2.1.45 Problem 45

Internal problem ID [10404]
Book : Second order enumerated odes
Section : section 1
Problem number : 45
Date solved : Monday, January 26, 2026 at 09:53:47 PM
CAS classification : [[_2nd_order, _missing_x]]

2.1.45.1 second order ode missing x

1141.074 (sec)

\begin{align*} y {y^{\prime \prime }}^{2}+y^{\prime }&=0 \\ \end{align*}

Entering second order ode missing \(x\) solverThis is missing independent variable second order ode. Solved by reduction of order by using substitution which makes the dependent variable \(y\) an independent variable. Using

\begin{align*} y' &= p \end{align*}

Then

\begin{align*} y'' &= \frac {dp}{dx}\\ &= \frac {dp}{dy}\frac {dy}{dx}\\ &= p \frac {dp}{dy} \end{align*}

Hence the ode becomes

\begin{align*} y p \left (y \right )^{2} \left (\frac {d}{d y}p \left (y \right )\right )^{2}+p \left (y \right ) = 0 \end{align*}

Which is now solved as first order ode for \(p(y)\).

2.1.45.2 Solved by factoring the differential equation

Time used: 1.710 (sec)

\begin{align*} y p^{2} {p^{\prime }}^{2}+p&=0 \\ \end{align*}

Writing the ode as

\begin{align*} \left (p\right )\left ({p^{\prime }}^{2} p y +1\right )&=0 \end{align*}

Therefore we need to solve the following equations

\begin{align*} \tag{1} p &= 0 \\ \tag{2} {p^{\prime }}^{2} p y +1 &= 0 \\ \end{align*}

Now each of the above equations is solved in turn.

Solving equation (1)

Entering zero order ode solverSolving for \(p\) from

\begin{align*} p = 0 \end{align*}

Solving gives

\begin{align*} p &= 0 \\ \end{align*}

Solving equation (2)

Entering first order ode homog type G solverMultiplying the right side of the ode, which is \(-\frac {1}{\sqrt {-p y}}\) by \(\frac {y}{p}\) gives

\begin{align*} p^{\prime } &= \left (\frac {y}{p}\right ) -\frac {1}{\sqrt {-p y}}\\ &= -\frac {y}{p \sqrt {-p y}}\\ &= F(y,p) \end{align*}

Since \(F \left (y , p\right )\) has \(p\), then let

\begin{align*} f_y&= y \left (\frac {\partial }{\partial y}F \left (y , p\right )\right )\\ &= \frac {y^{2}}{2 \left (-p y \right )^{{3}/{2}}}\\ f_p&= p \left (\frac {\partial }{\partial p}F \left (y , p\right )\right )\\ &= -\frac {3 y^{2}}{2 \left (-p y \right )^{{3}/{2}}}\\ \alpha &= \frac {f_y}{f_p} \\ &=-{\frac {1}{3}} \end{align*}

Since \(\alpha \) is independent of \(y,p\) then this is Homogeneous type G.

Let

\begin{align*} p&=\frac {z}{y^ \alpha }\\ &=\frac {z}{\frac {1}{y^{{1}/{3}}}} \end{align*}

Substituting the above back into \(F(y,p)\) gives

\begin{align*} F \left (z \right ) &=\frac {1}{\left (-z \right )^{{3}/{2}}} \end{align*}

We see that \(F \left (z \right )\) does not depend on \(y\) nor on \(p\). If this was not the case, then this method will not work.

Therefore, the implicit solution is given by

\begin{align*} \ln \left (y \right )- c_1 - \int ^{p y^\alpha } \frac {1}{z \left (\alpha + F(z)\right ) } \,dz & = 0 \end{align*}

Which gives

\[ \ln \left (y \right )-c_1 +\int _{}^{\frac {p}{y^{{1}/{3}}}}\frac {1}{z \left (\frac {1}{3}-\frac {1}{\left (-z \right )^{{3}/{2}}}\right )}d z = 0 \]

The value of the above is

\[ \ln \left (y \right )-c_1 +2 \ln \left (\left (-\frac {p}{y^{{1}/{3}}}\right )^{{3}/{2}}-3\right ) = 0 \]

Multiplying the right side of the ode, which is \(\frac {1}{\sqrt {-p y}}\) by \(\frac {y}{p}\) gives

\begin{align*} p^{\prime } &= \left (\frac {y}{p}\right ) \frac {1}{\sqrt {-p y}}\\ &= \frac {y}{p \sqrt {-p y}}\\ &= F(y,p) \end{align*}

