5.33 problem 30

Internal problem ID [1007]
Internal file name [OUTPUT/1008_Sunday_June_05_2022_01_56_41_AM_39474354/index.tex]

Book: Elementary differential equations with boundary value problems. William F. Trench. Brooks/Cole 2001
Section: Chapter 2, First order equations. Transformation of Nonlinear Equations into Separable Equations. Section 2.4 Page 68
Problem number: 30.
ODE order: 1.
ODE degree: 1.

The type(s) of ODE detected by this program : "dAlembert", "homogeneousTypeD2", "exactWithIntegrationFactor", "first_order_ode_lie_symmetry_calculated"

Maple gives the following as the ode type

[[_homogeneous, `class A`], _rational, _dAlembert]

\[ \boxed {y^{\prime }-\frac {y^{3}+2 x y^{2}+x^{2} y+x^{3}}{x \left (x +y\right )^{2}}=0} \]

5.33.1 Solving as homogeneousTypeD2 ode

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*} u^{\prime }\left (x \right ) x +u \left (x \right )-\frac {u \left (x \right )^{3} x^{3}+2 x^{3} u \left (x \right )^{2}+x^{3} u \left (x \right )+x^{3}}{x \left (x +u \left (x \right ) x \right )^{2}} = 0 \end {align*}

In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= \frac {1}{x \left (u +1\right )^{2}} \end {align*}

Where \(f(x)=\frac {1}{x}\) and \(g(u)=\frac {1}{\left (u +1\right )^{2}}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{\left (u +1\right )^{2}}} \,du &= \frac {1}{x} \,d x \\ \int { \frac {1}{\frac {1}{\left (u +1\right )^{2}}} \,du} &= \int {\frac {1}{x} \,d x} \\ \frac {\left (u +1\right )^{3}}{3}&=\ln \left (x \right )+c_{2} \\ \end{align*} The solution is \[ \frac {\left (u \left (x \right )+1\right )^{3}}{3}-\ln \left (x \right )-c_{2} = 0 \] Replacing \(u(x)\) in the above solution by \(\frac {y}{x}\) results in the solution for \(y\) in implicit form \begin {align*} \frac {\left (1+\frac {y}{x}\right )^{3}}{3}-\ln \left (x \right )-c_{2} = 0\\ \frac {\left (x +y\right )^{3}}{3 x^{3}}-\ln \left (x \right )-c_{2} = 0 \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} \frac {\left (x +y\right )^{3}}{3 x^{3}}-\ln \left (x \right )-c_{2} &= 0 \\ \end{align*}

Figure 201: Slope field plot

Verification of solutions

\[ \frac {\left (x +y\right )^{3}}{3 x^{3}}-\ln \left (x \right )-c_{2} = 0 \] Verified OK.

5.33.2 Solving as first order ode lie symmetry calculated ode

Writing the ode as \begin {align*} y^{\prime }&=\frac {x^{3}+x^{2} y +2 x \,y^{2}+y^{3}}{x \left (x +y \right )^{2}}\\ 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 {\left (x^{3}+x^{2} y +2 x \,y^{2}+y^{3}\right ) \left (b_{3}-a_{2}\right )}{x \left (x +y \right )^{2}}-\frac {\left (x^{3}+x^{2} y +2 x \,y^{2}+y^{3}\right )^{2} a_{3}}{x^{2} \left (x +y \right )^{4}}-\left (\frac {3 x^{2}+2 y x +2 y^{2}}{x \left (x +y \right )^{2}}-\frac {x^{3}+x^{2} y +2 x \,y^{2}+y^{3}}{x^{2} \left (x +y \right )^{2}}-\frac {2 \left (x^{3}+x^{2} y +2 x \,y^{2}+y^{3}\right )}{x \left (x +y \right )^{3}}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )-\left (\frac {x^{2}+4 y x +3 y^{2}}{x \left (x +y \right )^{2}}-\frac {2 \left (x^{3}+x^{2} y +2 x \,y^{2}+y^{3}\right )}{x \left (x +y \right )^{3}}\right ) \left (x b_{2}+y b_{3}+b_{1}\right ) = 0 \end{equation} Putting the above in normal form gives \[ -\frac {x^{6} a_{2}+x^{6} a_{3}-2 x^{6} b_{2}-x^{6} b_{3}+4 x^{5} y a_{2}+2 x^{5} y a_{3}-2 x^{5} y b_{2}-4 x^{5} y b_{3}+3 x^{4} y^{2} a_{2}+6 x^{4} y^{2} a_{3}-3 x^{4} y^{2} b_{3}+4 x^{3} y^{3} a_{3}-x^{5} b_{1}+x^{4} y a_{1}+2 x^{4} y b_{1}-2 x^{3} y^{2} a_{1}+6 x^{3} y^{2} b_{1}-6 x^{2} y^{3} a_{1}+4 x^{2} y^{3} b_{1}-4 x \,y^{4} a_{1}+x \,y^{4} b_{1}-y^{5} a_{1}}{x^{2} \left (x +y \right )^{4}} = 0 \] Setting the numerator to zero gives \begin{equation} \tag{6E} -x^{6} a_{2}-x^{6} a_{3}+2 x^{6} b_{2}+x^{6} b_{3}-4 x^{5} y a_{2}-2 x^{5} y a_{3}+2 x^{5} y b_{2}+4 x^{5} y b_{3}-3 x^{4} y^{2} a_{2}-6 x^{4} y^{2} a_{3}+3 x^{4} y^{2} b_{3}-4 x^{3} y^{3} a_{3}+x^{5} b_{1}-x^{4} y a_{1}-2 x^{4} y b_{1}+2 x^{3} y^{2} a_{1}-6 x^{3} y^{2} b_{1}+6 x^{2} y^{3} a_{1}-4 x^{2} y^{3} b_{1}+4 x \,y^{4} a_{1}-x \,y^{4} b_{1}+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} -a_{2} v_{1}^{6}-4 a_{2} v_{1}^{5} v_{2}-3 a_{2} v_{1}^{4} v_{2}^{2}-a_{3} v_{1}^{6}-2 a_{3} v_{1}^{5} v_{2}-6 a_{3} v_{1}^{4} v_{2}^{2}-4 a_{3} v_{1}^{3} v_{2}^{3}+2 b_{2} v_{1}^{6}+2 b_{2} v_{1}^{5} v_{2}+b_{3} v_{1}^{6}+4 b_{3} v_{1}^{5} v_{2}+3 b_{3} v_{1}^{4} v_{2}^{2}-a_{1} v_{1}^{4} v_{2}+2 a_{1} v_{1}^{3} v_{2}^{2}+6 a_{1} v_{1}^{2} v_{2}^{3}+4 a_{1} v_{1} v_{2}^{4}+a_{1} v_{2}^{5}+b_{1} v_{1}^{5}-2 b_{1} v_{1}^{4} v_{2}-6 b_{1} v_{1}^{3} v_{2}^{2}-4 b_{1} v_{1}^{2} v_{2}^{3}-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} \left (-a_{2}-a_{3}+2 b_{2}+b_{3}\right ) v_{1}^{6}+\left (-4 a_{2}-2 a_{3}+2 b_{2}+4 b_{3}\right ) v_{1}^{5} v_{2}+b_{1} v_{1}^{5}+\left (-3 a_{2}-6 a_{3}+3 b_{3}\right ) v_{1}^{4} v_{2}^{2}+\left (-a_{1}-2 b_{1}\right ) v_{1}^{4} v_{2}-4 a_{3} v_{1}^{3} v_{2}^{3}+\left (2 a_{1}-6 b_{1}\right ) v_{1}^{3} v_{2}^{2}+\left (6 a_{1}-4 b_{1}\right ) v_{1}^{2} v_{2}^{3}+\left (4 a_{1}-b_{1}\right ) v_{1} v_{2}^{4}+a_{1} v_{2}^{5} = 0 \end{equation} Setting each coefficients in (8E) to zero gives the following equations to solve \begin {align*} a_{1}&=0\\ b_{1}&=0\\ -4 a_{3}&=0\\ -a_{1}-2 b_{1}&=0\\ 2 a_{1}-6 b_{1}&=0\\ 4 a_{1}-b_{1}&=0\\ 6 a_{1}-4 b_{1}&=0\\ -3 a_{2}-6 a_{3}+3 b_{3}&=0\\ -4 a_{2}-2 a_{3}+2 b_{2}+4 b_{3}&=0\\ -a_{2}-a_{3}+2 b_{2}+b_{3}&=0 \end {align*}

