Internal problem ID [6858]
Internal file name [OUTPUT/6105_Friday_July_29_2022_03_09_39_AM_93402856/index.tex]
Book: Elementary differential equations. By Earl D. Rainville, Phillip E. Bedient. Macmilliam
Publishing Co. NY. 6th edition. 1981.
Section: CHAPTER 16. Nonlinear equations. Section 101. Independent variable missing.
EXERCISES Page 324
Problem number: 41.
ODE order: 2.
ODE degree: 3.
The type(s) of ODE detected by this program : "second_order_ode_missing_y"
Maple gives the following as the ode type
[[_2nd_order, _missing_y]]
\[ \boxed {{y^{\prime \prime }}^{3}-12 y^{\prime } \left (x y^{\prime \prime }-2 y^{\prime }\right )=0} \]
This is second order ode with missing dependent variable \(y\). Let \begin {align*} p(x) &= y^{\prime } \end {align*}
Then \begin {align*} p'(x) &= y^{\prime \prime } \end {align*}
Hence the ode becomes \begin {align*} \left (-12 p \left (x \right ) x +{p^{\prime }\left (x \right )}^{2}\right ) p^{\prime }\left (x \right )+24 p \left (x \right )^{2} = 0 \end {align*}
Which is now solve for \(p(x)\) as first order ode. Solving the given ode for \(p^{\prime }\left (x \right )\) results in \(3\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} p^{\prime }\left (x \right )&=\left (-12 p \left (x \right )^{2}+4 \sqrt {-4 p \left (x \right )^{3} x^{3}+9 p \left (x \right )^{4}}\right )^{\frac {1}{3}}+\frac {4 p \left (x \right ) x}{\left (-12 p \left (x \right )^{2}+4 \sqrt {-4 p \left (x \right )^{3} x^{3}+9 p \left (x \right )^{4}}\right )^{\frac {1}{3}}} \tag {1} \\ p^{\prime }\left (x \right )&=-\frac {\left (-12 p \left (x \right )^{2}+4 \sqrt {-4 p \left (x \right )^{3} x^{3}+9 p \left (x \right )^{4}}\right )^{\frac {1}{3}}}{2}-\frac {2 p \left (x \right ) x}{\left (-12 p \left (x \right )^{2}+4 \sqrt {-4 p \left (x \right )^{3} x^{3}+9 p \left (x \right )^{4}}\right )^{\frac {1}{3}}}+i \sqrt {3}\, \left (\frac {\left (-12 p \left (x \right )^{2}+4 \sqrt {-4 p \left (x \right )^{3} x^{3}+9 p \left (x \right )^{4}}\right )^{\frac {1}{3}}}{2}-\frac {2 p \left (x \right ) x}{\left (-12 p \left (x \right )^{2}+4 \sqrt {-4 p \left (x \right )^{3} x^{3}+9 p \left (x \right )^{4}}\right )^{\frac {1}{3}}}\right ) \tag {2} \\ p^{\prime }\left (x \right )&=-\frac {\left (-12 p \left (x \right )^{2}+4 \sqrt {-4 p \left (x \right )^{3} x^{3}+9 p \left (x \right )^{4}}\right )^{\frac {1}{3}}}{2}-\frac {2 p \left (x \right ) x}{\left (-12 p \left (x \right )^{2}+4 \sqrt {-4 p \left (x \right )^{3} x^{3}+9 p \left (x \right )^{4}}\right )^{\frac {1}{3}}}-i \sqrt {3}\, \left (\frac {\left (-12 p \left (x \right )^{2}+4 \sqrt {-4 p \left (x \right )^{3} x^{3}+9 p \left (x \right )^{4}}\right )^{\frac {1}{3}}}{2}-\frac {2 p \left (x \right ) x}{\left (-12 p \left (x \right )^{2}+4 \sqrt {-4 p \left (x \right )^{3} x^{3}+9 p \left (x \right )^{4}}\right )^{\frac {1}{3}}}\right ) \tag {3} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Writing the ode as \begin {align*} p^{\prime }\left (x \right )&=\frac {\left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}}+4 p x}{\left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {1}{3}}}\\ p^{\prime }\left (x \right )&= \omega \left ( x,p\right ) \end {align*}
The condition of Lie symmetry is the linearized PDE given by \begin {align*} \eta _{x}+\omega \left ( \eta _{p}-\xi _{x}\right ) -\omega ^{2}\xi _{p}-\omega _{x}\xi -\omega _{p}\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 &= p a_{3}+x a_{2}+a_{1} \\ \tag{2E} \eta &= p b_{3}+x b_{2}+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 (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}}+4 p x \right ) \left (b_{3}-a_{2}\right )}{\left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {1}{3}}}-\frac {{\left (\left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}}+4 p x \right )}^{2} a_{3}}{\left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}}}-\left (\frac {-\frac {16 p^{3} x^{2}}{\left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {1}{3}} \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}}+4 p}{\left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {1}{3}}}+\frac {8 \left (\left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}}+4 p x \right ) p^{3} x^{2}}{\left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {4}{3}} \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}}\right ) \left (p a_{3}+x a_{2}+a_{1}\right )-\left (\frac {\frac {-16 p +\frac {2 \left (6 p^{2} \left (-4 x^{3}+9 p \right )+18 p^{3}\right )}{3 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}}}{\left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {1}{3}}}+4 x}{\left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {1}{3}}}-\frac {\left (\left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}}+4 p x \right ) \left (-24 p +\frac {6 p^{2} \left (-4 x^{3}+9 p \right )+18 p^{3}}{\sqrt {p^{3} \left (-4 x^{3}+9 p \right )}}\right )}{3 \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {4}{3}}}\right ) \left (p b_{3}+x b_{2}+b_{1}\right ) = 0 \end{equation} Putting the above in normal form gives \[ -\frac {-96 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\, p^{3} x a_{2}-8 \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}} p^{4} x^{2} a_{3}-8 \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}} p^{3} x^{3} a_{2}-8 \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}} p^{3} x^{3} b_{3}-8 \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}} p^{2} x^{4} b_{2}-8 \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}} p^{3} x^{2} a_{1}-8 \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}} p^{2} x^{3} b_{1}+24 \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}} p^{3} x b_{2}-8 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\, \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}} p^{2} b_{3}+32 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\, p^{3} x b_{3}-16 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\, p^{2} x^{2} b_{2}-8 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\, \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}} p b_{1}-16 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\, p^{2} x b_{1}+16 \left (p^{3} \left (-4 x^{3}+9 p \right )\right )^{\frac {3}{2}} a_{3}-720 p^{6} a_{3}+144 p^{5} a_{1}+96 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\, p^{4} a_{3}-48 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\, p^{3} a_{1}+288 p^{5} x a_{2}+352 p^{5} x^{3} a_{3}-96 p^{4} x^{4} a_{2}+32 p^{4} x^{4} b_{3}-32 p^{3} x^{5} b_{2}-32 p^{4} x^{3} a_{1}-32 p^{3} x^{4} b_{1}-96 p^{5} x b_{3}+48 p^{4} x^{2} b_{2}+48 p^{4} x b_{1}+\sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\, \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {5}{3}} a_{2}-\sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\, \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {5}{3}} b_{3}+24 \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}} p^{4} b_{3}-b_{2} \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {4}{3}} \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}+24 \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}} p^{3} b_{1}+8 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\, \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {4}{3}} p x a_{3}+16 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\, \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}} p^{2} x^{2} a_{3}-8 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\, \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}} p x b_{2}}{\left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {4}{3}} \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}} = 0 \] Setting the numerator to zero gives \begin{equation} \tag{6E} 96 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\, p^{3} x a_{2}+8 \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}} p^{4} x^{2} a_{3}+8 \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}} p^{3} x^{3} a_{2}+8 \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}} p^{3} x^{3} b_{3}+8 \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}} p^{2} x^{4} b_{2}+8 \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}} p^{3} x^{2} a_{1}+8 \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}} p^{2} x^{3} b_{1}-24 \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}} p^{3} x b_{2}+8 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\, \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}} p^{2} b_{3}-32 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\, p^{3} x b_{3}+16 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\, p^{2} x^{2} b_{2}+8 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\, \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}} p b_{1}+16 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\, p^{2} x b_{1}-16 \left (p^{3} \left (-4 x^{3}+9 p \right )\right )^{\frac {3}{2}} a_{3}+720 p^{6} a_{3}-144 p^{5} a_{1}-96 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\, p^{4} a_{3}+48 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\, p^{3} a_{1}-288 p^{5} x a_{2}-352 p^{5} x^{3} a_{3}+96 p^{4} x^{4} a_{2}-32 p^{4} x^{4} b_{3}+32 p^{3} x^{5} b_{2}+32 p^{4} x^{3} a_{1}+32 p^{3} x^{4} b_{1}+96 p^{5} x b_{3}-48 p^{4} x^{2} b_{2}-48 p^{4} x b_{1}-\sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\, \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {5}{3}} a_{2}+\sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\, \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {5}{3}} b_{3}-24 \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}} p^{4} b_{3}+b_{2} \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {4}{3}} \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}-24 \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}} p^{3} b_{1}-8 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\, \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {4}{3}} p x a_{3}-16 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\, \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}} p^{2} x^{2} a_{3}+8 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\, \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}} p x b_{2} = 0 \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 \(\{p, x\}\) in them. \[ \left \{p, x, \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}, \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {1}{3}}, \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}}\right \} \] The following substitution is now made to be able to collect on all terms with \(\{p, x\}\) in them \[ \left \{p = v_{1}, x = v_{2}, \sqrt {p^{3} \left (-4 x^{3}+9 p \right )} = v_{3}, \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {1}{3}} = v_{4}, \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}} = v_{5}\right \} \] The above PDE (6E) now becomes \begin{equation} \tag{7E} -4 v_{1} \left (-32 v_{4} v_{1}^{3} v_{2}^{4} a_{3}-24 v_{1}^{3} v_{2}^{4} a_{2}+88 v_{1}^{4} v_{2}^{3} a_{3}-8 v_{1}^{2} v_{2}^{5} b_{2}+8 v_{1}^{3} v_{2}^{4} b_{3}-8 v_{1}^{3} v_{2}^{3} a_{1}-6 v_{5} v_{1}^{2} v_{2}^{3} a_{2}+72 v_{4} v_{1}^{4} v_{2} a_{3}-2 v_{5} v_{1}^{3} v_{2}^{2} a_{3}-16 v_{3} v_{1}^{2} v_{2}^{3} a_{3}-8 v_{1}^{2} v_{2}^{4} b_{1}+4 v_{4} v_{1}^{2} v_{2}^{3} b_{2}-2 v_{5} v_{1} v_{2}^{4} b_{2}+2 v_{5} v_{1}^{2} v_{2}^{3} b_{3}-2 v_{5} v_{1}^{2} v_{2}^{2} a_{1}+72 v_{1}^{4} v_{2} a_{2}-180 v_{1}^{5} a_{3}-24 v_{3} v_{4} v_{1}^{2} v_{2} a_{3}+4 v_{3} v_{5} v_{1} v_{2}^{2} a_{3}-2 v_{5} v_{1} v_{2}^{3} b_{1}+12 v_{1}^{3} v_{2}^{2} b_{2}-24 v_{1}^{4} v_{2} b_{3}+36 v_{1}^{4} a_{1}+9 v_{5} v_{1}^{3} a_{2}-24 v_{3} v_{1}^{2} v_{2} a_{2}+60 v_{3} v_{1}^{3} a_{3}+12 v_{1}^{3} v_{2} b_{1}-9 v_{4} v_{1}^{3} b_{2}+6 v_{5} v_{1}^{2} v_{2} b_{2}-4 v_{3} v_{1} v_{2}^{2} b_{2}-3 v_{5} v_{1}^{3} b_{3}+8 v_{3} v_{1}^{2} v_{2} b_{3}-12 v_{3} v_{1}^{2} a_{1}-3 v_{3} v_{5} v_{1} a_{2}+6 v_{5} v_{1}^{2} b_{1}-4 v_{3} v_{1} v_{2} b_{1}+3 v_{3} v_{4} v_{1} b_{2}-2 v_{3} v_{5} v_{2} b_{2}+v_{3} v_{5} v_{1} b_{3}-2 v_{3} v_{5} b_{1}\right ) = 0 \end{equation} Collecting the above on the terms \(v_i\) introduced, and these are \[ \{v_{1}, v_{2}, v_{3}, v_{4}, v_{5}\} \] Equation (7E) now becomes \begin{equation} \tag{8E} 32 b_{1} v_{2}^{4} v_{1}^{3}+48 a_{1} v_{3} v_{1}^{3}-24 b_{1} v_{5} v_{1}^{3}-352 a_{3} v_{2}^{3} v_{1}^{5}+\left (-288 a_{2}+96 b_{3}\right ) v_{2} v_{1}^{5}+\left (96 a_{2}-32 b_{3}\right ) v_{2}^{4} v_{1}^{4}+32 a_{1} v_{2}^{3} v_{1}^{4}-48 b_{2} v_{2}^{2} v_{1}^{4}-48 b_{1} v_{2} v_{1}^{4}-240 a_{3} v_{3} v_{1}^{4}+36 b_{2} v_{4} v_{1}^{4}+\left (-36 a_{2}+12 b_{3}\right ) v_{5} v_{1}^{4}+32 b_{2} v_{2}^{5} v_{1}^{3}-144 a_{1} v_{1}^{5}+720 a_{3} v_{1}^{6}-16 a_{3} v_{2}^{2} v_{3} v_{5} v_{1}^{2}+8 b_{2} v_{2} v_{3} v_{5} v_{1}+96 a_{3} v_{2} v_{3} v_{4} v_{1}^{3}+8 a_{3} v_{2}^{2} v_{5} v_{1}^{4}+64 a_{3} v_{2}^{3} v_{3} v_{1}^{3}-16 b_{2} v_{2}^{3} v_{4} v_{1}^{3}+\left (24 a_{2}-8 b_{3}\right ) v_{2}^{3} v_{5} v_{1}^{3}+8 a_{1} v_{2}^{2} v_{5} v_{1}^{3}+\left (96 a_{2}-32 b_{3}\right ) v_{2} v_{3} v_{1}^{3}-24 b_{2} v_{2} v_{5} v_{1}^{3}+8 b_{2} v_{2}^{4} v_{5} v_{1}^{2}+8 b_{1} v_{2}^{3} v_{5} v_{1}^{2}+16 b_{2} v_{2}^{2} v_{3} v_{1}^{2}+16 b_{1} v_{2} v_{3} v_{1}^{2}-12 b_{2} v_{3} v_{4} v_{1}^{2}+\left (12 a_{2}-4 b_{3}\right ) v_{3} v_{5} v_{1}^{2}+8 b_{1} v_{3} v_{5} v_{1}-288 a_{3} v_{2} v_{4} v_{1}^{5}+128 a_{3} v_{2}^{4} v_{4} v_{1}^{4} = 0 \end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve \begin {align*} -144 a_{1}&=0\\ 8 a_{1}&=0\\ 32 a_{1}&=0\\ 48 a_{1}&=0\\ -352 a_{3}&=0\\ -288 a_{3}&=0\\ -240 a_{3}&=0\\ -16 a_{3}&=0\\ 8 a_{3}&=0\\ 64 a_{3}&=0\\ 96 a_{3}&=0\\ 128 a_{3}&=0\\ 720 a_{3}&=0\\ -48 b_{1}&=0\\ -24 b_{1}&=0\\ 8 b_{1}&=0\\ 16 b_{1}&=0\\ 32 b_{1}&=0\\ -48 b_{2}&=0\\ -24 b_{2}&=0\\ -16 b_{2}&=0\\ -12 b_{2}&=0\\ 8 b_{2}&=0\\ 16 b_{2}&=0\\ 32 b_{2}&=0\\ 36 b_{2}&=0\\ -288 a_{2}+96 b_{3}&=0\\ -36 a_{2}+12 b_{3}&=0\\ 12 a_{2}-4 b_{3}&=0\\ 24 a_{2}-8 b_{3}&=0\\ 96 a_{2}-32 b_{3}&=0 \end {align*}
Solving the above equations for the unknowns gives \begin {align*} a_{1}&=0\\ a_{2}&=a_{2}\\ a_{3}&=0\\ b_{1}&=0\\ b_{2}&=0\\ b_{3}&=3 a_{2} \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 &= 3 p \\ \end{align*} The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( x,p\right ) \to \left ( R,S \right )\) 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 p}{\eta } = dS \tag {1} \end {align*}
The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial p}\right ) S(x,p) = 1\). 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 {dp}{dx} &= \frac {\eta }{\xi }\\ &= \frac {3 p}{x}\\ &= \frac {3 p}{x} \end {align*}
This is easily solved to give \begin {align*} p \left (x \right ) = c_{1} x^{3} \end {align*}
Where now the coordinate \(R\) is taken as the constant of integration. Hence \begin {align*} R &= \frac {p}{x^{3}} \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,p) S_{p} }{ R_{x} + \omega (x,p) R_{p} }\tag {2} \end {align*}
Where in the above \(R_{x},R_{p},S_{x},S_{p}\) are all partial derivatives and \(\omega (x,p)\) is the right hand side of the original ode given by \begin {align*} \omega (x,p) &= \frac {\left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}}+4 p x}{\left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {1}{3}}} \end {align*}
Evaluating all the partial derivatives gives \begin {align*} R_{x} &= -\frac {3 p}{x^{4}}\\ R_{p} &= \frac {1}{x^{3}}\\ S_{x} &= \frac {1}{x}\\ S_{p} &= 0 \end {align*}