Since \(F \left (y , p\right )\) has \(p\), then let

\begin{align*} f_y&= y \left (\frac {\partial }{\partial y}F \left (y , p\right )\right )\\ &= -\frac {y^{2}}{2 \left (-p y \right )^{{3}/{2}}}\\ f_p&= p \left (\frac {\partial }{\partial p}F \left (y , p\right )\right )\\ &= \frac {3 y^{2}}{2 \left (-p y \right )^{{3}/{2}}}\\ \alpha &= \frac {f_y}{f_p} \\ &=-{\frac {1}{3}} \end{align*}

Since \(\alpha \) is independent of \(y,p\) then this is Homogeneous type G.

Let

\begin{align*} p&=\frac {z}{y^ \alpha }\\ &=\frac {z}{\frac {1}{y^{{1}/{3}}}} \end{align*}

Substituting the above back into \(F(y,p)\) gives

\begin{align*} F \left (z \right ) &=-\frac {1}{\left (-z \right )^{{3}/{2}}} \end{align*}

We see that \(F \left (z \right )\) does not depend on \(y\) nor on \(p\). If this was not the case, then this method will not work.

Therefore, the implicit solution is given by

\begin{align*} \ln \left (y \right )- c_1 - \int ^{p y^\alpha } \frac {1}{z \left (\alpha + F(z)\right ) } \,dz & = 0 \end{align*}

Which gives

\[ \ln \left (y \right )-c_2 +\int _{}^{\frac {p}{y^{{1}/{3}}}}\frac {1}{z \left (\frac {1}{3}+\frac {1}{\left (-z \right )^{{3}/{2}}}\right )}d z = 0 \]

The value of the above is

\[ \ln \left (y \right )-c_2 +2 \ln \left (\left (-\frac {p}{y^{{1}/{3}}}\right )^{{3}/{2}}+3\right ) = 0 \]

Summary of solutions found

\begin{align*} \ln \left (y \right )-c_1 +2 \ln \left (\left (-\frac {p}{y^{{1}/{3}}}\right )^{{3}/{2}}-3\right ) &= 0 \\ \ln \left (y \right )-c_2 +2 \ln \left (\left (-\frac {p}{y^{{1}/{3}}}\right )^{{3}/{2}}+3\right ) &= 0 \\ p &= 0 \\ \end{align*}

For solution (1) found earlier, since \(p=y^{\prime }\) then the new first order ode to solve is

\begin{align*} \ln \left (y\right )-c_1 +2 \ln \left (\left (-\frac {y^{\prime }}{y^{{1}/{3}}}\right )^{{3}/{2}}-3\right ) = 0 \end{align*}

Solving for the derivative gives these ODE’s to solve

\begin{align*} \tag{1} y^{\prime }&=-{\left (-\frac {\left ({\mathrm e}^{-\frac {\ln \left (y\right )}{2}+\frac {c_1}{2}}+3\right )^{{1}/{3}}}{2}-\frac {i \sqrt {3}\, \left ({\mathrm e}^{-\frac {\ln \left (y\right )}{2}+\frac {c_1}{2}}+3\right )^{{1}/{3}}}{2}\right )}^{2} y^{{1}/{3}} \\ \tag{2} y^{\prime }&=-{\left (-\frac {\left ({\mathrm e}^{-\frac {\ln \left (y\right )}{2}+\frac {c_1}{2}}+3\right )^{{1}/{3}}}{2}+\frac {i \sqrt {3}\, \left ({\mathrm e}^{-\frac {\ln \left (y\right )}{2}+\frac {c_1}{2}}+3\right )^{{1}/{3}}}{2}\right )}^{2} y^{{1}/{3}} \\ \tag{3} y^{\prime }&=-\left ({\mathrm e}^{-\frac {\ln \left (y\right )}{2}+\frac {c_1}{2}}+3\right )^{{2}/{3}} y^{{1}/{3}} \\ \end{align*}

Now each of the above is solved separately.