Solving the above equations for the unknowns gives \begin {align*} a_{1}&=0\\ a_{2}&=b_{3}\\ a_{3}&=0\\ b_{1}&=0\\ b_{2}&=0\\ b_{3}&=b_{3} \end {align*}

Substituting the above solution in the anstaz (1E,2E) (using \(1\) as arbitrary value for any unknown in the RHS) gives \begin{align*} \xi &= x \\ \eta &= y \\ \end{align*} 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 {x^{3}+x^{2} y +2 x \,y^{2}+y^{3}}{x \left (x +y \right )^{2}}\right ) \left (x\right ) \\ &= -\frac {x^{3}}{x^{2}+2 y x +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 {x^{3}}{x^{2}+2 y x +y^{2}}}} dy \end {align*}

Which results in \begin {align*} S&= -\frac {\left (x +y \right )^{3}}{3 x^{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 {x^{3}+x^{2} y +2 x \,y^{2}+y^{3}}{x \left (x +y \right )^{2}} \end {align*}

Evaluating all the partial derivatives gives \begin {align*} R_{x} &= 1\\ R_{y} &= 0\\ S_{x} &= \frac {y \left (x +y \right )^{2}}{x^{4}}\\ S_{y} &= -\frac {\left (x +y \right )^{2}}{x^{3}} \end {align*}

Substituting all the above in (2) and simplifying gives the ode in canonical coordinates. \begin {align*} \frac {dS}{dR} &= -\frac {1}{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}{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 ) = -\ln \left (R \right )+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 {\left (x +y\right )^{3}}{3 x^{3}} = -\ln \left (x \right )+c_{1} \end {align*}

Which simplifies to \begin {align*} -\frac {\left (x +y\right )^{3}}{3 x^{3}} = -\ln \left (x \right )+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 \((R,S)\)
\( \frac {dy}{dx} = \frac {x^{3}+x^{2} y +2 x \,y^{2}+y^{3}}{x \left (x +y \right )^{2}}\)		\( \frac {d S}{d R} = -\frac {1}{R}\)
	\(\!\begin {aligned} R&= x\\ S&= -\frac {\left (x +y \right )^{3}}{3 x^{3}} \end {aligned} \)

Summary

The solution(s) found are the following \begin{align*} \tag{1} -\frac {\left (x +y\right )^{3}}{3 x^{3}} &= -\ln \left (x \right )+c_{1} \\ \end{align*}

Figure 202: Slope field plot

Verification of solutions

\[ -\frac {\left (x +y\right )^{3}}{3 x^{3}} = -\ln \left (x \right )+c_{1} \] Verified OK.

5.33.3 Solving as exact ode

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 (x \left (x +y \right )^{2}\right )\mathop {\mathrm {d}y} &= \left (x^{3}+x^{2} y +2 x \,y^{2}+y^{3}\right )\mathop {\mathrm {d}x}\\ \left (-x^{3}-x^{2} y -2 x \,y^{2}-y^{3}\right )\mathop {\mathrm {d}x} + \left (x \left (x +y \right )^{2}\right )\mathop {\mathrm {d}y} &= 0 \tag {2A} \end {align*}

Comparing (1A) and (2A) shows that \begin {align*} M(x,y) &= -x^{3}-x^{2} y -2 x \,y^{2}-y^{3}\\ N(x,y) &= x \left (x +y \right )^{2} \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 (-x^{3}-x^{2} y -2 x \,y^{2}-y^{3}\right )\\ &= -\left (x +3 y \right ) \left (x +y \right ) \end {align*}

And \begin {align*} \frac {\partial N}{\partial x} &= \frac {\partial }{\partial x} \left (x \left (x +y \right )^{2}\right )\\ &= 3 x^{2}+4 y x +y^{2} \end {align*}

Since \(\frac {\partial M}{\partial y} \neq \frac {\partial N}{\partial x}\), then the ODE is not exact. Since the ODE is not exact, we will try to find an integrating factor to make it exact. Let \begin {align*} A &= \frac {1}{N} \left (\frac {\partial M}{\partial y} - \frac {\partial N}{\partial x} \right ) \\ &=\frac {1}{x \left (x +y \right )^{2}}\left ( \left ( -x^{2}-4 y x -3 y^{2}\right ) - \left (\left (x +y \right )^{2}+2 x \left (x +y \right ) \right ) \right ) \\ &=-\frac {4}{x} \end {align*}

