Internal problem ID [4281]
Internal file name [OUTPUT/3774_Sunday_June_05_2022_10_51_02_AM_35366318/index.tex]
Book: Ordinary differential equations and their solutions. By George Moseley Murphy.
1960
Section: Various 35
Problem number: 1061.
ODE order: 1.
ODE degree: 3.
The type(s) of ODE detected by this program : "first_order_ode_lie_symmetry_calculated"
Maple gives the following as the ode type
[[_1st_order, _with_linear_symmetries]]
\[ \boxed {x {y^{\prime }}^{3}-2 y {y^{\prime }}^{2}=-4 x^{2}} \] 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 (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}}}{3 x}+\frac {4 y^{2}}{3 x \left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}}}+\frac {2 y}{3 x} \tag {1} \\ y^{\prime }&=-\frac {\left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}}}{6 x}-\frac {2 y^{2}}{3 x \left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}}}+\frac {2 y}{3 x}+\frac {i \sqrt {3}\, \left (\frac {\left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}}}{3 x}-\frac {4 y^{2}}{3 x \left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}}}\right )}{2} \tag {2} \\ y^{\prime }&=-\frac {\left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}}}{6 x}-\frac {2 y^{2}}{3 x \left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}}}+\frac {2 y}{3 x}-\frac {i \sqrt {3}\, \left (\frac {\left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}}}{3 x}-\frac {4 y^{2}}{3 x \left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {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 (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {2}{3}}+2 y \left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}}+4 y^{2}}{3 x \left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\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. \[ \left \{x, y, \left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}}, \left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {2}{3}}, \sqrt {81 x^{4}-24 y^{3}}\right \} \] The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them \[ \left \{x = v_{1}, y = v_{2}, \left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}} = v_{3}, \left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {2}{3}} = v_{4}, \sqrt {81 x^{4}-24 y^{3}} = v_{5}\right \} \] The above PDE (6E) now becomes \begin{equation} \tag{7E} 52488 v_{1}^{10} a_{3}+7776 v_{1}^{7} v_{2}^{2} a_{2}-5832 v_{5} v_{1}^{8} a_{3}-17496 v_{1}^{6} v_{2}^{3} a_{3}-2916 v_{1}^{6} v_{3} v_{2}^{2} a_{3}-11664 v_{1}^{8} v_{2} b_{2}+1458 v_{1}^{8} v_{3} b_{2}-5832 v_{1}^{7} v_{2}^{2} b_{3}+13608 v_{1}^{6} v_{2}^{2} a_{1}+2916 v_{1}^{6} v_{3} v_{2} a_{1}-1944 v_{4} v_{1}^{7} a_{2}-864 v_{5} v_{1}^{5} v_{2}^{2} a_{2}-1152 v_{1}^{3} v_{2}^{5} a_{2}-2430 v_{4} v_{1}^{6} v_{2} a_{3}+1080 v_{5} v_{1}^{4} v_{2}^{3} a_{3}+324 v_{1}^{4} v_{5} v_{3} v_{2}^{2} a_{3}+1728 