2.7.1 problem 1 (eq 98)
Internal
problem
ID
[18255]
Book
:
Elementary
Differential
Equations.
By
Thornton
C.
Fry.
D
Van
Nostrand.
NY.
First
Edition
(1929)
Section
:
Chapter
V.
Singular
solutions.
section
36.
Problems
at
page
99
Problem
number
:
1
(eq
98)
Date
solved
:
Monday, December 23, 2024 at 09:19:33 PM
CAS
classification
:
[[_1st_order, _with_linear_symmetries]]
Solve
\begin{align*} 4 y {y^{\prime }}^{3}-2 {y^{\prime }}^{2} x^{2}+4 x y y^{\prime }+x^{3}&=16 y^{2} \end{align*}
Solving for the derivative gives these ODE’s to solve
\begin{align*}
\tag{1} y^{\prime }&=\frac {\left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y\right )^{{1}/{3}}}{6 y}-\frac {x \left (-x^{3}+12 y^{2}\right )}{6 y \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y\right )^{{1}/{3}}}+\frac {x^{2}}{6 y} \\
\tag{2} y^{\prime }&=-\frac {\left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y\right )^{{1}/{3}}}{12 y}+\frac {x \left (-x^{3}+12 y^{2}\right )}{12 y \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y\right )^{{1}/{3}}}+\frac {x^{2}}{6 y}+\frac {i \sqrt {3}\, \left (\frac {\left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y\right )^{{1}/{3}}}{6 y}+\frac {x \left (-x^{3}+12 y^{2}\right )}{6 y \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y\right )^{{1}/{3}}}\right )}{2} \\
\tag{3} y^{\prime }&=-\frac {\left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y\right )^{{1}/{3}}}{12 y}+\frac {x \left (-x^{3}+12 y^{2}\right )}{12 y \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y\right )^{{1}/{3}}}+\frac {x^{2}}{6 y}-\frac {i \sqrt {3}\, \left (\frac {\left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y\right )^{{1}/{3}}}{6 y}+\frac {x \left (-x^{3}+12 y^{2}\right )}{6 y \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y\right )^{{1}/{3}}}\right )}{2} \\
\end{align*}
Now each of the above is solved
separately.
Solving Eq. (1)
Solving for \(y'\) gives
\begin{align*}
\tag{1} y' &= -\frac {-x^{4}+12 x y^{2}-x^{2} \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y\right )^{{1}/{3}}-\left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y\right )^{{2}/{3}}}{6 y \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y\right )^{{1}/{3}}} \\
\end{align*}
Each of the above ode’s is now solved An ode \(y^{\prime }=f(x,y)\) is isobaric if
\[ f(t x, t^m y) = t^{m-1} f(x,y)\tag {1} \]
Where here
\[ f(x,y) = -\frac {-x^{4}+12 x y^{2}-x^{2} \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y\right )^{{1}/{3}}-\left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y\right )^{{2}/{3}}}{6 y \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y\right )^{{1}/{3}}}\tag {2} \]
\(m\)
is the order of isobaric. Substituting (2) into (1) and solving for \(m\) gives
\[ m = {\frac {3}{2}} \]
Since the ode is
isobaric of order \(m={\frac {3}{2}}\), then the substitution
\begin{align*} y&=u x^m \\ &=u \,x^{{3}/{2}} \end{align*}
Converts the ODE to a separable in \(u \left (x \right )\). Performing this substitution gives
\[ \frac {3 \sqrt {x}\, u \left (x \right )}{2}+x^{{3}/{2}} u^{\prime }\left (x \right ) = -\frac {-x^{4}+12 x^{4} u \left (x \right )^{2}-x^{2} \left (x^{6}-45 x^{6} u \left (x \right )^{2}+432 x^{6} u \left (x \right )^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 x^{9} u \left (x \right )^{2}-1376 x^{9} u \left (x \right )^{4}+6912 x^{9} u \left (x \right )^{6}}\, x^{{3}/{2}} u \left (x \right )\right )^{{1}/{3}}-\left (x^{6}-45 x^{6} u \left (x \right )^{2}+432 x^{6} u \left (x \right )^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 x^{9} u \left (x \right )^{2}-1376 x^{9} u \left (x \right )^{4}+6912 x^{9} u \left (x \right )^{6}}\, x^{{3}/{2}} u \left (x \right )\right )^{{2}/{3}}}{6 x^{{3}/{2}} u \left (x \right ) \left (x^{6}-45 x^{6} u \left (x \right )^{2}+432 x^{6} u \left (x \right )^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 x^{9} u \left (x \right )^{2}-1376 x^{9} u \left (x \right )^{4}+6912 x^{9} u \left (x \right )^{6}}\, x^{{3}/{2}} u \left (x \right )\right )^{{1}/{3}}} \]
The ode \(u^{\prime }\left (x \right ) = \frac {\left (-9 \,3^{{1}/{3}} {\left (\sqrt {3}\, \left (432 \sqrt {3}\, u \left (x \right )^{4}-45 \sqrt {3}\, u \left (x \right )^{2}+9 u \left (x \right ) \sqrt {\left (27 u \left (x \right )^{2}-2\right ) \left (4 u \left (x \right )-1\right )^{2} \left (4 u \left (x \right )+1\right )^{2}}+\sqrt {3}\right )\right )}^{{1}/{3}} u \left (x \right )^{2}-12 u \left (x \right )^{2} 3^{{2}/{3}}+{\left (\sqrt {3}\, \left (432 \sqrt {3}\, u \left (x \right )^{4}-45 \sqrt {3}\, u \left (x \right )^{2}+9 u \left (x \right ) \sqrt {\left (27 u \left (x \right )^{2}-2\right ) \left (4 u \left (x \right )-1\right )^{2} \left (4 u \left (x \right )+1\right )^{2}}+\sqrt {3}\right )\right )}^{{2}/{3}}+3^{{1}/{3}} {\left (\sqrt {3}\, \left (432 \sqrt {3}\, u \left (x \right )^{4}-45 \sqrt {3}\, u \left (x \right )^{2}+9 u \left (x \right ) \sqrt {\left (27 u \left (x \right )^{2}-2\right ) \left (4 u \left (x \right )-1\right )^{2} \left (4 u \left (x \right )+1\right )^{2}}+\sqrt {3}\right )\right )}^{{1}/{3}}+3^{{2}/{3}}\right ) 3^{{2}/{3}}}{18 {\left (\sqrt {3}\, \left (432 \sqrt {3}\, u \left (x \right )^{4}-45 \sqrt {3}\, u \left (x \right )^{2}+9 u \left (x \right ) \sqrt {\left (27 u \left (x \right )^{2}-2\right ) \left (4 u \left (x \right )-1\right )^{2} \left (4 u \left (x \right )+1\right )^{2}}+\sqrt {3}\right )\right )}^{{1}/{3}} u \left (x \right ) x}\) is
separable as it can be written as
\begin{align*} u^{\prime }\left (x \right )&= \frac {\left (-9 \,3^{{1}/{3}} {\left (\sqrt {3}\, \left (432 \sqrt {3}\, u \left (x \right )^{4}-45 \sqrt {3}\, u \left (x \right )^{2}+9 u \left (x \right ) \sqrt {\left (27 u \left (x \right )^{2}-2\right ) \left (4 u \left (x \right )-1\right )^{2} \left (4 u \left (x \right )+1\right )^{2}}+\sqrt {3}\right )\right )}^{{1}/{3}} u \left (x \right )^{2}-12 u \left (x \right )^{2} 3^{{2}/{3}}+{\left (\sqrt {3}\, \left (432 \sqrt {3}\, u \left (x \right )^{4}-45 \sqrt {3}\, u \left (x \right )^{2}+9 u \left (x \right ) \sqrt {\left (27 u \left (x \right )^{2}-2\right ) \left (4 u \left (x \right )-1\right )^{2} \left (4 u \left (x \right )+1\right )^{2}}+\sqrt {3}\right )\right )}^{{2}/{3}}+3^{{1}/{3}} {\left (\sqrt {3}\, \left (432 \sqrt {3}\, u \left (x \right )^{4}-45 \sqrt {3}\, u \left (x \right )^{2}+9 u \left (x \right ) \sqrt {\left (27 u \left (x \right )^{2}-2\right ) \left (4 u \left (x \right )-1\right )^{2} \left (4 u \left (x \right )+1\right )^{2}}+\sqrt {3}\right )\right )}^{{1}/{3}}+3^{{2}/{3}}\right ) 3^{{2}/{3}}}{18 {\left (\sqrt {3}\, \left (432 \sqrt {3}\, u \left (x \right )^{4}-45 \sqrt {3}\, u \left (x \right )^{2}+9 u \left (x \right ) \sqrt {\left (27 u \left (x \right )^{2}-2\right ) \left (4 u \left (x \right )-1\right )^{2} \left (4 u \left (x \right )+1\right )^{2}}+\sqrt {3}\right )\right )}^{{1}/{3}} u \left (x \right ) x}\\ &= f(x) g(u) \end{align*}