Since \(A\) does not depend on \(y\), then it can be used to find an integrating factor. The integrating factor \(\mu \) is \begin {align*} \mu &= e^{ \int A \mathop {\mathrm {d}x} } \\ &= e^{\int -\frac {4}{x}\mathop {\mathrm {d}x} } \end {align*}

The result of integrating gives \begin {align*} \mu &= e^{-4 \ln \left (x \right ) } \\ &= \frac {1}{x^{4}} \end {align*}

\(M\) and \(N\) are multiplied by this integrating factor, giving new \(M\) and new \(N\) which are called \(\overline {M}\) and \(\overline {N}\) for now so not to confuse them with the original \(M\) and \(N\). \begin {align*} \overline {M} &=\mu M \\ &= \frac {1}{x^{4}}\left (-x^{3}-x^{2} y -2 x \,y^{2}-y^{3}\right ) \\ &= \frac {-x^{3}-x^{2} y -2 x \,y^{2}-y^{3}}{x^{4}} \end {align*}

And \begin {align*} \overline {N} &=\mu N \\ &= \frac {1}{x^{4}}\left (x \left (x +y \right )^{2}\right ) \\ &= \frac {\left (x +y \right )^{2}}{x^{3}} \end {align*}

Now a modified ODE is ontained from the original ODE, which is exact and can be solved. The modified ODE is \begin {align*} \overline {M} + \overline {N} \frac { \mathop {\mathrm {d}y}}{\mathop {\mathrm {d}x}} &= 0 \\ \left (\frac {-x^{3}-x^{2} y -2 x \,y^{2}-y^{3}}{x^{4}}\right ) + \left (\frac {\left (x +y \right )^{2}}{x^{3}}\right ) \frac { \mathop {\mathrm {d}y}}{\mathop {\mathrm {d}x}} &= 0 \end {align*}

The following equations are now set up to solve for the function \(\phi \left (x,y\right )\) \begin {align*} \frac {\partial \phi }{\partial x } &= \overline {M}\tag {1} \\ \frac {\partial \phi }{\partial y } &= \overline {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 \overline {M}\mathop {\mathrm {d}x} \\ \int \frac {\partial \phi }{\partial x} \mathop {\mathrm {d}x} &= \int \frac {-x^{3}-x^{2} y -2 x \,y^{2}-y^{3}}{x^{4}}\mathop {\mathrm {d}x} \\ \tag{3} \phi &= \frac {y^{2}}{x^{2}}+\frac {y}{x}+\frac {y^{3}}{3 x^{3}}-\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 {2 y}{x^{2}}+\frac {1}{x}+\frac {y^{2}}{x^{3}}+f'(y) \\ &=\frac {\left (x +y \right )^{2}}{x^{3}}+f'(y) \\ \end{align*} But equation (2) says that \(\frac {\partial \phi }{\partial y} = \frac {\left (x +y \right )^{2}}{x^{3}}\). Therefore equation (4) becomes \begin{equation} \tag{5} \frac {\left (x +y \right )^{2}}{x^{3}} = \frac {\left (x +y \right )^{2}}{x^{3}}+f'(y) \end{equation} Solving equation (5) for \( f'(y)\) gives \[ f'(y) = 0 \] Therefore \[ f(y) = c_{1} \] Where \(c_{1}\) is constant of integration. Substituting this result for \(f(y)\) into equation (3) gives \(\phi \) \[ \phi = \frac {y^{2}}{x^{2}}+\frac {y}{x}+\frac {y^{3}}{3 x^{3}}-\ln \left (x \right )+ c_{1} \] 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 \[ c_{1} = \frac {y^{2}}{x^{2}}+\frac {y}{x}+\frac {y^{3}}{3 x^{3}}-\ln \left (x \right ) \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} \frac {y^{2}}{x^{2}}+\frac {y}{x}+\frac {y^{3}}{3 x^{3}}-\ln \left (x \right ) &= c_{1} \\ \end{align*}

Figure 203: Slope field plot

Verification of solutions

\[ \frac {y^{2}}{x^{2}}+\frac {y}{x}+\frac {y^{3}}{3 x^{3}}-\ln \left (x \right ) = c_{1} \] Verified OK.

5.33.4 Solving as dAlembert ode

Let \(p=y^{\prime }\) the ode becomes \begin {align*} p -\frac {x^{3}+x^{2} y +2 x \,y^{2}+y^{3}}{x \left (x +y \right )^{2}} = 0 \end {align*}

Solving for \(y\) from the above results in \begin {align*} y &= \left (\frac {\left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}}{6}-\frac {6 \left (-\frac {2}{9} p -\frac {1}{9}-\frac {1}{9} p^{2}\right )}{\left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}}+\frac {p}{3}-\frac {2}{3}\right ) x\tag {1A}\\ y &= \left (-\frac {\left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}}{12}+\frac {-\frac {2}{3} p -\frac {1}{3}-\frac {1}{3} p^{2}}{\left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}}+\frac {p}{3}-\frac {2}{3}+\frac {i \sqrt {3}\, \left (\frac {\left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}}{6}+\frac {-\frac {4}{3} p -\frac {2}{3}-\frac {2}{3} p^{2}}{\left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}}\right )}{2}\right ) x\tag {2A}\\ y &= \left (-\frac {\left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}}{12}+\frac {-\frac {2}{3} p -\frac {1}{3}-\frac {1}{3} p^{2}}{\left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}}+\frac {p}{3}-\frac {2}{3}-\frac {i \sqrt {3}\, \left (\frac {\left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}}{6}+\frac {-\frac {4}{3} p -\frac {2}{3}-\frac {2}{3} p^{2}}{\left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\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 (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}}{6}+\frac {2 \left (p +1\right )^{2}}{3 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}}+\frac {p}{3}-\frac {2}{3}\\ g &= 0 \end {align*}

Hence (2) becomes \begin {align*} \frac {2 p}{3}-\frac {\left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}}{6}-\frac {2 \left (p +1\right )^{2}}{3 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}}+\frac {2}{3} = x \left (\frac {48 p +24+24 p^{2}+\frac {-216 p^{2}-432 p -216}{\sqrt {-12 p^{3}-36 p^{2}-36 p +69}}}{18 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}}+\frac {\frac {4 p}{3}+\frac {4}{3}}{\left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}}-\frac {2 \left (p +1\right )^{2} \left (48 p +24+24 p^{2}+\frac {-216 p^{2}-432 p -216}{\sqrt {-12 p^{3}-36 p^{2}-36 p +69}}\right )}{9 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{4}/{3}}}+\frac {1}{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 {2 p}{3}-\frac {\left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}}{6}-\frac {2 \left (p +1\right )^{2}}{3 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}}+\frac {2}{3} = 0 \end {align*}