Substituting all the above in (2) and simplifying gives the ode in canonical coordinates. \begin {align*} \frac {dS}{dR} &= \frac {\left (-12 p^{2}+4 \sqrt {-4 p^{3} x^{3}+9 p^{4}}\right )^{\frac {1}{3}} x^{3}}{\left (-12 p^{2}+4 \sqrt {-4 p^{3} x^{3}+9 p^{4}}\right )^{\frac {2}{3}} x +4 p \left (x^{2}-\frac {3 \left (-12 p^{2}+4 \sqrt {-4 p^{3} x^{3}+9 p^{4}}\right )^{\frac {1}{3}}}{4}\right )}\tag {2A} \end {align*}
We now need to express the RHS as function of \(R\) only. This is done by solving for \(x,p\) in terms of \(R,S\) from the result obtained earlier and simplifying. This gives \begin {align*} \frac {dS}{dR} &= \frac {\left (1+i \sqrt {3}\right ) 2^{\frac {2}{3}} \left (-\sqrt {9 R -4}+3 \sqrt {R}\right )^{\frac {1}{3}}}{\sqrt {R}\, \left (8+2 \left (i \sqrt {3}-1\right ) 2^{\frac {1}{3}} \left (-\sqrt {9 R -4}+3 \sqrt {R}\right )^{\frac {2}{3}}+3 \left (-i \sqrt {3}-1\right ) \sqrt {R}\, 2^{\frac {2}{3}} \left (-\sqrt {9 R -4}+3 \sqrt {R}\right )^{\frac {1}{3}}\right )} \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 {\left (1+i \sqrt {3}\right ) \left (-4 \sqrt {9 R -4}+12 \sqrt {R}\right )^{\frac {1}{3}}}{\left (2 \,2^{\frac {1}{3}} \left (\left (-\sqrt {9 R -4}+3 \sqrt {R}\right )^{2}\right )^{\frac {1}{3}}+3 i \sqrt {R}\, \sqrt {3}\, \left (-4 \sqrt {9 R -4}+12 \sqrt {R}\right )^{\frac {1}{3}}-2 i \sqrt {3}\, 2^{\frac {1}{3}} \left (\left (-\sqrt {9 R -4}+3 \sqrt {R}\right )^{2}\right )^{\frac {1}{3}}+3 \sqrt {R}\, \left (-4 \sqrt {9 R -4}+12 \sqrt {R}\right )^{\frac {1}{3}}-8\right ) \sqrt {R}}d R +c_{1}\tag {4} \end {align*}
To complete the solution, we just need to transform (4) back to \(x,p\) coordinates. This results in \begin {align*} \ln \left (x \right ) = \int _{}^{\frac {p \left (x \right )}{x^{3}}}-\frac {\left (1+i \sqrt {3}\right ) \left (-4 \sqrt {9 \textit {\_a} -4}+12 \sqrt {\textit {\_a}}\right )^{\frac {1}{3}}}{\left (2 \,2^{\frac {1}{3}} \left (\left (-\sqrt {9 \textit {\_a} -4}+3 \sqrt {\textit {\_a}}\right )^{2}\right )^{\frac {1}{3}}+3 i \sqrt {\textit {\_a}}\, \sqrt {3}\, \left (-4 \sqrt {9 \textit {\_a} -4}+12 \sqrt {\textit {\_a}}\right )^{\frac {1}{3}}-2 i \sqrt {3}\, 2^{\frac {1}{3}} \left (\left (-\sqrt {9 \textit {\_a} -4}+3 \sqrt {\textit {\_a}}\right )^{2}\right )^{\frac {1}{3}}+3 \sqrt {\textit {\_a}}\, \left (-4 \sqrt {9 \textit {\_a} -4}+12 \sqrt {\textit {\_a}}\right )^{\frac {1}{3}}-8\right ) \sqrt {\textit {\_a}}}d \textit {\_a} +c_{1} \end {align*}
Which simplifies to \begin {align*} 2^{\frac {2}{3}} \left (1+i \sqrt {3}\right ) \left (\int _{}^{\frac {p \left (x \right )}{x^{3}}}-\frac {\left (-\sqrt {9 \textit {\_a} -4}+3 \sqrt {\textit {\_a}}\right )^{\frac {1}{3}}}{\sqrt {\textit {\_a}}\, \left (\left (2 i \sqrt {3}-2\right ) 2^{\frac {1}{3}} \left (\sqrt {9 \textit {\_a} -4}-3 \sqrt {\textit {\_a}}\right )^{\frac {2}{3}}+8-3 \sqrt {\textit {\_a}}\, 2^{\frac {2}{3}} \left (1+i \sqrt {3}\right ) \left (-\sqrt {9 \textit {\_a} -4}+3 \sqrt {\textit {\_a}}\right )^{\frac {1}{3}}\right )}d \textit {\_a} \right )+\ln \left (x \right )-c_{1} = 0 \end {align*}
Solving equation (2)
Writing the ode as \begin {align*} p^{\prime }\left (x \right )&=-\frac {2 \left (2 i \sqrt {3}\, p x -2 p x +\left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}}\right )}{\left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {1}{3}} \left (1+i \sqrt {3}\right )}\\ p^{\prime }\left (x \right )&= \omega \left ( x,p\right ) \end {align*}
The condition of Lie symmetry is the linearized PDE given by \begin {align*} \eta _{x}+\omega \left ( \eta _{p}-\xi _{x}\right ) -\omega ^{2}\xi _{p}-\omega _{x}\xi -\omega _{p}\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 &= p a_{3}+x a_{2}+a_{1} \\ \tag{2E} \eta &= p b_{3}+x b_{2}+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 {2 \left (2 i \sqrt {3}\, p x -2 p x +\left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}}\right ) \left (b_{3}-a_{2}\right )}{\left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {1}{3}} \left (1+i \sqrt {3}\right )}-\frac {4 {\left (2 i \sqrt {3}\, p x -2 p x +\left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}}\right )}^{2} a_{3}}{\left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}} \left (1+i \sqrt {3}\right )^{2}}-\left (-\frac {2 \left (2 i \sqrt {3}\, p -2 p -\frac {16 p^{3} x^{2}}{\left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {1}{3}} \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}}\right )}{\left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {1}{3}} \left (1+i \sqrt {3}\right )}-\frac {16 \left (2 i \sqrt {3}\, p x -2 p x +\left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}}\right ) p^{3} x^{2}}{\left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {4}{3}} \left (1+i \sqrt {3}\right ) \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}}\right ) \left (p a_{3}+x a_{2}+a_{1}\right )-\left (-\frac {2 \left (2 i x \sqrt {3}-2 x +\frac {-16 p +\frac {2 \left (6 p^{2} \left (-4 x^{3}+9 p \right )+18 p^{3}\right )}{3 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}}}{\left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {1}{3}}}\right )}{\left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {1}{3}} \left (1+i \sqrt {3}\right )}+\frac {2 \left (2 i \sqrt {3}\, p x -2 p x +\left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}}\right ) \left (-24 p +\frac {6 p^{2} \left (-4 x^{3}+9 p \right )+18 p^{3}}{\sqrt {p^{3} \left (-4 x^{3}+9 p \right )}}\right )}{3 \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {4}{3}} \left (1+i \sqrt {3}\right )}\right ) \left (p b_{3}+x b_{2}+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 \(\{p, x\}\) in them. \[ \left \{p, x, \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}, \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {1}{3}}, \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}}\right \} \] The following substitution is now made to be able to collect on all terms with \(\{p, x\}\) in them \[ \left \{p = v_{1}, x = v_{2}, \sqrt {p^{3} \left (-4 x^{3}+9 p \right )} = v_{3}, \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {1}{3}} = v_{4}, \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}} = v_{5}\right \} \] The above PDE (6E) now becomes \begin{equation} \tag{7E} 8 v_{1} \left (360 v_{1}^{5} a_{3}-72 v_{1}^{4} a_{1}-24 v_{1}^{3} v_{2}^{2} b_{2}-24 v_{1}^{3} v_{2} b_{1}-144 v_{1}^{4} v_{2} a_{2}-176 v_{1}^{4} v_{2}^{3} a_{3}+48 v_{1}^{3} v_{2}^{4} a_{2}-16 v_{1}^{3} v_{2}^{4} b_{3}+16 v_{1}^{2} v_{2}^{5} b_{2}+16 v_{1}^{3} v_{2}^{3} a_{1}+16 v_{1}^{2} v_{2}^{4} b_{1}+48 v_{1}^{4} v_{2} b_{3}+9 v_{5} v_{1}^{3} a_{2}-3 v_{5} v_{1}^{3} b_{3}-120 v_{3} v_{1}^{3} a_{3}+6 v_{5} v_{1}^{2} b_{1}-9 v_{4} v_{1}^{3} b_{2}-2 v_{3} v_{5} b_{1}+24 v_{3} v_{1}^{2} a_{1}+4 i \sqrt {3}\, v_{3} v_{5} v_{1} v_{2}^{2} a_{3}+24 i \sqrt {3}\, v_{3} v_{4} v_{1}^{2} v_{2} a_{3}-2 i \sqrt {3}\, v_{3} v_{5} b_{1}+4 v_{3} v_{5} v_{1} v_{2}^{2} a_{3}-24 v_{3} v_{4} v_{1}^{2} v_{2} a_{3}+9 i \sqrt {3}\, v_{5} v_{1}^{3} a_{2}-3 i \sqrt {3}\, v_{5} v_{1}^{3} b_{3}+6 i \sqrt {3}\, v_{5} v_{1}^{2} b_{1}+9 i \sqrt {3}\, v_{4} v_{1}^{3} b_{2}-4 i \sqrt {3}\, v_{4} v_{1}^{2} v_{2}^{3} b_{2}+6 i \sqrt {3}\, v_{5} v_{1}^{2} v_{2} b_{2}-3 i \sqrt {3}\, v_{3} v_{5} v_{1} a_{2}-2 i \sqrt {3}\, v_{3} v_{5} v_{2} b_{2}-3 i \sqrt {3}\, v_{3} v_{4} v_{1} b_{2}+i \sqrt {3}\, v_{3} v_{5} v_{1} b_{3}+32 i \sqrt {3}\, v_{4} v_{1}^{3} v_{2}^{4} a_{3}-2 i \sqrt {3}\, v_{5} v_{1}^{3} v_{2}^{2} a_{3}-6 i \sqrt {3}\, v_{5} v_{1}^{2} v_{2}^{3} a_{2}+2 i \sqrt {3}\, v_{5} v_{1}^{2} v_{2}^{3} b_{3}-2 i \sqrt {3}\, v_{5} v_{1} v_{2}^{4} b_{2}-2 i \sqrt {3}\, v_{5} v_{1}^{2} v_{2}^{2} a_{1}-2 i \sqrt {3}\, v_{5} v_{1} v_{2}^{3} b_{1}-72 i \sqrt {3}\, v_{4} v_{1}^{4} v_{2} a_{3}-2 v_{5} v_{1}^{3} v_{2}^{2} a_{3}-6 v_{5} v_{1}^{2} v_{2}^{3} a_{2}+2 v_{5} v_{1}^{2} v_{2}^{3} b_{3}-2 v_{5} v_{1} v_{2}^{4} b_{2}+32 v_{3} v_{1}^{2} v_{2}^{3} a_{3}-2 v_{5} v_{1}^{2} v_{2}^{2} a_{1}-2 v_{5} v_{1} v_{2}^{3} b_{1}+72 v_{4} v_{1}^{4} v_{2} a_{3}+4 v_{4} v_{1}^{2} v_{2}^{3} b_{2}+6 v_{5} v_{1}^{2} v_{2} b_{2}-3 v_{3} v_{5} v_{1} a_{2}+v_{3} v_{5} v_{1} b_{3}-2 v_{3} v_{5} v_{2} b_{2}+48 v_{3} v_{1}^{2} v_{2} a_{2}-16 v_{3} v_{1}^{2} v_{2} b_{3}+8 v_{3} v_{1} v_{2}^{2} b_{2}+3 v_{3} v_{4} v_{1} b_{2}+8 v_{3} v_{1} v_{2} b_{1}-32 v_{4} v_{1}^{3} v_{2}^{4} a_{3}\right ) = 0 \end{equation} Collecting the above on the terms \(v_i\) introduced, and these are \[ \{v_{1}, v_{2}, v_{3}, v_{4}, v_{5}\} \] Equation (7E) now becomes \begin{equation} \tag{8E} \left (48 i \sqrt {3}\, b_{1}+48 b_{1}\right ) v_{5} v_{1}^{3}+\left (72 i \sqrt {3}\, b_{2}-72 b_{2}\right ) v_{4} v_{1}^{4}+128 b_{2} v_{2}^{5} v_{1}^{3}+128 b_{1} v_{2}^{4} v_{1}^{3}+192 a_{1} v_{3} v_{1}^{3}-1408 a_{3} v_{2}^{3} v_{1}^{5}+\left (-1152 a_{2}+384 b_{3}\right ) v_{2} v_{1}^{5}+\left (384 a_{2}-128 b_{3}\right ) v_{2}^{4} v_{1}^{4}+128 a_{1} v_{2}^{3} v_{1}^{4}-192 b_{2} v_{2}^{2} v_{1}^{4}-192 b_{1} v_{2} v_{1}^{4}-960 a_{3} v_{3} v_{1}^{4}+\left (256 i \sqrt {3}\, a_{3}-256 a_{3}\right ) v_{2}^{4} v_{4} v_{1}^{4}+\left (-16 i \sqrt {3}\, a_{3}-16 a_{3}\right ) v_{2}^{2} v_{5} v_{1}^{4}+\left (-32 i \sqrt {3}\, b_{2}+32 b_{2}\right ) v_{2}^{3} v_{4} v_{1}^{3}+\left (-48 i \sqrt {3}\, a_{2}+16 i \sqrt {3}\, b_{3}-48 a_{2}+16 b_{3}\right ) v_{2}^{3} v_{5} v_{1}^{3}+\left (-16 i \sqrt {3}\, a_{1}-16 a_{1}\right ) v_{2}^{2} v_{5} v_{1}^{3}+\left (48 i \sqrt {3}\, b_{2}+48 b_{2}\right ) v_{2} v_{5} v_{1}^{3}+64 b_{2} v_{2}^{2} v_{3} v_{1}^{2}+64 b_{1} v_{2} v_{3} v_{1}^{2}+\left (-16 i \sqrt {3}\, b_{2}-16 b_{2}\right ) v_{2}^{4} v_{5} v_{1}^{2}+\left (-16 i \sqrt {3}\, b_{1}-16 b_{1}\right ) v_{2}^{3} v_{5} v_{1}^{2}+\left (-24 i \sqrt {3}\, b_{2}+24 b_{2}\right ) v_{3} v_{4} v_{1}^{2}+\left (-24 i \sqrt {3}\, a_{2}+8 i \sqrt {3}\, b_{3}-24 a_{2}+8 b_{3}\right ) v_{3} v_{5} v_{1}^{2}+\left (-16 i \sqrt {3}\, b_{1}-16 b_{1}\right ) v_{3} v_{5} v_{1}+\left (-576 i \sqrt {3}\, a_{3}+576 a_{3}\right ) v_{2} v_{4} v_{1}^{5}+256 a_{3} v_{2}^{3} v_{3} v_{1}^{3}+\left (384 a_{2}-128 b_{3}\right ) v_{2} v_{3} v_{1}^{3}+\left (72 i \sqrt {3}\, a_{2}-24 i \sqrt {3}\, b_{3}+72 a_{2}-24 b_{3}\right ) v_{5} v_{1}^{4}+2880 a_{3} v_{1}^{6}-576 a_{1} v_{1}^{5}+\left (192 i \sqrt {3}\, a_{3}-192 a_{3}\right ) v_{2} v_{3} v_{4} v_{1}^{3}+\left (32 i \sqrt {3}\, a_{3}+32 a_{3}\right ) v_{2}^{2} v_{3} v_{5} v_{1}^{2}+\left (-16 i \sqrt {3}\, b_{2}-16 b_{2}\right ) v_{2} v_{3} v_{5} v_{1} = 0 \end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve \begin {align*} -576 a_{1}&=0\\ 128 a_{1}&=0\\ 192 a_{1}&=0\\ -1408 a_{3}&=0\\ -960 a_{3}&=0\\ 256 a_{3}&=0\\ 2880 a_{3}&=0\\ -192 b_{1}&=0\\ 64 b_{1}&=0\\ 128 b_{1}&=0\\ -192 b_{2}&=0\\ 64 b_{2}&=0\\ 128 b_{2}&=0\\ -1152 a_{2}+384 b_{3}&=0\\ 384 a_{2}-128 b_{3}&=0\\ -576 i \sqrt {3}\, a_{3}+576 a_{3}&=0\\ -32 i \sqrt {3}\, b_{2}+32 b_{2}&=0\\ -24 i \sqrt {3}\, b_{2}+24 b_{2}&=0\\ -16 i \sqrt {3}\, a_{1}-16 a_{1}&=0\\ -16 i \sqrt {3}\, a_{3}-16 a_{3}&=0\\ -16 i \sqrt {3}\, b_{1}-16 b_{1}&=0\\ -16 i \sqrt {3}\, b_{2}-16 b_{2}&=0\\ 32 i \sqrt {3}\, a_{3}+32 a_{3}&=0\\ 48 i \sqrt {3}\, b_{1}+48 b_{1}&=0\\ 48 i \sqrt {3}\, b_{2}+48 b_{2}&=0\\ 72 i \sqrt {3}\, b_{2}-72 b_{2}&=0\\ 192 i \sqrt {3}\, a_{3}-192 a_{3}&=0\\ 256 i \sqrt {3}\, a_{3}-256 a_{3}&=0\\ -48 i \sqrt {3}\, a_{2}+16 i \sqrt {3}\, b_{3}-48 a_{2}+16 b_{3}&=0\\ -24 i \sqrt {3}\, a_{2}+8 i \sqrt {3}\, b_{3}-24 a_{2}+8 b_{3}&=0\\ 72 i \sqrt {3}\, a_{2}-24 i \sqrt {3}\, b_{3}+72 a_{2}-24 b_{3}&=0 \end {align*}
Solving the above equations for the unknowns gives \begin {align*} a_{1}&=0\\ a_{2}&=\frac {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 {x}{3} \\ \eta &= p \\ \end{align*} The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( x,p\right ) \to \left ( R,S \right )\) 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 p}{\eta } = dS \tag {1} \end {align*}
The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial p}\right ) S(x,p) = 1\). 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*} p^{\prime }\left (x \right )&=-\frac {2 \left (2 i \sqrt {3}\, p x +2 p x -\left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}}\right )}{\left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {1}{3}} \left (i \sqrt {3}-1\right )}\\ p^{\prime }\left (x \right )&= \omega \left ( x,p\right ) \end {align*}
The condition of Lie symmetry is the linearized PDE given by \begin {align*} \eta _{x}+\omega \left ( \eta _{p}-\xi _{x}\right ) -\omega ^{2}\xi _{p}-\omega _{x}\xi -\omega _{p}\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 &= p a_{3}+x a_{2}+a_{1} \\ \tag{2E} \eta &= p b_{3}+x b_{2}+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 {2 \left (2 i \sqrt {3}\, p x +2 p x -\left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}}\right ) \left (b_{3}-a_{2}\right )}{\left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {1}{3}} \left (i \sqrt {3}-1\right )}-\frac {4 {\left (2 i \sqrt {3}\, p x +2 p x -\left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}}\right )}^{2} a_{3}}{\left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}} \left (i \sqrt {3}-1\right )^{2}}-\left (-\frac {2 \left (2 i \sqrt {3}\, p +2 p +\frac {16 p^{3} x^{2}}{\left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {1}{3}} \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}}\right )}{\left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {1}{3}} \left (i \sqrt {3}-1\right )}-\frac {16 \left (2 i \sqrt {3}\, p x +2 p x -\left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}}\right ) p^{3} x^{2}}{\left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {4}{3}} \left (i \sqrt {3}-1\right ) \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}}\right ) \left (p a_{3}+x a_{2}+a_{1}\right )-\left (-\frac {2 \left (2 i x \sqrt {3}+2 x -\frac {2 \left (-24 p +\frac {6 p^{2} \left (-4 x^{3}+9 p \right )+18 p^{3}}{\sqrt {p^{3} \left (-4 x^{3}+9 p \right )}}\right )}{3 \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {1}{3}}}\right )}{\left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {1}{3}} \left (i \sqrt {3}-1\right )}+\frac {2 \left (2 i \sqrt {3}\, p x +2 p x -\left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}}\right ) \left (-24 p +\frac {6 p^{2} \left (-4 x^{3}+9 p \right )+18 p^{3}}{\sqrt {p^{3} \left (-4 x^{3}+9 p \right )}}\right )}{3 \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {4}{3}} \left (i \sqrt {3}-1\right )}\right ) \left (p b_{3}+x b_{2}+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 \(\{p, x\}\) in them. \[ \left \{p, x, \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}, \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {1}{3}}, \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}}\right \} \] The following substitution is now made to be able to collect on all terms with \(\{p, x\}\) in them \[ \left \{p = v_{1}, x = v_{2}, \sqrt {p^{3} \left (-4 x^{3}+9 p \right )} = v_{3}, \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {1}{3}} = v_{4}, \left (-12 p^{2}+4 \sqrt {p^{3} \left (-4 x^{3}+9 p \right )}\right )^{\frac {2}{3}} = v_{5}\right \} \] The above PDE (6E) now becomes \begin{equation} \tag{7E} -8 v_{1} \left (32 i \sqrt {3}\, v_{4} v_{1}^{3} v_{2}^{4} a_{3}-2 i \sqrt {3}\, v_{5} v_{1}^{3} v_{2}^{2} a_{3}-6 i \sqrt {3}\, v_{5} v_{1}^{2} v_{2}^{3} a_{2}+2 i \sqrt {3}\, v_{5} v_{1}^{2} v_{2}^{3} b_{3}-2 i \sqrt {3}\, v_{5} v_{1} v_{2}^{4} b_{2}-2 i \sqrt {3}\, v_{5} v_{1}^{2} v_{2}^{2} a_{1}-2 i \sqrt {3}\, v_{5} v_{1} v_{2}^{3} b_{1}-72 i \sqrt {3}\, v_{4} v_{1}^{4} v_{2} a_{3}-4 i \sqrt {3}\, v_{4} v_{1}^{2} v_{2}^{3} b_{2}+6 i \sqrt {3}\, v_{5} v_{1}^{2} v_{2} b_{2}-3 i \sqrt {3}\, v_{3} v_{5} v_{1} a_{2}-2 i \sqrt {3}\, v_{3} v_{5} v_{2} b_{2}-3 i \sqrt {3}\, v_{3} v_{4} v_{1} b_{2}+72 v_{1}^{4} a_{1}-360 v_{1}^{5} a_{3}+4 i \sqrt {3}\, v_{3} v_{5} v_{1} v_{2}^{2} a_{3}+24 i \sqrt {3}\, v_{3} v_{4} v_{1}^{2} v_{2} a_{3}+9 v_{4} v_{1}^{3} b_{2}+2 v_{3} v_{5} b_{1}-24 v_{3} v_{1}^{2} a_{1}+24 v_{1}^{3} v_{2}^{2} b_{2}+24 v_{1}^{3} v_{2} b_{1}+144 v_{1}^{4} v_{2} a_{2}+176 v_{1}^{4} v_{2}^{3} a_{3}-48 v_{1}^{3} v_{2}^{4} a_{2}+16 v_{1}^{3} v_{2}^{4} b_{3}-16 v_{1}^{2} v_{2}^{5} b_{2}-16 v_{1}^{3} v_{2}^{3} a_{1}-16 v_{1}^{2} v_{2}^{4} b_{1}-48 v_{1}^{4} v_{2} b_{3}-9 v_{5} v_{1}^{3} a_{2}+3 v_{5} v_{1}^{3} b_{3}+120 v_{3} v_{1}^{3} a_{3}-6 v_{5} v_{1}^{2} b_{1}-2 i \sqrt {3}\, v_{3} v_{5} b_{1}+9 i \sqrt {3}\, v_{5} v_{1}^{3} a_{2}-3 i \sqrt {3}\, v_{5} v_{1}^{3} b_{3}+6 i \sqrt {3}\, v_{5} v_{1}^{2} b_{1}+9 i \sqrt {3}\, v_{4} v_{1}^{3} b_{2}-4 v_{3} v_{5} v_{1} v_{2}^{2} a_{3}+24 v_{3} v_{4} v_{1}^{2} v_{2} a_{3}+i \sqrt {3}\, v_{3} v_{5} v_{1} b_{3}-8 v_{3} v_{1} v_{2} b_{1}+32 v_{4} v_{1}^{3} v_{2}^{4} a_{3}+2 v_{5} v_{1}^{3} v_{2}^{2} a_{3}+6 v_{5} v_{1}^{2} v_{2}^{3} a_{2}-2 v_{5} v_{1}^{2} v_{2}^{3} b_{3}+2 v_{5} v_{1} v_{2}^{4} b_{2}-32 v_{3} v_{1}^{2} v_{2}^{3} a_{3}+2 v_{5} v_{1}^{2} v_{2}^{2} a_{1}+2 v_{5} v_{1} v_{2}^{3} b_{1}-72 v_{4} v_{1}^{4} v_{2} a_{3}-4 v_{4} v_{1}^{2} v_{2}^{3} b_{2}-6 v_{5} v_{1}^{2} v_{2} b_{2}+3 v_{3} v_{5} v_{1} a_{2}-v_{3} v_{5} v_{1} b_{3}+2 v_{3} v_{5} v_{2} b_{2}+16 v_{3} v_{1}^{2} v_{2} b_{3}-8 v_{3} v_{1} v_{2}^{2} b_{2}-3 v_{3} v_{4} v_{1} b_{2}-48 v_{3} v_{1}^{2} v_{2} a_{2}\right ) = 0 \end{equation} Collecting the above on the terms \(v_i\) introduced, and these are \[ \{v_{1}, v_{2}, v_{3}, v_{4}, v_{5}\} \] Equation (7E) now becomes \begin{equation} \tag{8E} 2880 a_{3} v_{1}^{6}-576 a_{1} v_{1}^{5}+\left (-72 i \sqrt {3}\, b_{2}-72 b_{2}\right ) v_{4} v_{1}^{4}+\left (-72 i \sqrt {3}\, a_{2}+24 i \sqrt {3}\, b_{3}+72 a_{2}-24 b_{3}\right ) v_{5} v_{1}^{4}+\left (-48 i \sqrt {3}\, b_{1}+48 b_{1}\right ) v_{5} v_{1}^{3}-1408 a_{3} v_{2}^{3} v_{1}^{5}+\left (-1152 a_{2}+384 b_{3}\right ) v_{2} v_{1}^{5}+\left (384 a_{2}-128 b_{3}\right ) v_{2}^{4} v_{1}^{4}+128 a_{1} v_{2}^{3} v_{1}^{4}-192 b_{2} v_{2}^{2} v_{1}^{4}-192 b_{1} v_{2} v_{1}^{4}-960 a_{3} v_{3} v_{1}^{4}+128 b_{2} v_{2}^{5} v_{1}^{3}+128 b_{1} v_{2}^{4} v_{1}^{3}+192 a_{1} v_{3} v_{1}^{3}+\left (-32 i \sqrt {3}\, a_{3}+32 a_{3}\right ) v_{2}^{2} v_{3} v_{5} v_{1}^{2}+\left (16 i \sqrt {3}\, b_{2}-16 b_{2}\right ) v_{2} v_{3} v_{5} v_{1}+\left (-192 i \sqrt {3}\, a_{3}-192 a_{3}\right ) v_{2} v_{3} v_{4} v_{1}^{3}+\left (32 i \sqrt {3}\, b_{2}+32 b_{2}\right ) v_{2}^{3} v_{4} v_{1}^{3}+\left (48 i \sqrt {3}\, a_{2}-16 i \sqrt {3}\, b_{3}-48 a_{2}+16 b_{3}\right ) v_{2}^{3} v_{5} v_{1}^{3}+\left (16 i \sqrt {3}\, a_{1}-16 a_{1}\right ) v_{2}^{2} v_{5} v_{1}^{3}+\left (-48 i \sqrt {3}\, b_{2}+48 b_{2}\right ) v_{2} v_{5} v_{1}^{3}+\left (16 i \sqrt {3}\, b_{2}-16 b_{2}\right ) v_{2}^{4} v_{5} v_{1}^{2}+\left (16 i \sqrt {3}\, b_{1}-16 b_{1}\right ) v_{2}^{3} v_{5} v_{1}^{2}+\left (24 i \sqrt {3}\, b_{2}+24 b_{2}\right ) v_{3} v_{4} v_{1}^{2}+\left (24 i \sqrt {3}\, a_{2}-8 i \sqrt {3}\, b_{3}-24 a_{2}+8 b_{3}\right ) v_{3} v_{5} v_{1}^{2}+\left (16 i \sqrt {3}\, b_{1}-16 b_{1}\right ) v_{3} v_{5} v_{1}+\left (576 i \sqrt {3}\, a_{3}+576 a_{3}\right ) v_{2} v_{4} v_{1}^{5}+\left (-256 i \sqrt {3}\, a_{3}-256 a_{3}\right ) v_{2}^{4} v_{4} v_{1}^{4}+\left (16 i \sqrt {3}\, a_{3}-16 a_{3}\right ) v_{2}^{2} v_{5} v_{1}^{4}+256 a_{3} v_{2}^{3} v_{3} v_{1}^{3}+\left (384 a_{2}-128 b_{3}\right ) v_{2} v_{3} v_{1}^{3}+64 b_{2} v_{2}^{2} v_{3} v_{1}^{2}+64 b_{1} v_{2} v_{3} v_{1}^{2} = 0 \end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve \begin {align*} -576 a_{1}&=0\\ 128 a_{1}&=0\\ 192 a_{1}&=0\\ -1408 a_{3}&=0\\ -960 a_{3}&=0\\ 256 a_{3}&=0\\ 2880 a_{3}&=0\\ -192 b_{1}&=0\\ 64 b_{1}&=0\\ 128 b_{1}&=0\\ -192 b_{2}&=0\\ 64 b_{2}&=0\\ 128 b_{2}&=0\\ -1152 a_{2}+384 b_{3}&=0\\ 384 a_{2}-128 b_{3}&=0\\ -256 i \sqrt {3}\, a_{3}-256 a_{3}&=0\\ -192 i \sqrt {3}\, a_{3}-192 a_{3}&=0\\ -72 i \sqrt {3}\, b_{2}-72 b_{2}&=0\\ -48 i \sqrt {3}\, b_{1}+48 b_{1}&=0\\ -48 i \sqrt {3}\, b_{2}+48 b_{2}&=0\\ -32 i \sqrt {3}\, a_{3}+32 a_{3}&=0\\ 16 i \sqrt {3}\, a_{1}-16 a_{1}&=0\\ 16 i \sqrt {3}\, a_{3}-16 a_{3}&=0\\ 16 i \sqrt {3}\, b_{1}-16 b_{1}&=0\\ 16 i \sqrt {3}\, b_{2}-16 b_{2}&=0\\ 24 i \sqrt {3}\, b_{2}+24 b_{2}&=0\\ 32 i \sqrt {3}\, b_{2}+32 b_{2}&=0\\ 576 i \sqrt {3}\, a_{3}+576 a_{3}&=0\\ -72 i \sqrt {3}\, a_{2}+24 i \sqrt {3}\, b_{3}+72 a_{2}-24 b_{3}&=0\\ 24 i \sqrt {3}\, a_{2}-8 i \sqrt {3}\, b_{3}-24 a_{2}+8 b_{3}&=0\\ 48 i \sqrt {3}\, a_{2}-16 i \sqrt {3}\, b_{3}-48 a_{2}+16 b_{3}&=0 \end {align*}
Solving the above equations for the unknowns gives \begin {align*} a_{1}&=0\\ a_{2}&=\frac {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 {x}{3} \\ \eta &= p \\ \end{align*} The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( x,p\right ) \to \left ( R,S \right )\) 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 p}{\eta } = dS \tag {1} \end {align*}
The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial p}\right ) S(x,p) = 1\). 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.