Where
\begin{align*} f(x) &= \frac {3^{{2}/{3}}}{18 x}\\ g(u) &= \frac {-9 \,3^{{1}/{3}} {\left (\sqrt {3}\, \left (432 \sqrt {3}\, u^{4}-45 \sqrt {3}\, u^{2}+9 u \sqrt {\left (27 u^{2}-2\right ) \left (4 u -1\right )^{2} \left (4 u +1\right )^{2}}+\sqrt {3}\right )\right )}^{{1}/{3}} u^{2}-12 u^{2} 3^{{2}/{3}}+{\left (\sqrt {3}\, \left (432 \sqrt {3}\, u^{4}-45 \sqrt {3}\, u^{2}+9 u \sqrt {\left (27 u^{2}-2\right ) \left (4 u -1\right )^{2} \left (4 u +1\right )^{2}}+\sqrt {3}\right )\right )}^{{2}/{3}}+3^{{1}/{3}} {\left (\sqrt {3}\, \left (432 \sqrt {3}\, u^{4}-45 \sqrt {3}\, u^{2}+9 u \sqrt {\left (27 u^{2}-2\right ) \left (4 u -1\right )^{2} \left (4 u +1\right )^{2}}+\sqrt {3}\right )\right )}^{{1}/{3}}+3^{{2}/{3}}}{{\left (\sqrt {3}\, \left (432 \sqrt {3}\, u^{4}-45 \sqrt {3}\, u^{2}+9 u \sqrt {\left (27 u^{2}-2\right ) \left (4 u -1\right )^{2} \left (4 u +1\right )^{2}}+\sqrt {3}\right )\right )}^{{1}/{3}} u} \end{align*}
Integrating gives
\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(x) \,dx}\\ \int { \frac {{\left (\sqrt {3}\, \left (432 \sqrt {3}\, u^{4}-45 \sqrt {3}\, u^{2}+9 u \sqrt {\left (27 u^{2}-2\right ) \left (4 u -1\right )^{2} \left (4 u +1\right )^{2}}+\sqrt {3}\right )\right )}^{{1}/{3}} u}{-9 \,3^{{1}/{3}} {\left (\sqrt {3}\, \left (432 \sqrt {3}\, u^{4}-45 \sqrt {3}\, u^{2}+9 u \sqrt {\left (27 u^{2}-2\right ) \left (4 u -1\right )^{2} \left (4 u +1\right )^{2}}+\sqrt {3}\right )\right )}^{{1}/{3}} u^{2}-12 u^{2} 3^{{2}/{3}}+{\left (\sqrt {3}\, \left (432 \sqrt {3}\, u^{4}-45 \sqrt {3}\, u^{2}+9 u \sqrt {\left (27 u^{2}-2\right ) \left (4 u -1\right )^{2} \left (4 u +1\right )^{2}}+\sqrt {3}\right )\right )}^{{2}/{3}}+3^{{1}/{3}} {\left (\sqrt {3}\, \left (432 \sqrt {3}\, u^{4}-45 \sqrt {3}\, u^{2}+9 u \sqrt {\left (27 u^{2}-2\right ) \left (4 u -1\right )^{2} \left (4 u +1\right )^{2}}+\sqrt {3}\right )\right )}^{{1}/{3}}+3^{{2}/{3}}}\,du} &= \int { \frac {3^{{2}/{3}}}{18 x} \,dx}\\ \int _{}^{u \left (x \right )}\frac {{\left (\sqrt {3}\, \left (432 \sqrt {3}\, \tau ^{4}-45 \sqrt {3}\, \tau ^{2}+9 \tau \sqrt {\left (27 \tau ^{2}-2\right ) \left (4 \tau -1\right )^{2} \left (4 \tau +1\right )^{2}}+\sqrt {3}\right )\right )}^{{1}/{3}} \tau }{-9 \,3^{{1}/{3}} {\left (\sqrt {3}\, \left (432 \sqrt {3}\, \tau ^{4}-45 \sqrt {3}\, \tau ^{2}+9 \tau \sqrt {\left (27 \tau ^{2}-2\right ) \left (4 \tau -1\right )^{2} \left (4 \tau +1\right )^{2}}+\sqrt {3}\right )\right )}^{{1}/{3}} \tau ^{2}-12 \tau ^{2} 3^{{2}/{3}}+{\left (\sqrt {3}\, \left (432 \sqrt {3}\, \tau ^{4}-45 \sqrt {3}\, \tau ^{2}+9 \tau \sqrt {\left (27 \tau ^{2}-2\right ) \left (4 \tau -1\right )^{2} \left (4 \tau +1\right )^{2}}+\sqrt {3}\right )\right )}^{{2}/{3}}+3^{{1}/{3}} {\left (\sqrt {3}\, \left (432 \sqrt {3}\, \tau ^{4}-45 \sqrt {3}\, \tau ^{2}+9 \tau \sqrt {\left (27 \tau ^{2}-2\right ) \left (4 \tau -1\right )^{2} \left (4 \tau +1\right )^{2}}+\sqrt {3}\right )\right )}^{{1}/{3}}+3^{{2}/{3}}}d \tau = 3^{{2}/{3}} \ln \left (x^{{1}/{18}}\right )+c_{1} \end{align*}
We now need to find the singular solutions, these are found by finding for what values \(g(u)\) is
zero, since we had to divide by this above. Solving \(g(u)=0\) or \(\frac {-9 \,3^{{1}/{3}} {\left (\sqrt {3}\, \left (432 \sqrt {3}\, u^{4}-45 \sqrt {3}\, u^{2}+9 u \sqrt {\left (27 u^{2}-2\right ) \left (4 u -1\right )^{2} \left (4 u +1\right )^{2}}+\sqrt {3}\right )\right )}^{{1}/{3}} u^{2}-12 u^{2} 3^{{2}/{3}}+{\left (\sqrt {3}\, \left (432 \sqrt {3}\, u^{4}-45 \sqrt {3}\, u^{2}+9 u \sqrt {\left (27 u^{2}-2\right ) \left (4 u -1\right )^{2} \left (4 u +1\right )^{2}}+\sqrt {3}\right )\right )}^{{2}/{3}}+3^{{1}/{3}} {\left (\sqrt {3}\, \left (432 \sqrt {3}\, u^{4}-45 \sqrt {3}\, u^{2}+9 u \sqrt {\left (27 u^{2}-2\right ) \left (4 u -1\right )^{2} \left (4 u +1\right )^{2}}+\sqrt {3}\right )\right )}^{{1}/{3}}+3^{{2}/{3}}}{{\left (\sqrt {3}\, \left (432 \sqrt {3}\, u^{4}-45 \sqrt {3}\, u^{2}+9 u \sqrt {\left (27 u^{2}-2\right ) \left (4 u -1\right )^{2} \left (4 u +1\right )^{2}}+\sqrt {3}\right )\right )}^{{1}/{3}} u}=0\) for \(u \left (x \right )\) gives
\begin{align*} u \left (x \right )&=1 \end{align*}
Now we go over each such singular solution and check if it verifies the ode itself and
any initial conditions given. If it does not then the singular solution will not be
used.
Therefore the solutions found are
\begin{align*} \int _{}^{u \left (x \right )}\frac {{\left (\sqrt {3}\, \left (432 \sqrt {3}\, \tau ^{4}-45 \sqrt {3}\, \tau ^{2}+9 \tau \sqrt {\left (27 \tau ^{2}-2\right ) \left (4 \tau -1\right )^{2} \left (4 \tau +1\right )^{2}}+\sqrt {3}\right )\right )}^{{1}/{3}} \tau }{-9 \,3^{{1}/{3}} {\left (\sqrt {3}\, \left (432 \sqrt {3}\, \tau ^{4}-45 \sqrt {3}\, \tau ^{2}+9 \tau \sqrt {\left (27 \tau ^{2}-2\right ) \left (4 \tau -1\right )^{2} \left (4 \tau +1\right )^{2}}+\sqrt {3}\right )\right )}^{{1}/{3}} \tau ^{2}-12 \tau ^{2} 3^{{2}/{3}}+{\left (\sqrt {3}\, \left (432 \sqrt {3}\, \tau ^{4}-45 \sqrt {3}\, \tau ^{2}+9 \tau \sqrt {\left (27 \tau ^{2}-2\right ) \left (4 \tau -1\right )^{2} \left (4 \tau +1\right )^{2}}+\sqrt {3}\right )\right )}^{{2}/{3}}+3^{{1}/{3}} {\left (\sqrt {3}\, \left (432 \sqrt {3}\, \tau ^{4}-45 \sqrt {3}\, \tau ^{2}+9 \tau \sqrt {\left (27 \tau ^{2}-2\right ) \left (4 \tau -1\right )^{2} \left (4 \tau +1\right )^{2}}+\sqrt {3}\right )\right )}^{{1}/{3}}+3^{{2}/{3}}}d \tau = 3^{{2}/{3}} \ln \left (x^{{1}/{18}}\right )+c_{1}\\ u \left (x \right ) = 1 \end{align*}
