Internal problem ID [11122]
Internal file name [OUTPUT/10108_Wednesday_November_23_2022_11_50_42_AM_71634107/index.tex]
Book: An elementary treatise on differential equations by Abraham Cohen. DC heath publishers.
1906
Section: Chapter 2, differential equations of the first order and the first degree. Article 8. Exact
differential equations. Page 11
Problem number: Ex 2.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "exact", "dAlembert", "homogeneousTypeD2", "first_order_ode_lie_symmetry_calculated"
Maple gives the following as the ode type
[[_homogeneous, `class A`], _exact, _rational, _dAlembert]
\[ \boxed {\frac {y^{2}-2 x^{2}}{x y^{2}-x^{3}}+\frac {\left (2 y^{2}-x^{2}\right ) y^{\prime }}{y^{3}-x^{2} y}=0} \]
Using the change of variables \(y = u \left (x \right ) x\) on the above ode results in new ode in \(u \left (x \right )\) \begin {align*} \frac {u \left (x \right )^{2} x^{2}-2 x^{2}}{x^{3} u \left (x \right )^{2}-x^{3}}+\frac {\left (2 u \left (x \right )^{2} x^{2}-x^{2}\right ) \left (u^{\prime }\left (x \right ) x +u \left (x \right )\right )}{u \left (x \right )^{3} x^{3}-x^{3} u \left (x \right )} = 0 \end {align*}
In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= -\frac {3 \left (u^{3}-u \right )}{x \left (2 u^{2}-1\right )} \end {align*}
Where \(f(x)=-\frac {3}{x}\) and \(g(u)=\frac {u^{3}-u}{2 u^{2}-1}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {u^{3}-u}{2 u^{2}-1}} \,du &= -\frac {3}{x} \,d x \\ \int { \frac {1}{\frac {u^{3}-u}{2 u^{2}-1}} \,du} &= \int {-\frac {3}{x} \,d x} \\ \ln \left (u \right )+\frac {\ln \left (u^{2}-1\right )}{2}&=-3 \ln \left (x \right )+c_{2} \\ \end{align*} Raising both side to exponential gives \begin {align*} {\mathrm e}^{\ln \left (u \right )+\frac {\ln \left (u^{2}-1\right )}{2}} &= {\mathrm e}^{-3 \ln \left (x \right )+c_{2}} \end {align*}
Which simplifies to \begin {align*} u \sqrt {u^{2}-1} &= \frac {c_{3}}{x^{3}} \end {align*}
The solution is \[ u \left (x \right ) \sqrt {u \left (x \right )^{2}-1} = \frac {c_{3}}{x^{3}} \] Replacing \(u(x)\) in the above solution by \(\frac {y}{x}\) results in the solution for \(y\) in implicit form \begin {align*} \frac {y \sqrt {\frac {y^{2}}{x^{2}}-1}}{x} = \frac {c_{3}}{x^{3}}\\ \frac {y \sqrt {\frac {y^{2}-x^{2}}{x^{2}}}}{x} = \frac {c_{3}}{x^{3}} \end {align*}
Which simplifies to \begin {align*} y \sqrt {-\frac {\left (x -y\right ) \left (x +y\right )}{x^{2}}} = \frac {c_{3}}{x^{2}} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y \sqrt {-\frac {\left (x -y\right ) \left (x +y\right )}{x^{2}}} &= \frac {c_{3}}{x^{2}} \\ \end{align*}
Verification of solutions
\[ y \sqrt {-\frac {\left (x -y\right ) \left (x +y\right )}{x^{2}}} = \frac {c_{3}}{x^{2}} \] Verified OK.
Writing the ode as \begin {align*} y^{\prime }&=-\frac {y \left (-2 x^{2}+y^{2}\right )}{x \left (-x^{2}+2 y^{2}\right )}\\ y^{\prime }&= \omega \left ( x,y\right ) \end {align*}
The condition of Lie symmetry is the linearized PDE given by \begin {align*} \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0\tag {A} \end {align*}
The type of this ode is not in the lookup table. To determine \(\xi ,\eta \) then (A) is solved using ansatz. Making bivariate polynomials of degree 1 to use as anstaz gives \begin{align*} \tag{1E} \xi &= x a_{2}+y a_{3}+a_{1} \\ \tag{2E} \eta &= x b_{2}+y b_{3}+b_{1} \\ \end{align*} Where the unknown coefficients are \[ \{a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}\} \] Substituting equations (1E,2E) and \(\omega \) into (A) gives \begin{equation} \tag{5E} b_{2}-\frac {y \left (-2 x^{2}+y^{2}\right ) \left (b_{3}-a_{2}\right )}{x \left (-x^{2}+2 y^{2}\right )}-\frac {y^{2} \left (-2 x^{2}+y^{2}\right )^{2} a_{3}}{x^{2} \left (-x^{2}+2 y^{2}\right )^{2}}-\left (\frac {4 y}{-x^{2}+2 y^{2}}+\frac {y \left (-2 x^{2}+y^{2}\right )}{x^{2} \left (-x^{2}+2 y^{2}\right )}-\frac {2 y \left (-2 x^{2}+y^{2}\right )}{\left (-x^{2}+2 y^{2}\right )^{2}}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )-\left (-\frac {-2 x^{2}+y^{2}}{x \left (-x^{2}+2 y^{2}\right )}-\frac {2 y^{2}}{x \left (-x^{2}+2 y^{2}\right )}+\frac {4 y^{2} \left (-2 x^{2}+y^{2}\right )}{x \left (-x^{2}+2 y^{2}\right )^{2}}\right ) \left (x b_{2}+y b_{3}+b_{1}\right ) = 0 \end{equation} Putting the above in normal form gives \[ \frac {3 x^{6} b_{2}-6 x^{4} y^{2} a_{3}-3 x^{4} y^{2} b_{2}-6 x^{3} y^{3} a_{2}+6 x^{3} y^{3} b_{3}+3 x^{2} y^{4} a_{3}+6 x^{2} y^{4} b_{2}-3 y^{6} a_{3}+2 x^{5} b_{1}-2 x^{4} y a_{1}+x^{3} y^{2} b_{1}-x^{2} y^{3} a_{1}+2 x \,y^{4} b_{1}-2 y^{5} a_{1}}{\left (x^{2}-2 y^{2}\right )^{2} x^{2}} = 0 \] Setting the numerator to zero gives \begin{equation} \tag{6E} 3 x^{6} b_{2}-6 x^{4} y^{2} a_{3}-3 x^{4} y^{2} b_{2}-6 x^{3} y^{3} a_{2}+6 x^{3} y^{3} b_{3}+3 x^{2} y^{4} a_{3}+6 x^{2} y^{4} b_{2}-3 y^{6} a_{3}+2 x^{5} b_{1}-2 x^{4} y a_{1}+x^{3} y^{2} b_{1}-x^{2} y^{3} a_{1}+2 x \,y^{4} b_{1}-2 y^{5} a_{1} = 0 \end{equation} Looking at the above PDE shows the following are all the terms with \(\{x, y\}\) in them. \[ \{x, y\} \] The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them \[ \{x = v_{1}, y = v_{2}\} \] The above PDE (6E) now becomes \begin{equation} \tag{7E} -6 a_{2} v_{1}^{3} v_{2}^{3}-6 a_{3} v_{1}^{4} v_{2}^{2}+3 a_{3} v_{1}^{2} v_{2}^{4}-3 a_{3} v_{2}^{6}+3 b_{2} v_{1}^{6}-3 b_{2} v_{1}^{4} v_{2}^{2}+6 b_{2} v_{1}^{2} v_{2}^{4}+6 b_{3} v_{1}^{3} v_{2}^{3}-2 a_{1} v_{1}^{4} v_{2}-a_{1} v_{1}^{2} v_{2}^{3}-2 a_{1} v_{2}^{5}+2 b_{1} v_{1}^{5}+b_{1} v_{1}^{3} v_{2}^{2}+2 b_{1} v_{1} v_{2}^{4} = 0 \end{equation} Collecting the above on the terms \(v_i\) introduced, and these are \[ \{v_{1}, v_{2}\} \] Equation (7E) now becomes \begin{equation} \tag{8E} 3 b_{2} v_{1}^{6}+2 b_{1} v_{1}^{5}+\left (-6 a_{3}-3 b_{2}\right ) v_{1}^{4} v_{2}^{2}-2 a_{1} v_{1}^{4} v_{2}+\left (-6 a_{2}+6 b_{3}\right ) v_{1}^{3} v_{2}^{3}+b_{1} v_{1}^{3} v_{2}^{2}+\left (3 a_{3}+6 b_{2}\right ) v_{1}^{2} v_{2}^{4}-a_{1} v_{1}^{2} v_{2}^{3}+2 b_{1} v_{1} v_{2}^{4}-3 a_{3} v_{2}^{6}-2 a_{1} v_{2}^{5} = 0 \end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve \begin {align*} b_{1}&=0\\ -2 a_{1}&=0\\ -a_{1}&=0\\ -3 a_{3}&=0\\ 2 b_{1}&=0\\ 3 b_{2}&=0\\ -6 a_{2}+6 b_{3}&=0\\ -6 a_{3}-3 b_{2}&=0\\ 3 a_{3}+6 b_{2}&=0 \end {align*}
Solving the above equations for the unknowns gives \begin {align*} a_{1}&=0\\ a_{2}&=b_{3}\\ a_{3}&=0\\ b_{1}&=0\\ b_{2}&=0\\ b_{3}&=b_{3} \end {align*}
Substituting the above solution in the anstaz (1E,2E) (using \(1\) as arbitrary value for any unknown in the RHS) gives \begin{align*} \xi &= x \\ \eta &= y \\ \end{align*} Shifting is now applied to make \(\xi =0\) in order to simplify the rest of the computation \begin {align*} \eta &= \eta - \omega \left (x,y\right ) \xi \\ &= y - \left (-\frac {y \left (-2 x^{2}+y^{2}\right )}{x \left (-x^{2}+2 y^{2}\right )}\right ) \left (x\right ) \\ &= \frac {3 x^{2} y -3 y^{3}}{x^{2}-2 y^{2}}\\ \xi &= 0 \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 \(S(R)\). Since \(\xi =0\) then in this special case \begin {align*} R = x \end {align*}
\(S\) is found from \begin {align*} S &= \int { \frac {1}{\eta }} dy\\ &= \int { \frac {1}{\frac {3 x^{2} y -3 y^{3}}{x^{2}-2 y^{2}}}} dy \end {align*}
Which results in \begin {align*} S&= \frac {\ln \left (x +y \right )}{6}+\frac {\ln \left (y -x \right )}{6}+\frac {\ln \left (y \right )}{3} \end {align*}
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 {y \left (-2 x^{2}+y^{2}\right )}{x \left (-x^{2}+2 y^{2}\right )} \end {align*}
Evaluating all the partial derivatives gives \begin {align*} R_{x} &= 1\\ R_{y} &= 0\\ S_{x} &= \frac {x}{3 x^{2}-3 y^{2}}\\ S_{y} &= \frac {x^{2}-2 y^{2}}{3 x^{2} y -3 y^{3}} \end {align*}
Substituting all the above in (2) and simplifying gives the ode in canonical coordinates. \begin {align*} \frac {dS}{dR} &= -\frac {1}{3 x}\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 {1}{3 R} \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 ) = -\frac {\ln \left (R \right )}{3}+c_{1}\tag {4} \end {align*}
To complete the solution, we just need to transform (4) back to \(x,y\) coordinates. This results in \begin {align*} \frac {\ln \left (x +y\right )}{6}+\frac {\ln \left (y-x \right )}{6}+\frac {\ln \left (y\right )}{3} = -\frac {\ln \left (x \right )}{3}+c_{1} \end {align*}
Which simplifies to \begin {align*} \frac {\ln \left (x +y\right )}{6}+\frac {\ln \left (y-x \right )}{6}+\frac {\ln \left (y\right )}{3} = -\frac {\ln \left (x \right )}{3}+c_{1} \end {align*}
The following diagram shows solution curves of the original ode and how they transform in the canonical coordinates space using the mapping shown.