No singular solution are found

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 {2 p \left (x \right )}{3}-\frac {\left (24 p \left (x \right )^{2}+24 p \left (x \right )-100+8 p \left (x \right )^{3}+12 \sqrt {-12 p \left (x \right )^{3}-36 p \left (x \right )^{2}-36 p \left (x \right )+69}\right )^{{1}/{3}}}{6}-\frac {2 \left (p \left (x \right )+1\right )^{2}}{3 \left (24 p \left (x \right )^{2}+24 p \left (x \right )-100+8 p \left (x \right )^{3}+12 \sqrt {-12 p \left (x \right )^{3}-36 p \left (x \right )^{2}-36 p \left (x \right )+69}\right )^{{1}/{3}}}+\frac {2}{3}}{x \left (\frac {48 p \left (x \right )+24+24 p \left (x \right )^{2}+\frac {-216 p \left (x \right )^{2}-432 p \left (x \right )-216}{\sqrt {-12 p \left (x \right )^{3}-36 p \left (x \right )^{2}-36 p \left (x \right )+69}}}{18 \left (24 p \left (x \right )^{2}+24 p \left (x \right )-100+8 p \left (x \right )^{3}+12 \sqrt {-12 p \left (x \right )^{3}-36 p \left (x \right )^{2}-36 p \left (x \right )+69}\right )^{{2}/{3}}}+\frac {\frac {4 p \left (x \right )}{3}+\frac {4}{3}}{\left (24 p \left (x \right )^{2}+24 p \left (x \right )-100+8 p \left (x \right )^{3}+12 \sqrt {-12 p \left (x \right )^{3}-36 p \left (x \right )^{2}-36 p \left (x \right )+69}\right )^{{1}/{3}}}-\frac {2 \left (p \left (x \right )+1\right )^{2} \left (48 p \left (x \right )+24+24 p \left (x \right )^{2}+\frac {-216 p \left (x \right )^{2}-432 p \left (x \right )-216}{\sqrt {-12 p \left (x \right )^{3}-36 p \left (x \right )^{2}-36 p \left (x \right )+69}}\right )}{9 \left (24 p \left (x \right )^{2}+24 p \left (x \right )-100+8 p \left (x \right )^{3}+12 \sqrt {-12 p \left (x \right )^{3}-36 p \left (x \right )^{2}-36 p \left (x \right )+69}\right )^{{4}/{3}}}+\frac {1}{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 {48 p +24+24 p^{2}+\frac {-216 p^{2}-432 p -216}{\sqrt {-12 p^{3}-36 p^{2}-36 p +69}}}{18 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}}+\frac {\frac {4 p}{3}+\frac {4}{3}}{\left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}}-\frac {2 \left (p +1\right )^{2} \left (48 p +24+24 p^{2}+\frac {-216 p^{2}-432 p -216}{\sqrt {-12 p^{3}-36 p^{2}-36 p +69}}\right )}{9 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{4}/{3}}}+\frac {1}{3}\right )}{\frac {2 p}{3}-\frac {\left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}}{6}-\frac {2 \left (p +1\right )^{2}}{3 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}}+\frac {2}{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 (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{4}/{3}} \sqrt {-12 p^{3}-36 p^{2}-36 p +69}+4 p^{2} \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}} \sqrt {-12 p^{3}-36 p^{2}-36 p +69}+16 p^{4} \sqrt {-12 p^{3}-36 p^{2}-36 p +69}+8 p \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}} \sqrt {-12 p^{3}-36 p^{2}-36 p +69}+64 p^{3} \sqrt {-12 p^{3}-36 p^{2}-36 p +69}-36 p^{2} \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}+144 p^{4}+48 \left (-12 p^{3}-36 p^{2}-36 p +69\right ) p +4 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}} \sqrt {-12 p^{3}-36 p^{2}-36 p +69}+96 p^{2} \sqrt {-12 p^{3}-36 p^{2}-36 p +69}-72 p \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}-864 p^{2}-1152 p +3456-368 p \sqrt {-12 p^{3}-36 p^{2}-36 p +69}-36 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}-416 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}}{2 \left (\left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}-4 p \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}+4 p^{2}-4 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}+8 p +4\right ) \left (6 p^{2}+6 p -25+2 p^{3}+3 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right ) \sqrt {-12 p^{3}-36 p^{2}-36 p +69}}\\ 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 (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{4}/{3}} \sqrt {-12 p^{3}-36 p^{2}-36 p +69}+4 p^{2} \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}} \sqrt {-12 p^{3}-36 p^{2}-36 p +69}+16 p^{4} \sqrt {-12 p^{3}-36 p^{2}-36 p +69}+8 p \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}} \sqrt {-12 p^{3}-36 p^{2}-36 p +69}+64 p^{3} \sqrt {-12 p^{3}-36 p^{2}-36 p +69}-36 p^{2} \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}+144 p^{4}+48 \left (-12 p^{3}-36 p^{2}-36 p +69\right ) p +4 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}} \sqrt {-12 p^{3}-36 p^{2}-36 p +69}+96 p^{2} \sqrt {-12 p^{3}-36 p^{2}-36 p +69}-72 p \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}-864 p^{2}-1152 p +3456-368 p \sqrt {-12 p^{3}-36 p^{2}-36 p +69}-36 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}-416 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )}{2 \left (\left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}-4 p \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}+4 p^{2}-4 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}+8 p +4\right ) \left (6 p^{2}+6 p -25+2 p^{3}+3 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right ) \sqrt {-12 p^{3}-36 p^{2}-36 p +69}} = 0 \end {align*}

