}-1\right )^{\frac {1}{3}}}{24 y x^{3}}+\frac {24 y^{2} x^{2}-1}{24 y x^{3} \left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 y^{2} x^{2}-1}\, y^{3} x^{3}+36 y^{2} x^{2}-1\right )^{\frac {1}{3}}}\right )}{2} \tag {2} \\ y^{\prime }&=-\frac {\left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 y^{2} x^{2}-1}\, y^{3} x^{3}+36 y^{2} x^{2}-1\right )^{\frac {1}{3}}}{48 y x^{3}}+\frac {24 y^{2} x^{2}-1}{48 y x^{3} \left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 y^{2} x^{2}-1}\, y^{3} x^{3}+36 y^{2} x^{2}-1\right )^{\frac {1}{3}}}+\frac {12 y^{2} x^{2}-1}{24 y x^{3}}-\frac {i \sqrt {3}\, \left (\frac {\left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 y^{2} x^{2}-1}\, y^{3} x^{3}+36 y^{2} x^{2}-1\right )^{\frac {1}{3}}}{24 y x^{3}}+\frac {24 y^{2} x^{2}-1}{24 y x^{3} \left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 y^{2} x^{2}-1}\, y^{3} x^{3}+36 y^{2} x^{2}-1\right )^{\frac {1}{3}}}\right )}{2} \tag {3} \end {align*}

Writing the ode as \begin {align*} y^{\prime }&=\frac {12 y^{2} x^{2} \left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 x^{2} y^{2}-1}\, y^{3} x^{3}+36 x^{2} y^{2}-1\right )^{\frac {1}{3}}-24 x^{2} y^{2}+\left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 x^{2} y^{2}-1}\, y^{3} x^{3}+36 x^{2} y^{2}-1\right )^{\frac {2}{3}}-\left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 x^{2} y^{2}-1}\, y^{3} x^{3}+36 x^{2} y^{2}-1\right )^{\frac {1}{3}}+1}{24 y \,x^{3} \left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 x^{2} y^{2}-1}\, y^{3} x^{3}+36 x^{2} y^{2}-1\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} \text {Expression too large to display} \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. \[ \left \{x, y, \left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 x^{2} y^{2}-1}\, y^{3} x^{3}+36 x^{2} y^{2}-1\right )^{\frac {1}{3}}, \left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 x^{2} y^{2}-1}\, y^{3} x^{3}+36 x^{2} y^{2}-1\right )^{\frac {2}{3}}, \sqrt {27 x^{2} y^{2}-1}\right \} \] The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them \[ \left \{x = v_{1}, y = v_{2}, \left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 x^{2} y^{2}-1}\, y^{3} x^{3}+36 x^{2} y^{2}-1\right )^{\frac {1}{3}} = v_{3}, \left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 x^{2} y^{2}-1}\, y^{3} x^{3}+36 x^{2} y^{2}-1\right )^{\frac {2}{3}} = v_{4}, \sqrt {27 x^{2} y^{2}-1} = v_{5}\right \} \] 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} \text {Expression too large to display} \end{equation} Setting each coeﬃcients in (8E) to zero gives the following equations to solve \begin {align*} -62208 a_{1}&=0\\ -57024 a_{1}&=0\\ -8640 a_{1}&=0\\ -2880 a_{1}&=0\\ -72 a_{1}&=0\\ 72 a_{1}&=0\\ 2016 a_{1}&=0\\ 3744 a_{1}&=0\\ 25920 a_{1}&=0\\ 290304 a_{1}&=0\\ -31104 a_{3}&=0\\ -3456 a_{3}&=0\\ -1584 a_{3}&=0\\ -108 a_{3}&=0\\ -3 a_{3}&=0\\ 3 a_{3}&=0\\ 72 a_{3}&=0\\ 144 a_{3}&=0\\ 504 a_{3}&=0\\ 5184 a_{3}&=0\\ 72576 a_{3}&=0\\ -5184 b_{1}&=0\\ -576 b_{1}&=0\\ -24 b_{1}&=0\\ 24 b_{1}&=0\\ 288 b_{1}&=0\\ 864 b_{1}&=0\\ 1728 b_{1}&=0\\ 41472 b_{1}&=0\\ 62208 b_{1}&=0\\ -62208 b_{2}&=0\\ -5184 b_{2}&=0\\ -1152 b_{2}&=0\\ -24 b_{2}&=0\\ 24 b_{2}&=0\\ 288 b_{2}&=0\\ 864 b_{2}&=0\\ 1728 b_{2}&=0\\ 15552 b_{2}&=0\\ 41472 b_{2}&=0\\ -870912 \sqrt {3}\, a_{1}&=0\\ -53568 \sqrt {3}\, a_{1}&=0\\ -2304 \sqrt {3}\, a_{1}&=0\\ -1152 \sqrt {3}\, a_{1}&=0\\ 1728 \sqrt {3}\, a_{1}&=0\\ 25920 \sqrt {3}\, a_{1}&=0\\ 95040 \sqrt {3}\, a_{1}&=0\\ 186624 \sqrt {3}\, a_{1}&=0\\ -217728 \sqrt {3}\, a_{3}&=0\\ -22464 \sqrt {3}\, a_{3}&=0\\ -3456 \sqrt {3}\, a_{3}&=0\\ -1944 \sqrt {3}\, a_{3}&=0\\ -96 \sqrt {3}\, a_{3}&=0\\ -48 \sqrt {3}\, a_{3}&=0\\ 72 \sqrt {3}\, a_{3}&=0\\ 720 \sqrt {3}\, a_{3}&=0\\ 3744 \sqrt {3}\, a_{3}&=0\\ 10368 \sqrt {3}\, a_{3}&=0\\ 93312 \sqrt {3}\, a_{3}&=0\\ -186624 \sqrt {3}\, b_{1}&=0\\ -124416 \sqrt {3}\, b_{1}&=0\\ -8640 \sqrt {3}\, b_{1}&=0\\ -5184 \sqrt {3}\, b_{1}&=0\\ -1152 \sqrt {3}\, b_{1}&=0\\ 576 \sqrt {3}\, b_{1}&=0\\ 36288 \sqrt {3}\, b_{1}&=0\\ -124416 \sqrt {3}\, b_{2}&=0\\ -22464 \sqrt {3}\, b_{2}&=0\\ -5184 \sqrt {3}\, b_{2}&=0\\ -1152 \sqrt {3}\, b_{2}&=0\\ 576 \sqrt {3}\, b_{2}&=0\\ 36288 \sqrt {3}\, b_{2}&=0\\ 186624 \sqrt {3}\, b_{2}&=0\\ -31104 a_{2}-31104 b_{3}&=0\\ -3456 a_{2}-3456 b_{3}&=0\\ -1728 a_{2}-1728 b_{3}&=0\\ -48 a_{2}-48 b_{3}&=0\\ 48 a_{2}+48 b_{3}&=0\\ 1152 a_{2}+1152 b_{3}&=0\\ 2304 a_{2}+2304 b_{3}&=0\\ 10368 a_{2}+10368 b_{3}&=0\\ 165888 a_{2}+165888 b_{3}&=0\\ -497664 \sqrt {3}\, a_{2}-497664 \sqrt {3}\, b_{3}&=0\\ -31104 \sqrt {3}\, a_{2}-31104 \sqrt {3}\, b_{3}&=0\\ -1728 \sqrt {3}\, a_{2}-1728 \sqrt {3}\, b_{3}&=0\\ -576 \sqrt {3}\, a_{2}-576 \sqrt {3}\, b_{3}&=0\\ 1152 \sqrt {3}\, a_{2}+1152 \sqrt {3}\, b_{3}&=0\\ 10368 \sqrt {3}\, a_{2}+10368 \sqrt {3}\, b_{3}&=0\\ 65664 \sqrt {3}\, a_{2}+65664 \sqrt {3}\, b_{3}&=0 \end {align*}