v_{1}^{2} v_{2}^{6} a_{3}+864 v_{1}^{2} v_{3} v_{2}^{5} a_{3}-11664 v_{1}^{7} v_{2} b_{1}-2916 v_{1}^{7} v_{3} b_{1}+1296 v_{5} v_{1}^{6} v_{2} b_{2}-162 v_{1}^{6} v_{5} v_{3} b_{2}+2592 v_{1}^{4} v_{2}^{4} b_{2}-432 v_{1}^{4} v_{3} v_{2}^{3} b_{2}+1458 v_{1}^{7} v_{4} b_{3}+648 v_{5} v_{1}^{5} v_{2}^{2} b_{3}+864 v_{1}^{3} v_{2}^{5} b_{3}-486 v_{4} v_{1}^{6} a_{1}-1512 v_{5} v_{1}^{4} v_{2}^{2} a_{1}-324 v_{1}^{4} v_{5} v_{3} v_{2} a_{1}-2880 v_{1}^{2} v_{2}^{5} a_{1}-864 v_{1}^{2} v_{3} v_{2}^{4} a_{1}+216 v_{5} v_{4} v_{1}^{5} a_{2}+288 v_{4} v_{1}^{3} v_{2}^{3} a_{2}+270 v_{5} v_{4} v_{1}^{4} v_{2} a_{3}+432 v_{4} v_{1}^{2} v_{2}^{4} a_{3}-96 v_{5} v_{2}^{6} a_{3}-48 v_{5} v_{3} v_{2}^{5} a_{3}+1296 v_{5} v_{1}^{5} v_{2} b_{1}+324 v_{1}^{5} v_{5} v_{3} b_{1}+2592 v_{1}^{3} v_{2}^{4} b_{1}+864 v_{1}^{3} v_{3} v_{2}^{3} b_{1}+216 v_{4} v_{1}^{4} v_{2}^{2} b_{2}-96 v_{5} v_{1}^{2} v_{2}^{4} b_{2}+24 v_{1}^{2} v_{5} v_{3} v_{2}^{3} b_{2}-162 v_{1}^{5} v_{5} v_{4} b_{3}-216 v_{4} v_{1}^{3} v_{2}^{3} b_{3}+54 v_{5} v_{4} v_{1}^{4} a_{1}-144 v_{4} v_{1}^{2} v_{2}^{3} a_{1}+96 v_{5} v_{2}^{5} a_{1}+48 v_{5} v_{3} v_{2}^{4} a_{1}-24 v_{5} v_{4} v_{2}^{4} a_{3}+216 v_{4} v_{1}^{3} v_{2}^{2} b_{1}-96 v_{5} v_{1} v_{2}^{4} b_{1}-48 v_{1} v_{5} v_{3} v_{2}^{3} b_{1}-24 v_{5} v_{4} v_{1}^{2} v_{2}^{2} b_{2}+24 v_{5} v_{4} v_{2}^{3} a_{1}-24 v_{5} v_{4} v_{1} v_{2}^{2} b_{1} = 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 (7776 a_{2}-5832 b_{3}\right ) v_{1}^{7} v_{2}^{2}+\left (-1944 a_{2}+1458 b_{3}\right ) v_{1}^{7} v_{4}+\left (-864 a_{2}+648 b_{3}\right ) v_{1}^{5} v_{2}^{2} v_{5}+\left (216 a_{2}-162 b_{3}\right ) v_{1}^{5} v_{4} v_{5}+\left (288 a_{2}-216 b_{3}\right ) v_{1}^{3} v_{2}^{3} v_{4}+1080 v_{5} v_{1}^{4} v_{2}^{3} a_{3}+216 v_{4} v_{1}^{4} v_{2}^{2} b_{2}+54 v_{5} v_{4} v_{1}^{4} a_{1}-24 v_{5} v_{4} v_{2}^{4} a_{3}-1512 v_{5} v_{1}^{4} v_{2}^{2} a_{1}-96 v_{5} v_{1}^{2} v_{2}^{4} b_{2}+216 v_{4} v_{1}^{3} v_{2}^{2} b_{1}-96 v_{5} v_{1} v_{2}^{4} b_{1}-2916 v_{1}^{6} v_{3} v_{2}^{2} a_{3}+2916 v_{1}^{6} v_{3} v_{2} a_{1}-432 v_{1}^{4} v_{3} v_{2}^{3} b_{2}+1296 v_{5} v_{1}^{6} v_{2} b_{2}+432 v_{4} v_{1}^{2} v_{2}^{4} a_{3}+1296 v_{5} v_{1}^{5} v_{2} b_{1}-144 v_{4} v_{1}^{2} v_{2}^{3} a_{1}-48 v_{5} v_{3} v_{2}^{5} a_{3}+24 v_{5} v_{4} v_{2}^{3} a_{1}+48 v_{5} v_{3} v_{2}^{4} a_{1}-162 v_{1}^{6} v_{5} v_{3} b_{2}+\left (-1152 a_{2}+864 b_{3}\right ) v_{1}^{3} v_{2}^{5}+324 v_{1}^{5} v_{5} v_{3} b_{1}+864 v_{1}^{2} v_{3} v_{2}^{5} a_{3}+864 v_{1}^{3} v_{3} v_{2}^{3} b_{1}-864 v_{1}^{2} v_{3} v_{2}^{4} a_{1}-2430 v_{4} v_{1}^{6} v_{2} a_{3}+324 v_{1}^{4} v_{5} v_{3} v_{2}^{2} a_{3}-48 v_{1} v_{5} v_{3} v_{2}^{3} b_{1}-324 v_{1}^{4} v_{5} v_{3} v_{2} a_{1}+24 v_{1}^{2} v_{5} v_{3} v_{2}^{3} b_{2}+270 v_{5} v_{4} v_{1}^{4} v_{2} a_{3}-24 v_{5} v_{4} v_{1}^{2} v_{2}^{2} b_{2}-24 v_{5} v_{4} v_{1} v_{2}^{2} b_{1}-17496 