Converting \(\int _{}^{u \left (x \right )}\frac {{\left (\sqrt {3}\, \left (432 \sqrt {3}\, \tau ^{4}-45 \sqrt {3}\, \tau ^{2}+9 \tau \sqrt {\left (27 \tau ^{2}-2\right ) \left (4 \tau -1\right )^{2} \left (4 \tau +1\right )^{2}}+\sqrt {3}\right )\right )}^{{1}/{3}} \tau }{-9 \,3^{{1}/{3}} {\left (\sqrt {3}\, \left (432 \sqrt {3}\, \tau ^{4}-45 \sqrt {3}\, \tau ^{2}+9 \tau \sqrt {\left (27 \tau ^{2}-2\right ) \left (4 \tau -1\right )^{2} \left (4 \tau +1\right )^{2}}+\sqrt {3}\right )\right )}^{{1}/{3}} \tau ^{2}-12 \tau ^{2} 3^{{2}/{3}}+{\left (\sqrt {3}\, \left (432 \sqrt {3}\, \tau ^{4}-45 \sqrt {3}\, \tau ^{2}+9 \tau \sqrt {\left (27 \tau ^{2}-2\right ) \left (4 \tau -1\right )^{2} \left (4 \tau +1\right )^{2}}+\sqrt {3}\right )\right )}^{{2}/{3}}+3^{{1}/{3}} {\left (\sqrt {3}\, \left (432 \sqrt {3}\, \tau ^{4}-45 \sqrt {3}\, \tau ^{2}+9 \tau \sqrt {\left (27 \tau ^{2}-2\right ) \left (4 \tau -1\right )^{2} \left (4 \tau +1\right )^{2}}+\sqrt {3}\right )\right )}^{{1}/{3}}+3^{{2}/{3}}}d \tau = \frac {3^{{2}/{3}} \ln \left (x \right )}{18}+c_{1}\) back to \(y\) gives
\begin{align*} \int _{}^{\frac {y}{x^{{3}/{2}}}}\frac {{\left (\sqrt {3}\, \left (432 \sqrt {3}\, \tau ^{4}-45 \sqrt {3}\, \tau ^{2}+9 \tau \sqrt {\left (27 \tau ^{2}-2\right ) \left (4 \tau -1\right )^{2} \left (4 \tau +1\right )^{2}}+\sqrt {3}\right )\right )}^{{1}/{3}} \tau }{-9 \,3^{{1}/{3}} {\left (\sqrt {3}\, \left (432 \sqrt {3}\, \tau ^{4}-45 \sqrt {3}\, \tau ^{2}+9 \tau \sqrt {\left (27 \tau ^{2}-2\right ) \left (4 \tau -1\right )^{2} \left (4 \tau +1\right )^{2}}+\sqrt {3}\right )\right )}^{{1}/{3}} \tau ^{2}-12 \tau ^{2} 3^{{2}/{3}}+{\left (\sqrt {3}\, \left (432 \sqrt {3}\, \tau ^{4}-45 \sqrt {3}\, \tau ^{2}+9 \tau \sqrt {\left (27 \tau ^{2}-2\right ) \left (4 \tau -1\right )^{2} \left (4 \tau +1\right )^{2}}+\sqrt {3}\right )\right )}^{{2}/{3}}+3^{{1}/{3}} {\left (\sqrt {3}\, \left (432 \sqrt {3}\, \tau ^{4}-45 \sqrt {3}\, \tau ^{2}+9 \tau \sqrt {\left (27 \tau ^{2}-2\right ) \left (4 \tau -1\right )^{2} \left (4 \tau +1\right )^{2}}+\sqrt {3}\right )\right )}^{{1}/{3}}+3^{{2}/{3}}}d \tau = \frac {3^{{2}/{3}} \ln \left (x \right )}{18}+c_{1} \end{align*}
Converting \(u \left (x \right ) = 1\) back to \(y\) gives
\begin{align*} \frac {y}{x^{{3}/{2}}} = 1 \end{align*}
Solving for \(y\) gives
\begin{align*}
\int _{}^{\frac {y}{x^{{3}/{2}}}}\frac {{\left (\sqrt {3}\, \left (432 \sqrt {3}\, \tau ^{4}-45 \sqrt {3}\, \tau ^{2}+9 \tau \sqrt {\left (27 \tau ^{2}-2\right ) \left (4 \tau -1\right )^{2} \left (4 \tau +1\right )^{2}}+\sqrt {3}\right )\right )}^{{1}/{3}} \tau }{-9 \,3^{{1}/{3}} {\left (\sqrt {3}\, \left (432 \sqrt {3}\, \tau ^{4}-45 \sqrt {3}\, \tau ^{2}+9 \tau \sqrt {\left (27 \tau ^{2}-2\right ) \left (4 \tau -1\right )^{2} \left (4 \tau +1\right )^{2}}+\sqrt {3}\right )\right )}^{{1}/{3}} \tau ^{2}-12 \tau ^{2} 3^{{2}/{3}}+{\left (\sqrt {3}\, \left (432 \sqrt {3}\, \tau ^{4}-45 \sqrt {3}\, \tau ^{2}+9 \tau \sqrt {\left (27 \tau ^{2}-2\right ) \left (4 \tau -1\right )^{2} \left (4 \tau +1\right )^{2}}+\sqrt {3}\right )\right )}^{{2}/{3}}+3^{{1}/{3}} {\left (\sqrt {3}\, \left (432 \sqrt {3}\, \tau ^{4}-45 \sqrt {3}\, \tau ^{2}+9 \tau \sqrt {\left (27 \tau ^{2}-2\right ) \left (4 \tau -1\right )^{2} \left (4 \tau +1\right )^{2}}+\sqrt {3}\right )\right )}^{{1}/{3}}+3^{{2}/{3}}}d \tau &= \frac {3^{{2}/{3}} \ln \left (x \right )}{18}+c_{1} \\
y &= x^{{3}/{2}} \\
\end{align*}
We now need to find the singular solutions, these are found by finding for
what values \((\frac {\left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y \right )^{{1}/{3}}}{6 y}-\frac {x \left (-x^{3}+12 y^{2}\right )}{6 y \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y \right )^{{1}/{3}}}+\frac {x^{2}}{6 y})\) is zero. These give
\begin{align*}
y&=\operatorname {RootOf}\left (-x^{4}-x^{2} \left (x^{6}-45 x^{3} \textit {\_Z}^{2}+432 \textit {\_Z}^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 \textit {\_Z}^{2} x^{6}-1376 \textit {\_Z}^{4} x^{3}+6912 \textit {\_Z}^{6}}\, \textit {\_Z} \right )^{{1}/{3}}+12 x \,\textit {\_Z}^{2}-\left (x^{6}-45 x^{3} \textit {\_Z}^{2}+432 \textit {\_Z}^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 \textit {\_Z}^{2} x^{6}-1376 \textit {\_Z}^{4} x^{3}+6912 \textit {\_Z}^{6}}\, \textit {\_Z} \right )^{{2}/{3}}\right ) \\
\end{align*}
Now we go over each such singular solution and check if it
verifies the ode itself and any initial conditions given. If it does not then the singular solution
will not be used.
The solution \(\operatorname {RootOf}\left (-x^{4}-x^{2} \left (x^{6}-45 x^{3} \textit {\_Z}^{2}+432 \textit {\_Z}^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 \textit {\_Z}^{2} x^{6}-1376 \textit {\_Z}^{4} x^{3}+6912 \textit {\_Z}^{6}}\, \textit {\_Z} \right )^{{1}/{3}}+12 x \,\textit {\_Z}^{2}-\left (x^{6}-45 x^{3} \textit {\_Z}^{2}+432 \textit {\_Z}^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 \textit {\_Z}^{2} x^{6}-1376 \textit {\_Z}^{4} x^{3}+6912 \textit {\_Z}^{6}}\, \textit {\_Z} \right )^{{2}/{3}}\right )\) will not be used
Solving Eq. (2)
Writing the ode as
\begin{align*} y^{\prime }&=\frac {-i \sqrt {3}\, x^{4}+12 i \sqrt {3}\, y^{2} x +i \sqrt {3}\, \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y \right )^{{2}/{3}}-x^{4}+12 x \,y^{2}+2 x^{2} \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y \right )^{{1}/{3}}-\left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y \right )^{{2}/{3}}}{12 y \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y \right )^{{1}/{3}}}\\ y^{\prime }&= \omega \left ( x,y\right ) \end{align*}
The condition of Lie symmetry is the linearized PDE given by
\begin{align*} \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0\tag {A} \end{align*}
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.
\[
\left \{x, y, \sqrt {-\left (2 x^{3}-27 y^{2}\right ) \left (x^{3}-16 y^{2}\right )^{2}}, \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-\left (2 x^{3}-27 y^{2}\right ) \left (x^{3}-16 y^{2}\right )^{2}}\, y \right )^{{1}/{3}}, \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-\left (2 x^{3}-27 y^{2}\right ) \left (x^{3}-16 y^{2}\right )^{2}}\, y \right )^{{2}/{3}}\right \}
\]
The following substitution is now made to be able to
collect on all terms with \(\{x, y\}\) in them
\[