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

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

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

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

Integrating gives \begin {align*} S &= \int { \frac {dx}{T}}\\ &= -\ln \left (x \right ) \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 {12 y^{2} x^{2} \left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 x^{2} y^{2}-1}\, y^{3} x^{3}+36 x^{2} y^{2}-1\right )^{\frac {1}{3}}-24 x^{2} y^{2}+\left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 x^{2} y^{2}-1}\, y^{3} x^{3}+36 x^{2} y^{2}-1\right )^{\frac {2}{3}}-\left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 x^{2} y^{2}-1}\, y^{3} x^{3}+36 x^{2} y^{2}-1\right )^{\frac {1}{3}}+1}{24 y \,x^{3} \left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 x^{2} y^{2}-1}\, y^{3} x^{3}+36 x^{2} y^{2}-1\right )^{\frac {1}{3}}} \end {align*}

Evaluating all the partial derivatives gives \begin {align*} R_{x} &= y\\ R_{y} &= x\\ S_{x} &= -\frac {1}{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 {24 y x \left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 x^{2} y^{2}-1}\, y^{3} x^{3}+36 x^{2} y^{2}-1\right )^{\frac {1}{3}}}{36 y^{2} x^{2} \left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 x^{2} y^{2}-1}\, y^{3} x^{3}+36 x^{2} y^{2}-1\right )^{\frac {1}{3}}-24 x^{2} y^{2}+\left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 x^{2} y^{2}-1}\, y^{3} x^{3}+36 x^{2} y^{2}-1\right )^{\frac {2}{3}}-\left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 x^{2} y^{2}-1}\, y^{3} x^{3}+36 x^{2} y^{2}-1\right )^{\frac {1}{3}}+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} &= -\frac {24 R \left (-216 R^{4}+24 \sqrt {3}\, \sqrt {27 R^{2}-1}\, R^{3}+36 R^{2}-1\right )^{\frac {1}{3}}}{36 R^{2} \left (-216 R^{4}+24 \sqrt {3}\, \sqrt {27 R^{2}-1}\, R^{3}+36 R^{2}-1\right )^{\frac {1}{3}}-24 R^{2}+\left (-216 R^{4}+24 \sqrt {3}\, \sqrt {27 R^{2}-1}\, R^{3}+36 R^{2}-1\right )^{\frac {2}{3}}-\left (-216 R^{4}+24 \sqrt {3}\, \sqrt {27 R^{2}-1}\, R^{3}+36 R^{2}-1\right )^{\frac {1}{3}}+1} \end {align*}

The above is a quadrature ode. This is the whole point of Lie symmetry method. It converts an ode, no matter how complicated it is, to one that can be solved by integration when the ode is in the canonical coordiates \(R,S\). Integrating the above gives \begin {align*} S \left (R \right ) = \int -\frac {24 R \left (-216 R^{4}+24 R^{3} \sqrt {81 R^{2}-3}+36 R^{2}-1\right )^{\frac {1}{3}}}{36 R^{2} \left (-216 R^{4}+24 R^{3} \sqrt {81 R^{2}-3}+36 R^{2}-1\right )^{\frac {1}{3}}-24 R^{2}+\left (-216 R^{4}+24 R^{3} \sqrt {81 R^{2}-3}+36 R^{2}-1\right )^{\frac {2}{3}}-\left (-216 R^{4}+24 R^{3} \sqrt {81 R^{2}-3}+36 R^{2}-1\right )^{\frac {1}{3}}+1}d R +c_{1}\tag {4} \end {align*}