|
Original ode in \(x,y\) coordinates |
Canonical coordinates transformation |
ODE in canonical coordinates \((R,S)\) |
| \( \frac {dy}{dx} = -\frac {y \left (-2 x^{2}+y^{2}\right )}{x \left (-x^{2}+2 y^{2}\right )}\) |
|
\( \frac {d S}{d R} = -\frac {1}{3 R}\) |
| |
\(\!\begin {aligned} R&= x\\ S&= \frac {\ln \left (x +y \right )}{6}+\frac {\ln \left (y -x \right )}{6}+\frac {\ln \left (y \right )}{3} \end {aligned} \) |
|
Summary
The solution(s) found are the following \begin{align*} \tag{1} \frac {\ln \left (x +y\right )}{6}+\frac {\ln \left (y-x \right )}{6}+\frac {\ln \left (y\right )}{3} &= -\frac {\ln \left (x \right )}{3}+c_{1} \\ \end{align*}
Verification of solutions
\[ \frac {\ln \left (x +y\right )}{6}+\frac {\ln \left (y-x \right )}{6}+\frac {\ln \left (y\right )}{3} = -\frac {\ln \left (x \right )}{3}+c_{1} \] Verified OK.
Entering Exact first order ODE solver. (Form one type)
To solve an ode of the form\begin {equation} M\left ( x,y\right ) +N\left ( x,y\right ) \frac {dy}{dx}=0\tag {A} \end {equation} We assume there exists a function \(\phi \left ( x,y\right ) =c\) where \(c\) is constant, that satisfies the ode. Taking derivative of \(\phi \) w.r.t. \(x\) gives\[ \frac {d}{dx}\phi \left ( x,y\right ) =0 \] Hence\begin {equation} \frac {\partial \phi }{\partial x}+\frac {\partial \phi }{\partial y}\frac {dy}{dx}=0\tag {B} \end {equation} Comparing (A,B) shows that\begin {align*} \frac {\partial \phi }{\partial x} & =M\\ \frac {\partial \phi }{\partial y} & =N \end {align*}
But since \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) then for the above to be valid, we require that\[ \frac {\partial M}{\partial y}=\frac {\partial N}{\partial x}\] If the above condition is satisfied, then the original ode is called exact. We still need to determine \(\phi \left ( x,y\right ) \) but at least we know now that we can do that since the condition \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) is satisfied. If this condition is not satisfied then this method will not work and we have to now look for an integrating factor to force this condition, which might or might not exist. The first step is to write the ODE in standard form to check for exactness, which is \[ M(x,y) \mathop {\mathrm {d}x}+ N(x,y) \mathop {\mathrm {d}y}=0 \tag {1A} \] Therefore \begin {align*} \left (\frac {-x^{2}+2 y^{2}}{-x^{2} y +y^{3}}\right )\mathop {\mathrm {d}y} &= \left (-\frac {-2 x^{2}+y^{2}}{-x^{3}+x \,y^{2}}\right )\mathop {\mathrm {d}x}\\ \left (\frac {-2 x^{2}+y^{2}}{-x^{3}+x \,y^{2}}\right )\mathop {\mathrm {d}x} + \left (\frac {-x^{2}+2 y^{2}}{-x^{2} y +y^{3}}\right )\mathop {\mathrm {d}y} &= 0 \tag {2A} \end {align*}
Comparing (1A) and (2A) shows that \begin {align*} M(x,y) &= \frac {-2 x^{2}+y^{2}}{-x^{3}+x \,y^{2}}\\ N(x,y) &= \frac {-x^{2}+2 y^{2}}{-x^{2} y +y^{3}} \end {align*}
The next step is to determine if the ODE is is exact or not. The ODE is exact when the following condition is satisfied \[ \frac {\partial M}{\partial y} = \frac {\partial N}{\partial x} \] Using result found above gives \begin {align*} \frac {\partial M}{\partial y} &= \frac {\partial }{\partial y} \left (\frac {-2 x^{2}+y^{2}}{-x^{3}+x \,y^{2}}\right )\\ &= \frac {2 y x}{\left (x^{2}-y^{2}\right )^{2}} \end {align*}
And \begin {align*} \frac {\partial N}{\partial x} &= \frac {\partial }{\partial x} \left (\frac {-x^{2}+2 y^{2}}{-x^{2} y +y^{3}}\right )\\ &= \frac {2 y x}{\left (x^{2}-y^{2}\right )^{2}} \end {align*}
Since \(\frac {\partial M}{\partial y}= \frac {\partial N}{\partial x}\), then the ODE is exact The following equations are now set up to solve for the function \(\phi \left (x,y\right )\) \begin {align*} \frac {\partial \phi }{\partial x } &= M\tag {1} \\ \frac {\partial \phi }{\partial y } &= N\tag {2} \end {align*}
Integrating (1) w.r.t. \(x\) gives \begin{align*} \int \frac {\partial \phi }{\partial x} \mathop {\mathrm {d}x} &= \int M\mathop {\mathrm {d}x} \\ \int \frac {\partial \phi }{\partial x} \mathop {\mathrm {d}x} &= \int \frac {-2 x^{2}+y^{2}}{-x^{3}+x \,y^{2}}\mathop {\mathrm {d}x} \\ \tag{3} \phi &= \frac {\ln \left (x +y \right )}{2}+\frac {\ln \left (x -y \right )}{2}+\ln \left (x \right )+ f(y) \\ \end{align*} Where \(f(y)\) is used for the constant of integration since \(\phi \) is a function of both \(x\) and \(y\). Taking derivative of equation (3) w.r.t \(y\) gives \begin{align*} \tag{4} \frac {\partial \phi }{\partial y} &= \frac {1}{2 x +2 y}-\frac {1}{2 \left (x -y \right )}+f'(y) \\ &=-\frac {y}{x^{2}-y^{2}}+f'(y) \\ \end{align*} But equation (2) says that \(\frac {\partial \phi }{\partial y} = \frac {-x^{2}+2 y^{2}}{-x^{2} y +y^{3}}\). Therefore equation (4) becomes \begin{equation} \tag{5} \frac {-x^{2}+2 y^{2}}{-x^{2} y +y^{3}} = -\frac {y}{x^{2}-y^{2}}+f'(y) \end{equation} Solving equation (5) for \( f'(y)\) gives \[ f'(y) = \frac {1}{y} \] Integrating the above w.r.t \(y\) gives \begin{align*} \int f'(y) \mathop {\mathrm {d}y} &= \int \left ( \frac {1}{y}\right ) \mathop {\mathrm {d}y} \\ f(y) &= \ln \left (y \right )+ c_{1} \\ \end{align*} Where \(c_{1}\) is constant of integration. Substituting result found above for \(f(y)\) into equation (3) gives \(\phi \) \[ \phi = \frac {\ln \left (x +y \right )}{2}+\frac {\ln \left (x -y \right )}{2}+\ln \left (x \right )+\ln \left (y \right )+ c_{1} \] But since \(\phi \) itself is a constant function, then let \(\phi =c_{2}\) where \(c_{2}\) is new constant and combining \(c_{1}\) and \(c_{2}\) constants into new constant \(c_{1}\) gives the solution as \[ c_{1} = \frac {\ln \left (x +y \right )}{2}+\frac {\ln \left (x -y \right )}{2}+\ln \left (x \right )+\ln \left (y \right ) \]
The solution(s) found are the following \begin{align*} \tag{1} \frac {\ln \left (x +y\right )}{2}+\frac {\ln \left (x -y\right )}{2}+\ln \left (x \right )+\ln \left (y\right ) &= c_{1} \\ \end{align*}
Verification of solutions
\[ \frac {\ln \left (x +y\right )}{2}+\frac {\ln \left (x -y\right )}{2}+\ln \left (x \right )+\ln \left (y\right ) = c_{1} \] Verified OK.