Solving Eq. (1)

Entering first order ode autonomous solverUnable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

\[ \int _{}^{y}-\frac {4}{\left ({\mathrm e}^{-\frac {\ln \left (\tau \right )}{2}+\frac {c_1}{2}}+3\right )^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2} \tau ^{{1}/{3}}}d \tau = x +c_9 \]

Solving Eq. (2)

Entering first order ode autonomous solverUnable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

\[ \int _{}^{y}-\frac {4}{\left ({\mathrm e}^{-\frac {\ln \left (\tau \right )}{2}+\frac {c_1}{2}}+3\right )^{{2}/{3}} \left (-1+i \sqrt {3}\right )^{2} \tau ^{{1}/{3}}}d \tau = x +\textit {\_C10} \]

Solving Eq. (3)

Entering first order ode autonomous solverUnable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

\[ \int _{}^{y}-\frac {1}{\left ({\mathrm e}^{-\frac {\ln \left (\tau \right )}{2}+\frac {c_1}{2}}+3\right )^{{2}/{3}} \tau ^{{1}/{3}}}d \tau = x +\textit {\_C11} \]

For solution (2) found earlier, since \(p=y^{\prime }\) then the new first order ode to solve is

\begin{align*} \ln \left (y\right )-c_2 +2 \ln \left (\left (-\frac {y^{\prime }}{y^{{1}/{3}}}\right )^{{3}/{2}}+3\right ) = 0 \end{align*}

Entering first order ode dAlembert solverLet \(p=y^{\prime }\) the ode becomes

\begin{align*} \ln \left (y \right )-c_2 +2 \ln \left (\left (-\frac {p}{y^{{1}/{3}}}\right )^{{3}/{2}}+3\right ) = 0 \end{align*}

Solving for \(y\) from the above results in

\begin{align*} \tag{1} y &= \frac {p^{3}+{\mathrm e}^{c_2}}{\frac {18 p \left (-p^{2}+\sqrt {-{\mathrm e}^{c_2} p}\right )}{p^{3}+{\mathrm e}^{c_2}}+9} \\ \tag{2} y &= \frac {p^{3}+{\mathrm e}^{c_2}}{\frac {18 \left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right )^{3} p \left (-p^{2}+\sqrt {-{\mathrm e}^{c_2} p}\right )}{p^{3}+{\mathrm e}^{c_2}}+9} \\ \tag{3} y &= \frac {p^{3}+{\mathrm e}^{c_2}}{\frac {18 \left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )^{3} p \left (-p^{2}+\sqrt {-{\mathrm e}^{c_2} p}\right )}{p^{3}+{\mathrm e}^{c_2}}+9} \\ \tag{4} y &= \frac {p^{3}+{\mathrm e}^{c_2}}{-\frac {18 p \left (p^{2}+\sqrt {-{\mathrm e}^{c_2} p}\right )}{p^{3}+{\mathrm e}^{c_2}}+9} \\ \tag{5} y &= \frac {p^{3}+{\mathrm e}^{c_2}}{-\frac {18 \left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right )^{3} p \left (p^{2}+\sqrt {-{\mathrm e}^{c_2} p}\right )}{p^{3}+{\mathrm e}^{c_2}}+9} \\ \tag{6} y &= \frac {p^{3}+{\mathrm e}^{c_2}}{-\frac {18 \left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )^{3} p \left (p^{2}+\sqrt {-{\mathrm e}^{c_2} p}\right )}{p^{3}+{\mathrm e}^{c_2}}+9} \\ \end{align*}

This has the form

\begin{align*} y=x f(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 &= 0\\ g &= \frac {\left (p^{3}+{\mathrm e}^{c_2}\right )^{2}}{-9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p +9 \,{\mathrm e}^{c_2}} \end{align*}

Hence (2) becomes

\begin{equation} \tag{2A} p = \left (\frac {6 p^{5}}{-9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p +9 \,{\mathrm e}^{c_2}}+\frac {6 p^{2} {\mathrm e}^{c_2}}{-9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p +9 \,{\mathrm e}^{c_2}}+\frac {27 p^{8}}{\left (-9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p +9 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {9 p^{7} {\mathrm e}^{c_2}}{\left (-9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p +9 \,{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p}}-\frac {18 p^{6} \sqrt {-{\mathrm e}^{c_2} p}}{\left (-9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p +9 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {54 \,{\mathrm e}^{c_2} p^{5}}{\left (-9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p +9 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {18 \,{\mathrm e}^{2 c_2} p^{4}}{\left (-9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p +9 \,{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p}}-\frac {36 \,{\mathrm e}^{c_2} p^{3} \sqrt {-{\mathrm e}^{c_2} p}}{\left (-9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p +9 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {27 \,{\mathrm e}^{2 c_2} p^{2}}{\left (-9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p +9 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {9 \,{\mathrm e}^{3 c_2} p}{\left (-9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p +9 \,{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p}}-\frac {18 \,{\mathrm e}^{2 c_2} \sqrt {-{\mathrm e}^{c_2} p}}{\left (-9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p +9 \,{\mathrm e}^{c_2}\right )^{2}}\right ) p^{\prime }\left (x \right ) \end{equation}