To complete the solution, we just need to transform (4) back to \(x,y\) coordinates. This results in \begin {align*} -\ln \left (x \right ) = \int _{}^{y x}-\frac {24 \textit {\_a} \left (-216 \textit {\_a}^{4}+24 \textit {\_a}^{3} \sqrt {81 \textit {\_a}^{2}-3}+36 \textit {\_a}^{2}-1\right )^{\frac {1}{3}}}{36 \textit {\_a}^{2} \left (-216 \textit {\_a}^{4}+24 \textit {\_a}^{3} \sqrt {81 \textit {\_a}^{2}-3}+36 \textit {\_a}^{2}-1\right )^{\frac {1}{3}}-24 \textit {\_a}^{2}+\left (-216 \textit {\_a}^{4}+24 \textit {\_a}^{3} \sqrt {81 \textit {\_a}^{2}-3}+36 \textit {\_a}^{2}-1\right )^{\frac {2}{3}}-\left (-216 \textit {\_a}^{4}+24 \textit {\_a}^{3} \sqrt {81 \textit {\_a}^{2}-3}+36 \textit {\_a}^{2}-1\right )^{\frac {1}{3}}+1}d \textit {\_a} +c_{1} \end {align*}

Which simpliﬁes to \begin {align*} -\ln \left (x \right ) = \int _{}^{y x}-\frac {24 \textit {\_a} \left (-216 \textit {\_a}^{4}+24 \textit {\_a}^{3} \sqrt {81 \textit {\_a}^{2}-3}+36 \textit {\_a}^{2}-1\right )^{\frac {1}{3}}}{36 \textit {\_a}^{2} \left (-216 \textit {\_a}^{4}+24 \textit {\_a}^{3} \sqrt {81 \textit {\_a}^{2}-3}+36 \textit {\_a}^{2}-1\right )^{\frac {1}{3}}-24 \textit {\_a}^{2}+\left (-216 \textit {\_a}^{4}+24 \textit {\_a}^{3} \sqrt {81 \textit {\_a}^{2}-3}+36 \textit {\_a}^{2}-1\right )^{\frac {2}{3}}-\left (-216 \textit {\_a}^{4}+24 \textit {\_a}^{3} \sqrt {81 \textit {\_a}^{2}-3}+36 \textit {\_a}^{2}-1\right )^{\frac {1}{3}}+1}d \textit {\_a} +c_{1} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} -\ln \left (x \right ) &= \int _{}^{y x}-\frac {24 \textit {\_a} \left (-216 \textit {\_a}^{4}+24 \textit {\_a}^{3} \sqrt {81 \textit {\_a}^{2}-3}+36 \textit {\_a}^{2}-1\right )^{\frac {1}{3}}}{36 \textit {\_a}^{2} \left (-216 \textit {\_a}^{4}+24 \textit {\_a}^{3} \sqrt {81 \textit {\_a}^{2}-3}+36 \textit {\_a}^{2}-1\right )^{\frac {1}{3}}-24 \textit {\_a}^{2}+\left (-216 \textit {\_a}^{4}+24 \textit {\_a}^{3} \sqrt {81 \textit {\_a}^{2}-3}+36 \textit {\_a}^{2}-1\right )^{\frac {2}{3}}-\left (-216 \textit {\_a}^{4}+24 \textit {\_a}^{3} \sqrt {81 \textit {\_a}^{2}-3}+36 \textit {\_a}^{2}-1\right )^{\frac {1}{3}}+1}d \textit {\_a} +c_{1} \\ \end{align*}

Veriﬁcation of solutions

\[ -\ln \left (x \right ) = \int _{}^{y x}-\frac {24 \textit {\_a} \left (-216 \textit {\_a}^{4}+24 \textit {\_a}^{3} \sqrt {81 \textit {\_a}^{2}-3}+36 \textit {\_a}^{2}-1\right )^{\frac {1}{3}}}{36 \textit {\_a}^{2} \left (-216 \textit {\_a}^{4}+24 \textit {\_a}^{3} \sqrt {81 \textit {\_a}^{2}-3}+36 \textit {\_a}^{2}-1\right )^{\frac {1}{3}}-24 \textit {\_a}^{2}+\left (-216 \textit {\_a}^{4}+24 \textit {\_a}^{3} \sqrt {81 \textit {\_a}^{2}-3}+36 \textit {\_a}^{2}-1\right )^{\frac {2}{3}}-\left (-216 \textit {\_a}^{4}+24 \textit {\_a}^{3} \sqrt {81 \textit {\_a}^{2}-3}+36 \textit {\_a}^{2}-1\right )^{\frac {1}{3}}+1}d \textit {\_a} +c_{1} \] Veriﬁed OK.