\left \{x = v_{1}, y = v_{2}, \sqrt {-\left (2 x^{3}-27 y^{2}\right ) \left (x^{3}-16 y^{2}\right )^{2}} = v_{3}, \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-\left (2 x^{3}-27 y^{2}\right ) \left (x^{3}-16 y^{2}\right )^{2}}\, y \right )^{{1}/{3}} = v_{4}, \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-\left (2 x^{3}-27 y^{2}\right ) \left (x^{3}-16 y^{2}\right )^{2}}\, y \right )^{{2}/{3}} = v_{5}\right \}
\]
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*} -20736 a_{1}&=0\\ -48 a_{1}&=0\\ 2160 a_{1}&=0\\ -7344 a_{3}&=0\\ -12 a_{3}&=0\\ 588 a_{3}&=0\\ 20736 a_{3}&=0\\ -1080 b_{1}&=0\\ 24 b_{1}&=0\\ 10368 b_{1}&=0\\ -936 b_{2}&=0\\ 24 b_{2}&=0\\ 3888 b_{2}&=0\\ 62208 b_{2}&=0\\ -995328 \sqrt {3}\, a_{1}&=0\\ -13104 \sqrt {3}\, a_{1}&=0\\ 288 \sqrt {3}\, a_{1}&=0\\ 198144 \sqrt {3}\, a_{1}&=0\\ -446976 \sqrt {3}\, a_{3}&=0\\ -3564 \sqrt {3}\, a_{3}&=0\\ 72 \sqrt {3}\, a_{3}&=0\\ 62640 \sqrt {3}\, a_{3}&=0\\ 995328 \sqrt {3}\, a_{3}&=0\\ -99072 \sqrt {3}\, b_{1}&=0\\ -144 \sqrt {3}\, b_{1}&=0\\ 6552 \sqrt {3}\, b_{1}&=0\\ 497664 \sqrt {3}\, b_{1}&=0\\ -96768 \sqrt {3}\, b_{2}&=0\\ -59760 \sqrt {3}\, b_{2}&=0\\ -144 \sqrt {3}\, b_{2}&=0\\ 5688 \sqrt {3}\, b_{2}&=0\\ 2985984 \sqrt {3}\, b_{2}&=0\\ -31104 a_{2}+20736 b_{3}&=0\\ -72 a_{2}+48 b_{3}&=0\\ 3240 a_{2}-2160 b_{3}&=0\\ -2985984 \sqrt {3}\, a_{1}-8957952 i a_{1}&=0\\ -173376 \sqrt {3}\, a_{1}-520128 i a_{1}&=0\\ -24768 \sqrt {3}\, a_{1}+74304 i a_{1}&=0\\ -180 \sqrt {3}\, a_{1}-540 i a_{1}&=0\\ -108 \sqrt {3}\, a_{1}+324 i a_{1}&=0\\ 3276 \sqrt {3}\, a_{1}-9828 i a_{1}&=0\\ 9396 \sqrt {3}\, a_{1}+28188 i a_{1}&=0\\ 1292544 \sqrt {3}\, a_{1}+3877632 i a_{1}&=0\\ -1492992 \sqrt {3}\, a_{2}+995328 \sqrt {3}\, b_{3}&=0\\ -19656 \sqrt {3}\, a_{2}+13104 \sqrt {3}\, b_{3}&=0\\ 432 \sqrt {3}\, a_{2}-288 \sqrt {3}\, b_{3}&=0\\ 297216 \sqrt {3}\, a_{2}-198144 \sqrt {3}\, b_{3}&=0\\ -10568448 \sqrt {3}\, a_{3}-31705344 i a_{3}&=0\\ -96240 \sqrt {3}\, a_{3}-288720 i a_{3}&=0\\ -13236 \sqrt {3}\, a_{3}+39708 i a_{3}&=0\\ -48 \sqrt {3}\, a_{3}-144 i a_{3}&=0\\ -24 \sqrt {3}\, a_{3}+72 i a_{3}&=0\\ 984 \sqrt {3}\, a_{3}-2952 i a_{3}&=0\\ 3372 \sqrt {3}\, a_{3}+10116 i a_{3}&=0\\ 58176 \sqrt {3}\, a_{3}-174528 i a_{3}&=0\\ 1405440 \sqrt {3}\, a_{3}+4216320 i a_{3}&=0\\ 32845824 \sqrt {3}\, a_{3}+98537472 i a_{3}&=0\\ -580608 \sqrt {3}\, b_{1}-1741824 i b_{1}&=0\\ -4884 \sqrt {3}\, b_{1}-14652 i b_{1}&=0\\ -1092 \sqrt {3}\, b_{1}+3276 i b_{1}&=0\\ 48 \sqrt {3}\, b_{1}-144 i b_{1}&=0\\ 96 \sqrt {3}\, b_{1}+288 i b_{1}&=0\\ 82944 \sqrt {3}\, b_{1}-248832 i b_{1}&=0\\ 85968 \sqrt {3}\, b_{1}+257904 i b_{1}&=0\\ 995328 \sqrt {3}\, b_{1}+2985984 i b_{1}&=0\\ -580608 \sqrt {3}\, b_{2}-1741824 i b_{2}&=0\\ -4884 \sqrt {3}\, b_{2}-14652 i b_{2}&=0\\ -1092 \sqrt {3}\, b_{2}+3276 i b_{2}&=0\\ 48 \sqrt {3}\, b_{2}-144 i b_{2}&=0\\ 96 \sqrt {3}\, b_{2}+288 i b_{2}&=0\\ 82944 \sqrt {3}\, b_{2}-248832 i b_{2}&=0\\ 85968 \sqrt {3}\, b_{2}+257904 i b_{2}&=0\\ 995328 \sqrt {3}\, b_{2}+2985984 i b_{2}&=0\\ -152064 i \sqrt {3}\, a_{3}-152064 a_{3}&=0\\ -62208 i \sqrt {3}\, a_{1}-62208 a_{1}&=0\\ -9936 i \sqrt {3}\, b_{1}-9936 b_{1}&=0\\ -9936 i \sqrt {3}\, b_{2}-9936 b_{2}&=0\\ -1728 i \sqrt {3}\, b_{1}+1728 b_{1}&=0\\ -1728 i \sqrt {3}\, b_{2}+1728 b_{2}&=0\\ -1476 i \sqrt {3}\, a_{1}-1476 a_{1}&=0\\ -1476 i \sqrt {3}\, a_{3}+1476 a_{3}&=0\\ -492 i \sqrt {3}\, a_{3}-492 a_{3}&=0\\ -180 i \sqrt {3}\, b_{1}+180 b_{1}&=0\\ -180 i \sqrt {3}\, b_{2}+180 b_{2}&=0\\ -24 i \sqrt {3}\, a_{1}+24 a_{1}&=0\\ -12 i \sqrt {3}\, b_{1}-12 b_{1}&=0\\ -12 i \sqrt {3}\, b_{2}-12 b_{2}&=0\\ -6 i \sqrt {3}\, a_{3}+6 a_{3}&=0\\ 6 i \sqrt {3}\, a_{3}+6 a_{3}&=0\\ 12 i \sqrt {3}\, b_{1}-12 b_{1}&=0\\ 12 i \sqrt {3}\, b_{2}-12 b_{2}&=0\\ 24 i \sqrt {3}\, a_{1}+24 a_{1}&=0\\ 204 i \sqrt {3}\, a_{3}-204 a_{3}&=0\\ 540 i \sqrt {3}\, a_{1}-540 a_{1}&=0\\ 756 i \sqrt {3}\, b_{1}+756 b_{1}&=0\\ 756 i \sqrt {3}\, b_{2}+756 b_{2}&=0\\ 13104 i \sqrt {3}\, a_{3}+13104 a_{3}&=0\\ 20736 i \sqrt {3}\, a_{1}+20736 a_{1}&=0\\ 20736 i \sqrt {3}\, b_{1}+20736 b_{1}&=0\\ 20736 i \sqrt {3}\, b_{2}+20736 b_{2}&=0\\ 684288 i \sqrt {3}\, a_{3}+684288 a_{3}&=0\\ -5971968 \sqrt {3}\, a_{2}+3981312 \sqrt {3}\, b_{3}-17915904 i a_{2}+11943936 i b_{3}&=0\\ -262224 \sqrt {3}\, a_{2}+174816 \sqrt {3}\, b_{3}-786672 i a_{2}+524448 i b_{3}&=0\\ -74304 \sqrt {3}\, a_{2}+49536 \sqrt {3}\, b_{3}+222912 i a_{2}-148608 i b_{3}&=0\\ -252 \sqrt {3}\, a_{2}+168 \sqrt {3}\, b_{3}-756 i a_{2}+504 i b_{3}&=0\\ -180 \sqrt {3}\, a_{2}+120 \sqrt {3}\, b_{3}+540 i a_{2}-360 i b_{3}&=0\\ 6552 \sqrt {3}\, a_{2}-4368 \sqrt {3}\, b_{3}-19656 i a_{2}+13104 i b_{3}&=0\\ 13536 \sqrt {3}\, a_{2}-9024 \sqrt {3}\, b_{3}+40608 i a_{2}-27072 i b_{3}&=0\\ 248832 \sqrt {3}\, a_{2}-165888 \sqrt {3}\, b_{3}-746496 i a_{2}+497664 i b_{3}&=0\\ 2135808 \sqrt {3}\, a_{2}-1423872 \sqrt {3}\, b_{3}+6407424 i a_{2}-4271616 i b_{3}&=0\\ -124416 i \sqrt {3}\, a_{2}+82944 i \sqrt {3}\, b_{3}-124416 a_{2}+82944 b_{3}&=0\\ -5184 i \sqrt {3}\, a_{2}+3456 i \sqrt {3}\, b_{3}+5184 a_{2}-3456 b_{3}&=0\\ -2160 i \sqrt {3}\, a_{2}+1440 i \sqrt {3}\, b_{3}-2160 a_{2}+1440 b_{3}&=0\\ -36 i \sqrt {3}\, a_{2}+24 i \sqrt {3}\, b_{3}+36 a_{2}-24 b_{3}&=0\\ 36 i \sqrt {3}\, a_{2}-24 i \sqrt {3}\, b_{3}+36 a_{2}-24 b_{3}&=0\\ 1080 i \sqrt {3}\, a_{2}-720 i \sqrt {3}\, b_{3}-1080 a_{2}+720 b_{3}&=0\\ 32400 i \sqrt {3}\, a_{2}-21600 i \sqrt {3}\, b_{3}+32400 a_{2}-21600 b_{3}&=0 \end{align*}
Solving the above equations for the unknowns gives
\begin{align*} a_{1}&=0\\ a_{2}&=\frac {2 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 {2 x}{3} \\
\eta &= y \\
\end{align*}
The next step is to determine the canonical coordinates \(R,S\). The
canonical coordinates map \(\left ( x,y\right ) \to \left ( R,S \right )\) where \(\left ( R,S \right )\) are the canonical coordinates which make the original ode
become a quadrature and hence solved by integration.
The characteristic pde which is used to find the canonical coordinates is
\begin{align*} \frac {d x}{\xi } &= \frac {d y}{\eta } = dS \tag {1} \end{align*}
The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial y}\right ) S(x,y) = 1\). Starting with the first pair of ode’s in (1)
gives an ode to solve for the independent variable \(R\) in the canonical coordinates, where \(S(R)\).