Let \(p=y^{\prime }\) the ode becomes \begin {align*} \frac {-2 x^{2}+y^{2}}{-x^{3}+x \,y^{2}}+\frac {\left (-x^{2}+2 y^{2}\right ) p}{-x^{2} y +y^{3}} = 0 \end {align*}
Solving for \(y\) from the above results in \begin {align*} y &= \left (\frac {\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}}{6}-\frac {6 \left (-\frac {2}{3}-\frac {4 p^{2}}{9}\right )}{\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}}-\frac {2 p}{3}\right ) x\tag {1A}\\ y &= \left (-\frac {\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}}{12}+\frac {-2-\frac {4 p^{2}}{3}}{\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}}-\frac {2 p}{3}+\frac {i \sqrt {3}\, \left (\frac {\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}}{6}+\frac {-4-\frac {8 p^{2}}{3}}{\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}}\right )}{2}\right ) x\tag {2A}\\ y &= \left (-\frac {\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}}{12}+\frac {-2-\frac {4 p^{2}}{3}}{\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}}-\frac {2 p}{3}-\frac {i \sqrt {3}\, \left (\frac {\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}}{6}+\frac {-4-\frac {8 p^{2}}{3}}{\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}}\right )}{2}\right ) x\tag {3A} \end {align*}
This has the form \begin {align*} y=xf(p)+g(p)\tag {*} \end {align*}
Where \(f,g\) are functions of \(p=y'(x)\). Each of the above ode’s is dAlembert ode which is now solved. Solving ode 1A Taking derivative of (*) w.r.t. \(x\) gives \begin {align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end {align*}
Comparing the form \(y=x f + g\) to (1A) shows that \begin {align*} f &= \frac {\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}}{6}+\frac {\frac {8 p^{2}}{3}+4}{\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}}-\frac {2 p}{3}\\ g &= 0 \end {align*}
Hence (2) becomes \begin {align*} \frac {5 p}{3}-\frac {\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}}{6}-\frac {4 \left (2 p^{2}+3\right )}{3 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}} = x \left (\frac {-36-192 p^{2}+\frac {-2304 p^{3}-2196 p}{\sqrt {-96 p^{4}-183 p^{2}-96}}}{18 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}}+\frac {16 p}{3 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}}-\frac {4 \left (2 p^{2}+3\right ) \left (-36-192 p^{2}+\frac {-2304 p^{3}-2196 p}{\sqrt {-96 p^{4}-183 p^{2}-96}}\right )}{9 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{4}/{3}}}-\frac {2}{3}\right ) p^{\prime }\left (x \right )\tag {2A} \end {align*}
The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives \begin {align*} \frac {5 p}{3}-\frac {\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}}{6}-\frac {4 \left (2 p^{2}+3\right )}{3 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}} = 0 \end {align*}
Solving for \(p\) from the above gives \begin {align*} p&=1 \end {align*}
Substituting these in (1A) gives \begin {align*} y&=\frac {x \left (-100+60 i \sqrt {15}\right )^{{2}/{3}}-4 \left (-100+60 i \sqrt {15}\right )^{{1}/{3}} x +40 x}{6 \left (-100+60 i \sqrt {15}\right )^{{1}/{3}}} \end {align*}
The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in \begin {align*} p^{\prime }\left (x \right ) = \frac {\frac {5 p \left (x \right )}{3}-\frac {\left (-36 p \left (x \right )-64 p \left (x \right )^{3}+12 \sqrt {-96 p \left (x \right )^{4}-183 p \left (x \right )^{2}-96}\right )^{{1}/{3}}}{6}-\frac {4 \left (2 p \left (x \right )^{2}+3\right )}{3 \left (-36 p \left (x \right )-64 p \left (x \right )^{3}+12 \sqrt {-96 p \left (x \right )^{4}-183 p \left (x \right )^{2}-96}\right )^{{1}/{3}}}}{x \left (\frac {-36-192 p \left (x \right )^{2}+\frac {-2304 p \left (x \right )^{3}-2196 p \left (x \right )}{\sqrt {-96 p \left (x \right )^{4}-183 p \left (x \right )^{2}-96}}}{18 \left (-36 p \left (x \right )-64 p \left (x \right )^{3}+12 \sqrt {-96 p \left (x \right )^{4}-183 p \left (x \right )^{2}-96}\right )^{{2}/{3}}}+\frac {16 p \left (x \right )}{3 \left (-36 p \left (x \right )-64 p \left (x \right )^{3}+12 \sqrt {-96 p \left (x \right )^{4}-183 p \left (x \right )^{2}-96}\right )^{{1}/{3}}}-\frac {4 \left (2 p \left (x \right )^{2}+3\right ) \left (-36-192 p \left (x \right )^{2}+\frac {-2304 p \left (x \right )^{3}-2196 p \left (x \right )}{\sqrt {-96 p \left (x \right )^{4}-183 p \left (x \right )^{2}-96}}\right )}{9 \left (-36 p \left (x \right )-64 p \left (x \right )^{3}+12 \sqrt {-96 p \left (x \right )^{4}-183 p \left (x \right )^{2}-96}\right )^{{4}/{3}}}-\frac {2}{3}\right )}\tag {3} \end {align*}
This ODE is now solved for \(p \left (x \right )\).
Inverting the above ode gives \begin {align*} \frac {d}{d p}x \left (p \right ) = \frac {x \left (p \right ) \left (\frac {-36-192 p^{2}+\frac {-2304 p^{3}-2196 p}{\sqrt {-96 p^{4}-183 p^{2}-96}}}{18 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}}+\frac {16 p}{3 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}}-\frac {4 \left (2 p^{2}+3\right ) \left (-36-192 p^{2}+\frac {-2304 p^{3}-2196 p}{\sqrt {-96 p^{4}-183 p^{2}-96}}\right )}{9 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{4}/{3}}}-\frac {2}{3}\right )}{\frac {5 p}{3}-\frac {\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}}{6}-\frac {4 \left (2 p^{2}+3\right )}{3 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}}}\tag {4} \end {align*}
This ODE is now solved for \(x \left (p \right )\).
Entering Linear first order ODE solver. In canonical form a linear first order is \begin {align*} \frac {d}{d p}x \left (p \right ) + p(p)x \left (p \right ) &= q(p) \end {align*}
Where here \begin {align*} p(p) &=-\frac {\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{4}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}+16 p^{2} \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}+192 p^{3} \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}+256 p^{4} \sqrt {-96 p^{4}-183 p^{2}-96}-3072 p^{5}+3 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}+183 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-96 \left (-96 p^{4}-183 p^{2}-96\right ) p -144 p^{2} \sqrt {-96 p^{4}-183 p^{2}-96}-7536 p^{3}-72 \sqrt {-96 p^{4}-183 p^{2}-96}-4392 p}{\left (-16 p^{3}+3 \sqrt {-96 p^{4}-183 p^{2}-96}-9 p \right ) \sqrt {-96 p^{4}-183 p^{2}-96}\, \left (\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-10 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}+16 p^{2}+24\right )}\\ q(p) &=0 \end {align*}
Hence the ode is \begin {align*} \frac {d}{d p}x \left (p \right )-\frac {x \left (p \right ) \left (\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{4}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}+16 p^{2} \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}+192 p^{3} \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}+256 p^{4} \sqrt {-96 p^{4}-183 p^{2}-96}-3072 p^{5}+3 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}+183 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-96 \left (-96 p^{4}-183 p^{2}-96\right ) p -144 p^{2} \sqrt {-96 p^{4}-183 p^{2}-96}-7536 p^{3}-72 \sqrt {-96 p^{4}-183 p^{2}-96}-4392 p \right )}{\left (-16 p^{3}+3 \sqrt {-96 p^{4}-183 p^{2}-96}-9 p \right ) \sqrt {-96 p^{4}-183 p^{2}-96}\, \left (\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-10 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}+16 p^{2}+24\right )} = 0 \end {align*}
The integrating factor \(\mu \) is \[ \mu = {\mathrm e}^{\int -\frac {\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{4}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}+16 p^{2} \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}+192 p^{3} \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}+256 p^{4} \sqrt {-96 p^{4}-183 p^{2}-96}-3072 p^{5}+3 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}+183 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-96 \left (-96 p^{4}-183 p^{2}-96\right ) p -144 p^{2} \sqrt {-96 p^{4}-183 p^{2}-96}-7536 p^{3}-72 \sqrt {-96 p^{4}-183 p^{2}-96}-4392 p}{\left (-16 p^{3}+3 \sqrt {-96 p^{4}-183 p^{2}-96}-9 p \right ) \sqrt {-96 p^{4}-183 p^{2}-96}\, \left (\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-10 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}+16 p^{2}+24\right )}d p} \] The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}p}} \mu x &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}p}} \left ({\mathrm e}^{\int -\frac {\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{4}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}+16 p^{2} \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}+192 p^{3} \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}+256 p^{4} \sqrt {-96 p^{4}-183 p^{2}-96}-3072 p^{5}+3 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}+183 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-96 \left (-96 p^{4}-183 p^{2}-96\right ) p -144 p^{2} \sqrt {-96 p^{4}-183 p^{2}-96}-7536 p^{3}-72 \sqrt {-96 p^{4}-183 p^{2}-96}-4392 p}{\left (-16 p^{3}+3 \sqrt {-96 p^{4}-183 p^{2}-96}-9 p \right ) \sqrt {-96 p^{4}-183 p^{2}-96}\, \left (\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-10 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}+16 p^{2}+24\right )}d p} x\right ) &= 0 \end {align*}
Integrating gives \begin {align*} {\mathrm e}^{\int -\frac {\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{4}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}+16 p^{2} \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}+192 p^{3} \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}+256 p^{4} \sqrt {-96 p^{4}-183 p^{2}-96}-3072 p^{5}+3 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}+183 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-96 \left (-96 p^{4}-183 p^{2}-96\right ) p -144 p^{2} \sqrt {-96 p^{4}-183 p^{2}-96}-7536 p^{3}-72 \sqrt {-96 p^{4}-183 p^{2}-96}-4392 p}{\left (-16 p^{3}+3 \sqrt {-96 p^{4}-183 p^{2}-96}-9 p \right ) \sqrt {-96 p^{4}-183 p^{2}-96}\, \left (\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-10 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}+16 p^{2}+24\right )}d p} x &= c_{2} \end {align*}
Dividing both sides by the integrating factor \(\mu ={\mathrm e}^{\int -\frac {\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{4}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}+16 p^{2} \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}+192 p^{3} \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}+256 p^{4} \sqrt {-96 p^{4}-183 p^{2}-96}-3072 p^{5}+3 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}+183 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-96 \left (-96 p^{4}-183 p^{2}-96\right ) p -144 p^{2} \sqrt {-96 p^{4}-183 p^{2}-96}-7536 p^{3}-72 \sqrt {-96 p^{4}-183 p^{2}-96}-4392 p}{\left (-16 p^{3}+3 \sqrt {-96 p^{4}-183 p^{2}-96}-9 p \right ) \sqrt {-96 p^{4}-183 p^{2}-96}\, \left (\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-10 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}+16 p^{2}+24\right )}d p}\) results in \begin {align*} x \left (p \right ) &= c_{2} {\mathrm e}^{\int \frac {\left (-16 p^{2} \sqrt {-96 p^{4}-183 p^{2}-96}-192 p^{3}-3 \sqrt {-96 p^{4}-183 p^{2}-96}-183 p \right ) \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}+1152 \left (1+\frac {\left (\frac {1}{9} p^{3}+\frac {1}{16} p \right ) \sqrt {-96 p^{4}-183 p^{2}-96}}{2}+p^{4}+\frac {61 p^{2}}{32}\right ) \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}+8 \left (-32 p^{4}+18 p^{2}+9\right ) \sqrt {-96 p^{4}-183 p^{2}-96}-6144 p^{5}-10032 p^{3}-4824 p}{\sqrt {-96 p^{4}-183 p^{2}-96}\, \left (9 p +16 p^{3}-3 \sqrt {-96 p^{4}-183 p^{2}-96}\right ) \left (\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-10 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}+16 p^{2}+24\right )}d p} \end {align*}
Since the solution \(x \left (p \right )\) has unresolved integral, unable to continue.