Writing the ode as \begin {align*} y^{\prime }&=\frac {24 i \sqrt {3}\, y^{2} x^{2}+24 y^{2} x^{2} \left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 x^{2} y^{2}-1}\, y^{3} x^{3}+36 x^{2} y^{2}-1\right )^{\frac {1}{3}}+i \sqrt {3}\, \left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 x^{2} y^{2}-1}\, y^{3} x^{3}+36 x^{2} y^{2}-1\right )^{\frac {2}{3}}+24 x^{2} y^{2}-i \sqrt {3}-\left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 x^{2} y^{2}-1}\, y^{3} x^{3}+36 x^{2} y^{2}-1\right )^{\frac {2}{3}}-2 \left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 x^{2} y^{2}-1}\, y^{3} x^{3}+36 x^{2} y^{2}-1\right )^{\frac {1}{3}}-1}{48 y \,x^{3} \left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 x^{2} y^{2}-1}\, y^{3} x^{3}+36 x^{2} y^{2}-1\right )^{\frac {1}{3}}}\\ 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} \text {Expression too large to display} \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. \[ \left \{x, y, \left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 x^{2} y^{2}-1}\, y^{3} x^{3}+36 x^{2} y^{2}-1\right )^{\frac {1}{3}}, \left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 x^{2} y^{2}-1}\, y^{3} x^{3}+36 x^{2} y^{2}-1\right )^{\frac {2}{3}}, \sqrt {27 x^{2} y^{2}-1}\right \} \] The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them \[ \left \{x = v_{1}, y = v_{2}, \left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 x^{2} y^{2}-1}\, y^{3} x^{3}+36 x^{2} y^{2}-1\right )^{\frac {1}{3}} = v_{3}, \left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 x^{2} y^{2}-1}\, y^{3} x^{3}+36 x^{2} y^{2}-1\right )^{\frac {2}{3}} = v_{4}, \sqrt {27 x^{2} y^{2}-1} = v_{5}\right \} \] 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} \text {Expression too large to display} \end{equation} Setting each coeﬃcients in (8E) to zero gives the following equations to solve \begin {align*} -248832 a_{1}&=0\\ -11520 a_{1}&=0\\ 288 a_{1}&=0\\ 103680 a_{1}&=0\\ -124416 a_{3}&=0\\ -432 a_{3}&=0\\ 12 a_{3}&=0\\ 2016 a_{3}&=0\\ 20736 a_{3}&=0\\ -20736 b_{1}&=0\\ -2304 b_{1}&=0\\ 96 b_{1}&=0\\ 248832 b_{1}&=0\\ -248832 b_{2}&=0\\ -4608 b_{2}&=0\\ 96 b_{2}&=0\\ 62208 b_{2}&=0\\ -214272 \sqrt {3}\, a_{1}&=0\\ 6912 \sqrt {3}\, a_{1}&=0\\ 746496 \sqrt {3}\, a_{1}&=0\\ -13824 \sqrt {3}\, a_{3}&=0\\ -7776 \sqrt {3}\, a_{3}&=0\\ 288 \sqrt {3}\, a_{3}&=0\\ 373248 \sqrt {3}\, a_{3}&=0\\ -746496 \sqrt {3}\, b_{1}&=0\\ -34560 \sqrt {3}\, b_{1}&=0\\ 2304 \sqrt {3}\, b_{1}&=0\\ -89856 \sqrt {3}\, b_{2}&=0\\ 2304 \sqrt {3}\, b_{2}&=0\\ 746496 \sqrt {3}\, b_{2}&=0\\ -6912 a_{2}-6912 b_{3}&=0\\ 192 a_{2}+192 b_{3}&=0\\ 41472 a_{2}+41472 b_{3}&=0\\ -124416 \sqrt {3}\, a_{2}-124416 \sqrt {3}\, b_{3}&=0\\ 4608 \sqrt {3}\, a_{2}+4608 \sqrt {3}\, b_{3}&=0\\ -570240 i a_{1}-190080 \sqrt {3}\, a_{1}&=0\\ -217728 i b_{1}-72576 \sqrt {3}\, b_{1}&=0\\ -217728 i b_{2}-72576 \sqrt {3}\, b_{2}&=0\\ -31104 i b_{1}+10368 \sqrt {3}\, b_{1}&=0\\ -31104 i b_{2}+10368 \sqrt {3}\, b_{2}&=0\\ -22464 i a_{3}-7488 \sqrt {3}\, a_{3}&=0\\ -6912 i a_{1}+2304 \sqrt {3}\, a_{1}&=0\\ -288 i a_{3}+96 \sqrt {3}\, a_{3}&=0\\ 576 i a_{3}+192 \sqrt {3}\, a_{3}&=0\\ 4320 i a_{3}-1440 \sqrt {3}\, a_{3}&=0\\ 6912 i b_{1}+2304 \sqrt {3}\, b_{1}&=0\\ 6912 i b_{2}+2304 \sqrt {3}\, b_{2}&=0\\ 13824 i a_{1}+4608 \sqrt {3}\, a_{1}&=0\\ 62208 i a_{3}-20736 \sqrt {3}\, a_{3}&=0\\ 134784 i a_{3}+44928 \sqrt {3}\, a_{3}&=0\\ 155520 i a_{1}-51840 \sqrt {3}\, a_{1}&=0\\ 746496 i b_{1}+248832 \sqrt {3}\, b_{1}&=0\\ 746496 i b_{2}+248832 \sqrt {3}\, b_{2}&=0\\ 1306368 i a_{3}+435456 \sqrt {3}\, a_{3}&=0\\ 5225472 i a_{1}+1741824 \sqrt {3}\, a_{1}&=0\\ -580608 i \sqrt {3}\, a_{1}-580608 a_{1}&=0\\ -145152 i \sqrt {3}\, a_{3}-145152 a_{3}&=0\\ -82944 i \sqrt {3}\, b_{1}-82944 b_{1}&=0\\ -82944 i \sqrt {3}\, b_{2}-82944 b_{2}&=0\\ -17280 i \sqrt {3}\, a_{1}+17280 a_{1}&=0\\ -7488 i \sqrt {3}\, a_{1}-7488 a_{1}&=0\\ -6912 i \sqrt {3}\, a_{3}+6912 a_{3}&=0\\ -1728 i \sqrt {3}\, b_{1}-1728 b_{1}&=0\\ -1728 i \sqrt {3}\, b_{2}-1728 b_{2}&=0\\ -288 i \sqrt {3}\, a_{3}-288 a_{3}&=0\\ -144 i \sqrt {3}\, a_{1}+144 a_{1}&=0\\ -48 i \sqrt {3}\, b_{1}+48 b_{1}&=0\\ -48 i \sqrt {3}\, b_{2}+48 b_{2}&=0\\ -6 i \sqrt {3}\, a_{3}+6 a_{3}&=0\\ 6 i \sqrt {3}\, a_{3}+6 a_{3}&=0\\ 48 i \sqrt {3}\, b_{1}+48 b_{1}&=0\\ 48 i \sqrt {3}\, b_{2}+48 b_{2}&=0\\ 144 i \sqrt {3}\, a_{1}+144 a_{1}&=0\\ 144 i \sqrt {3}\, a_{3}-144 a_{3}&=0\\ 288 i \sqrt {3}\, a_{3}-288 a_{3}&=0\\ 576 i \sqrt {3}\, b_{1}-576 b_{1}&=0\\ 576 i \sqrt {3}\, b_{2}-576 b_{2}&=0\\ 3168 i \sqrt {3}\, a_{3}+3168 a_{3}&=0\\ 3456 i \sqrt {3}\, b_{1}-3456 b_{1}&=0\\ 3456 i \sqrt {3}\, b_{2}-3456 b_{2}&=0\\ 4032 i \sqrt {3}\, a_{1}-4032 a_{1}&=0\\ 6912 i \sqrt {3}\, a_{3}+6912 a_{3}&=0\\ 10368 i \sqrt {3}\, b_{1}+10368 b_{1}&=0\\ 10368 i \sqrt {3}\, b_{2}+10368 b_{2}&=0\\ 114048 i \sqrt {3}\, a_{1}+114048 a_{1}&=0\\ -393984 i a_{2}-393984 i b_{3}-131328 \sqrt {3}\, a_{2}-131328 \sqrt {3}\, b_{3}&=0\\ -3456 i a_{2}-3456 i b_{3}+1152 \sqrt {3}\, a_{2}+1152 \sqrt {3}\, b_{3}&=0\\ 10368 i a_{2}+10368 i b_{3}+3456 \sqrt {3}\, a_{2}+3456 \sqrt {3}\, b_{3}&=0\\ 62208 i a_{2}+62208 i b_{3}-20736 \sqrt {3}\, a_{2}-20736 \sqrt {3}\, b_{3}&=0\\ 2985984 i a_{2}+2985984 i b_{3}+995328 \sqrt {3}\, a_{2}+995328 \sqrt {3}\, b_{3}&=0\\ -331776 i \sqrt {3}\, a_{2}-331776 i \sqrt {3}\, b_{3}-331776 a_{2}-331776 b_{3}&=0\\ -6912 i \sqrt {3}\, a_{2}-6912 i \sqrt {3}\, b_{3}+6912 a_{2}+6912 b_{3}&=0\\ -4608 i \sqrt {3}\, a_{2}-4608 i \sqrt {3}\, b_{3}-4608 a_{2}-4608 b_{3}&=0\\ -96 i \sqrt {3}\, a_{2}-96 i \sqrt {3}\, b_{3}+96 a_{2}+96 b_{3}&=0\\ 96 i \sqrt {3}\, a_{2}+96 i \sqrt {3}\, b_{3}+96 a_{2}+96 b_{3}&=0\\ 2304 i \sqrt {3}\, a_{2}+2304 i \sqrt {3}\, b_{3}-2304 a_{2}-2304 b_{3}&=0\\ 62208 i \sqrt {3}\, a_{2}+62208 i \sqrt {3}\, b_{3}+62208 a_{2}+62208 b_{3}&=0 \end {align*}