The integrating factor \(\mu \) is \[ \mu = {\mathrm e}^{\int \frac {\left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{4}/{3}} \sqrt {-12 p^{3}-36 p^{2}-36 p +69}+4 p^{2} \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}} \sqrt {-12 p^{3}-36 p^{2}-36 p +69}+16 p^{4} \sqrt {-12 p^{3}-36 p^{2}-36 p +69}+8 p \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}} \sqrt {-12 p^{3}-36 p^{2}-36 p +69}+64 p^{3} \sqrt {-12 p^{3}-36 p^{2}-36 p +69}-36 p^{2} \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}+144 p^{4}+48 \left (-12 p^{3}-36 p^{2}-36 p +69\right ) p +4 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}} \sqrt {-12 p^{3}-36 p^{2}-36 p +69}+96 p^{2} \sqrt {-12 p^{3}-36 p^{2}-36 p +69}-72 p \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}-864 p^{2}-1152 p +3456-368 p \sqrt {-12 p^{3}-36 p^{2}-36 p +69}-36 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}-416 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}}{2 \left (\left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}-4 p \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}+4 p^{2}-4 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}+8 p +4\right ) \left (6 p^{2}+6 p -25+2 p^{3}+3 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right ) \sqrt {-12 p^{3}-36 p^{2}-36 p +69}}d p} \] 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 (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{4}/{3}} \sqrt {-12 p^{3}-36 p^{2}-36 p +69}+4 p^{2} \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}} \sqrt {-12 p^{3}-36 p^{2}-36 p +69}+16 p^{4} \sqrt {-12 p^{3}-36 p^{2}-36 p +69}+8 p \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}} \sqrt {-12 p^{3}-36 p^{2}-36 p +69}+64 p^{3} \sqrt {-12 p^{3}-36 p^{2}-36 p +69}-36 p^{2} \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}+144 p^{4}+48 \left (-12 p^{3}-36 p^{2}-36 p +69\right ) p +4 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}} \sqrt {-12 p^{3}-36 p^{2}-36 p +69}+96 p^{2} \sqrt {-12 p^{3}-36 p^{2}-36 p +69}-72 p \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}-864 p^{2}-1152 p +3456-368 p \sqrt {-12 p^{3}-36 p^{2}-36 p +69}-36 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}-416 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}}{2 \left (\left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}-4 p \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}+4 p^{2}-4 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}+8 p +4\right ) \left (6 p^{2}+6 p -25+2 p^{3}+3 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right ) \sqrt {-12 p^{3}-36 p^{2}-36 p +69}}d p} x\right ) &= 0 \end {align*}

Integrating gives \begin {align*} {\mathrm e}^{\int \frac {\left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{4}/{3}} \sqrt {-12 p^{3}-36 p^{2}-36 p +69}+4 p^{2} \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}} \sqrt {-12 p^{3}-36 p^{2}-36 p +69}+16 p^{4} \sqrt {-12 p^{3}-36 p^{2}-36 p +69}+8 p \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}} \sqrt {-12 p^{3}-36 p^{2}-36 p +69}+64 p^{3} \sqrt {-12 p^{3}-36 p^{2}-36 p +69}-36 p^{2} \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}+144 p^{4}+48 \left (-12 p^{3}-36 p^{2}-36 p +69\right ) p +4 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}} \sqrt {-12 p^{3}-36 p^{2}-36 p +69}+96 p^{2} \sqrt {-12 p^{3}-36 p^{2}-36 p +69}-72 p \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}-864 p^{2}-1152 p +3456-368 p \sqrt {-12 p^{3}-36 p^{2}-36 p +69}-36 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}-416 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}}{2 \left (\left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}-4 p \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}+4 p^{2}-4 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}+8 p +4\right ) \left (6 p^{2}+6 p -25+2 p^{3}+3 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right ) \sqrt {-12 p^{3}-36 p^{2}-36 p +69}}d p} x &= c_{2} \end {align*}

Dividing both sides by the integrating factor \(\mu ={\mathrm e}^{\int \frac {\left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{4}/{3}} \sqrt {-12 p^{3}-36 p^{2}-36 p +69}+4 p^{2} \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}} \sqrt {-12 p^{3}-36 p^{2}-36 p +69}+16 p^{4} \sqrt {-12 p^{3}-36 p^{2}-36 p +69}+8 p \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}} \sqrt {-12 p^{3}-36 p^{2}-36 p +69}+64 p^{3} \sqrt {-12 p^{3}-36 p^{2}-36 p +69}-36 p^{2} \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}+144 p^{4}+48 \left (-12 p^{3}-36 p^{2}-36 p +69\right ) p +4 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}} \sqrt {-12 p^{3}-36 p^{2}-36 p +69}+96 p^{2} \sqrt {-12 p^{3}-36 p^{2}-36 p +69}-72 p \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}-864 p^{2}-1152 p +3456-368 p \sqrt {-12 p^{3}-36 p^{2}-36 p +69}-36 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}-416 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}}{2 \left (\left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}-4 p \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}+4 p^{2}-4 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}+8 p +4\right ) \left (6 p^{2}+6 p -25+2 p^{3}+3 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right ) \sqrt {-12 p^{3}-36 p^{2}-36 p +69}}d p}\) results in \begin {align*} x \left (p \right ) &= c_{2} {\mathrm e}^{-2 \left (\int \frac {\left (p +1\right )^{2} \left (\sqrt {-12 p^{3}-36 p^{2}-36 p +69}-9\right ) \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}+4 \left (\frac {207}{4}+\frac {\left (p^{3}+3 p^{2}+3 p -\frac {25}{2}\right ) \sqrt {-12 p^{3}-36 p^{2}-36 p +69}}{2}-9 p^{3}-27 p^{2}-27 p \right ) \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}+4 \left (\left (p^{3}+3 p^{2}+3 p -26\right ) \sqrt {-12 p^{3}-36 p^{2}-36 p +69}-27 p^{3}-81 p^{2}-81 p +216\right ) \left (p +1\right )}{\left (\left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}-4 p \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}+4 p^{2}-4 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}+8 p +4\right ) \left (6 p^{2}+6 p -25+2 p^{3}+3 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right ) \sqrt {-12 p^{3}-36 p^{2}-36 p +69}}d p \right )} \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 {\left (-\frac {i \sqrt {3}}{4}+\frac {1}{4}\right ) \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}+\left (-p +2\right ) \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}+\left (1+i \sqrt {3}\right ) \left (p +1\right )^{2}}{3 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}}\\ g &= 0 \end {align*}

Hence (2) becomes \begin {align*} p +\frac {\left (-\frac {i \sqrt {3}}{4}+\frac {1}{4}\right ) \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}+\left (-p +2\right ) \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}+\left (1+i \sqrt {3}\right ) \left (p +1\right )^{2}}{3 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}} = x \left (-\frac {\frac {2 \left (-\frac {i \sqrt {3}}{4}+\frac {1}{4}\right ) \left (48 p +24+24 p^{2}+\frac {-216 p^{2}-432 p -216}{\sqrt {-12 p^{3}-36 p^{2}-36 p +69}}\right )}{3 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}}-\left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}+\frac {\left (-p +2\right ) \left (48 p +24+24 p^{2}+\frac {-216 p^{2}-432 p -216}{\sqrt {-12 p^{3}-36 p^{2}-36 p +69}}\right )}{3 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}}+2 \left (1+i \sqrt {3}\right ) \left (p +1\right )}{3 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}}+\frac {\left (\left (-\frac {i \sqrt {3}}{4}+\frac {1}{4}\right ) \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}+\left (-p +2\right ) \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}+\left (1+i \sqrt {3}\right ) \left (p +1\right )^{2}\right ) \left (48 p +24+24 p^{2}+\frac {-216 p^{2}-432 p -216}{\sqrt {-12 p^{3}-36 p^{2}-36 p +69}}\right )}{9 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\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 {\left (-\frac {i \sqrt {3}}{4}+\frac {1}{4}\right ) \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}+\left (-p +2\right ) \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}+\left (1+i \sqrt {3}\right ) \left (p +1\right )^{2}}{3 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}} = 0 \end {align*}