The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives

\begin{align*} p = 0 \end{align*}

Solving the above for \(p\) results in

\begin{align*} p_{1} &=0 \end{align*}

Substituting these in (1A) and keeping singular solution that verifies the ode gives

\begin{align*} y = \frac {{\mathrm e}^{c_2}}{9} \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{equation} \tag{3} p^{\prime }\left (x \right ) = \frac {p \left (x \right )}{\frac {6 p \left (x \right )^{5}}{-9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )+9 \,{\mathrm e}^{c_2}}+\frac {6 p \left (x \right )^{2} {\mathrm e}^{c_2}}{-9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )+9 \,{\mathrm e}^{c_2}}+\frac {27 p \left (x \right )^{8}}{\left (-9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )+9 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {9 p \left (x \right )^{7} {\mathrm e}^{c_2}}{\left (-9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )+9 \,{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}}-\frac {18 p \left (x \right )^{6} \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}}{\left (-9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )+9 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {54 \,{\mathrm e}^{c_2} p \left (x \right )^{5}}{\left (-9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )+9 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {18 \,{\mathrm e}^{2 c_2} p \left (x \right )^{4}}{\left (-9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )+9 \,{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}}-\frac {36 \,{\mathrm e}^{c_2} p \left (x \right )^{3} \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}}{\left (-9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )+9 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {27 \,{\mathrm e}^{2 c_2} p \left (x \right )^{2}}{\left (-9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )+9 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {9 \,{\mathrm e}^{3 c_2} p \left (x \right )}{\left (-9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )+9 \,{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}}-\frac {18 \,{\mathrm e}^{2 c_2} \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}}{\left (-9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )+9 \,{\mathrm e}^{c_2}\right )^{2}}} \end{equation}

This ODE is now solved for \(p \left (x \right )\). No inversion is needed.

Integrating gives

\begin{align*} \int -\frac {\sqrt {-{\mathrm e}^{c_2} p}\, p^{7}+3 \,{\mathrm e}^{c_2} p^{6}-2 \sqrt {-{\mathrm e}^{c_2} p}\, {\mathrm e}^{c_2} p^{4}+2 \,{\mathrm e}^{2 c_2} p^{3}-3 \sqrt {-{\mathrm e}^{c_2} p}\, {\mathrm e}^{2 c_2} p -{\mathrm e}^{3 c_2}}{3 \left (-p^{3}+2 \sqrt {-{\mathrm e}^{c_2} p}\, p +{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p}}d p &= dx\\ \frac {2 \,{\mathrm e}^{-2 c_2} \left (-\frac {{\mathrm e}^{2 c_2} p^{2}}{4}-{\mathrm e}^{2 c_2} \sqrt {-{\mathrm e}^{c_2} p}\right )}{3}&= x +\textit {\_C12} \end{align*}

Singular solutions are found by solving

\begin{align*} -\frac {3 \left (-p^{3}+2 \sqrt {-{\mathrm e}^{c_2} p}\, p +{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p}}{\sqrt {-{\mathrm e}^{c_2} p}\, p^{7}+3 \,{\mathrm e}^{c_2} p^{6}-2 \sqrt {-{\mathrm e}^{c_2} p}\, {\mathrm e}^{c_2} p^{4}+2 \,{\mathrm e}^{2 c_2} p^{3}-3 \sqrt {-{\mathrm e}^{c_2} p}\, {\mathrm e}^{2 c_2} p -{\mathrm e}^{3 c_2}}&= 0 \end{align*}

for \(p \left (x \right )\). This is because of dividing by the above earlier. This gives the following singular solution(s), which also has to satisfy the given ODE.