Solving the above equations for the unknowns gives \begin {align*} a_{1}&=0\\ a_{2}&=-b_{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 {24 i \sqrt {3}\, y^{2} x^{2}-24 y^{2} x^{2} \left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 x^{2} y^{2}-1}\, y^{3} x^{3}+36 x^{2} y^{2}-1\right )^{\frac {1}{3}}+i \sqrt {3}\, \left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 x^{2} y^{2}-1}\, y^{3} x^{3}+36 x^{2} y^{2}-1\right )^{\frac {2}{3}}-24 x^{2} y^{2}-i \sqrt {3}+\left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 x^{2} y^{2}-1}\, y^{3} x^{3}+36 x^{2} y^{2}-1\right )^{\frac {2}{3}}+2 \left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 x^{2} y^{2}-1}\, y^{3} x^{3}+36 x^{2} y^{2}-1\right )^{\frac {1}{3}}+1}{48 y \,x^{3} \left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 x^{2} y^{2}-1}\, y^{3} x^{3}+36 x^{2} y^{2}-1\right )^{\frac {1}{3}}}\\ 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} \text {Expression too large to display} \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. \[ \left \{x, y, \left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 x^{2} y^{2}-1}\, y^{3} x^{3}+36 x^{2} y^{2}-1\right )^{\frac {1}{3}}, \left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 x^{2} y^{2}-1}\, y^{3} x^{3}+36 x^{2} y^{2}-1\right )^{\frac {2}{3}}, \sqrt {27 x^{2} y^{2}-1}\right \} \] The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them \[ \left \{x = v_{1}, y = v_{2}, \left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 x^{2} y^{2}-1}\, y^{3} x^{3}+36 x^{2} y^{2}-1\right )^{\frac {1}{3}} = v_{3}, \left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 x^{2} y^{2}-1}\, y^{3} x^{3}+36 x^{2} y^{2}-1\right )^{\frac {2}{3}} = v_{4}, \sqrt {27 x^{2} y^{2}-1} = v_{5}\right \} \] 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} \text {Expression too large to display} \end{equation} Setting each coeﬃcients in (8E) to zero gives the following equations to solve \begin {align*} -248832 a_{1}&=0\\ -11520 a_{1}&=0\\ 288 a_{1}&=0\\ 103680 a_{1}&=0\\ -124416 a_{3}&=0\\ -432 a_{3}&=0\\ 12 a_{3}&=0\\ 2016 a_{3}&=0\\ 20736 a_{3}&=0\\ -20736 b_{1}&=0\\ -2304 b_{1}&=0\\ 96 b_{1}&=0\\ 248832 b_{1}&=0\\ -248832 b_{2}&=0\\ -4608 b_{2}&=0\\ 96 b_{2}&=0\\ 62208 b_{2}&=0\\ -214272 \sqrt {3}\, a_{1}&=0\\ 6912 \sqrt {3}\, a_{1}&=0\\ 746496 \sqrt {3}\, a_{1}&=0\\ -13824 \sqrt {3}\, a_{3}&=0\\ -7776 \sqrt {3}\, a_{3}&=0\\ 288 \sqrt {3}\, a_{3}&=0\\ 373248 \sqrt {3}\, a_{3}&=0\\ -746496 \sqrt {3}\, b_{1}&=0\\ -34560 \sqrt {3}\, b_{1}&=0\\ 2304 \sqrt {3}\, b_{1}&=0\\ -89856 \sqrt {3}\, b_{2}&=0\\ 2304 \sqrt {3}\, b_{2}&=0\\ 746496 \sqrt {3}\, b_{2}&=0\\ -6912 a_{2}-6912 b_{3}&=0\\ 192 a_{2}+192 b_{3}&=0\\ 41472 a_{2}+41472 b_{3}&=0\\ -190080 \sqrt {3}\, a_{1}+570240 i a_{1}&=0\\ -51840 \sqrt {3}\, a_{1}-155520 i a_{1}&=0\\ 2304 \sqrt {3}\, a_{1}+6912 i a_{1}&=0\\ 4608 \sqrt {3}\, a_{1}-13824 i a_{1}&=0\\ 1741824 \sqrt {3}\, a_{1}-5225472 i a_{1}&=0\\ -124416 \sqrt {3}\, a_{2}-124416 \sqrt {3}\, b_{3}&=0\\ 4608 \sqrt {3}\, a_{2}+4608 \sqrt {3}\, b_{3}&=0\\ -20736 \sqrt {3}\, a_{3}-62208 i a_{3}&=0\\ -7488 \sqrt {3}\, a_{3}+22464 i a_{3}&=0\\ -1440 \sqrt {3}\, a_{3}-4320 i a_{3}&=0\\ 96 \sqrt {3}\, a_{3}+288 i a_{3}&=0\\ 192 \sqrt {3}\, a_{3}-576 i a_{3}&=0\\ 44928 \sqrt {3}\, a_{3}-134784 i a_{3}&=0\\ 435456 \sqrt {3}\, a_{3}-1306368 i a_{3}&=0\\ -72576 \sqrt {3}\, b_{1}+217728 i b_{1}&=0\\ 2304 \sqrt {3}\, b_{1}-6912 i b_{1}&=0\\ 10368 \sqrt {3}\, b_{1}+31104 i b_{1}&=0\\ 248832 \sqrt {3}\, b_{1}-746496 i b_{1}&=0\\ -72576 \sqrt {3}\, b_{2}+217728 i b_{2}&=0\\ 2304 \sqrt {3}\, b_{2}-6912 i b_{2}&=0\\ 10368 \sqrt {3}\, b_{2}+31104 i b_{2}&=0\\ 248832 \sqrt {3}\, b_{2}-746496 i b_{2}&=0\\ -114048 i \sqrt {3}\, a_{1}+114048 a_{1}&=0\\ -10368 i \sqrt {3}\, b_{1}+10368 b_{1}&=0\\ -10368 i \sqrt {3}\, b_{2}+10368 b_{2}&=0\\ -6912 i \sqrt {3}\, a_{3}+6912 a_{3}&=0\\ -4032 i \sqrt {3}\, a_{1}-4032 a_{1}&=0\\ -3456 i \sqrt {3}\, b_{1}-3456 b_{1}&=0\\ -3456 i \sqrt {3}\, b_{2}-3456 b_{2}&=0\\ -3168 i \sqrt {3}\, a_{3}+3168 a_{3}&=0\\ -576 i \sqrt {3}\, b_{1}-576 b_{1}&=0\\ -576 i \sqrt {3}\, b_{2}-576 b_{2}&=0\\ -288 i \sqrt {3}\, a_{3}-288 a_{3}&=0\\ -144 i \sqrt {3}\, a_{1}+144 a_{1}&=0\\ -144 i \sqrt {3}\, a_{3}-144 a_{3}&=0\\ -48 i \sqrt {3}\, b_{1}+48 b_{1}&=0\\ -48 i \sqrt {3}\, b_{2}+48 b_{2}&=0\\ -6 i \sqrt {3}\, a_{3}+6 a_{3}&=0\\ 6 i \sqrt {3}\, a_{3}+6 a_{3}&=0\\ 48 i \sqrt {3}\, b_{1}+48 b_{1}&=0\\ 48 i \sqrt {3}\, b_{2}+48 b_{2}&=0\\ 144 i \sqrt {3}\, a_{1}+144 a_{1}&=0\\ 288 i \sqrt {3}\, a_{3}-288 a_{3}&=0\\ 1728 i \sqrt {3}\, b_{1}-1728 b_{1}&=0\\ 1728 i \sqrt {3}\, b_{2}-1728 b_{2}&=0\\ 6912 i \sqrt {3}\, a_{3}+6912 a_{3}&=0\\ 7488 i \sqrt {3}\, a_{1}-7488 a_{1}&=0\\ 17280 i \sqrt {3}\, a_{1}+17280 a_{1}&=0\\ 82944 i \sqrt {3}\, b_{1}-82944 b_{1}&=0\\ 82944 i \sqrt {3}\, b_{2}-82944 b_{2}&=0\\ 145152 i \sqrt {3}\, a_{3}-145152 a_{3}&=0\\ 580608 i \sqrt {3}\, a_{1}-580608 a_{1}&=0\\ -131328 \sqrt {3}\, a_{2}-131328 \sqrt {3}\, b_{3}+393984 i a_{2}+393984 i b_{3}&=0\\ -20736 \sqrt {3}\, a_{2}-20736 \sqrt {3}\, b_{3}-62208 i a_{2}-62208 i b_{3}&=0\\ 1152 \sqrt {3}\, a_{2}+1152 \sqrt {3}\, b_{3}+3456 i a_{2}+3456 i b_{3}&=0\\ 3456 \sqrt {3}\, a_{2}+3456 \sqrt {3}\, b_{3}-10368 i a_{2}-10368 i b_{3}&=0\\ 995328 \sqrt {3}\, a_{2}+995328 \sqrt {3}\, b_{3}-2985984 i a_{2}-2985984 i b_{3}&=0\\ -62208 i \sqrt {3}\, a_{2}-62208 i \sqrt {3}\, b_{3}+62208 a_{2}+62208 b_{3}&=0\\ -2304 i \sqrt {3}\, a_{2}-2304 i \sqrt {3}\, b_{3}-2304 a_{2}-2304 b_{3}&=0\\ -96 i \sqrt {3}\, a_{2}-96 i \sqrt {3}\, b_{3}+96 a_{2}+96 b_{3}&=0\\ 96 i \sqrt {3}\, a_{2}+96 i \sqrt {3}\, b_{3}+96 a_{2}+96 b_{3}&=0\\ 4608 i \sqrt {3}\, a_{2}+4608 i \sqrt {3}\, b_{3}-4608 a_{2}-4608 b_{3}&=0\\ 6912 i \sqrt {3}\, a_{2}+6912 i \sqrt {3}\, b_{3}+6912 a_{2}+6912 b_{3}&=0\\ 331776 i \sqrt {3}\, a_{2}+331776 i \sqrt {3}\, b_{3}-331776 a_{2}-331776 b_{3}&=0 \end {align*}