For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} 2^{\frac {2}{3}} \left (1+i \sqrt {3}\right ) \left (\int _{}^{\frac {y^{\prime }}{x^{3}}}-\frac {\left (-\sqrt {9 \textit {\_a} -4}+3 \sqrt {\textit {\_a}}\right )^{\frac {1}{3}}}{\sqrt {\textit {\_a}}\, \left (\left (2 i \sqrt {3}-2\right ) 2^{\frac {1}{3}} \left (\sqrt {9 \textit {\_a} -4}-3 \sqrt {\textit {\_a}}\right )^{\frac {2}{3}}+8-3 \sqrt {\textit {\_a}}\, 2^{\frac {2}{3}} \left (1+i \sqrt {3}\right ) \left (-\sqrt {9 \textit {\_a} -4}+3 \sqrt {\textit {\_a}}\right )^{\frac {1}{3}}\right )}d \textit {\_a} \right )+\ln \left (x \right )-c_{1} = 0 \end {align*}
Integrating both sides gives \begin {align*} y = \int \operatorname {RootOf}\left (-i \ln \left (x \right ) \sqrt {3}\, 2^{\frac {1}{3}}+i c_{1} \sqrt {3}\, 2^{\frac {1}{3}}+\ln \left (x \right ) 2^{\frac {1}{3}}-c_{1} 2^{\frac {1}{3}}+8 \left (\int _{}^{\frac {\textit {\_Z}}{x^{3}}}-\frac {i \left (-\sqrt {9 \textit {\_a} -4}+3 \sqrt {\textit {\_a}}\right )^{\frac {1}{3}}}{\left (3 \sqrt {\textit {\_a}}\, \sqrt {3}\, 2^{\frac {2}{3}} \left (-\sqrt {9 \textit {\_a} -4}+3 \sqrt {\textit {\_a}}\right )^{\frac {1}{3}}-2 \sqrt {3}\, 2^{\frac {1}{3}} \left (\sqrt {9 \textit {\_a} -4}-3 \sqrt {\textit {\_a}}\right )^{\frac {2}{3}}-3 i \sqrt {\textit {\_a}}\, 2^{\frac {2}{3}} \left (-\sqrt {9 \textit {\_a} -4}+3 \sqrt {\textit {\_a}}\right )^{\frac {1}{3}}-2 i 2^{\frac {1}{3}} \left (\sqrt {9 \textit {\_a} -4}-3 \sqrt {\textit {\_a}}\right )^{\frac {2}{3}}+8 i\right ) \sqrt {\textit {\_a}}}d \textit {\_a} \right )\right )d x +c_{4} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \int \operatorname {RootOf}\left (-i \ln \left (x \right ) \sqrt {3}\, 2^{\frac {1}{3}}+i c_{1} \sqrt {3}\, 2^{\frac {1}{3}}+\ln \left (x \right ) 2^{\frac {1}{3}}-c_{1} 2^{\frac {1}{3}}+8 \left (\int _{}^{\frac {\textit {\_Z}}{x^{3}}}-\frac {i \left (-\sqrt {9 \textit {\_a} -4}+3 \sqrt {\textit {\_a}}\right )^{\frac {1}{3}}}{\left (3 \sqrt {\textit {\_a}}\, \sqrt {3}\, 2^{\frac {2}{3}} \left (-\sqrt {9 \textit {\_a} -4}+3 \sqrt {\textit {\_a}}\right )^{\frac {1}{3}}-2 \sqrt {3}\, 2^{\frac {1}{3}} \left (\sqrt {9 \textit {\_a} -4}-3 \sqrt {\textit {\_a}}\right )^{\frac {2}{3}}-3 i \sqrt {\textit {\_a}}\, 2^{\frac {2}{3}} \left (-\sqrt {9 \textit {\_a} -4}+3 \sqrt {\textit {\_a}}\right )^{\frac {1}{3}}-2 i 2^{\frac {1}{3}} \left (\sqrt {9 \textit {\_a} -4}-3 \sqrt {\textit {\_a}}\right )^{\frac {2}{3}}+8 i\right ) \sqrt {\textit {\_a}}}d \textit {\_a} \right )\right )d x +c_{4} \\ \end{align*}
Verification of solutions
\[ y = \int \operatorname {RootOf}\left (-i \ln \left (x \right ) \sqrt {3}\, 2^{\frac {1}{3}}+i c_{1} \sqrt {3}\, 2^{\frac {1}{3}}+\ln \left (x \right ) 2^{\frac {1}{3}}-c_{1} 2^{\frac {1}{3}}+8 \left (\int _{}^{\frac {\textit {\_Z}}{x^{3}}}-\frac {i \left (-\sqrt {9 \textit {\_a} -4}+3 \sqrt {\textit {\_a}}\right )^{\frac {1}{3}}}{\left (3 \sqrt {\textit {\_a}}\, \sqrt {3}\, 2^{\frac {2}{3}} \left (-\sqrt {9 \textit {\_a} -4}+3 \sqrt {\textit {\_a}}\right )^{\frac {1}{3}}-2 \sqrt {3}\, 2^{\frac {1}{3}} \left (\sqrt {9 \textit {\_a} -4}-3 \sqrt {\textit {\_a}}\right )^{\frac {2}{3}}-3 i \sqrt {\textit {\_a}}\, 2^{\frac {2}{3}} \left (-\sqrt {9 \textit {\_a} -4}+3 \sqrt {\textit {\_a}}\right )^{\frac {1}{3}}-2 i 2^{\frac {1}{3}} \left (\sqrt {9 \textit {\_a} -4}-3 \sqrt {\textit {\_a}}\right )^{\frac {2}{3}}+8 i\right ) \sqrt {\textit {\_a}}}d \textit {\_a} \right )\right )d x +c_{4} \] Warning, solution could not be verified
Maple trace
`Methods for second order ODEs: *** Sublevel 2 *** Methods for second order ODEs: Successful isolation of d^2y/dx^2: 3 solutions were found. Trying to solve each resulting ODE. *** Sublevel 3 *** Methods for second order ODEs: --- Trying classification methods --- trying 2nd order Liouville trying 2nd order WeierstrassP trying 2nd order JacobiSN differential order: 2; trying a linearization to 3rd order trying 2nd order ODE linearizable_by_differentiation trying 2nd order, 2 integrating factors of the form mu(x,y) trying differential order: 2; missing variables `, `-> Computing symmetries using: way = 3 -> Calling odsolve with the ODE`, diff(_b(_a), _a) = (4*_b(_a)*_a+(-12*_b(_a)^2+4*(_b(_a)^3*(-4*_a^3+9*_b(_a)))^(1/2))^(2/3))/ symmetry methods on request `, `1st order, trying reduction of order with given symmetries:`[_a, 3*_b]
✓ Solution by Maple
Time used: 0.5 (sec). Leaf size: 174
dsolve(diff(y(x),x$2)^3=12*diff(y(x),x)*(x*diff(y(x),x$2)-2*diff(y(x),x)),y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \frac {x^{4}}{9}+c_{1} \\ y \left (x \right ) &= c_{1} \\ y \left (x \right ) &= \int \operatorname {RootOf}\left (-6 \ln \left (x \right )-\left (\int _{}^{\textit {\_Z}}\frac {3 \textit {\_f} \sqrt {\frac {1}{\textit {\_f} \left (9 \textit {\_f} -4\right )}}\, 2^{\frac {1}{3}} \left (\left (3 \sqrt {\frac {1}{\textit {\_f} \left (9 \textit {\_f} -4\right )}}\, \textit {\_f} +1\right )^{2} \left (9 \textit {\_f} -4\right )^{4}\right )^{\frac {1}{3}}-2 \,2^{\frac {2}{3}} \left (\left (3 \sqrt {\frac {1}{\textit {\_f} \left (9 \textit {\_f} -4\right )}}\, \textit {\_f} +1\right ) \left (9 \textit {\_f} -4\right )^{2}\right )^{\frac {1}{3}}-2^{\frac {1}{3}} \left (\left (3 \sqrt {\frac {1}{\textit {\_f} \left (9 \textit {\_f} -4\right )}}\, \textit {\_f} +1\right )^{2} \left (9 \textit {\_f} -4\right )^{4}\right )^{\frac {1}{3}}+18 \textit {\_f} -8}{\textit {\_f} \left (9 \textit {\_f} -4\right )}d \textit {\_f} \right )+6 c_{1} \right ) x^{3}d x +c_{2} \\ \end{align*}
✗ Solution by Mathematica
Time used: 0.0 (sec). Leaf size: 0
DSolve[(y''[x])^3==12*y'[x]*(x*y''[x]-2*y'[x]),y[x],x,IncludeSingularSolutions -> True]
Not solved