\begin{align*} p \left (x \right ) = 0 \end{align*}

Substituing the above solution for \(p\) in (2A) gives

\begin{align*} y &= \frac {{\left (-\operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C12} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x +4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{6} {\mathrm e}^{-3 c_2}+{\mathrm e}^{c_2}\right )}^{2}}{9 \operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C12} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x +4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{6} {\mathrm e}^{-3 c_2}-18 \sqrt {\operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C12} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x +4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{2}}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C12} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x +4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{2} {\mathrm e}^{-c_2}+9 \,{\mathrm e}^{c_2}} \\ y &= \frac {{\mathrm e}^{c_2}}{9} \\ \end{align*}

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 &= 0\\ g &= \frac {\left (p^{3}+{\mathrm e}^{c_2}\right )^{2}}{-9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p +9 \,{\mathrm e}^{c_2}} \end{align*}

Hence (2) becomes

\begin{equation} \tag{2A} p = \left (\frac {6 p^{5}}{-9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p +9 \,{\mathrm e}^{c_2}}+\frac {6 p^{2} {\mathrm e}^{c_2}}{-9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p +9 \,{\mathrm e}^{c_2}}+\frac {27 p^{8}}{\left (-9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p +9 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {9 p^{7} {\mathrm e}^{c_2}}{\left (-9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p +9 \,{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p}}-\frac {18 p^{6} \sqrt {-{\mathrm e}^{c_2} p}}{\left (-9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p +9 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {54 \,{\mathrm e}^{c_2} p^{5}}{\left (-9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p +9 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {18 \,{\mathrm e}^{2 c_2} p^{4}}{\left (-9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p +9 \,{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p}}-\frac {36 \,{\mathrm e}^{c_2} p^{3} \sqrt {-{\mathrm e}^{c_2} p}}{\left (-9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p +9 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {27 \,{\mathrm e}^{2 c_2} p^{2}}{\left (-9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p +9 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {9 \,{\mathrm e}^{3 c_2} p}{\left (-9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p +9 \,{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p}}-\frac {18 \,{\mathrm e}^{2 c_2} \sqrt {-{\mathrm e}^{c_2} p}}{\left (-9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p +9 \,{\mathrm e}^{c_2}\right )^{2}}\right ) p^{\prime }\left (x \right ) \end{equation}

The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives

\begin{align*} p = 0 \end{align*}

Solving the above for \(p\) results in

\begin{align*} p_{1} &=0 \end{align*}

Substituting these in (1A) and keeping singular solution that verifies the ode gives

\begin{align*} y = \frac {{\mathrm e}^{c_2}}{9} \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{equation} \tag{3} p^{\prime }\left (x \right ) = \frac {p \left (x \right )}{\frac {6 p \left (x \right )^{5}}{-9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )+9 \,{\mathrm e}^{c_2}}+\frac {6 p \left (x \right )^{2} {\mathrm e}^{c_2}}{-9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )+9 \,{\mathrm e}^{c_2}}+\frac {27 p \left (x \right )^{8}}{\left (-9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )+9 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {9 p \left (x \right )^{7} {\mathrm e}^{c_2}}{\left (-9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )+9 \,{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}}-\frac {18 p \left (x \right )^{6} \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}}{\left (-9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )+9 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {54 \,{\mathrm e}^{c_2} p \left (x \right )^{5}}{\left (-9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )+9 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {18 \,{\mathrm e}^{2 c_2} p \left (x \right )^{4}}{\left (-9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )+9 \,{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}}-\frac {36 \,{\mathrm e}^{c_2} p \left (x \right )^{3} \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}}{\left (-9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )+9 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {27 \,{\mathrm e}^{2 c_2} p \left (x \right )^{2}}{\left (-9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )+9 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {9 \,{\mathrm e}^{3 c_2} p \left (x \right )}{\left (-9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )+9 \,{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}}-\frac {18 \,{\mathrm e}^{2 c_2} \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}}{\left (-9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )+9 \,{\mathrm e}^{c_2}\right )^{2}}} \end{equation}

This ODE is now solved for \(p \left (x \right )\). No inversion is needed.