v_{1}^{6} v_{2}^{3} a_{3}+13608 v_{1}^{6} v_{2}^{2} a_{1}+2592 v_{1}^{4} v_{2}^{4} b_{2}+2592 v_{1}^{3} v_{2}^{4} b_{1}-486 v_{4} v_{1}^{6} a_{1}+1458 v_{1}^{8} v_{3} b_{2}-2916 v_{1}^{7} v_{3} b_{1}+1728 v_{1}^{2} v_{2}^{6} a_{3}-2880 v_{1}^{2} v_{2}^{5} a_{1}-5832 v_{5} v_{1}^{8} a_{3}-96 v_{5} v_{2}^{6} a_{3}+96 v_{5} v_{2}^{5} a_{1}-11664 v_{1}^{8} v_{2} b_{2}-11664 v_{1}^{7} v_{2} b_{1}+52488 v_{1}^{10} a_{3} = 0 \end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve \begin {align*} -2880 a_{1}&=0\\ -1512 a_{1}&=0\\ -864 a_{1}&=0\\ -486 a_{1}&=0\\ -324 a_{1}&=0\\ -144 a_{1}&=0\\ 24 a_{1}&=0\\ 48 a_{1}&=0\\ 54 a_{1}&=0\\ 96 a_{1}&=0\\ 2916 a_{1}&=0\\ 13608 a_{1}&=0\\ -17496 a_{3}&=0\\ -5832 a_{3}&=0\\ -2916 a_{3}&=0\\ -2430 a_{3}&=0\\ -96 a_{3}&=0\\ -48 a_{3}&=0\\ -24 a_{3}&=0\\ 270 a_{3}&=0\\ 324 a_{3}&=0\\ 432 a_{3}&=0\\ 864 a_{3}&=0\\ 1080 a_{3}&=0\\ 1728 a_{3}&=0\\ 52488 a_{3}&=0\\ -11664 b_{1}&=0\\ -2916 b_{1}&=0\\ -96 b_{1}&=0\\ -48 b_{1}&=0\\ -24 b_{1}&=0\\ 216 b_{1}&=0\\ 324 b_{1}&=0\\ 864 b_{1}&=0\\ 1296 b_{1}&=0\\ 2592 b_{1}&=0\\ -11664 b_{2}&=0\\ -432 b_{2}&=0\\ -162 b_{2}&=0\\ -96 b_{2}&=0\\ -24 b_{2}&=0\\ 24 b_{2}&=0\\ 216 b_{2}&=0\\ 1296 b_{2}&=0\\ 1458 b_{2}&=0\\ 2592 b_{2}&=0\\ -1944 a_{2}+1458 b_{3}&=0\\ -1152 a_{2}+864 b_{3}&=0\\ -864 a_{2}+648 b_{3}&=0\\ 216 a_{2}-162 b_{3}&=0\\ 288 a_{2}-216 b_{3}&=0\\ 7776 a_{2}-5832 b_{3}&=0 \end {align*}
Solving the above equations for the unknowns gives \begin {align*} a_{1}&=0\\ a_{2}&=\frac {3 b_{3}}{4}\\ 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 {3 x}{4} \\ \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 {3 x}{4}}\\ &= \frac {4 y}{3 x} \end {align*}
This is easily solved to give \begin {align*} y = c_{1} x^{\frac {4}{3}} \end {align*}
Where now the coordinate \(R\) is taken as the constant of integration. Hence \begin {align*} R &= \frac {y}{x^{\frac {4}{3}}} \end {align*}
And \(S\) is found from \begin {align*} dS &= \frac {dx}{\xi } \\ &= \frac {dx}{\frac {3 x}{4}} \end {align*}
Integrating gives \begin {align*} S &= \int { \frac {dx}{T}}\\ &= \frac {4 \ln \left (x \right )}{3} \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 (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {2}{3}}+2 y \left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}}+4 y^{2}}{3 x \left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}}} \end {align*}
Evaluating all the partial derivatives gives \begin {align*} R_{x} &= -\frac {4 y}{3 x^{\frac {7}{3}}}\\ R_{y} &= \frac {1}{x^{\frac {4}{3}}}\\ S_{x} &= \frac {4}{3 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 {4 x^{\frac {4}{3}} \left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}}}{\left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {2}{3}}-2 y \left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}}+4 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 {4 \,2^{\frac {1}{3}} \left (-27+4 R^{3}+3 i \sqrt {3}\, \sqrt {8 R^{3}-27}\right )^{\frac {1}{3}}}{\left (-27+4 R^{3}+3 i \sqrt {3}\, \sqrt {8 R^{3}-27}\right )^{\frac {2}{3}} 2^{\frac {2}{3}}-2 \left (-27+4 R^{3}+3 i \sqrt {3}\, \sqrt {8 R^{3}-27}\right )^{\frac {1}{3}} 2^{\frac {1}{3}} R +4 R^{2}} \end {align*}