No singular solution are found

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 {\left (-\frac {i \sqrt {3}}{4}+\frac {1}{4}\right ) \left (24 p \left (x \right )^{2}+24 p \left (x \right )-100+8 p \left (x \right )^{3}+12 \sqrt {-12 p \left (x \right )^{3}-36 p \left (x \right )^{2}-36 p \left (x \right )+69}\right )^{{2}/{3}}+\left (-p \left (x \right )+2\right ) \left (24 p \left (x \right )^{2}+24 p \left (x \right )-100+8 p \left (x \right )^{3}+12 \sqrt {-12 p \left (x \right )^{3}-36 p \left (x \right )^{2}-36 p \left (x \right )+69}\right )^{{1}/{3}}+\left (1+i \sqrt {3}\right ) \left (p \left (x \right )+1\right )^{2}}{3 \left (24 p \left (x \right )^{2}+24 p \left (x \right )-100+8 p \left (x \right )^{3}+12 \sqrt {-12 p \left (x \right )^{3}-36 p \left (x \right )^{2}-36 p \left (x \right )+69}\right )^{{1}/{3}}}}{x \left (-\frac {\frac {2 \left (-\frac {i \sqrt {3}}{4}+\frac {1}{4}\right ) \left (48 p \left (x \right )+24+24 p \left (x \right )^{2}+\frac {-216 p \left (x \right )^{2}-432 p \left (x \right )-216}{\sqrt {-12 p \left (x \right )^{3}-36 p \left (x \right )^{2}-36 p \left (x \right )+69}}\right )}{3 \left (24 p \left (x \right )^{2}+24 p \left (x \right )-100+8 p \left (x \right )^{3}+12 \sqrt {-12 p \left (x \right )^{3}-36 p \left (x \right )^{2}-36 p \left (x \right )+69}\right )^{{1}/{3}}}-\left (24 p \left (x \right )^{2}+24 p \left (x \right )-100+8 p \left (x \right )^{3}+12 \sqrt {-12 p \left (x \right )^{3}-36 p \left (x \right )^{2}-36 p \left (x \right )+69}\right )^{{1}/{3}}+\frac {\left (-p \left (x \right )+2\right ) \left (48 p \left (x \right )+24+24 p \left (x \right )^{2}+\frac {-216 p \left (x \right )^{2}-432 p \left (x \right )-216}{\sqrt {-12 p \left (x \right )^{3}-36 p \left (x \right )^{2}-36 p \left (x \right )+69}}\right )}{3 \left (24 p \left (x \right )^{2}+24 p \left (x \right )-100+8 p \left (x \right )^{3}+12 \sqrt {-12 p \left (x \right )^{3}-36 p \left (x \right )^{2}-36 p \left (x \right )+69}\right )^{{2}/{3}}}+2 \left (1+i \sqrt {3}\right ) \left (p \left (x \right )+1\right )}{3 \left (24 p \left (x \right )^{2}+24 p \left (x \right )-100+8 p \left (x \right )^{3}+12 \sqrt {-12 p \left (x \right )^{3}-36 p \left (x \right )^{2}-36 p \left (x \right )+69}\right )^{{1}/{3}}}+\frac {\left (\left (-\frac {i \sqrt {3}}{4}+\frac {1}{4}\right ) \left (24 p \left (x \right )^{2}+24 p \left (x \right )-100+8 p \left (x \right )^{3}+12 \sqrt {-12 p \left (x \right )^{3}-36 p \left (x \right )^{2}-36 p \left (x \right )+69}\right )^{{2}/{3}}+\left (-p \left (x \right )+2\right ) \left (24 p \left (x \right )^{2}+24 p \left (x \right )-100+8 p \left (x \right )^{3}+12 \sqrt {-12 p \left (x \right )^{3}-36 p \left (x \right )^{2}-36 p \left (x \right )+69}\right )^{{1}/{3}}+\left (1+i \sqrt {3}\right ) \left (p \left (x \right )+1\right )^{2}\right ) \left (48 p \left (x \right )+24+24 p \left (x \right )^{2}+\frac {-216 p \left (x \right )^{2}-432 p \left (x \right )-216}{\sqrt {-12 p \left (x \right )^{3}-36 p \left (x \right )^{2}-36 p \left (x \right )+69}}\right )}{9 \left (24 p \left (x \right )^{2}+24 p \left (x \right )-100+8 p \left (x \right )^{3}+12 \sqrt {-12 p \left (x \right )^{3}-36 p \left (x \right )^{2}-36 p \left (x \right )+69}\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 {2 \left (-\frac {i \sqrt {3}}{4}+\frac {1}{4}\right ) \left (48 p +24+24 p^{2}+\frac {-216 p^{2}-432 p -216}{\sqrt {-12 p^{3}-36 p^{2}-36 p +69}}\right )}{3 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}}-\left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}+\frac {\left (-p +2\right ) \left (48 p +24+24 p^{2}+\frac {-216 p^{2}-432 p -216}{\sqrt {-12 p^{3}-36 p^{2}-36 p +69}}\right )}{3 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}}+2 \left (1+i \sqrt {3}\right ) \left (p +1\right )}{3 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}}+\frac {\left (\left (-\frac {i \sqrt {3}}{4}+\frac {1}{4}\right ) \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}+\left (-p +2\right ) \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}+\left (1+i \sqrt {3}\right ) \left (p +1\right )^{2}\right ) \left (48 p +24+24 p^{2}+\frac {-216 p^{2}-432 p -216}{\sqrt {-12 p^{3}-36 p^{2}-36 p +69}}\right )}{9 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{4}/{3}}}\right )}{p +\frac {\left (-\frac {i \sqrt {3}}{4}+\frac {1}{4}\right ) \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}+\left (-p +2\right ) \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}+\left (1+i \sqrt {3}\right ) \left (p +1\right )^{2}}{3 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\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) &=\text {Expression too large to display}\\ q(p) &=0 \end {align*}