Solving ode 2A Taking derivative of (*) w.r.t. \(x\) gives \begin {align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end {align*}
Comparing the form \(y=x f + g\) to (1A) shows that \begin {align*} f &= \frac {\frac {\left (i \sqrt {3}-1\right ) \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}}{4}-2 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}+4 \left (-1-i \sqrt {3}\right ) \left (p^{2}+\frac {3}{2}\right )}{3 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}}\\ g &= 0 \end {align*}
Hence (2) becomes \begin {align*} p -\frac {\frac {\left (i \sqrt {3}-1\right ) \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}}{4}-2 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}+4 \left (-1-i \sqrt {3}\right ) \left (p^{2}+\frac {3}{2}\right )}{3 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}} = x \left (\frac {\frac {\left (i \sqrt {3}-1\right ) \left (-36-192 p^{2}+\frac {-2304 p^{3}-2196 p}{\sqrt {-96 p^{4}-183 p^{2}-96}}\right )}{6 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}}-2 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}-\frac {2 p \left (-36-192 p^{2}+\frac {-2304 p^{3}-2196 p}{\sqrt {-96 p^{4}-183 p^{2}-96}}\right )}{3 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}}+8 \left (-1-i \sqrt {3}\right ) p}{3 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}}-\frac {\left (\frac {\left (i \sqrt {3}-1\right ) \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}}{4}-2 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}+4 \left (-1-i \sqrt {3}\right ) \left (p^{2}+\frac {3}{2}\right )\right ) \left (-36-192 p^{2}+\frac {-2304 p^{3}-2196 p}{\sqrt {-96 p^{4}-183 p^{2}-96}}\right )}{9 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{4}/{3}}}\right ) p^{\prime }\left (x \right )\tag {2A} \end {align*}
The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives \begin {align*} p -\frac {\frac {\left (i \sqrt {3}-1\right ) \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}}{4}-2 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}+4 \left (-1-i \sqrt {3}\right ) \left (p^{2}+\frac {3}{2}\right )}{3 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}} = 0 \end {align*}
Solving for \(p\) from the above gives \begin {align*} p&=-1 \end {align*}
Substituting these in (1A) gives \begin {align*} y&=\frac {i x \left (100+60 i \sqrt {15}\right )^{{2}/{3}} \sqrt {3}-40 i x \sqrt {3}-x \left (100+60 i \sqrt {15}\right )^{{2}/{3}}+8 \left (100+60 i \sqrt {15}\right )^{{1}/{3}} x -40 x}{12 \left (100+60 i \sqrt {15}\right )^{{1}/{3}}} \end {align*}
The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in \begin {align*} p^{\prime }\left (x \right ) = \frac {p \left (x \right )-\frac {\frac {\left (i \sqrt {3}-1\right ) \left (-36 p \left (x \right )-64 p \left (x \right )^{3}+12 \sqrt {-96 p \left (x \right )^{4}-183 p \left (x \right )^{2}-96}\right )^{{2}/{3}}}{4}-2 p \left (x \right ) \left (-36 p \left (x \right )-64 p \left (x \right )^{3}+12 \sqrt {-96 p \left (x \right )^{4}-183 p \left (x \right )^{2}-96}\right )^{{1}/{3}}+4 \left (-1-i \sqrt {3}\right ) \left (p \left (x \right )^{2}+\frac {3}{2}\right )}{3 \left (-36 p \left (x \right )-64 p \left (x \right )^{3}+12 \sqrt {-96 p \left (x \right )^{4}-183 p \left (x \right )^{2}-96}\right )^{{1}/{3}}}}{x \left (\frac {\frac {\left (i \sqrt {3}-1\right ) \left (-36-192 p \left (x \right )^{2}+\frac {-2304 p \left (x \right )^{3}-2196 p \left (x \right )}{\sqrt {-96 p \left (x \right )^{4}-183 p \left (x \right )^{2}-96}}\right )}{6 \left (-36 p \left (x \right )-64 p \left (x \right )^{3}+12 \sqrt {-96 p \left (x \right )^{4}-183 p \left (x \right )^{2}-96}\right )^{{1}/{3}}}-2 \left (-36 p \left (x \right )-64 p \left (x \right )^{3}+12 \sqrt {-96 p \left (x \right )^{4}-183 p \left (x \right )^{2}-96}\right )^{{1}/{3}}-\frac {2 p \left (x \right ) \left (-36-192 p \left (x \right )^{2}+\frac {-2304 p \left (x \right )^{3}-2196 p \left (x \right )}{\sqrt {-96 p \left (x \right )^{4}-183 p \left (x \right )^{2}-96}}\right )}{3 \left (-36 p \left (x \right )-64 p \left (x \right )^{3}+12 \sqrt {-96 p \left (x \right )^{4}-183 p \left (x \right )^{2}-96}\right )^{{2}/{3}}}+8 \left (-1-i \sqrt {3}\right ) p \left (x \right )}{3 \left (-36 p \left (x \right )-64 p \left (x \right )^{3}+12 \sqrt {-96 p \left (x \right )^{4}-183 p \left (x \right )^{2}-96}\right )^{{1}/{3}}}-\frac {\left (\frac {\left (i \sqrt {3}-1\right ) \left (-36 p \left (x \right )-64 p \left (x \right )^{3}+12 \sqrt {-96 p \left (x \right )^{4}-183 p \left (x \right )^{2}-96}\right )^{{2}/{3}}}{4}-2 p \left (x \right ) \left (-36 p \left (x \right )-64 p \left (x \right )^{3}+12 \sqrt {-96 p \left (x \right )^{4}-183 p \left (x \right )^{2}-96}\right )^{{1}/{3}}+4 \left (-1-i \sqrt {3}\right ) \left (p \left (x \right )^{2}+\frac {3}{2}\right )\right ) \left (-36-192 p \left (x \right )^{2}+\frac {-2304 p \left (x \right )^{3}-2196 p \left (x \right )}{\sqrt {-96 p \left (x \right )^{4}-183 p \left (x \right )^{2}-96}}\right )}{9 \left (-36 p \left (x \right )-64 p \left (x \right )^{3}+12 \sqrt {-96 p \left (x \right )^{4}-183 p \left (x \right )^{2}-96}\right )^{{4}/{3}}}\right )}\tag {3} \end {align*}
This ODE is now solved for \(p \left (x \right )\).
Inverting the above ode gives \begin {align*} \frac {d}{d p}x \left (p \right ) = \frac {x \left (p \right ) \left (\frac {\frac {\left (i \sqrt {3}-1\right ) \left (-36-192 p^{2}+\frac {-2304 p^{3}-2196 p}{\sqrt {-96 p^{4}-183 p^{2}-96}}\right )}{6 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}}-2 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}-\frac {2 p \left (-36-192 p^{2}+\frac {-2304 p^{3}-2196 p}{\sqrt {-96 p^{4}-183 p^{2}-96}}\right )}{3 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}}+8 \left (-1-i \sqrt {3}\right ) p}{3 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}}-\frac {\left (\frac {\left (i \sqrt {3}-1\right ) \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}}{4}-2 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}+4 \left (-1-i \sqrt {3}\right ) \left (p^{2}+\frac {3}{2}\right )\right ) \left (-36-192 p^{2}+\frac {-2304 p^{3}-2196 p}{\sqrt {-96 p^{4}-183 p^{2}-96}}\right )}{9 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{4}/{3}}}\right )}{p -\frac {\frac {\left (i \sqrt {3}-1\right ) \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}}{4}-2 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}+4 \left (-1-i \sqrt {3}\right ) \left (p^{2}+\frac {3}{2}\right )}{3 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}}}\tag {4} \end {align*}
This ODE is now solved for \(x \left (p \right )\).