The above is a quadrature ode. This is the whole point of Lie symmetry method. It converts an ode, no matter how complicated it is, to one that can be solved by integration when the ode is in the canonical coordiates \(R,S\). Integrating the above gives \begin {align*} S \left (R \right ) = \int \frac {4 \left (-54+8 R^{3}+6 i \sqrt {24 R^{3}-81}\right )^{\frac {1}{3}}}{4^{\frac {1}{3}} {\left (\left (-27+4 R^{3}+3 i \sqrt {24 R^{3}-81}\right )^{2}\right )}^{\frac {1}{3}}-2 R \left (-54+8 R^{3}+6 i \sqrt {24 R^{3}-81}\right )^{\frac {1}{3}}+4 R^{2}}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 {4 \ln \left (x \right )}{3} = \int _{}^{\frac {y}{x^{\frac {4}{3}}}}\frac {4 \left (-54+8 \textit {\_a}^{3}+6 i \sqrt {24 \textit {\_a}^{3}-81}\right )^{\frac {1}{3}}}{4^{\frac {1}{3}} {\left (\left (-27+4 \textit {\_a}^{3}+3 i \sqrt {24 \textit {\_a}^{3}-81}\right )^{2}\right )}^{\frac {1}{3}}-2 \textit {\_a} \left (-54+8 \textit {\_a}^{3}+6 i \sqrt {24 \textit {\_a}^{3}-81}\right )^{\frac {1}{3}}+4 \textit {\_a}^{2}}d \textit {\_a} +c_{1} \end {align*}
Which simplifies to \begin {align*} \frac {4 \ln \left (x \right )}{3}-4 \left (\int _{}^{\frac {y}{x^{\frac {4}{3}}}}\frac {\left (-54+8 \textit {\_a}^{3}+6 i \sqrt {24 \textit {\_a}^{3}-81}\right )^{\frac {1}{3}}}{4 \textit {\_a}^{2}-2 \textit {\_a} \left (-54+8 \textit {\_a}^{3}+6 i \sqrt {24 \textit {\_a}^{3}-81}\right )^{\frac {1}{3}}+\left (-54+8 \textit {\_a}^{3}+6 i \sqrt {24 \textit {\_a}^{3}-81}\right )^{\frac {2}{3}}}d \textit {\_a} \right )-c_{1} = 0 \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \frac {4 \ln \left (x \right )}{3}-4 \left (\int _{}^{\frac {y}{x^{\frac {4}{3}}}}\frac {\left (-54+8 \textit {\_a}^{3}+6 i \sqrt {24 \textit {\_a}^{3}-81}\right )^{\frac {1}{3}}}{4 \textit {\_a}^{2}-2 \textit {\_a} \left (-54+8 \textit {\_a}^{3}+6 i \sqrt {24 \textit {\_a}^{3}-81}\right )^{\frac {1}{3}}+\left (-54+8 \textit {\_a}^{3}+6 i \sqrt {24 \textit {\_a}^{3}-81}\right )^{\frac {2}{3}}}d \textit {\_a} \right )-c_{1} &= 0 \\ \end{align*}
Verification of solutions
\[ \frac {4 \ln \left (x \right )}{3}-4 \left (\int _{}^{\frac {y}{x^{\frac {4}{3}}}}\frac {\left (-54+8 \textit {\_a}^{3}+6 i \sqrt {24 \textit {\_a}^{3}-81}\right )^{\frac {1}{3}}}{4 \textit {\_a}^{2}-2 \textit {\_a} \left (-54+8 \textit {\_a}^{3}+6 i \sqrt {24 \textit {\_a}^{3}-81}\right )^{\frac {1}{3}}+\left (-54+8 \textit {\_a}^{3}+6 i \sqrt {24 \textit {\_a}^{3}-81}\right )^{\frac {2}{3}}}d \textit {\_a} \right )-c_{1} = 0 \] Verified OK.