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

1.87.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y \left (y-2 x y^{\prime }\right )^{3}-{y^{\prime }}^{2}=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 (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 y^{2} x^{2}-1}\, y^{3} x^{3}+36 y^{2} x^{2}-1\right )^{\frac {1}{3}}}{24 y x^{3}}-\frac {24 y^{2} x^{2}-1}{24 y x^{3} \left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 y^{2} x^{2}-1}\, y^{3} x^{3}+36 y^{2} x^{2}-1\right )^{\frac {1}{3}}}+\frac {12 y^{2} x^{2}-1}{24 y x^{3}}, y^{\prime }=-\frac {\left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 y^{2} x^{2}-1}\, y^{3} x^{3}+36 y^{2} x^{2}-1\right )^{\frac {1}{3}}}{48 y x^{3}}+\frac {24 y^{2} x^{2}-1}{48 y x^{3} \left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 y^{2} x^{2}-1}\, y^{3} x^{3}+36 y^{2} x^{2}-1\right )^{\frac {1}{3}}}+\frac {12 y^{2} x^{2}-1}{24 y x^{3}}-\frac {\mathrm {I} \sqrt {3}\, \left (\frac {\left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 y^{2} x^{2}-1}\, y^{3} x^{3}+36 y^{2} x^{2}-1\right )^{\frac {1}{3}}}{24 y x^{3}}+\frac {24 y^{2} x^{2}-1}{24 y x^{3} \left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 y^{2} x^{2}-1}\, y^{3} x^{3}+36 y^{2} x^{2}-1\right )^{\frac {1}{3}}}\right )}{2}, y^{\prime }=-\frac {\left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 y^{2} x^{2}-1}\, y^{3} x^{3}+36 y^{2} x^{2}-1\right )^{\frac {1}{3}}}{48 y x^{3}}+\frac {24 y^{2} x^{2}-1}{48 y x^{3} \left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 y^{2} x^{2}-1}\, y^{3} x^{3}+36 y^{2} x^{2}-1\right )^{\frac {1}{3}}}+\frac {12 y^{2} x^{2}-1}{24 y x^{3}}+\frac {\mathrm {I} \sqrt {3}\, \left (\frac {\left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 y^{2} x^{2}-1}\, y^{3} x^{3}+36 y^{2} x^{2}-1\right )^{\frac {1}{3}}}{24 y x^{3}}+\frac {24 y^{2} x^{2}-1}{24 y x^{3} \left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 y^{2} x^{2}-1}\, y^{3} x^{3}+36 y^{2} x^{2}-1\right )^{\frac {1}{3}}}\right )}{2}\right ] \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {\left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 y^{2} x^{2}-1}\, y^{3} x^{3}+36 y^{2} x^{2}-1\right )^{\frac {1}{3}}}{24 y x^{3}}-\frac {24 y^{2} x^{2}-1}{24 y x^{3} \left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 y^{2} x^{2}-1}\, y^{3} x^{3}+36 y^{2} x^{2}-1\right )^{\frac {1}{3}}}+\frac {12 y^{2} x^{2}-1}{24 y x^{3}} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {\left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 y^{2} x^{2}-1}\, y^{3} x^{3}+36 y^{2} x^{2}-1\right )^{\frac {1}{3}}}{48 y x^{3}}+\frac {24 y^{2} x^{2}-1}{48 y x^{3} \left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 y^{2} x^{2}-1}\, y^{3} x^{3}+36 y^{2} x^{2}-1\right )^{\frac {1}{3}}}+\frac {12 y^{2} x^{2}-1}{24 y x^{3}}-\frac {\mathrm {I} \sqrt {3}\, \left (\frac {\left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 y^{2} x^{2}-1}\, y^{3} x^{3}+36 y^{2} x^{2}-1\right )^{\frac {1}{3}}}{24 y x^{3}}+\frac {24 y^{2} x^{2}-1}{24 y x^{3} \left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 y^{2} x^{2}-1}\, y^{3} x^{3}+36 y^{2} x^{2}-1\right )^{\frac {1}{3}}}\right )}{2} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {\left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 y^{2} x^{2}-1}\, y^{3} x^{3}+36 y^{2} x^{2}-1\right )^{\frac {1}{3}}}{48 y x^{3}}+\frac {24 y^{2} x^{2}-1}{48 y x^{3} \left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 y^{2} x^{2}-1}\, y^{3} x^{3}+36 y^{2} x^{2}-1\right )^{\frac {1}{3}}}+\frac {12 y^{2} x^{2}-1}{24 y x^{3}}+\frac {\mathrm {I} \sqrt {3}\, \left (\frac {\left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 y^{2} x^{2}-1}\, y^{3} x^{3}+36 y^{2} x^{2}-1\right )^{\frac {1}{3}}}{24 y x^{3}}+\frac {24 y^{2} x^{2}-1}{24 y x^{3} \left (-216 y^{4} x^{4}+24 \sqrt {3}\, \sqrt {27 y^{2} x^{2}-1}\, y^{3} x^{3}+36 y^{2} x^{2}-1\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 homogeneous G 
      <- homogeneous successful 
   ------------------- 
   * Tackling next ODE. 
      *** Sublevel 3 *** 
      Methods for first order ODEs: 
      --- Trying classification methods --- 
      trying homogeneous types: 
      trying homogeneous G 
      <- homogeneous successful 
   ------------------- 
   * Tackling next ODE. 
      *** Sublevel 3 *** 
      Methods for first order ODEs: 
      --- Trying classification methods --- 
      trying homogeneous types: 
      trying homogeneous G 
      <- homogeneous successful`