Entering Linear first order ODE solver. In canonical form a linear first order is \begin {align*} \frac {d}{d p}x \left (p \right ) + p(p)x \left (p \right ) &= q(p) \end {align*}
Where here \begin {align*} p(p) &=-\frac {183 i \sqrt {3}\, p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}+144 i p^{2} \sqrt {3}\, \sqrt {-96 p^{4}-183 p^{2}-96}+192 i p^{3} \sqrt {3}\, \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}+72 i \sqrt {3}\, \sqrt {-96 p^{4}-183 p^{2}-96}+2 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{4}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}+4392 i \sqrt {3}\, p +16 i p^{2} \sqrt {3}\, \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-16 p^{2} \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-192 p^{3} \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}+3072 i p^{5} \sqrt {3}-256 i p^{4} \sqrt {3}\, \sqrt {-96 p^{4}-183 p^{2}-96}+96 i \left (-96 p^{4}-183 p^{2}-96\right ) \sqrt {3}\, p -256 p^{4} \sqrt {-96 p^{4}-183 p^{2}-96}+3072 p^{5}-3 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-183 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}+3 i \sqrt {3}\, \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}+7536 i p^{3} \sqrt {3}+96 \left (-96 p^{4}-183 p^{2}-96\right ) p +144 p^{2} \sqrt {-96 p^{4}-183 p^{2}-96}+7536 p^{3}+72 \sqrt {-96 p^{4}-183 p^{2}-96}+4392 p}{\left (-16 p^{3}+3 \sqrt {-96 p^{4}-183 p^{2}-96}-9 p \right ) \sqrt {-96 p^{4}-183 p^{2}-96}\, \left (i \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {3}-16 i \sqrt {3}\, p^{2}-\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-20 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}-24 i \sqrt {3}-16 p^{2}-24\right )}\\ q(p) &=0 \end {align*}
Hence the ode is \begin {align*} \frac {d}{d p}x \left (p \right )-\frac {x \left (p \right ) \left (183 i \sqrt {3}\, p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}+144 i p^{2} \sqrt {3}\, \sqrt {-96 p^{4}-183 p^{2}-96}+192 i p^{3} \sqrt {3}\, \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}+72 i \sqrt {3}\, \sqrt {-96 p^{4}-183 p^{2}-96}+2 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{4}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}+4392 i \sqrt {3}\, p +16 i p^{2} \sqrt {3}\, \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-16 p^{2} \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-192 p^{3} \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}+3072 i p^{5} \sqrt {3}-256 i p^{4} \sqrt {3}\, \sqrt {-96 p^{4}-183 p^{2}-96}+96 i \left (-96 p^{4}-183 p^{2}-96\right ) \sqrt {3}\, p -256 p^{4} \sqrt {-96 p^{4}-183 p^{2}-96}+3072 p^{5}-3 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-183 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}+3 i \sqrt {3}\, \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}+7536 i p^{3} \sqrt {3}+96 \left (-96 p^{4}-183 p^{2}-96\right ) p +144 p^{2} \sqrt {-96 p^{4}-183 p^{2}-96}+7536 p^{3}+72 \sqrt {-96 p^{4}-183 p^{2}-96}+4392 p \right )}{\left (-16 p^{3}+3 \sqrt {-96 p^{4}-183 p^{2}-96}-9 p \right ) \sqrt {-96 p^{4}-183 p^{2}-96}\, \left (i \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {3}-16 i \sqrt {3}\, p^{2}-\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-20 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}-24 i \sqrt {3}-16 p^{2}-24\right )} = 0 \end {align*}
The integrating factor \(\mu \) is \[ \mu = {\mathrm e}^{\int -\frac {183 i \sqrt {3}\, p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}+144 i p^{2} \sqrt {3}\, \sqrt {-96 p^{4}-183 p^{2}-96}+192 i p^{3} \sqrt {3}\, \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}+72 i \sqrt {3}\, \sqrt {-96 p^{4}-183 p^{2}-96}+2 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{4}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}+4392 i \sqrt {3}\, p +16 i p^{2} \sqrt {3}\, \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-16 p^{2} \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-192 p^{3} \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}+3072 i p^{5} \sqrt {3}-256 i p^{4} \sqrt {3}\, \sqrt {-96 p^{4}-183 p^{2}-96}+96 i \left (-96 p^{4}-183 p^{2}-96\right ) \sqrt {3}\, p -256 p^{4} \sqrt {-96 p^{4}-183 p^{2}-96}+3072 p^{5}-3 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-183 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}+3 i \sqrt {3}\, \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}+7536 i p^{3} \sqrt {3}+96 \left (-96 p^{4}-183 p^{2}-96\right ) p +144 p^{2} \sqrt {-96 p^{4}-183 p^{2}-96}+7536 p^{3}+72 \sqrt {-96 p^{4}-183 p^{2}-96}+4392 p}{\left (-16 p^{3}+3 \sqrt {-96 p^{4}-183 p^{2}-96}-9 p \right ) \sqrt {-96 p^{4}-183 p^{2}-96}\, \left (i \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {3}-16 i \sqrt {3}\, p^{2}-\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-20 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}-24 i \sqrt {3}-16 p^{2}-24\right )}d p} \] The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}p}} \mu x &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}p}} \left ({\mathrm e}^{\int -\frac {183 i \sqrt {3}\, p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}+144 i p^{2} \sqrt {3}\, \sqrt {-96 p^{4}-183 p^{2}-96}+192 i p^{3} \sqrt {3}\, \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}+72 i \sqrt {3}\, \sqrt {-96 p^{4}-183 p^{2}-96}+2 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{4}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}+4392 i \sqrt {3}\, p +16 i p^{2} \sqrt {3}\, \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-16 p^{2} \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-192 p^{3} \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}+3072 i p^{5} \sqrt {3}-256 i p^{4} \sqrt {3}\, \sqrt {-96 p^{4}-183 p^{2}-96}+96 i \left (-96 p^{4}-183 p^{2}-96\right ) \sqrt {3}\, p -256 p^{4} \sqrt {-96 p^{4}-183 p^{2}-96}+3072 p^{5}-3 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-183 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}+3 i \sqrt {3}\, \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}+7536 i p^{3} \sqrt {3}+96 \left (-96 p^{4}-183 p^{2}-96\right ) p +144 p^{2} \sqrt {-96 p^{4}-183 p^{2}-96}+7536 p^{3}+72 \sqrt {-96 p^{4}-183 p^{2}-96}+4392 p}{\left (-16 p^{3}+3 \sqrt {-96 p^{4}-183 p^{2}-96}-9 p \right ) \sqrt {-96 p^{4}-183 p^{2}-96}\, \left (i \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {3}-16 i \sqrt {3}\, p^{2}-\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-20 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}-24 i \sqrt {3}-16 p^{2}-24\right )}d p} x\right ) &= 0 \end {align*}
Integrating gives \begin {align*} {\mathrm e}^{\int -\frac {183 i \sqrt {3}\, p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}+144 i p^{2} \sqrt {3}\, \sqrt {-96 p^{4}-183 p^{2}-96}+192 i p^{3} \sqrt {3}\, \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}+72 i \sqrt {3}\, \sqrt {-96 p^{4}-183 p^{2}-96}+2 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{4}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}+4392 i \sqrt {3}\, p +16 i p^{2} \sqrt {3}\, \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-16 p^{2} \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-192 p^{3} \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}+3072 i p^{5} \sqrt {3}-256 i p^{4} \sqrt {3}\, \sqrt {-96 p^{4}-183 p^{2}-96}+96 i \left (-96 p^{4}-183 p^{2}-96\right ) \sqrt {3}\, p -256 p^{4} \sqrt {-96 p^{4}-183 p^{2}-96}+3072 p^{5}-3 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-183 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}+3 i \sqrt {3}\, \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}+7536 i p^{3} \sqrt {3}+96 \left (-96 p^{4}-183 p^{2}-96\right ) p +144 p^{2} \sqrt {-96 p^{4}-183 p^{2}-96}+7536 p^{3}+72 \sqrt {-96 p^{4}-183 p^{2}-96}+4392 p}{\left (-16 p^{3}+3 \sqrt {-96 p^{4}-183 p^{2}-96}-9 p \right ) \sqrt {-96 p^{4}-183 p^{2}-96}\, \left (i \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {3}-16 i \sqrt {3}\, p^{2}-\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-20 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}-24 i \sqrt {3}-16 p^{2}-24\right )}d p} x &= c_{4} \end {align*}
Dividing both sides by the integrating factor \(\mu ={\mathrm e}^{\int -\frac {183 i \sqrt {3}\, p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}+144 i p^{2} \sqrt {3}\, \sqrt {-96 p^{4}-183 p^{2}-96}+192 i p^{3} \sqrt {3}\, \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}+72 i \sqrt {3}\, \sqrt {-96 p^{4}-183 p^{2}-96}+2 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{4}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}+4392 i \sqrt {3}\, p +16 i p^{2} \sqrt {3}\, \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-16 p^{2} \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-192 p^{3} \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}+3072 i p^{5} \sqrt {3}-256 i p^{4} \sqrt {3}\, \sqrt {-96 p^{4}-183 p^{2}-96}+96 i \left (-96 p^{4}-183 p^{2}-96\right ) \sqrt {3}\, p -256 p^{4} \sqrt {-96 p^{4}-183 p^{2}-96}+3072 p^{5}-3 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-183 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}+3 i \sqrt {3}\, \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}+7536 i p^{3} \sqrt {3}+96 \left (-96 p^{4}-183 p^{2}-96\right ) p +144 p^{2} \sqrt {-96 p^{4}-183 p^{2}-96}+7536 p^{3}+72 \sqrt {-96 p^{4}-183 p^{2}-96}+4392 p}{\left (-16 p^{3}+3 \sqrt {-96 p^{4}-183 p^{2}-96}-9 p \right ) \sqrt {-96 p^{4}-183 p^{2}-96}\, \left (i \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {3}-16 i \sqrt {3}\, p^{2}-\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-20 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}-24 i \sqrt {3}-16 p^{2}-24\right )}d p}\) results in \begin {align*} x \left (p \right ) &= c_{4} {\mathrm e}^{\int -\frac {6144 \left (-\frac {\left (i \sqrt {3}-1\right ) \left (\left (\frac {p^{2}}{12}+\frac {1}{64}\right ) \sqrt {-96 p^{4}-183 p^{2}-96}+p^{3}+\frac {61 p}{64}\right ) \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}}{32}+\left (\left (\frac {1}{48} p^{3}+\frac {3}{256} p \right ) \sqrt {-96 p^{4}-183 p^{2}-96}+\frac {3}{8}+\frac {3 p^{4}}{8}+\frac {183 p^{2}}{256}\right ) \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}+\left (\left (\frac {1}{24} p^{4}-\frac {3}{128} p^{2}-\frac {3}{256}\right ) \sqrt {-96 p^{4}-183 p^{2}-96}+p^{5}+\frac {209 p^{3}}{128}+\frac {201 p}{256}\right ) \left (1+i \sqrt {3}\right )\right )}{\sqrt {-96 p^{4}-183 p^{2}-96}\, \left (9 p +16 p^{3}-3 \sqrt {-96 p^{4}-183 p^{2}-96}\right ) \left (-i \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {3}+16 i \sqrt {3}\, p^{2}+\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}+20 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}+24 i \sqrt {3}+16 p^{2}+24\right )}d p} \end {align*}
Since the solution \(x \left (p \right )\) has unresolved integral, unable to continue.