Solving equation (2)
Writing the ode as \begin {align*} y^{\prime }&=-\frac {2 \left (-i y \sqrt {3}\, \left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}}+2 i \sqrt {3}\, y^{2}+\left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {2}{3}}-y \left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}}-2 y^{2}\right )}{3 x \left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}} \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} \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. \[ \left \{x, y, \left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}}, \left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {2}{3}}, \sqrt {81 x^{4}-24 y^{3}}\right \} \] The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them \[ \left \{x = v_{1}, y = v_{2}, \left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}} = v_{3}, \left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {2}{3}} = v_{4}, \sqrt {81 x^{4}-24 y^{3}} = v_{5}\right \} \] 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} \text {Expression too large to display} \end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve \begin {align*} -11520 a_{1}&=0\\ -6048 a_{1}&=0\\ 384 a_{1}&=0\\ 54432 a_{1}&=0\\ -69984 a_{3}&=0\\ -23328 a_{3}&=0\\ -384 a_{3}&=0\\ 4320 a_{3}&=0\\ 6912 a_{3}&=0\\ 209952 a_{3}&=0\\ -46656 b_{1}&=0\\ -384 b_{1}&=0\\ 5184 b_{1}&=0\\ 10368 b_{1}&=0\\ -46656 b_{2}&=0\\ -384 b_{2}&=0\\ 5184 b_{2}&=0\\ 10368 b_{2}&=0\\ -4608 a_{2}+3456 b_{3}&=0\\ -3456 a_{2}+2592 b_{3}&=0\\ 31104 a_{2}-23328 b_{3}&=0\\ -5832 i \sqrt {3}\, a_{3}+5832 a_{3}&=0\\ -5832 i \sqrt {3}\, b_{1}+5832 b_{1}&=0\\ -1728 i \sqrt {3}\, a_{1}+1728 a_{1}&=0\\ -864 i \sqrt {3}\, a_{3}-864 a_{3}&=0\\ -864 i \sqrt {3}\, b_{2}+864 b_{2}&=0\\ -648 i \sqrt {3}\, a_{1}+648 a_{1}&=0\\ -540 i \sqrt {3}\, a_{3}-540 a_{3}&=0\\ -432 i \sqrt {3}\, b_{1}-432 b_{1}&=0\\ -432 i \sqrt {3}\, b_{2}-432 b_{2}&=0\\ -324 i \sqrt {3}\, b_{2}+324 b_{2}&=0\\ -108 i \sqrt {3}\, a_{1}-108 a_{1}&=0\\ -96 i \sqrt {3}\, a_{3}+96 a_{3}&=0\\ -96 i \sqrt {3}\, b_{1}+96 b_{1}&=0\\ -48 i \sqrt {3}\, a_{1}-48 a_{1}&=0\\ 48 i \sqrt {3}\, a_{3}+48 a_{3}&=0\\ 48 i \sqrt {3}\, b_{1}+48 b_{1}&=0\\ 48 i \sqrt {3}\, b_{2}-48 b_{2}&=0\\ 48 i \sqrt {3}\, b_{2}+48 b_{2}&=0\\ 96 i \sqrt {3}\, a_{1}-96 a_{1}&=0\\ 288 i \sqrt {3}\, a_{1}+288 a_{1}&=0\\ 648 i \sqrt {3}\, a_{3}-648 a_{3}&=0\\ 648 i \sqrt {3}\, b_{1}-648 b_{1}&=0\\ 972 i \sqrt {3}\, a_{1}+972 a_{1}&=0\\ 1728 i \sqrt {3}\, a_{3}-1728 a_{3}&=0\\ 1728 i \sqrt {3}\, b_{1}-1728 b_{1}&=0\\ 2916 i \sqrt {3}\, b_{2}-2916 b_{2}&=0\\ 4860 i \sqrt {3}\, a_{3}+4860 a_{3}&=0\\ 5832 i \sqrt {3}\, a_{1}-5832 a_{1}&=0\\ -576 i \sqrt {3}\, a_{2}+432 i \sqrt {3}\, b_{3}-576 a_{2}+432 b_{3}&=0\\ -432 i \sqrt {3}\, a_{2}+324 i \sqrt {3}\, b_{3}-432 a_{2}+324 b_{3}&=0\\ 3888 i \sqrt {3}\, a_{2}-2916 i \sqrt {3}\, b_{3}+3888 a_{2}-2916 b_{3}&=0 \end {align*}