\begin{align*} y \left (x \right ) &= -\frac {\sqrt {3}}{9 x} \\ y \left (x \right ) &= \frac {\sqrt {3}}{9 x} \\ y \left (x \right ) &= 0 \\ y \left (x \right ) &= \frac {\operatorname {RootOf}\left (-\ln \left (x \right )+c_{1} +24 \left (\int _{}^{\textit {\_Z}}\frac {\left (-216 \textit {\_a}^{4}+24 \textit {\_a}^{3} \sqrt {81 \textit {\_a}^{2}-3}+36 \textit {\_a}^{2}-1\right )^{\frac {1}{3}} \textit {\_a}}{36 \left (-216 \textit {\_a}^{4}+24 \textit {\_a}^{3} \sqrt {81 \textit {\_a}^{2}-3}+36 \textit {\_a}^{2}-1\right )^{\frac {1}{3}} \textit {\_a}^{2}+\left (-216 \textit {\_a}^{4}+24 \textit {\_a}^{3} \sqrt {81 \textit {\_a}^{2}-3}+36 \textit {\_a}^{2}-1\right )^{\frac {2}{3}}-24 \textit {\_a}^{2}-\left (-216 \textit {\_a}^{4}+24 \textit {\_a}^{3} \sqrt {81 \textit {\_a}^{2}-3}+36 \textit {\_a}^{2}-1\right )^{\frac {1}{3}}+1}d \textit {\_a} \right )\right )}{x} \\ y \left (x \right ) &= \frac {\operatorname {RootOf}\left (-\ln \left (x \right )+c_{1} -48 \left (\int _{}^{\textit {\_Z}}\frac {\left (-216 \textit {\_a}^{4}+24 \textit {\_a}^{3} \sqrt {81 \textit {\_a}^{2}-3}+36 \textit {\_a}^{2}-1\right )^{\frac {1}{3}} \textit {\_a}}{i \left (-216 \textit {\_a}^{4}+24 \textit {\_a}^{3} \sqrt {81 \textit {\_a}^{2}-3}+36 \textit {\_a}^{2}-1\right )^{\frac {2}{3}} \sqrt {3}+24 i \sqrt {3}\, \textit {\_a}^{2}-72 \left (-216 \textit {\_a}^{4}+24 \textit {\_a}^{3} \sqrt {81 \textit {\_a}^{2}-3}+36 \textit {\_a}^{2}-1\right )^{\frac {1}{3}} \textit {\_a}^{2}-i \sqrt {3}+\left (-216 \textit {\_a}^{4}+24 \textit {\_a}^{3} \sqrt {81 \textit {\_a}^{2}-3}+36 \textit {\_a}^{2}-1\right )^{\frac {2}{3}}-24 \textit {\_a}^{2}+2 \left (-216 \textit {\_a}^{4}+24 \textit {\_a}^{3} \sqrt {81 \textit {\_a}^{2}-3}+36 \textit {\_a}^{2}-1\right )^{\frac {1}{3}}+1}d \textit {\_a} \right )\right )}{x} \\ y \left (x \right ) &= \frac {\operatorname {RootOf}\left (-\ln \left (x \right )+c_{1} +48 \left (\int _{}^{\textit {\_Z}}\frac {\left (-216 \textit {\_a}^{4}+24 \textit {\_a}^{3} \sqrt {81 \textit {\_a}^{2}-3}+36 \textit {\_a}^{2}-1\right )^{\frac {1}{3}} \textit {\_a}}{i \left (-216 \textit {\_a}^{4}+24 \textit {\_a}^{3} \sqrt {81 \textit {\_a}^{2}-3}+36 \textit {\_a}^{2}-1\right )^{\frac {2}{3}} \sqrt {3}+24 i \sqrt {3}\, \textit {\_a}^{2}+72 \left (-216 \textit {\_a}^{4}+24 \textit {\_a}^{3} \sqrt {81 \textit {\_a}^{2}-3}+36 \textit {\_a}^{2}-1\right )^{\frac {1}{3}} \textit {\_a}^{2}-i \sqrt {3}-\left (-216 \textit {\_a}^{4}+24 \textit {\_a}^{3} \sqrt {81 \textit {\_a}^{2}-3}+36 \textit {\_a}^{2}-1\right )^{\frac {2}{3}}+24 \textit {\_a}^{2}-2 \left (-216 \textit {\_a}^{4}+24 \textit {\_a}^{3} \sqrt {81 \textit {\_a}^{2}-3}+36 \textit {\_a}^{2}-1\right )^{\frac {1}{3}}-1}d \textit {\_a} \right )\right )}{x} \\ \end{align*}

1.87 problem 90

1.87.1 Maple step by step solution