Solving ode 3A Taking derivative of (*) w.r.t. \(x\) gives \begin {align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end {align*}
Comparing the form \(y=x f + g\) to (1A) shows that \begin {align*} f &= \frac {16 i \sqrt {3}\, p^{2}-i \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {3}+24 i \sqrt {3}-16 p^{2}-8 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}-\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-24}{12 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}}\\ g &= 0 \end {align*}
Hence (2) becomes \begin {align*} p -\frac {16 i \sqrt {3}\, p^{2}-i \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {3}+24 i \sqrt {3}-16 p^{2}-8 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}-\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-24}{12 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}} = x \left (\frac {32 i \sqrt {3}\, p -\frac {2 i \sqrt {3}\, \left (-36-192 p^{2}+\frac {-2304 p^{3}-2196 p}{\sqrt {-96 p^{4}-183 p^{2}-96}}\right )}{3 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}}-32 p -8 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}-\frac {8 p \left (-36-192 p^{2}+\frac {-2304 p^{3}-2196 p}{\sqrt {-96 p^{4}-183 p^{2}-96}}\right )}{3 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}}-\frac {2 \left (-36-192 p^{2}+\frac {-2304 p^{3}-2196 p}{\sqrt {-96 p^{4}-183 p^{2}-96}}\right )}{3 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}}}{12 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}}-\frac {\left (16 i \sqrt {3}\, p^{2}-i \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {3}+24 i \sqrt {3}-16 p^{2}-8 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}-\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-24\right ) \left (-36-192 p^{2}+\frac {-2304 p^{3}-2196 p}{\sqrt {-96 p^{4}-183 p^{2}-96}}\right )}{36 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{4}/{3}}}\right ) p^{\prime }\left (x \right )\tag {2A} \end {align*}
The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives \begin {align*} p -\frac {16 i \sqrt {3}\, p^{2}-i \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {3}+24 i \sqrt {3}-16 p^{2}-8 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}-\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-24}{12 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}} = 0 \end {align*}
Solving for \(p\) from the above gives \begin {align*} p&=0 \end {align*}
Substituting these in (1A) gives \begin {align*} y&=0 \end {align*}
The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in \begin {align*} p^{\prime }\left (x \right ) = \frac {p \left (x \right )-\frac {16 i p \left (x \right )^{2} \sqrt {3}-i \left (-36 p \left (x \right )-64 p \left (x \right )^{3}+12 \sqrt {-96 p \left (x \right )^{4}-183 p \left (x \right )^{2}-96}\right )^{{2}/{3}} \sqrt {3}+24 i \sqrt {3}-16 p \left (x \right )^{2}-8 p \left (x \right ) \left (-36 p \left (x \right )-64 p \left (x \right )^{3}+12 \sqrt {-96 p \left (x \right )^{4}-183 p \left (x \right )^{2}-96}\right )^{{1}/{3}}-\left (-36 p \left (x \right )-64 p \left (x \right )^{3}+12 \sqrt {-96 p \left (x \right )^{4}-183 p \left (x \right )^{2}-96}\right )^{{2}/{3}}-24}{12 \left (-36 p \left (x \right )-64 p \left (x \right )^{3}+12 \sqrt {-96 p \left (x \right )^{4}-183 p \left (x \right )^{2}-96}\right )^{{1}/{3}}}}{x \left (\frac {32 i \sqrt {3}\, p \left (x \right )-\frac {2 i \sqrt {3}\, \left (-36-192 p \left (x \right )^{2}+\frac {-2304 p \left (x \right )^{3}-2196 p \left (x \right )}{\sqrt {-96 p \left (x \right )^{4}-183 p \left (x \right )^{2}-96}}\right )}{3 \left (-36 p \left (x \right )-64 p \left (x \right )^{3}+12 \sqrt {-96 p \left (x \right )^{4}-183 p \left (x \right )^{2}-96}\right )^{{1}/{3}}}-32 p \left (x \right )-8 \left (-36 p \left (x \right )-64 p \left (x \right )^{3}+12 \sqrt {-96 p \left (x \right )^{4}-183 p \left (x \right )^{2}-96}\right )^{{1}/{3}}-\frac {8 p \left (x \right ) \left (-36-192 p \left (x \right )^{2}+\frac {-2304 p \left (x \right )^{3}-2196 p \left (x \right )}{\sqrt {-96 p \left (x \right )^{4}-183 p \left (x \right )^{2}-96}}\right )}{3 \left (-36 p \left (x \right )-64 p \left (x \right )^{3}+12 \sqrt {-96 p \left (x \right )^{4}-183 p \left (x \right )^{2}-96}\right )^{{2}/{3}}}-\frac {2 \left (-36-192 p \left (x \right )^{2}+\frac {-2304 p \left (x \right )^{3}-2196 p \left (x \right )}{\sqrt {-96 p \left (x \right )^{4}-183 p \left (x \right )^{2}-96}}\right )}{3 \left (-36 p \left (x \right )-64 p \left (x \right )^{3}+12 \sqrt {-96 p \left (x \right )^{4}-183 p \left (x \right )^{2}-96}\right )^{{1}/{3}}}}{12 \left (-36 p \left (x \right )-64 p \left (x \right )^{3}+12 \sqrt {-96 p \left (x \right )^{4}-183 p \left (x \right )^{2}-96}\right )^{{1}/{3}}}-\frac {\left (16 i p \left (x \right )^{2} \sqrt {3}-i \left (-36 p \left (x \right )-64 p \left (x \right )^{3}+12 \sqrt {-96 p \left (x \right )^{4}-183 p \left (x \right )^{2}-96}\right )^{{2}/{3}} \sqrt {3}+24 i \sqrt {3}-16 p \left (x \right )^{2}-8 p \left (x \right ) \left (-36 p \left (x \right )-64 p \left (x \right )^{3}+12 \sqrt {-96 p \left (x \right )^{4}-183 p \left (x \right )^{2}-96}\right )^{{1}/{3}}-\left (-36 p \left (x \right )-64 p \left (x \right )^{3}+12 \sqrt {-96 p \left (x \right )^{4}-183 p \left (x \right )^{2}-96}\right )^{{2}/{3}}-24\right ) \left (-36-192 p \left (x \right )^{2}+\frac {-2304 p \left (x \right )^{3}-2196 p \left (x \right )}{\sqrt {-96 p \left (x \right )^{4}-183 p \left (x \right )^{2}-96}}\right )}{36 \left (-36 p \left (x \right )-64 p \left (x \right )^{3}+12 \sqrt {-96 p \left (x \right )^{4}-183 p \left (x \right )^{2}-96}\right )^{{4}/{3}}}\right )}\tag {3} \end {align*}
This ODE is now solved for \(p \left (x \right )\).
Inverting the above ode gives \begin {align*} \frac {d}{d p}x \left (p \right ) = \frac {x \left (p \right ) \left (\frac {32 i \sqrt {3}\, p -\frac {2 i \sqrt {3}\, \left (-36-192 p^{2}+\frac {-2304 p^{3}-2196 p}{\sqrt {-96 p^{4}-183 p^{2}-96}}\right )}{3 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}}-32 p -8 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}-\frac {8 p \left (-36-192 p^{2}+\frac {-2304 p^{3}-2196 p}{\sqrt {-96 p^{4}-183 p^{2}-96}}\right )}{3 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}}-\frac {2 \left (-36-192 p^{2}+\frac {-2304 p^{3}-2196 p}{\sqrt {-96 p^{4}-183 p^{2}-96}}\right )}{3 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}}}{12 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}}-\frac {\left (16 i \sqrt {3}\, p^{2}-i \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {3}+24 i \sqrt {3}-16 p^{2}-8 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}-\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-24\right ) \left (-36-192 p^{2}+\frac {-2304 p^{3}-2196 p}{\sqrt {-96 p^{4}-183 p^{2}-96}}\right )}{36 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{4}/{3}}}\right )}{p -\frac {16 i \sqrt {3}\, p^{2}-i \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {3}+24 i \sqrt {3}-16 p^{2}-8 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}-\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-24}{12 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}}}\tag {4} \end {align*}
This ODE is now solved for \(x \left (p \right )\).