Therefore
\begin{align*} \frac {dy}{dx} &= \frac {\eta }{\xi }\\ &= \frac {y}{\frac {2 x}{3}}\\ &= \frac {3 y}{2 x} \end{align*}
This is easily solved to give
\begin{align*} y = c_{1} x^{{3}/{2}} \end{align*}
Where now the coordinate \(R\) is taken as the constant of integration. Hence
\begin{align*} R &= \frac {y}{x^{{3}/{2}}} \end{align*}
And \(S\) is found from
\begin{align*} dS &= \frac {dx}{\xi } \\ &= \frac {dx}{\frac {2 x}{3}} \end{align*}
Integrating gives
\begin{align*} S &= \int { \frac {dx}{T}}\\ &= \frac {3 \ln \left (x \right )}{2} \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 {-i \sqrt {3}\, x^{4}+12 i \sqrt {3}\, y^{2} x +i \sqrt {3}\, \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y \right )^{{2}/{3}}-x^{4}+12 x \,y^{2}+2 x^{2} \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y \right )^{{1}/{3}}-\left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y \right )^{{2}/{3}}}{12 y \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y \right )^{{1}/{3}}} \end{align*}
Evaluating all the partial derivatives gives
\begin{align*} R_{x} &= -\frac {3 y}{2 x^{{5}/{2}}}\\ R_{y} &= \frac {1}{x^{{3}/{2}}}\\ S_{x} &= \frac {3}{2 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 {18 x^{{3}/{2}} \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 \left (x^{3}-\frac {27 y^{2}}{2}\right ) \left (x^{3}-16 y^{2}\right )^{2}}\, y \right )^{{1}/{3}} y}{\left (-i \sqrt {3}\, x +x \right ) \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 \left (x^{3}-\frac {27 y^{2}}{2}\right ) \left (x^{3}-16 y^{2}\right )^{2}}\, y \right )^{{2}/{3}}+\left (-2 x^{3}+18 y^{2}\right ) \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 \left (x^{3}-\frac {27 y^{2}}{2}\right ) \left (x^{3}-16 y^{2}\right )^{2}}\, y \right )^{{1}/{3}}+x^{2} \left (1+i \sqrt {3}\right ) \left (x^{3}-12 y^{2}\right )}\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 {18 R {\left (\left (48 R^{3}-3 R \right ) \sqrt {3}\, \sqrt {27 R^{2}-2}+432 R^{4}-45 R^{2}+1\right )}^{{1}/{3}}}{\left (-1+i \sqrt {3}\right ) {\left (\left (48 R^{3}-3 R \right ) \sqrt {3}\, \sqrt {27 R^{2}-2}+432 R^{4}-45 R^{2}+1\right )}^{{2}/{3}}+\left (-18 R^{2}+2\right ) {\left (\left (48 R^{3}-3 R \right ) \sqrt {3}\, \sqrt {27 R^{2}-2}+432 R^{4}-45 R^{2}+1\right )}^{{1}/{3}}+12 \left (1+i \sqrt {3}\right ) \left (R^{2}-\frac {1}{12}\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 \(R,S\).
Since the ode has the form \(\frac {d}{d R}S \left (R \right )=f(R)\), then we only need to integrate \(f(R)\).
\begin{align*} \int {dS} &= \int {\frac {18 R {\left (\left (48 R^{3}-3 R \right ) \sqrt {3}\, \sqrt {27 R^{2}-2}+432 R^{4}-45 R^{2}+1\right )}^{{1}/{3}}}{i {\left (\left (48 R^{3}-3 R \right ) \sqrt {3}\, \sqrt {27 R^{2}-2}+432 R^{4}-45 R^{2}+1\right )}^{{2}/{3}} \sqrt {3}+12 i \sqrt {3}\, R^{2}-18 {\left (\left (48 R^{3}-3 R \right ) \sqrt {3}\, \sqrt {27 R^{2}-2}+432 R^{4}-45 R^{2}+1\right )}^{{1}/{3}} R^{2}-i \sqrt {3}-{\left (\left (48 R^{3}-3 R \right ) \sqrt {3}\, \sqrt {27 R^{2}-2}+432 R^{4}-45 R^{2}+1\right )}^{{2}/{3}}+12 R^{2}+2 {\left (\left (48 R^{3}-3 R \right ) \sqrt {3}\, \sqrt {27 R^{2}-2}+432 R^{4}-45 R^{2}+1\right )}^{{1}/{3}}-1}\, dR}\\ S \left (R \right ) &= \int \frac {18 R {\left (\left (48 R^{3}-3 R \right ) \sqrt {3}\, \sqrt {27 R^{2}-2}+432 R^{4}-45 R^{2}+1\right )}^{{1}/{3}}}{i {\left (\left (48 R^{3}-3 R \right ) \sqrt {3}\, \sqrt {27 R^{2}-2}+432 R^{4}-45 R^{2}+1\right )}^{{2}/{3}} \sqrt {3}+12 i \sqrt {3}\, R^{2}-18 {\left (\left (48 R^{3}-3 R \right ) \sqrt {3}\, \sqrt {27 R^{2}-2}+432 R^{4}-45 R^{2}+1\right )}^{{1}/{3}} R^{2}-i \sqrt {3}-{\left (\left (48 R^{3}-3 R \right ) \sqrt {3}\, \sqrt {27 R^{2}-2}+432 R^{4}-45 R^{2}+1\right )}^{{2}/{3}}+12 R^{2}+2 {\left (\left (48 R^{3}-3 R \right ) \sqrt {3}\, \sqrt {27 R^{2}-2}+432 R^{4}-45 R^{2}+1\right )}^{{1}/{3}}-1}d R + c_3 \end{align*}
\begin{align*} S \left (R \right )&= \int \frac {18 R {\left (\left (48 R^{3}-3 R \right ) \sqrt {3}\, \sqrt {27 R^{2}-2}+432 R^{4}-45 R^{2}+1\right )}^{{1}/{3}}}{i {\left (\left (48 R^{3}-3 R \right ) \sqrt {3}\, \sqrt {27 R^{2}-2}+432 R^{4}-45 R^{2}+1\right )}^{{2}/{3}} \sqrt {3}+12 i \sqrt {3}\, R^{2}-18 {\left (\left (48 R^{3}-3 R \right ) \sqrt {3}\, \sqrt {27 R^{2}-2}+432 R^{4}-45 R^{2}+1\right )}^{{1}/{3}} R^{2}-i \sqrt {3}-{\left (\left (48 R^{3}-3 R \right ) \sqrt {3}\, \sqrt {27 R^{2}-2}+432 R^{4}-45 R^{2}+1\right )}^{{2}/{3}}+12 R^{2}+2 {\left (\left (48 R^{3}-3 R \right ) \sqrt {3}\, \sqrt {27 R^{2}-2}+432 R^{4}-45 R^{2}+1\right )}^{{1}/{3}}-1}d R +c_3 \end{align*}
To complete the solution, we just need to transform the above back to \(x,y\) coordinates. This
results in
\begin{align*} \frac {3 \ln \left (x \right )}{2} = \int _{}^{\frac {y}{x^{{3}/{2}}}}\frac {18 \textit {\_a} {\left (\left (48 \textit {\_a}^{3}-3 \textit {\_a} \right ) \sqrt {3}\, \sqrt {27 \textit {\_a}^{2}-2}+432 \textit {\_a}^{4}-45 \textit {\_a}^{2}+1\right )}^{{1}/{3}}}{i {\left (\left (48 \textit {\_a}^{3}-3 \textit {\_a} \right ) \sqrt {3}\, \sqrt {27 \textit {\_a}^{2}-2}+432 \textit {\_a}^{4}-45 \textit {\_a}^{2}+1\right )}^{{2}/{3}} \sqrt {3}+12 i \sqrt {3}\, \textit {\_a}^{2}-18 {\left (\left (48 \textit {\_a}^{3}-3 \textit {\_a} \right ) \sqrt {3}\, \sqrt {27 \textit {\_a}^{2}-2}+432 \textit {\_a}^{4}-45 \textit {\_a}^{2}+1\right )}^{{1}/{3}} \textit {\_a}^{2}-i \sqrt {3}-{\left (\left (48 \textit {\_a}^{3}-3 \textit {\_a} \right ) \sqrt {3}\, \sqrt {27 \textit {\_a}^{2}-2}+432 \textit {\_a}^{4}-45 \textit {\_a}^{2}+1\right )}^{{2}/{3}}+12 \textit {\_a}^{2}+2 {\left (\left (48 \textit {\_a}^{3}-3 \textit {\_a} \right ) \sqrt {3}\, \sqrt {27 \textit {\_a}^{2}-2}+432 \textit {\_a}^{4}-45 \textit {\_a}^{2}+1\right )}^{{1}/{3}}-1}d \textit {\_a} +c_3 \end{align*}
We now need to find the singular solutions, these are found by finding for what values \((-\frac {\left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y \right )^{{1}/{3}}}{12 y}+\frac {x \left (-x^{3}+12 y^{2}\right )}{12 y \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y \right )^{{1}/{3}}}+\frac {x^{2}}{6 y}+\frac {i \sqrt {3}\, \left (\frac {\left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y \right )^{{1}/{3}}}{6 y}+\frac {x \left (-x^{3}+12 y^{2}\right )}{6 y \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y \right )^{{1}/{3}}}\right )}{2})\) is
zero. These give
\begin{align*}
y&=\operatorname {RootOf}\left (-i \sqrt {3}\, x^{4}+12 i \sqrt {3}\, \textit {\_Z}^{2} x +i \sqrt {3}\, \left (x^{6}-45 x^{3} \textit {\_Z}^{2}+432 \textit {\_Z}^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 \textit {\_Z}^{2} x^{6}-1376 \textit {\_Z}^{4} x^{3}+6912 \textit {\_Z}^{6}}\, \textit {\_Z} \right )^{{2}/{3}}-x^{4}+12 x \,\textit {\_Z}^{2}+2 x^{2} \left (x^{6}-45 x^{3} \textit {\_Z}^{2}+432 \textit {\_Z}^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 \textit {\_Z}^{2} x^{6}-1376 \textit {\_Z}^{4} x^{3}+6912 \textit {\_Z}^{6}}\, \textit {\_Z} \right )^{{1}/{3}}-\left (x^{6}-45 x^{3} \textit {\_Z}^{2}+432 \textit {\_Z}^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 \textit {\_Z}^{2} x^{6}-1376 \textit {\_Z}^{4} x^{3}+6912 \textit {\_Z}^{6}}\, \textit {\_Z} \right )^{{2}/{3}}\right ) \\
\end{align*}
Now we go over each such singular solution and check if it verifies the ode
itself and any initial conditions given. If it does not then the singular solution will not be
used.