Solving the above equations for the unknowns gives \begin {align*} a_{1}&=0\\ a_{2}&=\frac {3 b_{3}}{4}\\ 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 {3 x}{4} \\ \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 {\frac {2 i y \sqrt {3}\, \left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}}}{3}-\frac {4 i \sqrt {3}\, y^{2}}{3}+\frac {2 \left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {2}{3}}}{3}-\frac {2 y \left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}}}{3}-\frac {4 y^{2}}{3}}{x \left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}} \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} \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. \[ \left \{x, y, \left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}}, \left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {2}{3}}, \sqrt {81 x^{4}-24 y^{3}}\right \} \] The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them \[ \left \{x = v_{1}, y = v_{2}, \left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}} = v_{3}, \left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {2}{3}} = v_{4}, \sqrt {81 x^{4}-24 y^{3}} = v_{5}\right \} \] 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} \text {Expression too large to display} \end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve \begin {align*} -11520 a_{1}&=0\\ -6048 a_{1}&=0\\ 384 a_{1}&=0\\ 54432 a_{1}&=0\\ -69984 a_{3}&=0\\ -23328 a_{3}&=0\\ -384 a_{3}&=0\\ 4320 a_{3}&=0\\ 6912 a_{3}&=0\\ 209952 a_{3}&=0\\ -46656 b_{1}&=0\\ -384 b_{1}&=0\\ 5184 b_{1}&=0\\ 10368 b_{1}&=0\\ -46656 b_{2}&=0\\ -384 b_{2}&=0\\ 5184 b_{2}&=0\\ 10368 b_{2}&=0\\ -4608 a_{2}+3456 b_{3}&=0\\ -3456 a_{2}+2592 b_{3}&=0\\ 31104 a_{2}-23328 b_{3}&=0\\ -5832 i \sqrt {3}\, a_{1}-5832 a_{1}&=0\\ -4860 i \sqrt {3}\, a_{3}+4860 a_{3}&=0\\ -2916 i \sqrt {3}\, b_{2}-2916 b_{2}&=0\\ -1728 i \sqrt {3}\, a_{3}-1728 a_{3}&=0\\ -1728 i \sqrt {3}\, b_{1}-1728 b_{1}&=0\\ -972 i \sqrt {3}\, a_{1}+972 a_{1}&=0\\ -648 i \sqrt {3}\, a_{3}-648 a_{3}&=0\\ -648 i \sqrt {3}\, b_{1}-648 b_{1}&=0\\ -288 i \sqrt {3}\, a_{1}+288 a_{1}&=0\\ -96 i \sqrt {3}\, a_{1}-96 a_{1}&=0\\ -48 i \sqrt {3}\, a_{3}+48 a_{3}&=0\\ -48 i \sqrt {3}\, b_{1}+48 b_{1}&=0\\ -48 i \sqrt {3}\, b_{2}-48 b_{2}&=0\\ -48 i \sqrt {3}\, b_{2}+48 b_{2}&=0\\ 48 i \sqrt {3}\, a_{1}-48 a_{1}&=0\\ 96 i \sqrt {3}\, a_{3}+96 a_{3}&=0\\ 96 i \sqrt {3}\, b_{1}+96 b_{1}&=0\\ 108 i \sqrt {3}\, a_{1}-108 a_{1}&=0\\ 324 i \sqrt {3}\, b_{2}+324 b_{2}&=0\\ 432 i \sqrt {3}\, b_{1}-432 b_{1}&=0\\ 432 i \sqrt {3}\, b_{2}-432 b_{2}&=0\\ 540 i \sqrt {3}\, a_{3}-540 a_{3}&=0\\ 648 i \sqrt {3}\, a_{1}+648 a_{1}&=0\\ 864 i \sqrt {3}\, a_{3}-864 a_{3}&=0\\ 864 i \sqrt {3}\, b_{2}+864 b_{2}&=0\\ 1728 i \sqrt {3}\, a_{1}+1728 a_{1}&=0\\ 5832 i \sqrt {3}\, a_{3}+5832 a_{3}&=0\\ 5832 i \sqrt {3}\, b_{1}+5832 b_{1}&=0\\ -3888 i \sqrt {3}\, a_{2}+2916 i \sqrt {3}\, b_{3}+3888 a_{2}-2916 b_{3}&=0\\ 432 i \sqrt {3}\, a_{2}-324 i \sqrt {3}\, b_{3}-432 a_{2}+324 b_{3}&=0\\ 576 i \sqrt {3}\, a_{2}-432 i \sqrt {3}\, b_{3}-576 a_{2}+432 b_{3}&=0 \end {align*}