Hence the ode is \begin {align*} \text {Expression too large to display} \end {align*}

The integrating factor \(\mu \) is \[ \mu = \text {Expression too large to display} \] 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 (\text {Expression too large to display}\right ) &= 0 \end {align*}

Integrating gives \begin {align*} \text {Expression too large to display} &= c_{4} \end {align*}

Dividing both sides by the integrating factor \(\mu =\text {Expression too large to display}\) results in \begin {align*} x \left (p \right ) &= c_{4} {\mathrm e}^{-2 \left (\int \frac {-\frac {\left (i \sqrt {3}-1\right ) \left (\sqrt {-12 p^{3}-36 p^{2}-36 p +69}-9\right ) \left (p +1\right )^{2} \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}}{4}+\left (\left (-p^{3}-3 p^{2}-3 p +\frac {25}{2}\right ) \sqrt {-12 p^{3}-36 p^{2}-36 p +69}-\frac {207}{2}+54 p +54 p^{2}+18 p^{3}\right ) \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}+\left (1+i \sqrt {3}\right ) \left (\left (p^{3}+3 p^{2}+3 p -26\right ) \sqrt {-12 p^{3}-36 p^{2}-36 p +69}-27 p^{3}-81 p^{2}-81 p +216\right ) \left (p +1\right )}{2 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\, \left (p^{3}+3 p^{2}+3 p +\frac {3 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}}{2}-\frac {25}{2}\right ) \left (\left (-\frac {i \sqrt {3}}{4}+\frac {1}{4}\right ) \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}+\left (2 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}+\left (1+i \sqrt {3}\right ) \left (p +1\right )\right ) \left (p +1\right )\right )}d p \right )} \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 {\frac {\left (-i \sqrt {3}-1\right ) \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}}{4}+\left (p -2\right ) \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}+\left (i \sqrt {3}-1\right ) \left (p +1\right )^{2}}{3 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}}\\ g &= 0 \end {align*}

Hence (2) becomes \begin {align*} p -\frac {\frac {\left (-i \sqrt {3}-1\right ) \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}}{4}+\left (p -2\right ) \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}+\left (i \sqrt {3}-1\right ) \left (p +1\right )^{2}}{3 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}} = x \left (\frac {\frac {\left (-i \sqrt {3}-1\right ) \left (48 p +24+24 p^{2}+\frac {-216 p^{2}-432 p -216}{\sqrt {-12 p^{3}-36 p^{2}-36 p +69}}\right )}{6 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}}+\left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}+\frac {\left (p -2\right ) \left (48 p +24+24 p^{2}+\frac {-216 p^{2}-432 p -216}{\sqrt {-12 p^{3}-36 p^{2}-36 p +69}}\right )}{3 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}}+2 \left (i \sqrt {3}-1\right ) \left (p +1\right )}{3 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}}-\frac {\left (\frac {\left (-i \sqrt {3}-1\right ) \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}}{4}+\left (p -2\right ) \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}+\left (i \sqrt {3}-1\right ) \left (p +1\right )^{2}\right ) \left (48 p +24+24 p^{2}+\frac {-216 p^{2}-432 p -216}{\sqrt {-12 p^{3}-36 p^{2}-36 p +69}}\right )}{9 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\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 (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}}{4}+\left (p -2\right ) \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}+\left (i \sqrt {3}-1\right ) \left (p +1\right )^{2}}{3 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}} = 0 \end {align*}

No singular solution are found

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 (24 p \left (x \right )^{2}+24 p \left (x \right )-100+8 p \left (x \right )^{3}+12 \sqrt {-12 p \left (x \right )^{3}-36 p \left (x \right )^{2}-36 p \left (x \right )+69}\right )^{{2}/{3}}}{4}+\left (p \left (x \right )-2\right ) \left (24 p \left (x \right )^{2}+24 p \left (x \right )-100+8 p \left (x \right )^{3}+12 \sqrt {-12 p \left (x \right )^{3}-36 p \left (x \right )^{2}-36 p \left (x \right )+69}\right )^{{1}/{3}}+\left (i \sqrt {3}-1\right ) \left (p \left (x \right )+1\right )^{2}}{3 \left (24 p \left (x \right )^{2}+24 p \left (x \right )-100+8 p \left (x \right )^{3}+12 \sqrt {-12 p \left (x \right )^{3}-36 p \left (x \right )^{2}-36 p \left (x \right )+69}\right )^{{1}/{3}}}}{x \left (\frac {\frac {\left (-i \sqrt {3}-1\right ) \left (48 p \left (x \right )+24+24 p \left (x \right )^{2}+\frac {-216 p \left (x \right )^{2}-432 p \left (x \right )-216}{\sqrt {-12 p \left (x \right )^{3}-36 p \left (x \right )^{2}-36 p \left (x \right )+69}}\right )}{6 \left (24 p \left (x \right )^{2}+24 p \left (x \right )-100+8 p \left (x \right )^{3}+12 \sqrt {-12 p \left (x \right )^{3}-36 p \left (x \right )^{2}-36 p \left (x \right )+69}\right )^{{1}/{3}}}+\left (24 p \left (x \right )^{2}+24 p \left (x \right )-100+8 p \left (x \right )^{3}+12 \sqrt {-12 p \left (x \right )^{3}-36 p \left (x \right )^{2}-36 p \left (x \right )+69}\right )^{{1}/{3}}+\frac {\left (p \left (x \right )-2\right ) \left (48 p \left (x \right )+24+24 p \left (x \right )^{2}+\frac {-216 p \left (x \right )^{2}-432 p \left (x \right )-216}{\sqrt {-12 p \left (x \right )^{3}-36 p \left (x \right )^{2}-36 p \left (x \right )+69}}\right )}{3 \left (24 p \left (x \right )^{2}+24 p \left (x \right )-100+8 p \left (x \right )^{3}+12 \sqrt {-12 p \left (x \right )^{3}-36 p \left (x \right )^{2}-36 p \left (x \right )+69}\right )^{{2}/{3}}}+2 \left (i \sqrt {3}-1\right ) \left (p \left (x \right )+1\right )}{3 \left (24 p \left (x \right )^{2}+24 p \left (x \right )-100+8 p \left (x \right )^{3}+12 \sqrt {-12 p \left (x \right )^{3}-36 p \left (x \right )^{2}-36 p \left (x \right )+69}\right )^{{1}/{3}}}-\frac {\left (\frac {\left (-i \sqrt {3}-1\right ) \left (24 p \left (x \right )^{2}+24 p \left (x \right )-100+8 p \left (x \right )^{3}+12 \sqrt {-12 p \left (x \right )^{3}-36 p \left (x \right )^{2}-36 p \left (x \right )+69}\right )^{{2}/{3}}}{4}+\left (p \left (x \right )-2\right ) \left (24 p \left (x \right )^{2}+24 p \left (x \right )-100+8 p \left (x \right )^{3}+12 \sqrt {-12 p \left (x \right )^{3}-36 p \left (x \right )^{2}-36 p \left (x \right )+69}\right )^{{1}/{3}}+\left (i \sqrt {3}-1\right ) \left (p \left (x \right )+1\right )^{2}\right ) \left (48 p \left (x \right )+24+24 p \left (x \right )^{2}+\frac {-216 p \left (x \right )^{2}-432 p \left (x \right )-216}{\sqrt {-12 p \left (x \right )^{3}-36 p \left (x \right )^{2}-36 p \left (x \right )+69}}\right )}{9 \left (24 p \left (x \right )^{2}+24 p \left (x \right )-100+8 p \left (x \right )^{3}+12 \sqrt {-12 p \left (x \right )^{3}-36 p \left (x \right )^{2}-36 p \left (x \right )+69}\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 (48 p +24+24 p^{2}+\frac {-216 p^{2}-432 p -216}{\sqrt {-12 p^{3}-36 p^{2}-36 p +69}}\right )}{6 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}}+\left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}+\frac {\left (p -2\right ) \left (48 p +24+24 p^{2}+\frac {-216 p^{2}-432 p -216}{\sqrt {-12 p^{3}-36 p^{2}-36 p +69}}\right )}{3 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}}+2 \left (i \sqrt {3}-1\right ) \left (p +1\right )}{3 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}}-\frac {\left (\frac {\left (-i \sqrt {3}-1\right ) \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}}{4}+\left (p -2\right ) \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}+\left (i \sqrt {3}-1\right ) \left (p +1\right )^{2}\right ) \left (48 p +24+24 p^{2}+\frac {-216 p^{2}-432 p -216}{\sqrt {-12 p^{3}-36 p^{2}-36 p +69}}\right )}{9 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{4}/{3}}}\right )}{p -\frac {\frac {\left (-i \sqrt {3}-1\right ) \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}}}{4}+\left (p -2\right ) \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}+\left (i \sqrt {3}-1\right ) \left (p +1\right )^{2}}{3 \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\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) &=\text {Expression too large to display}\\ q(p) &=0 \end {align*}