Entering Linear first order ODE solver. In canonical form a linear first order is \begin {align*} \frac {d}{d p}x \left (p \right ) + p(p)x \left (p \right ) &= q(p) \end {align*}
Where here \begin {align*} p(p) &=\frac {-183 i \sqrt {3}\, p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-144 i p^{2} \sqrt {3}\, \sqrt {-96 p^{4}-183 p^{2}-96}-192 i p^{3} \sqrt {3}\, \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-72 i \sqrt {3}\, \sqrt {-96 p^{4}-183 p^{2}-96}+2 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{4}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-4392 i \sqrt {3}\, p -16 i p^{2} \sqrt {3}\, \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-16 p^{2} \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-192 p^{3} \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-3072 i p^{5} \sqrt {3}+256 i p^{4} \sqrt {3}\, \sqrt {-96 p^{4}-183 p^{2}-96}-96 i \left (-96 p^{4}-183 p^{2}-96\right ) \sqrt {3}\, p -256 p^{4} \sqrt {-96 p^{4}-183 p^{2}-96}+3072 p^{5}-3 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-183 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-3 i \sqrt {3}\, \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-7536 i p^{3} \sqrt {3}+96 \left (-96 p^{4}-183 p^{2}-96\right ) p +144 p^{2} \sqrt {-96 p^{4}-183 p^{2}-96}+7536 p^{3}+72 \sqrt {-96 p^{4}-183 p^{2}-96}+4392 p}{\left (-16 p^{3}+3 \sqrt {-96 p^{4}-183 p^{2}-96}-9 p \right ) \sqrt {-96 p^{4}-183 p^{2}-96}\, \left (i \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {3}-16 i \sqrt {3}\, p^{2}+\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}+20 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}-24 i \sqrt {3}+16 p^{2}+24\right )}\\ q(p) &=0 \end {align*}
Hence the ode is \begin {align*} \frac {d}{d p}x \left (p \right )+\frac {x \left (p \right ) \left (-183 i \sqrt {3}\, p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-144 i p^{2} \sqrt {3}\, \sqrt {-96 p^{4}-183 p^{2}-96}-192 i p^{3} \sqrt {3}\, \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-72 i \sqrt {3}\, \sqrt {-96 p^{4}-183 p^{2}-96}+2 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{4}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-4392 i \sqrt {3}\, p -16 i p^{2} \sqrt {3}\, \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-16 p^{2} \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-192 p^{3} \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-3072 i p^{5} \sqrt {3}+256 i p^{4} \sqrt {3}\, \sqrt {-96 p^{4}-183 p^{2}-96}-96 i \left (-96 p^{4}-183 p^{2}-96\right ) \sqrt {3}\, p -256 p^{4} \sqrt {-96 p^{4}-183 p^{2}-96}+3072 p^{5}-3 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-183 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-3 i \sqrt {3}\, \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-7536 i p^{3} \sqrt {3}+96 \left (-96 p^{4}-183 p^{2}-96\right ) p +144 p^{2} \sqrt {-96 p^{4}-183 p^{2}-96}+7536 p^{3}+72 \sqrt {-96 p^{4}-183 p^{2}-96}+4392 p \right )}{\left (-16 p^{3}+3 \sqrt {-96 p^{4}-183 p^{2}-96}-9 p \right ) \sqrt {-96 p^{4}-183 p^{2}-96}\, \left (i \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {3}-16 i \sqrt {3}\, p^{2}+\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}+20 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}-24 i \sqrt {3}+16 p^{2}+24\right )} = 0 \end {align*}
The integrating factor \(\mu \) is \[ \mu = {\mathrm e}^{\int \frac {-183 i \sqrt {3}\, p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-144 i p^{2} \sqrt {3}\, \sqrt {-96 p^{4}-183 p^{2}-96}-192 i p^{3} \sqrt {3}\, \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-72 i \sqrt {3}\, \sqrt {-96 p^{4}-183 p^{2}-96}+2 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{4}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-4392 i \sqrt {3}\, p -16 i p^{2} \sqrt {3}\, \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-16 p^{2} \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-192 p^{3} \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-3072 i p^{5} \sqrt {3}+256 i p^{4} \sqrt {3}\, \sqrt {-96 p^{4}-183 p^{2}-96}-96 i \left (-96 p^{4}-183 p^{2}-96\right ) \sqrt {3}\, p -256 p^{4} \sqrt {-96 p^{4}-183 p^{2}-96}+3072 p^{5}-3 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-183 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-3 i \sqrt {3}\, \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-7536 i p^{3} \sqrt {3}+96 \left (-96 p^{4}-183 p^{2}-96\right ) p +144 p^{2} \sqrt {-96 p^{4}-183 p^{2}-96}+7536 p^{3}+72 \sqrt {-96 p^{4}-183 p^{2}-96}+4392 p}{\left (-16 p^{3}+3 \sqrt {-96 p^{4}-183 p^{2}-96}-9 p \right ) \sqrt {-96 p^{4}-183 p^{2}-96}\, \left (i \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {3}-16 i \sqrt {3}\, p^{2}+\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}+20 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}-24 i \sqrt {3}+16 p^{2}+24\right )}d p} \] The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}p}} \mu x &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}p}} \left ({\mathrm e}^{\int \frac {-183 i \sqrt {3}\, p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-144 i p^{2} \sqrt {3}\, \sqrt {-96 p^{4}-183 p^{2}-96}-192 i p^{3} \sqrt {3}\, \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-72 i \sqrt {3}\, \sqrt {-96 p^{4}-183 p^{2}-96}+2 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{4}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-4392 i \sqrt {3}\, p -16 i p^{2} \sqrt {3}\, \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-16 p^{2} \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-192 p^{3} \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-3072 i p^{5} \sqrt {3}+256 i p^{4} \sqrt {3}\, \sqrt {-96 p^{4}-183 p^{2}-96}-96 i \left (-96 p^{4}-183 p^{2}-96\right ) \sqrt {3}\, p -256 p^{4} \sqrt {-96 p^{4}-183 p^{2}-96}+3072 p^{5}-3 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-183 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-3 i \sqrt {3}\, \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-7536 i p^{3} \sqrt {3}+96 \left (-96 p^{4}-183 p^{2}-96\right ) p +144 p^{2} \sqrt {-96 p^{4}-183 p^{2}-96}+7536 p^{3}+72 \sqrt {-96 p^{4}-183 p^{2}-96}+4392 p}{\left (-16 p^{3}+3 \sqrt {-96 p^{4}-183 p^{2}-96}-9 p \right ) \sqrt {-96 p^{4}-183 p^{2}-96}\, \left (i \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {3}-16 i \sqrt {3}\, p^{2}+\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}+20 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}-24 i \sqrt {3}+16 p^{2}+24\right )}d p} x\right ) &= 0 \end {align*}
Integrating gives \begin {align*} {\mathrm e}^{\int \frac {-183 i \sqrt {3}\, p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-144 i p^{2} \sqrt {3}\, \sqrt {-96 p^{4}-183 p^{2}-96}-192 i p^{3} \sqrt {3}\, \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-72 i \sqrt {3}\, \sqrt {-96 p^{4}-183 p^{2}-96}+2 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{4}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-4392 i \sqrt {3}\, p -16 i p^{2} \sqrt {3}\, \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-16 p^{2} \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-192 p^{3} \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-3072 i p^{5} \sqrt {3}+256 i p^{4} \sqrt {3}\, \sqrt {-96 p^{4}-183 p^{2}-96}-96 i \left (-96 p^{4}-183 p^{2}-96\right ) \sqrt {3}\, p -256 p^{4} \sqrt {-96 p^{4}-183 p^{2}-96}+3072 p^{5}-3 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-183 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-3 i \sqrt {3}\, \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-7536 i p^{3} \sqrt {3}+96 \left (-96 p^{4}-183 p^{2}-96\right ) p +144 p^{2} \sqrt {-96 p^{4}-183 p^{2}-96}+7536 p^{3}+72 \sqrt {-96 p^{4}-183 p^{2}-96}+4392 p}{\left (-16 p^{3}+3 \sqrt {-96 p^{4}-183 p^{2}-96}-9 p \right ) \sqrt {-96 p^{4}-183 p^{2}-96}\, \left (i \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {3}-16 i \sqrt {3}\, p^{2}+\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}+20 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}-24 i \sqrt {3}+16 p^{2}+24\right )}d p} x &= c_{6} \end {align*}
Dividing both sides by the integrating factor \(\mu ={\mathrm e}^{\int \frac {-183 i \sqrt {3}\, p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-144 i p^{2} \sqrt {3}\, \sqrt {-96 p^{4}-183 p^{2}-96}-192 i p^{3} \sqrt {3}\, \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-72 i \sqrt {3}\, \sqrt {-96 p^{4}-183 p^{2}-96}+2 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{4}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-4392 i \sqrt {3}\, p -16 i p^{2} \sqrt {3}\, \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-16 p^{2} \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-192 p^{3} \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-3072 i p^{5} \sqrt {3}+256 i p^{4} \sqrt {3}\, \sqrt {-96 p^{4}-183 p^{2}-96}-96 i \left (-96 p^{4}-183 p^{2}-96\right ) \sqrt {3}\, p -256 p^{4} \sqrt {-96 p^{4}-183 p^{2}-96}+3072 p^{5}-3 \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-183 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-3 i \sqrt {3}\, \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {-96 p^{4}-183 p^{2}-96}-7536 i p^{3} \sqrt {3}+96 \left (-96 p^{4}-183 p^{2}-96\right ) p +144 p^{2} \sqrt {-96 p^{4}-183 p^{2}-96}+7536 p^{3}+72 \sqrt {-96 p^{4}-183 p^{2}-96}+4392 p}{\left (-16 p^{3}+3 \sqrt {-96 p^{4}-183 p^{2}-96}-9 p \right ) \sqrt {-96 p^{4}-183 p^{2}-96}\, \left (i \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {3}-16 i \sqrt {3}\, p^{2}+\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}+20 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}-24 i \sqrt {3}+16 p^{2}+24\right )}d p}\) results in \begin {align*} x \left (p \right ) &= c_{6} {\mathrm e}^{\int -\frac {6144 \left (-\frac {\left (1+i \sqrt {3}\right ) \left (\left (\frac {p^{2}}{12}+\frac {1}{64}\right ) \sqrt {-96 p^{4}-183 p^{2}-96}+p^{3}+\frac {61 p}{64}\right ) \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}}{32}+\left (\left (-\frac {1}{48} p^{3}-\frac {3}{256} p \right ) \sqrt {-96 p^{4}-183 p^{2}-96}-\frac {3}{8}-\frac {3 p^{4}}{8}-\frac {183 p^{2}}{256}\right ) \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}+\left (\left (\frac {1}{24} p^{4}-\frac {3}{128} p^{2}-\frac {3}{256}\right ) \sqrt {-96 p^{4}-183 p^{2}-96}+p^{5}+\frac {209 p^{3}}{128}+\frac {201 p}{256}\right ) \left (i \sqrt {3}-1\right )\right )}{\sqrt {-96 p^{4}-183 p^{2}-96}\, \left (9 p +16 p^{3}-3 \sqrt {-96 p^{4}-183 p^{2}-96}\right ) \left (-i \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}} \sqrt {3}+16 i \sqrt {3}\, p^{2}-\left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{2}/{3}}-20 p \left (-36 p -64 p^{3}+12 \sqrt {-96 p^{4}-183 p^{2}-96}\right )^{{1}/{3}}+24 i \sqrt {3}-16 p^{2}-24\right )}d p} \end {align*}
Since the solution \(x \left (p \right )\) has unresolved integral, unable to continue.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {x \left (-100+60 i \sqrt {15}\right )^{{2}/{3}}-4 \left (-100+60 i \sqrt {15}\right )^{{1}/{3}} x +40 x}{6 \left (-100+60 i \sqrt {15}\right )^{{1}/{3}}} \\ \tag{2} y &= \frac {i x \left (100+60 i \sqrt {15}\right )^{{2}/{3}} \sqrt {3}-40 i x \sqrt {3}-x \left (100+60 i \sqrt {15}\right )^{{2}/{3}}+8 \left (100+60 i \sqrt {15}\right )^{{1}/{3}} x -40 x}{12 \left (100+60 i \sqrt {15}\right )^{{1}/{3}}} \\ \tag{3} y &= 0 \\ \end{align*}
Verification of solutions
\[ y = \frac {x \left (-100+60 i \sqrt {15}\right )^{{2}/{3}}-4 \left (-100+60 i \sqrt {15}\right )^{{1}/{3}} x +40 x}{6 \left (-100+60 i \sqrt {15}\right )^{{1}/{3}}} \] Verified OK.