Solving the above equations for the unknowns gives \begin {align*} a_{1}&=0\\ a_{2}&=\frac {3 b_{3}}{4}\\ 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 {3 x}{4} \\ \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} \\ {} & {} & x {y^{\prime }}^{3}-2 y {y^{\prime }}^{2}=-4 x^{2} \\ \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 (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}}}{3 x}+\frac {4 y^{2}}{3 x \left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}}}+\frac {2 y}{3 x}, y^{\prime }=-\frac {\left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}}}{6 x}-\frac {2 y^{2}}{3 x \left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}}}+\frac {2 y}{3 x}-\frac {\mathrm {I} \sqrt {3}\, \left (\frac {\left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}}}{3 x}-\frac {4 y^{2}}{3 x \left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}}}\right )}{2}, y^{\prime }=-\frac {\left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}}}{6 x}-\frac {2 y^{2}}{3 x \left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}}}+\frac {2 y}{3 x}+\frac {\mathrm {I} \sqrt {3}\, \left (\frac {\left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}}}{3 x}-\frac {4 y^{2}}{3 x \left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}}}\right )}{2}\right ] \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {\left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}}}{3 x}+\frac {4 y^{2}}{3 x \left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}}}+\frac {2 y}{3 x} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {\left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}}}{6 x}-\frac {2 y^{2}}{3 x \left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}}}+\frac {2 y}{3 x}-\frac {\mathrm {I} \sqrt {3}\, \left (\frac {\left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}}}{3 x}-\frac {4 y^{2}}{3 x \left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}}}\right )}{2} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {\left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}}}{6 x}-\frac {2 y^{2}}{3 x \left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}}}+\frac {2 y}{3 x}+\frac {\mathrm {I} \sqrt {3}\, \left (\frac {\left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\right )^{\frac {1}{3}}}{3 x}-\frac {4 y^{2}}{3 x \left (-54 x^{4}+8 y^{3}+6 \sqrt {81 x^{4}-24 y^{3}}\, x^{2}\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 dAlembert 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 -> Calling odsolve with the ODE`, diff(y(x), x) = (-3*x^5-3*(x^4+32*y(x))^(1/2)*x^3-32*x*y(x))/(16*x^2+8*(x^4+32*y(x))^(1/2)), y( Methods for first order ODEs: --- Trying classification methods --- trying homogeneous types: trying homogeneous G 1st order, trying the canonical coordinates of the invariance group <- 1st order, canonical coordinates successful <- homogeneous successful -> Calling odsolve with the ODE`, diff(y(x), x) = (-3*x^5+3*(x^4+32*y(x))^(1/2)*x^3-32*x*y(x))/(16*x^2-8*(x^4+32*y(x))^(1/2)), y( Methods for first order ODEs: --- Trying classification methods --- trying homogeneous types: trying homogeneous G 1st order, trying the canonical coordinates of the invariance group <- 1st order, canonical coordinates successful <- homogeneous successful <- 1st order, parametric methods successful`
✓ Solution by Maple
Time used: 0.156 (sec). Leaf size: 800