Hence the ode is \begin {align*} \text {Expression too large to display} \end {align*}

The integrating factor \(\mu \) is \[ \mu = \text {Expression too large to display} \] 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 (\text {Expression too large to display}\right ) &= 0 \end {align*}

Integrating gives \begin {align*} \text {Expression too large to display} &= c_{6} \end {align*}

Dividing both sides by the integrating factor \(\mu =\text {Expression too large to display}\) results in \begin {align*} x \left (p \right ) &= c_{6} {\mathrm e}^{-2 \left (\int \frac {\frac {\left (\sqrt {-12 p^{3}-36 p^{2}-36 p +69}-9\right ) \left (p +1\right )^{2} \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}} \left (i-\sqrt {3}\right )}{4}-i \left (\left (p^{3}+3 p^{2}+3 p -\frac {25}{2}\right ) \sqrt {-12 p^{3}-36 p^{2}-36 p +69}+\frac {207}{2}-54 p -54 p^{2}-18 p^{3}\right ) \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}+\left (\left (p^{3}+3 p^{2}+3 p -26\right ) \sqrt {-12 p^{3}-36 p^{2}-36 p +69}-27 p^{3}-81 p^{2}-81 p +216\right ) \left (p +1\right ) \left (\sqrt {3}+i\right )}{2 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\, \left (\frac {\left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{2}/{3}} \left (i-\sqrt {3}\right )}{4}+\left (p +1\right ) \left (2 i \left (24 p^{2}+24 p -100+8 p^{3}+12 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}\right )^{{1}/{3}}+\left (p +1\right ) \left (\sqrt {3}+i\right )\right )\right ) \left (p^{3}+3 p^{2}+3 p +\frac {3 \sqrt {-12 p^{3}-36 p^{2}-36 p +69}}{2}-\frac {25}{2}\right )}d p \right )} \end {align*}

Since the solution \(x \left (p \right )\) has unresolved integral, unable to continue.

5.33.5 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-\frac {y^{3}+2 x y^{2}+x^{2} y+x^{3}}{x \left (x +y\right )^{2}}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y^{3}+2 x y^{2}+x^{2} y+x^{3}}{x \left (x +y\right )^{2}} \end {array} \]

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: 0.0 (sec). Leaf size: 80

dsolve(diff(y(x),x)=(y(x)^3+2*x*y(x)^2+x^2*y(x)+x^3)/(x*(y(x)+x)^2),y(x), singsol=all)
 

\begin{align*} y \left (x \right ) &= x \left (-1+\left (3 \ln \left (x \right )+3 c_{1} \right )^{\frac {1}{3}}\right ) \\ y \left (x \right ) &= -\frac {x \left (i \sqrt {3}\, \left (3 \ln \left (x \right )+3 c_{1} \right )^{\frac {1}{3}}+\left (3 \ln \left (x \right )+3 c_{1} \right )^{\frac {1}{3}}+2\right )}{2} \\ y \left (x \right ) &= \frac {x \left (i \sqrt {3}-1\right ) \left (3 \ln \left (x \right )+3 c_{1} \right )^{\frac {1}{3}}}{2}-x \\ \end{align*}

✓ Solution by Mathematica

Time used: 1.409 (sec). Leaf size: 109

DSolve[y'[x]==(y[x]^3+2*x*y[x]^2+x^2*y[x]+x^3)/(x*(y[x]+x)^2),y[x],x,IncludeSingularSolutions -> True]
 

\begin{align*} y(x)\to -x+\sqrt [3]{x^3 (3 \log (x)+1+3 c_1)} \\ y(x)\to -x+\frac {1}{2} i \left (\sqrt {3}+i\right ) \sqrt [3]{x^3 (3 \log (x)+1+3 c_1)} \\ y(x)\to -x-\frac {1}{2} \left (1+i \sqrt {3}\right ) \sqrt [3]{x^3 (3 \log (x)+1+3 c_1)} \\ \end{align*}