\[ y = \frac {i x \left (100+60 i \sqrt {15}\right )^{{2}/{3}} \sqrt {3}-40 i x \sqrt {3}-x \left (100+60 i \sqrt {15}\right )^{{2}/{3}}+8 \left (100+60 i \sqrt {15}\right )^{{1}/{3}} x -40 x}{12 \left (100+60 i \sqrt {15}\right )^{{1}/{3}}} \] Verified OK.
\[ y = 0 \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {y^{2}-2 x^{2}}{x y^{2}-x^{3}}+\frac {\left (2 y^{2}-x^{2}\right ) y^{\prime }}{y^{3}-x^{2} y}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \square & {} & \textrm {Check if ODE is exact}\hspace {3pt} \\ {} & \circ & \textrm {ODE is exact if the lhs is the total derivative of a}\hspace {3pt} C^{2}\hspace {3pt}\textrm {function}\hspace {3pt} \\ {} & {} & F^{\prime }\left (x , y\right )=0 \\ {} & \circ & \textrm {Compute derivative of lhs}\hspace {3pt} \\ {} & {} & F^{\prime }\left (x , y\right )+\left (\frac {\partial }{\partial y}F \left (x , y\right )\right ) y^{\prime }=0 \\ {} & \circ & \textrm {Evaluate derivatives}\hspace {3pt} \\ {} & {} & \frac {2 y}{-x^{3}+x \,y^{2}}-\frac {2 \left (-2 x^{2}+y^{2}\right ) y x}{\left (-x^{3}+x \,y^{2}\right )^{2}}=-\frac {2 x}{-x^{2} y +y^{3}}+\frac {2 \left (-x^{2}+2 y^{2}\right ) y x}{\left (-x^{2} y +y^{3}\right )^{2}} \\ {} & \circ & \textrm {Simplify}\hspace {3pt} \\ {} & {} & \frac {2 y x}{\left (x^{2}-y^{2}\right )^{2}}=\frac {2 y x}{\left (x^{2}-y^{2}\right )^{2}} \\ {} & \circ & \textrm {Condition met, ODE is exact}\hspace {3pt} \\ \bullet & {} & \textrm {Exact ODE implies solution will be of this form}\hspace {3pt} \\ {} & {} & \left [F \left (x , y\right )=c_{1} , M \left (x , y\right )=F^{\prime }\left (x , y\right ), N \left (x , y\right )=\frac {\partial }{\partial y}F \left (x , y\right )\right ] \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} F \left (x , y\right )\hspace {3pt}\textrm {by integrating}\hspace {3pt} M \left (x , y\right )\hspace {3pt}\textrm {with respect to}\hspace {3pt} x \\ {} & {} & F \left (x , y\right )=\int \frac {-2 x^{2}+y^{2}}{-x^{3}+x \,y^{2}}d x +\textit {\_F1} \left (y \right ) \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & F \left (x , y\right )=\frac {\ln \left (x +y \right )}{2}+\frac {\ln \left (x -y \right )}{2}+\ln \left (x \right )+\textit {\_F1} \left (y \right ) \\ \bullet & {} & \textrm {Take derivative of}\hspace {3pt} F \left (x , y\right )\hspace {3pt}\textrm {with respect to}\hspace {3pt} y \\ {} & {} & N \left (x , y\right )=\frac {\partial }{\partial y}F \left (x , y\right ) \\ \bullet & {} & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & \frac {-x^{2}+2 y^{2}}{-x^{2} y +y^{3}}=\frac {1}{2 \left (x +y \right )}-\frac {1}{2 \left (x -y \right )}+\frac {d}{d y}\textit {\_F1} \left (y \right ) \\ \bullet & {} & \textrm {Isolate for}\hspace {3pt} \frac {d}{d y}\textit {\_F1} \left (y \right ) \\ {} & {} & \frac {d}{d y}\textit {\_F1} \left (y \right )=\frac {-x^{2}+2 y^{2}}{-x^{2} y +y^{3}}-\frac {1}{2 \left (x +y \right )}+\frac {1}{2 \left (x -y \right )} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} \textit {\_F1} \left (y \right ) \\ {} & {} & \textit {\_F1} \left (y \right )=-\frac {\ln \left (x -y \right )}{2}+\frac {\ln \left (y -x \right )}{2}+\ln \left (y \right ) \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \textit {\_F1} \left (y \right )\hspace {3pt}\textrm {into equation for}\hspace {3pt} F \left (x , y\right ) \\ {} & {} & F \left (x , y\right )=\frac {\ln \left (x +y \right )}{2}+\ln \left (x \right )+\frac {\ln \left (y -x \right )}{2}+\ln \left (y \right ) \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} F \left (x , y\right )\hspace {3pt}\textrm {into the solution of the ODE}\hspace {3pt} \\ {} & {} & \frac {\ln \left (x +y \right )}{2}+\ln \left (x \right )+\frac {\ln \left (y -x \right )}{2}+\ln \left (y \right )=c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=-\frac {x \mathit {RootOf}\left (2 x^{5} \textit {\_Z}^{6}+\left ({\mathrm e}^{c_{1}}\right )^{2} \textit {\_Z}^{8}-5 x^{4} \textit {\_Z}^{4}+4 x^{3} \textit {\_Z}^{2}-x^{2}\right )^{2}-1}{\mathit {RootOf}\left (2 x^{5} \textit {\_Z}^{6}+\left ({\mathrm e}^{c_{1}}\right )^{2} \textit {\_Z}^{8}-5 x^{4} \textit {\_Z}^{4}+4 x^{3} \textit {\_Z}^{2}-x^{2}\right )^{2}} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli trying separable trying inverse linear trying homogeneous types: trying homogeneous D <- homogeneous successful`
✓ Solution by Maple
Time used: 0.266 (sec). Leaf size: 71
dsolve((y(x)^2-2*x^2)/(x*y(x)^2-x^3)+ (2*y(x)^2-x^2)/(y(x)^3-x^2*y(x))*diff(y(x),x)=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \frac {\sqrt {\frac {2 c_{1} x^{3}-2 \sqrt {c_{1}^{2} x^{6}+4}}{c_{1} x^{3}}}\, x}{2} \\ y \left (x \right ) &= \frac {\sqrt {2}\, \sqrt {\frac {c_{1} x^{3}+\sqrt {c_{1}^{2} x^{6}+4}}{c_{1} x^{3}}}\, x}{2} \\ \end{align*}
✓ Solution by Mathematica
Time used: 15.598 (sec). Leaf size: 277
DSolve[(y[x]^2-2*x^2)/(x*y[x]^2-x^3)+ (2*y[x]^2-x^2)/(y[x]^3-x^2*y[x])*y'[x]==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {\sqrt {x^2-\frac {\sqrt {x^6-4 e^{2 c_1}}}{x}}}{\sqrt {2}} \\ y(x)\to \frac {\sqrt {x^2-\frac {\sqrt {x^6-4 e^{2 c_1}}}{x}}}{\sqrt {2}} \\ y(x)\to -\frac {\sqrt {\frac {x^3+\sqrt {x^6-4 e^{2 c_1}}}{x}}}{\sqrt {2}} \\ y(x)\to \frac {\sqrt {\frac {x^3+\sqrt {x^6-4 e^{2 c_1}}}{x}}}{\sqrt {2}} \\ y(x)\to -\frac {\sqrt {x^2-\frac {\sqrt {x^6}}{x}}}{\sqrt {2}} \\ y(x)\to \frac {\sqrt {x^2-\frac {\sqrt {x^6}}{x}}}{\sqrt {2}} \\ y(x)\to -\frac {\sqrt {\frac {\sqrt {x^6}+x^3}{x}}}{\sqrt {2}} \\ y(x)\to \frac {\sqrt {\frac {\sqrt {x^6}+x^3}{x}}}{\sqrt {2}} \\ \end{align*}