dsolve(x*diff(y(x),x)^3-2*y(x)*diff(y(x),x)^2+4*x^2 = 0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \frac {3 x^{\frac {4}{3}}}{2} \\ y \left (x \right ) &= -\frac {3 x^{\frac {4}{3}} \left (1+i \sqrt {3}\right )}{4} \\ y \left (x \right ) &= \frac {3 x^{\frac {4}{3}} \left (-1+i \sqrt {3}\right )}{4} \\ y \left (x \right ) &= \frac {c_{1}^{3}+128 x^{2}}{32 c_{1}} \\ y \left (x \right ) &= \frac {c_{1}^{3}-128 x^{2}}{32 c_{1}} \\ y \left (x \right ) &= \frac {c_{1} \left (-1728 x^{2}+c_{1}^{3}+24 \sqrt {6}\, \sqrt {-x^{2} \left (c_{1}^{3}-864 x^{2}\right )}\right )^{\frac {1}{3}}}{96}+\frac {c_{1}^{3}}{96 \left (-1728 x^{2}+c_{1}^{3}+24 \sqrt {6}\, \sqrt {-x^{2} \left (c_{1}^{3}-864 x^{2}\right )}\right )^{\frac {1}{3}}}+\frac {c_{1}^{2}}{96} \\ y \left (x \right ) &= \frac {c_{1} \left (c_{1}^{3}+24 \sqrt {6}\, \sqrt {x^{2} \left (c_{1}^{3}+864 x^{2}\right )}+1728 x^{2}\right )^{\frac {1}{3}}}{96}+\frac {c_{1}^{3}}{96 \left (c_{1}^{3}+24 \sqrt {6}\, \sqrt {x^{2} \left (c_{1}^{3}+864 x^{2}\right )}+1728 x^{2}\right )^{\frac {1}{3}}}+\frac {c_{1}^{2}}{96} \\ y \left (x \right ) &= \frac {\left (c_{1} -\left (-1728 x^{2}+c_{1}^{3}+24 \sqrt {6}\, \sqrt {-c_{1}^{3} x^{2}+864 x^{4}}\right )^{\frac {1}{3}}\right ) c_{1} \left (i \left (\left (-1728 x^{2}+c_{1}^{3}+24 \sqrt {6}\, \sqrt {-c_{1}^{3} x^{2}+864 x^{4}}\right )^{\frac {1}{3}}+c_{1} \right ) \sqrt {3}-c_{1} +\left (-1728 x^{2}+c_{1}^{3}+24 \sqrt {6}\, \sqrt {-c_{1}^{3} x^{2}+864 x^{4}}\right )^{\frac {1}{3}}\right )}{192 \left (-1728 x^{2}+c_{1}^{3}+24 \sqrt {6}\, \sqrt {-c_{1}^{3} x^{2}+864 x^{4}}\right )^{\frac {1}{3}}} \\ y \left (x \right ) &= \frac {\left (-1+i \sqrt {3}\right ) c_{1} \left (-1728 x^{2}+c_{1}^{3}+24 \sqrt {3}\, \sqrt {2}\, \sqrt {-c_{1}^{3} x^{2}+864 x^{4}}\right )^{\frac {1}{3}}}{192}-\frac {\left (i \sqrt {3}\, c_{1} +c_{1} -2 \left (-1728 x^{2}+c_{1}^{3}+24 \sqrt {3}\, \sqrt {2}\, \sqrt {-c_{1}^{3} x^{2}+864 x^{4}}\right )^{\frac {1}{3}}\right ) c_{1}^{2}}{192 \left (-1728 x^{2}+c_{1}^{3}+24 \sqrt {3}\, \sqrt {2}\, \sqrt {-c_{1}^{3} x^{2}+864 x^{4}}\right )^{\frac {1}{3}}} \\ y \left (x \right ) &= \frac {\left (c_{1} -\left (c_{1}^{3}+24 \sqrt {6}\, \sqrt {c_{1}^{3} x^{2}+864 x^{4}}+1728 x^{2}\right )^{\frac {1}{3}}\right ) c_{1} \left (i \left (\left (c_{1}^{3}+24 \sqrt {6}\, \sqrt {c_{1}^{3} x^{2}+864 x^{4}}+1728 x^{2}\right )^{\frac {1}{3}}+c_{1} \right ) \sqrt {3}-c_{1} +\left (c_{1}^{3}+24 \sqrt {6}\, \sqrt {c_{1}^{3} x^{2}+864 x^{4}}+1728 x^{2}\right )^{\frac {1}{3}}\right )}{192 \left (c_{1}^{3}+24 \sqrt {6}\, \sqrt {c_{1}^{3} x^{2}+864 x^{4}}+1728 x^{2}\right )^{\frac {1}{3}}} \\ y \left (x \right ) &= \frac {\left (-1+i \sqrt {3}\right ) c_{1} \left (c_{1}^{3}+24 \sqrt {3}\, \sqrt {2}\, \sqrt {c_{1}^{3} x^{2}+864 x^{4}}+1728 x^{2}\right )^{\frac {1}{3}}}{192}-\frac {\left (i \sqrt {3}\, c_{1} +c_{1} -2 \left (c_{1}^{3}+24 \sqrt {3}\, \sqrt {2}\, \sqrt {c_{1}^{3} x^{2}+864 x^{4}}+1728 x^{2}\right )^{\frac {1}{3}}\right ) c_{1}^{2}}{192 \left (c_{1}^{3}+24 \sqrt {3}\, \sqrt {2}\, \sqrt {c_{1}^{3} x^{2}+864 x^{4}}+1728 x^{2}\right )^{\frac {1}{3}}} \\ \end{align*}
✓ Solution by Mathematica
Time used: 169.538 (sec). Leaf size: 15120
DSolve[x (y'[x])^3 - 2 y[x](y'[x])^2 + 4 x^2==0,y[x],x,IncludeSingularSolutions -> True]
Too large to display