Integrating gives

\begin{align*} \int -\frac {\sqrt {-{\mathrm e}^{c_2} p}\, p^{7}+3 \,{\mathrm e}^{c_2} p^{6}-2 \sqrt {-{\mathrm e}^{c_2} p}\, {\mathrm e}^{c_2} p^{4}+2 \,{\mathrm e}^{2 c_2} p^{3}-3 \sqrt {-{\mathrm e}^{c_2} p}\, {\mathrm e}^{2 c_2} p -{\mathrm e}^{3 c_2}}{3 \left (-p^{3}+2 \sqrt {-{\mathrm e}^{c_2} p}\, p +{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p}}d p &= dx\\ \frac {2 \,{\mathrm e}^{-2 c_2} \left (-\frac {{\mathrm e}^{2 c_2} p^{2}}{4}-{\mathrm e}^{2 c_2} \sqrt {-{\mathrm e}^{c_2} p}\right )}{3}&= x +\textit {\_C13} \end{align*}

Singular solutions are found by solving

\begin{align*} -\frac {3 \left (-p^{3}+2 \sqrt {-{\mathrm e}^{c_2} p}\, p +{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p}}{\sqrt {-{\mathrm e}^{c_2} p}\, p^{7}+3 \,{\mathrm e}^{c_2} p^{6}-2 \sqrt {-{\mathrm e}^{c_2} p}\, {\mathrm e}^{c_2} p^{4}+2 \,{\mathrm e}^{2 c_2} p^{3}-3 \sqrt {-{\mathrm e}^{c_2} p}\, {\mathrm e}^{2 c_2} p -{\mathrm e}^{3 c_2}}&= 0 \end{align*}

for \(p \left (x \right )\). This is because of dividing by the above earlier. This gives the following singular solution(s), which also has to satisfy the given ODE.

\begin{align*} p \left (x \right ) = 0 \end{align*}

Substituing the above solution for \(p\) in (2A) gives

\begin{align*} y &= \frac {{\left (-\operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C13} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x +4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{6} {\mathrm e}^{-3 c_2}+{\mathrm e}^{c_2}\right )}^{2}}{9 \operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C13} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x +4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{6} {\mathrm e}^{-3 c_2}-18 \sqrt {\operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C13} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x +4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{2}}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C13} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x +4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{2} {\mathrm e}^{-c_2}+9 \,{\mathrm e}^{c_2}} \\ y &= \frac {{\mathrm e}^{c_2}}{9} \\ \end{align*}

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 &= 0\\ g &= \frac {\left (p^{3}+{\mathrm e}^{c_2}\right )^{2}}{-9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p +9 \,{\mathrm e}^{c_2}} \end{align*}

Hence (2) becomes

\begin{equation} \tag{2A} p = \left (\frac {6 p^{5}}{-9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p +9 \,{\mathrm e}^{c_2}}+\frac {6 p^{2} {\mathrm e}^{c_2}}{-9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p +9 \,{\mathrm e}^{c_2}}+\frac {27 p^{8}}{\left (-9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p +9 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {9 p^{7} {\mathrm e}^{c_2}}{\left (-9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p +9 \,{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p}}-\frac {18 p^{6} \sqrt {-{\mathrm e}^{c_2} p}}{\left (-9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p +9 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {54 \,{\mathrm e}^{c_2} p^{5}}{\left (-9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p +9 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {18 \,{\mathrm e}^{2 c_2} p^{4}}{\left (-9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p +9 \,{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p}}-\frac {36 \,{\mathrm e}^{c_2} p^{3} \sqrt {-{\mathrm e}^{c_2} p}}{\left (-9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p +9 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {27 \,{\mathrm e}^{2 c_2} p^{2}}{\left (-9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p +9 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {9 \,{\mathrm e}^{3 c_2} p}{\left (-9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p +9 \,{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p}}-\frac {18 \,{\mathrm e}^{2 c_2} \sqrt {-{\mathrm e}^{c_2} p}}{\left (-9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p +9 \,{\mathrm e}^{c_2}\right )^{2}}\right ) p^{\prime }\left (x \right ) \end{equation}

The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives

\begin{align*} p = 0 \end{align*}

Solving the above for \(p\) results in

\begin{align*} p_{1} &=0 \end{align*}

Substituting these in (1A) and keeping singular solution that verifies the ode gives