The solution \(\operatorname {RootOf}\left (-i \sqrt {3}\, x^{4}+12 i \sqrt {3}\, \textit {\_Z}^{2} x +i \sqrt {3}\, \left (x^{6}-45 x^{3} \textit {\_Z}^{2}+432 \textit {\_Z}^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 \textit {\_Z}^{2} x^{6}-1376 \textit {\_Z}^{4} x^{3}+6912 \textit {\_Z}^{6}}\, \textit {\_Z} \right )^{{2}/{3}}-x^{4}+12 x \,\textit {\_Z}^{2}+2 x^{2} \left (x^{6}-45 x^{3} \textit {\_Z}^{2}+432 \textit {\_Z}^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 \textit {\_Z}^{2} x^{6}-1376 \textit {\_Z}^{4} x^{3}+6912 \textit {\_Z}^{6}}\, \textit {\_Z} \right )^{{1}/{3}}-\left (x^{6}-45 x^{3} \textit {\_Z}^{2}+432 \textit {\_Z}^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 \textit {\_Z}^{2} x^{6}-1376 \textit {\_Z}^{4} x^{3}+6912 \textit {\_Z}^{6}}\, \textit {\_Z} \right )^{{2}/{3}}\right )\) will not be used
Solving Eq. (3)
Writing the ode as
\begin{align*} y^{\prime }&=-\frac {-i \sqrt {3}\, x^{4}+12 i \sqrt {3}\, y^{2} x +i \sqrt {3}\, \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y \right )^{{2}/{3}}+x^{4}-12 x \,y^{2}-2 x^{2} \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y \right )^{{1}/{3}}+\left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y \right )^{{2}/{3}}}{12 y \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y \right )^{{1}/{3}}}\\ y^{\prime }&= \omega \left ( x,y\right ) \end{align*}
The condition of Lie symmetry is the linearized PDE given by
\begin{align*} \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0\tag {A} \end{align*}
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.
\[
\left \{x, y, \sqrt {-\left (2 x^{3}-27 y^{2}\right ) \left (x^{3}-16 y^{2}\right )^{2}}, \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-\left (2 x^{3}-27 y^{2}\right ) \left (x^{3}-16 y^{2}\right )^{2}}\, y \right )^{{1}/{3}}, \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-\left (2 x^{3}-27 y^{2}\right ) \left (x^{3}-16 y^{2}\right )^{2}}\, y \right )^{{2}/{3}}\right \}
\]
The following substitution is now made to be able to
collect on all terms with \(\{x, y\}\) in them
\[
\left \{x = v_{1}, y = v_{2}, \sqrt {-\left (2 x^{3}-27 y^{2}\right ) \left (x^{3}-16 y^{2}\right )^{2}} = v_{3}, \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-\left (2 x^{3}-27 y^{2}\right ) \left (x^{3}-16 y^{2}\right )^{2}}\, y \right )^{{1}/{3}} = v_{4}, \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-\left (2 x^{3}-27 y^{2}\right ) \left (x^{3}-16 y^{2}\right )^{2}}\, y \right )^{{2}/{3}} = v_{5}\right \}
\]
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*} -20736 a_{1}&=0\\ -48 a_{1}&=0\\ 2160 a_{1}&=0\\ -7344 a_{3}&=0\\ -12 a_{3}&=0\\ 588 a_{3}&=0\\ 20736 a_{3}&=0\\ -1080 b_{1}&=0\\ 24 b_{1}&=0\\ 10368 b_{1}&=0\\ -936 b_{2}&=0\\ 24 b_{2}&=0\\ 3888 b_{2}&=0\\ 62208 b_{2}&=0\\ -995328 \sqrt {3}\, a_{1}&=0\\ -13104 \sqrt {3}\, a_{1}&=0\\ 288 \sqrt {3}\, a_{1}&=0\\ 198144 \sqrt {3}\, a_{1}&=0\\ -446976 \sqrt {3}\, a_{3}&=0\\ -3564 \sqrt {3}\, a_{3}&=0\\ 72 \sqrt {3}\, a_{3}&=0\\ 62640 \sqrt {3}\, a_{3}&=0\\ 995328 \sqrt {3}\, a_{3}&=0\\ -99072 \sqrt {3}\, b_{1}&=0\\ -144 \sqrt {3}\, b_{1}&=0\\ 6552 \sqrt {3}\, b_{1}&=0\\ 497664 \sqrt {3}\, b_{1}&=0\\ -96768 \sqrt {3}\, b_{2}&=0\\ -59760 \sqrt {3}\, b_{2}&=0\\ -144 \sqrt {3}\, b_{2}&=0\\ 5688 \sqrt {3}\, b_{2}&=0\\ 2985984 \sqrt {3}\, b_{2}&=0\\ -31104 a_{2}+20736 b_{3}&=0\\ -72 a_{2}+48 b_{3}&=0\\ 3240 a_{2}-2160 b_{3}&=0\\ -2985984 \sqrt {3}\, a_{1}+8957952 i a_{1}&=0\\ -173376 \sqrt {3}\, a_{1}+520128 i a_{1}&=0\\ -24768 \sqrt {3}\, a_{1}-74304 i a_{1}&=0\\ -180 \sqrt {3}\, a_{1}+540 i a_{1}&=0\\ -108 \sqrt {3}\, a_{1}-324 i a_{1}&=0\\ 3276 \sqrt {3}\, a_{1}+9828 i a_{1}&=0\\ 9396 \sqrt {3}\, a_{1}-28188 i a_{1}&=0\\ 1292544 \sqrt {3}\, a_{1}-3877632 i a_{1}&=0\\ -1492992 \sqrt {3}\, a_{2}+995328 \sqrt {3}\, b_{3}&=0\\ -19656 \sqrt {3}\, a_{2}+13104 \sqrt {3}\, b_{3}&=0\\ 432 \sqrt {3}\, a_{2}-288 \sqrt {3}\, b_{3}&=0\\ 297216 \sqrt {3}\, a_{2}-198144 \sqrt {3}\, b_{3}&=0\\ -10568448 \sqrt {3}\, a_{3}+31705344 i a_{3}&=0\\ -96240 \sqrt {3}\, a_{3}+288720 i a_{3}&=0\\ -13236 \sqrt {3}\, a_{3}-39708 i a_{3}&=0\\ -48 \sqrt {3}\, a_{3}+144 i a_{3}&=0\\ -24 \sqrt {3}\, a_{3}-72 i a_{3}&=0\\ 984 \sqrt {3}\, a_{3}+2952 i a_{3}&=0\\ 3372 \sqrt {3}\, a_{3}-10116 i a_{3}&=0\\ 58176 \sqrt {3}\, a_{3}+174528 i a_{3}&=0\\ 1405440 \sqrt {3}\, a_{3}-4216320 i a_{3}&=0\\ 32845824 \sqrt {3}\, a_{3}-98537472 i a_{3}&=0\\ -580608 \sqrt {3}\, b_{1}+1741824 i b_{1}&=0\\ -4884 \sqrt {3}\, b_{1}+14652 i b_{1}&=0\\ -1092 \sqrt {3}\, b_{1}-3276 i b_{1}&=0\\ 48 \sqrt {3}\, b_{1}+144 i b_{1}&=0\\ 96 \sqrt {3}\, b_{1}-288 i b_{1}&=0\\ 82944 \sqrt {3}\, b_{1}+248832 i b_{1}&=0\\ 85968 \sqrt {3}\, b_{1}-257904 i b_{1}&=0\\ 995328 \sqrt {3}\, b_{1}-2985984 i b_{1}&=0\\ -580608 \sqrt {3}\, b_{2}+1741824 i b_{2}&=0\\ -4884 \sqrt {3}\, b_{2}+14652 i b_{2}&=0\\ -1092 \sqrt {3}\, b_{2}-3276 i b_{2}&=0\\ 48 \sqrt {3}\, b_{2}+144 i b_{2}&=0\\ 96 \sqrt {3}\, b_{2}-288 i b_{2}&=0\\ 82944 \sqrt {3}\, b_{2}+248832 i b_{2}&=0\\ 85968 \sqrt {3}\, b_{2}-257904 i b_{2}&=0\\ 995328 \sqrt {3}\, b_{2}-2985984 i b_{2}&=0\\ -684288 i \sqrt {3}\, a_{3}+684288 a_{3}&=0\\ -20736 i \sqrt {3}\, a_{1}+20736 a_{1}&=0\\ -20736 i \sqrt {3}\, b_{1}+20736 b_{1}&=0\\ -20736 i \sqrt {3}\, b_{2}+20736 b_{2}&=0\\ -13104 i \sqrt {3}\, a_{3}+13104 a_{3}&=0\\ -756 i \sqrt {3}\, b_{1}+756 b_{1}&=0\\ -756 i \sqrt {3}\, b_{2}+756 b_{2}&=0\\ -540 i \sqrt {3}\, a_{1}-540 a_{1}&=0\\ -204 i \sqrt {3}\, a_{3}-204 a_{3}&=0\\ -24 i \sqrt {3}\, a_{1}+24 a_{1}&=0\\ -12 i \sqrt {3}\, b_{1}-12 b_{1}&=0\\ -12 i \sqrt {3}\, b_{2}-12 b_{2}&=0\\ -6 i \sqrt {3}\, a_{3}+6 a_{3}&=0\\ 6 i \sqrt {3}\, a_{3}+6 a_{3}&=0\\ 12 i \sqrt {3}\, b_{1}-12 b_{1}&=0\\ 12 i \sqrt {3}\, b_{2}-12 b_{2}&=0\\ 24 i \sqrt {3}\, a_{1}+24 a_{1}&=0\\ 180 i \sqrt {3}\, b_{1}+180 b_{1}&=0\\ 180 i \sqrt {3}\, b_{2}+180 b_{2}&=0\\ 492 i \sqrt {3}\, a_{3}-492 a_{3}&=0\\ 1476 i \sqrt {3}\, a_{1}-1476 a_{1}&=0\\ 1476 i \sqrt {3}\, a_{3}+1476 a_{3}&=0\\ 1728 i \sqrt {3}\, b_{1}+1728 b_{1}&=0\\ 1728 i \sqrt {3}\, b_{2}+1728 b_{2}&=0\\ 9936 i \sqrt {3}\, b_{1}-9936 b_{1}&=0\\ 9936 i \sqrt {3}\, b_{2}-9936 b_{2}&=0\\ 62208 i \sqrt {3}\, a_{1}-62208 a_{1}&=0\\ 152064 i \sqrt {3}\, a_{3}-152064 a_{3}&=0\\ -5971968 \sqrt {3}\, a_{2}+3981312 \sqrt {3}\, b_{3}+17915904 i a_{2}-11943936 i b_{3}&=0\\ -262224 \sqrt {3}\, a_{2}+174816 \sqrt {3}\, b_{3}+786672 i a_{2}-524448 i b_{3}&=0\\ -74304 \sqrt {3}\, a_{2}+49536 \sqrt {3}\, b_{3}-222912 i a_{2}+148608 i b_{3}&=0\\ -252 \sqrt {3}\, a_{2}+168 \sqrt {3}\, b_{3}+756 i a_{2}-504 i b_{3}&=0\\ -180 \sqrt {3}\, a_{2}+120 \sqrt {3}\, b_{3}-540 i a_{2}+360 i b_{3}&=0\\ 6552 \sqrt {3}\, a_{2}-4368 \sqrt {3}\, b_{3}+19656 i a_{2}-13104 i b_{3}&=0\\ 13536 \sqrt {3}\, a_{2}-9024 \sqrt {3}\, b_{3}-40608 i a_{2}+27072 i b_{3}&=0\\ 248832 \sqrt {3}\, a_{2}-165888 \sqrt {3}\, b_{3}+746496 i a_{2}-497664 i b_{3}&=0\\ 2135808 \sqrt {3}\, a_{2}-1423872 \sqrt {3}\, b_{3}-6407424 i a_{2}+4271616 i b_{3}&=0\\ -32400 i \sqrt {3}\, a_{2}+21600 i \sqrt {3}\, b_{3}+32400 a_{2}-21600 b_{3}&=0\\ -1080 i \sqrt {3}\, a_{2}+720 i \sqrt {3}\, b_{3}-1080 a_{2}+720 b_{3}&=0\\ -36 i \sqrt {3}\, a_{2}+24 i \sqrt {3}\, b_{3}+36 a_{2}-24 b_{3}&=0\\ 36 i \sqrt {3}\, a_{2}-24 i \sqrt {3}\, b_{3}+36 a_{2}-24 b_{3}&=0\\ 2160 i \sqrt {3}\, a_{2}-1440 i \sqrt {3}\, b_{3}-2160 a_{2}+1440 b_{3}&=0\\ 5184 i \sqrt {3}\, a_{2}-3456 i \sqrt {3}\, b_{3}+5184 a_{2}-3456 b_{3}&=0\\ 124416 i \sqrt {3}\, a_{2}-82944 i \sqrt {3}\, b_{3}-124416 a_{2}+82944 b_{3}&=0 \end{align*}
Solving the above equations for the unknowns gives
\begin{align*} a_{1}&=0\\ a_{2}&=\frac {2 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 {2 x}{3} \\
\eta &= y \\
\end{align*}
The next step is to determine the canonical coordinates \(R,S\). The
canonical coordinates map \(\left ( x,y\right ) \to \left ( R,S \right )\) where \(\left ( R,S \right )\) are the canonical coordinates which make the original ode
become a quadrature and hence solved by integration.
The characteristic pde which is used to find the canonical coordinates is
\begin{align*} \frac {d x}{\xi } &= \frac {d y}{\eta } = dS \tag {1} \end{align*}
The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial y}\right ) S(x,y) = 1\). Starting with the first pair of ode’s in (1)
gives an ode to solve for the independent variable \(R\) in the canonical coordinates, where \(S(R)\).
Therefore
\begin{align*} \frac {dy}{dx} &= \frac {\eta }{\xi }\\ &= \frac {y}{\frac {2 x}{3}}\\ &= \frac {3 y}{2 x} \end{align*}
This is easily solved to give
\begin{align*} y = c_{1} x^{{3}/{2}} \end{align*}
Where now the coordinate \(R\) is taken as the constant of integration. Hence
\begin{align*} R &= \frac {y}{x^{{3}/{2}}} \end{align*}
And \(S\) is found from
\begin{align*} dS &= \frac {dx}{\xi } \\ &= \frac {dx}{\frac {2 x}{3}} \end{align*}
Integrating gives
\begin{align*} S &= \int { \frac {dx}{T}}\\ &= \frac {3 \ln \left (x \right )}{2} \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 {-i \sqrt {3}\, x^{4}+12 i \sqrt {3}\, y^{2} x +i \sqrt {3}\, \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y \right )^{{2}/{3}}+x^{4}-12 x \,y^{2}-2 x^{2} \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y \right )^{{1}/{3}}+\left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y \right )^{{2}/{3}}}{12 y \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y \right )^{{1}/{3}}} \end{align*}
Evaluating all the partial derivatives gives
\begin{align*} R_{x} &= -\frac {3 y}{2 x^{{5}/{2}}}\\ R_{y} &= \frac {1}{x^{{3}/{2}}}\\ S_{x} &= \frac {3}{2 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 {18 x^{{3}/{2}} \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 \left (x^{3}-\frac {27 y^{2}}{2}\right ) \left (x^{3}-16 y^{2}\right )^{2}}\, y \right )^{{1}/{3}} y}{x \left (-i \sqrt {3}-1\right ) \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 \left (x^{3}-\frac {27 y^{2}}{2}\right ) \left (x^{3}-16 y^{2}\right )^{2}}\, y \right )^{{2}/{3}}+2 \left (x^{3}-9 y^{2}\right ) \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 \left (x^{3}-\frac {27 y^{2}}{2}\right ) \left (x^{3}-16 y^{2}\right )^{2}}\, y \right )^{{1}/{3}}+x^{2} \left (x^{3}-12 y^{2}\right ) \left (-1+i \sqrt {3}\right )}\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 {18 R {\left (\left (48 R^{3}-3 R \right ) \sqrt {3}\, \sqrt {27 R^{2}-2}+432 R^{4}-45 R^{2}+1\right )}^{{1}/{3}}}{\left (1+i \sqrt {3}\right ) {\left (\left (48 R^{3}-3 R \right ) \sqrt {3}\, \sqrt {27 R^{2}-2}+432 R^{4}-45 R^{2}+1\right )}^{{2}/{3}}+\left (18 R^{2}-2\right ) {\left (\left (48 R^{3}-3 R \right ) \sqrt {3}\, \sqrt {27 R^{2}-2}+432 R^{4}-45 R^{2}+1\right )}^{{1}/{3}}+12 \left (-1+i \sqrt {3}\right ) \left (R^{2}-\frac {1}{12}\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 \(R,S\).
Since the ode has the form \(\frac {d}{d R}S \left (R \right )=f(R)\), then we only need to integrate \(f(R)\).