\begin{align*} y = \frac {{\mathrm e}^{c_2}}{9} \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{equation} \tag{3} p^{\prime }\left (x \right ) = \frac {p \left (x \right )}{\frac {6 p \left (x \right )^{5}}{-9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )+9 \,{\mathrm e}^{c_2}}+\frac {6 p \left (x \right )^{2} {\mathrm e}^{c_2}}{-9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )+9 \,{\mathrm e}^{c_2}}+\frac {27 p \left (x \right )^{8}}{\left (-9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )+9 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {9 p \left (x \right )^{7} {\mathrm e}^{c_2}}{\left (-9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )+9 \,{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}}-\frac {18 p \left (x \right )^{6} \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}}{\left (-9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )+9 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {54 \,{\mathrm e}^{c_2} p \left (x \right )^{5}}{\left (-9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )+9 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {18 \,{\mathrm e}^{2 c_2} p \left (x \right )^{4}}{\left (-9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )+9 \,{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}}-\frac {36 \,{\mathrm e}^{c_2} p \left (x \right )^{3} \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}}{\left (-9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )+9 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {27 \,{\mathrm e}^{2 c_2} p \left (x \right )^{2}}{\left (-9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )+9 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {9 \,{\mathrm e}^{3 c_2} p \left (x \right )}{\left (-9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )+9 \,{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}}-\frac {18 \,{\mathrm e}^{2 c_2} \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}}{\left (-9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )+9 \,{\mathrm e}^{c_2}\right )^{2}}} \end{equation}

This ODE is now solved for \(p \left (x \right )\). No inversion is needed.

Integrating gives

\begin{align*} \int -\frac {\sqrt {-{\mathrm e}^{c_2} p}\, p^{7}+3 \,{\mathrm e}^{c_2} p^{6}-2 \sqrt {-{\mathrm e}^{c_2} p}\, {\mathrm e}^{c_2} p^{4}+2 \,{\mathrm e}^{2 c_2} p^{3}-3 \sqrt {-{\mathrm e}^{c_2} p}\, {\mathrm e}^{2 c_2} p -{\mathrm e}^{3 c_2}}{3 \left (-p^{3}+2 \sqrt {-{\mathrm e}^{c_2} p}\, p +{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p}}d p &= dx\\ \frac {2 \,{\mathrm e}^{-2 c_2} \left (-\frac {{\mathrm e}^{2 c_2} p^{2}}{4}-{\mathrm e}^{2 c_2} \sqrt {-{\mathrm e}^{c_2} p}\right )}{3}&= x +\textit {\_C14} \end{align*}

Singular solutions are found by solving

\begin{align*} -\frac {3 \left (-p^{3}+2 \sqrt {-{\mathrm e}^{c_2} p}\, p +{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p}}{\sqrt {-{\mathrm e}^{c_2} p}\, p^{7}+3 \,{\mathrm e}^{c_2} p^{6}-2 \sqrt {-{\mathrm e}^{c_2} p}\, {\mathrm e}^{c_2} p^{4}+2 \,{\mathrm e}^{2 c_2} p^{3}-3 \sqrt {-{\mathrm e}^{c_2} p}\, {\mathrm e}^{2 c_2} p -{\mathrm e}^{3 c_2}}&= 0 \end{align*}

for \(p \left (x \right )\). This is because of dividing by the above earlier. This gives the following singular solution(s), which also has to satisfy the given ODE.

\begin{align*} p \left (x \right ) = 0 \end{align*}

Substituing the above solution for \(p\) in (2A) gives

\begin{align*} y &= \frac {{\left (-\operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C14} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x +4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{6} {\mathrm e}^{-3 c_2}+{\mathrm e}^{c_2}\right )}^{2}}{9 \operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C14} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x +4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{6} {\mathrm e}^{-3 c_2}-18 \sqrt {\operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C14} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x +4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{2}}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C14} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x +4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{2} {\mathrm e}^{-c_2}+9 \,{\mathrm e}^{c_2}} \\ y &= \frac {{\mathrm e}^{c_2}}{9} \\ \end{align*}

Solving ode 4A

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 &= 0\\ g &= -\frac {\left (p^{3}+{\mathrm e}^{c_2}\right )^{2}}{9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p -9 \,{\mathrm e}^{c_2}} \end{align*}