\begin{align*} \int {dS} &= \int {-\frac {18 R {\left (\left (48 R^{3}-3 R \right ) \sqrt {3}\, \sqrt {27 R^{2}-2}+432 R^{4}-45 R^{2}+1\right )}^{{1}/{3}}}{i \sqrt {3}\, {\left (\left (48 R^{3}-3 R \right ) \sqrt {3}\, \sqrt {27 R^{2}-2}+432 R^{4}-45 R^{2}+1\right )}^{{2}/{3}}+12 i \sqrt {3}\, R^{2}+18 {\left (\left (48 R^{3}-3 R \right ) \sqrt {3}\, \sqrt {27 R^{2}-2}+432 R^{4}-45 R^{2}+1\right )}^{{1}/{3}} R^{2}-i \sqrt {3}+{\left (\left (48 R^{3}-3 R \right ) \sqrt {3}\, \sqrt {27 R^{2}-2}+432 R^{4}-45 R^{2}+1\right )}^{{2}/{3}}-12 R^{2}-2 {\left (\left (48 R^{3}-3 R \right ) \sqrt {3}\, \sqrt {27 R^{2}-2}+432 R^{4}-45 R^{2}+1\right )}^{{1}/{3}}+1}\, dR}\\ S \left (R \right ) &= \int -\frac {18 R {\left (\left (48 R^{3}-3 R \right ) \sqrt {3}\, \sqrt {27 R^{2}-2}+432 R^{4}-45 R^{2}+1\right )}^{{1}/{3}}}{i \sqrt {3}\, {\left (\left (48 R^{3}-3 R \right ) \sqrt {3}\, \sqrt {27 R^{2}-2}+432 R^{4}-45 R^{2}+1\right )}^{{2}/{3}}+12 i \sqrt {3}\, R^{2}+18 {\left (\left (48 R^{3}-3 R \right ) \sqrt {3}\, \sqrt {27 R^{2}-2}+432 R^{4}-45 R^{2}+1\right )}^{{1}/{3}} R^{2}-i \sqrt {3}+{\left (\left (48 R^{3}-3 R \right ) \sqrt {3}\, \sqrt {27 R^{2}-2}+432 R^{4}-45 R^{2}+1\right )}^{{2}/{3}}-12 R^{2}-2 {\left (\left (48 R^{3}-3 R \right ) \sqrt {3}\, \sqrt {27 R^{2}-2}+432 R^{4}-45 R^{2}+1\right )}^{{1}/{3}}+1}d R + c_5 \end{align*}
\begin{align*} S \left (R \right )&= \int -\frac {18 R {\left (\left (48 R^{3}-3 R \right ) \sqrt {3}\, \sqrt {27 R^{2}-2}+432 R^{4}-45 R^{2}+1\right )}^{{1}/{3}}}{i \sqrt {3}\, {\left (\left (48 R^{3}-3 R \right ) \sqrt {3}\, \sqrt {27 R^{2}-2}+432 R^{4}-45 R^{2}+1\right )}^{{2}/{3}}+12 i \sqrt {3}\, R^{2}+18 {\left (\left (48 R^{3}-3 R \right ) \sqrt {3}\, \sqrt {27 R^{2}-2}+432 R^{4}-45 R^{2}+1\right )}^{{1}/{3}} R^{2}-i \sqrt {3}+{\left (\left (48 R^{3}-3 R \right ) \sqrt {3}\, \sqrt {27 R^{2}-2}+432 R^{4}-45 R^{2}+1\right )}^{{2}/{3}}-12 R^{2}-2 {\left (\left (48 R^{3}-3 R \right ) \sqrt {3}\, \sqrt {27 R^{2}-2}+432 R^{4}-45 R^{2}+1\right )}^{{1}/{3}}+1}d R +c_5 \end{align*}
To complete the solution, we just need to transform the above back to \(x,y\) coordinates. This
results in
\begin{align*} \frac {3 \ln \left (x \right )}{2} = \int _{}^{\frac {y}{x^{{3}/{2}}}}-\frac {18 \textit {\_a} {\left (\left (48 \textit {\_a}^{3}-3 \textit {\_a} \right ) \sqrt {3}\, \sqrt {27 \textit {\_a}^{2}-2}+432 \textit {\_a}^{4}-45 \textit {\_a}^{2}+1\right )}^{{1}/{3}}}{i {\left (\left (48 \textit {\_a}^{3}-3 \textit {\_a} \right ) \sqrt {3}\, \sqrt {27 \textit {\_a}^{2}-2}+432 \textit {\_a}^{4}-45 \textit {\_a}^{2}+1\right )}^{{2}/{3}} \sqrt {3}+12 i \sqrt {3}\, \textit {\_a}^{2}+18 {\left (\left (48 \textit {\_a}^{3}-3 \textit {\_a} \right ) \sqrt {3}\, \sqrt {27 \textit {\_a}^{2}-2}+432 \textit {\_a}^{4}-45 \textit {\_a}^{2}+1\right )}^{{1}/{3}} \textit {\_a}^{2}-i \sqrt {3}+{\left (\left (48 \textit {\_a}^{3}-3 \textit {\_a} \right ) \sqrt {3}\, \sqrt {27 \textit {\_a}^{2}-2}+432 \textit {\_a}^{4}-45 \textit {\_a}^{2}+1\right )}^{{2}/{3}}-12 \textit {\_a}^{2}-2 {\left (\left (48 \textit {\_a}^{3}-3 \textit {\_a} \right ) \sqrt {3}\, \sqrt {27 \textit {\_a}^{2}-2}+432 \textit {\_a}^{4}-45 \textit {\_a}^{2}+1\right )}^{{1}/{3}}+1}d \textit {\_a} +c_5 \end{align*}
We now need to find the singular solutions, these are found by finding for what values \((-\frac {\left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y \right )^{{1}/{3}}}{12 y}+\frac {x \left (-x^{3}+12 y^{2}\right )}{12 y \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y \right )^{{1}/{3}}}+\frac {x^{2}}{6 y}-\frac {i \sqrt {3}\, \left (\frac {\left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y \right )^{{1}/{3}}}{6 y}+\frac {x \left (-x^{3}+12 y^{2}\right )}{6 y \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y \right )^{{1}/{3}}}\right )}{2})\) is
zero. These give
\begin{align*}
y&=\operatorname {RootOf}\left (-i \sqrt {3}\, x^{4}+12 i \sqrt {3}\, \textit {\_Z}^{2} x +i \sqrt {3}\, \left (x^{6}-45 x^{3} \textit {\_Z}^{2}+432 \textit {\_Z}^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 \textit {\_Z}^{2} x^{6}-1376 \textit {\_Z}^{4} x^{3}+6912 \textit {\_Z}^{6}}\, \textit {\_Z} \right )^{{2}/{3}}+x^{4}-2 x^{2} \left (x^{6}-45 x^{3} \textit {\_Z}^{2}+432 \textit {\_Z}^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 \textit {\_Z}^{2} x^{6}-1376 \textit {\_Z}^{4} x^{3}+6912 \textit {\_Z}^{6}}\, \textit {\_Z} \right )^{{1}/{3}}-12 x \,\textit {\_Z}^{2}+\left (x^{6}-45 x^{3} \textit {\_Z}^{2}+432 \textit {\_Z}^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 \textit {\_Z}^{2} x^{6}-1376 \textit {\_Z}^{4} x^{3}+6912 \textit {\_Z}^{6}}\, \textit {\_Z} \right )^{{2}/{3}}\right ) \\
\end{align*}
Now we go over each such singular solution and check if it verifies the ode
itself and any initial conditions given. If it does not then the singular solution will not be
used.
The solution \(\operatorname {RootOf}\left (-i \sqrt {3}\, x^{4}+12 i \sqrt {3}\, \textit {\_Z}^{2} x +i \sqrt {3}\, \left (x^{6}-45 x^{3} \textit {\_Z}^{2}+432 \textit {\_Z}^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 \textit {\_Z}^{2} x^{6}-1376 \textit {\_Z}^{4} x^{3}+6912 \textit {\_Z}^{6}}\, \textit {\_Z} \right )^{{2}/{3}}+x^{4}-2 x^{2} \left (x^{6}-45 x^{3} \textit {\_Z}^{2}+432 \textit {\_Z}^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 \textit {\_Z}^{2} x^{6}-1376 \textit {\_Z}^{4} x^{3}+6912 \textit {\_Z}^{6}}\, \textit {\_Z} \right )^{{1}/{3}}-12 x \,\textit {\_Z}^{2}+\left (x^{6}-45 x^{3} \textit {\_Z}^{2}+432 \textit {\_Z}^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 \textit {\_Z}^{2} x^{6}-1376 \textit {\_Z}^{4} x^{3}+6912 \textit {\_Z}^{6}}\, \textit {\_Z} \right )^{{2}/{3}}\right )\) will not be used
Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 4 y {y^{\prime }}^{3}-2 {y^{\prime }}^{2} x^{2}+4 x y y^{\prime }+x^{3}=16 y^{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 (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y\right )^{{1}/{3}}}{6 y}-\frac {x \left (-x^{3}+12 y^{2}\right )}{6 y \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y\right )^{{1}/{3}}}+\frac {x^{2}}{6 y}, y^{\prime }=-\frac {\left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y\right )^{{1}/{3}}}{12 y}+\frac {x \left (-x^{3}+12 y^{2}\right )}{12 y \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y\right )^{{1}/{3}}}+\frac {x^{2}}{6 y}-\frac {\mathrm {I} \sqrt {3}\, \left (\frac {\left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y\right )^{{1}/{3}}}{6 y}+\frac {x \left (-x^{3}+12 y^{2}\right )}{6 y \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y\right )^{{1}/{3}}}\right )}{2}, y^{\prime }=-\frac {\left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y\right )^{{1}/{3}}}{12 y}+\frac {x \left (-x^{3}+12 y^{2}\right )}{12 y \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y\right )^{{1}/{3}}}+\frac {x^{2}}{6 y}+\frac {\mathrm {I} \sqrt {3}\, \left (\frac {\left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y\right )^{{1}/{3}}}{6 y}+\frac {x \left (-x^{3}+12 y^{2}\right )}{6 y \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y\right )^{{1}/{3}}}\right )}{2}\right ] \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {\left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y\right )^{{1}/{3}}}{6 y}-\frac {x \left (-x^{3}+12 y^{2}\right )}{6 y \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y\right )^{{1}/{3}}}+\frac {x^{2}}{6 y} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {\left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y\right )^{{1}/{3}}}{12 y}+\frac {x \left (-x^{3}+12 y^{2}\right )}{12 y \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y\right )^{{1}/{3}}}+\frac {x^{2}}{6 y}-\frac {\mathrm {I} \sqrt {3}\, \left (\frac {\left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y\right )^{{1}/{3}}}{6 y}+\frac {x \left (-x^{3}+12 y^{2}\right )}{6 y \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y\right )^{{1}/{3}}}\right )}{2} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {\left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y\right )^{{1}/{3}}}{12 y}+\frac {x \left (-x^{3}+12 y^{2}\right )}{12 y \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y\right )^{{1}/{3}}}+\frac {x^{2}}{6 y}+\frac {\mathrm {I} \sqrt {3}\, \left (\frac {\left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y\right )^{{1}/{3}}}{6 y}+\frac {x \left (-x^{3}+12 y^{2}\right )}{6 y \left (x^{6}-45 x^{3} y^{2}+432 y^{4}+3 \sqrt {3}\, \sqrt {-2 x^{9}+91 y^{2} x^{6}-1376 x^{3} y^{4}+6912 y^{6}}\, y\right )^{{1}/{3}}}\right )}{2} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\mathit {workingODE} , \mathit {workingODE} , \mathit {workingODE}\right \} \end {array} \]
Maple dsolve solution
Solving time : 393.754 (sec)
Leaf size : maple_leaf_size
dsolve(4*y(x)*diff(y(x),x)^3-2*diff(y(x),x)^2*x^2+4*x*y(x)*diff(y(x),x)+x^3 = 16*y(x)^2,
y(x),singsol=all)
\[ \text {No solution found} \]
Mathematica DSolve solution
Solving time : 35.094 (sec)
Leaf size : 49162
DSolve[{4*y[x]*D[y[x],x]^3-2*x^2*D[y[x],x]^2+4*x*y[x]*D[y[x],x]+x^3==16*y[x]^2,{}},
y[x],x,IncludeSingularSolutions->True]
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