Hence (2) becomes

\begin{equation} \tag{2A} p = \left (-\frac {6 p^{5}}{9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p -9 \,{\mathrm e}^{c_2}}-\frac {6 p^{2} {\mathrm e}^{c_2}}{9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p -9 \,{\mathrm e}^{c_2}}+\frac {27 p^{8}}{\left (9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p -9 \,{\mathrm e}^{c_2}\right )^{2}}-\frac {9 p^{7} {\mathrm e}^{c_2}}{\left (9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p -9 \,{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p}}+\frac {18 p^{6} \sqrt {-{\mathrm e}^{c_2} p}}{\left (9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p -9 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {54 \,{\mathrm e}^{c_2} p^{5}}{\left (9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p -9 \,{\mathrm e}^{c_2}\right )^{2}}-\frac {18 \,{\mathrm e}^{2 c_2} p^{4}}{\left (9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p -9 \,{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p}}+\frac {36 \,{\mathrm e}^{c_2} p^{3} \sqrt {-{\mathrm e}^{c_2} p}}{\left (9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p -9 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {27 \,{\mathrm e}^{2 c_2} p^{2}}{\left (9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p -9 \,{\mathrm e}^{c_2}\right )^{2}}-\frac {9 \,{\mathrm e}^{3 c_2} p}{\left (9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p -9 \,{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p}}+\frac {18 \,{\mathrm e}^{2 c_2} \sqrt {-{\mathrm e}^{c_2} p}}{\left (9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p -9 \,{\mathrm e}^{c_2}\right )^{2}}\right ) p^{\prime }\left (x \right ) \end{equation}

The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives

\begin{align*} p = 0 \end{align*}

Solving the above for \(p\) results in

\begin{align*} p_{1} &=0 \end{align*}

Substituting these in (1A) and keeping singular solution that verifies the ode gives

\begin{align*} y = \frac {{\mathrm e}^{c_2}}{9} \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{equation} \tag{3} p^{\prime }\left (x \right ) = \frac {p \left (x \right )}{-\frac {6 p \left (x \right )^{5}}{9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )-9 \,{\mathrm e}^{c_2}}-\frac {6 p \left (x \right )^{2} {\mathrm e}^{c_2}}{9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )-9 \,{\mathrm e}^{c_2}}+\frac {27 p \left (x \right )^{8}}{\left (9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )-9 \,{\mathrm e}^{c_2}\right )^{2}}-\frac {9 p \left (x \right )^{7} {\mathrm e}^{c_2}}{\left (9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )-9 \,{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}}+\frac {18 p \left (x \right )^{6} \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}}{\left (9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )-9 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {54 \,{\mathrm e}^{c_2} p \left (x \right )^{5}}{\left (9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )-9 \,{\mathrm e}^{c_2}\right )^{2}}-\frac {18 \,{\mathrm e}^{2 c_2} p \left (x \right )^{4}}{\left (9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )-9 \,{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}}+\frac {36 \,{\mathrm e}^{c_2} p \left (x \right )^{3} \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}}{\left (9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )-9 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {27 \,{\mathrm e}^{2 c_2} p \left (x \right )^{2}}{\left (9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )-9 \,{\mathrm e}^{c_2}\right )^{2}}-\frac {9 \,{\mathrm e}^{3 c_2} p \left (x \right )}{\left (9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )-9 \,{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}}+\frac {18 \,{\mathrm e}^{2 c_2} \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}}{\left (9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )-9 \,{\mathrm e}^{c_2}\right )^{2}}} \end{equation}

This ODE is now solved for \(p \left (x \right )\). No inversion is needed.

Integrating gives

\begin{align*} \int \frac {-\sqrt {-{\mathrm e}^{c_2} p}\, p^{7}+3 \,{\mathrm e}^{c_2} p^{6}+2 \sqrt {-{\mathrm e}^{c_2} p}\, {\mathrm e}^{c_2} p^{4}+2 \,{\mathrm e}^{2 c_2} p^{3}+3 \sqrt {-{\mathrm e}^{c_2} p}\, {\mathrm e}^{2 c_2} p -{\mathrm e}^{3 c_2}}{3 \left (-p^{3}-2 \sqrt {-{\mathrm e}^{c_2} p}\, p +{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p}}d p &= dx\\ \frac {2 \,{\mathrm e}^{-2 c_2} \left (-\frac {{\mathrm e}^{2 c_2} p^{2}}{4}+{\mathrm e}^{2 c_2} \sqrt {-{\mathrm e}^{c_2} p}\right )}{3}&= x +\textit {\_C15} \end{align*}

Singular solutions are found by solving

\begin{align*} \frac {3 \left (-p^{3}-2 \sqrt {-{\mathrm e}^{c_2} p}\, p +{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p}}{-\sqrt {-{\mathrm e}^{c_2} p}\, p^{7}+3 \,{\mathrm e}^{c_2} p^{6}+2 \sqrt {-{\mathrm e}^{c_2} p}\, {\mathrm e}^{c_2} p^{4}+2 \,{\mathrm e}^{2 c_2} p^{3}+3 \sqrt {-{\mathrm e}^{c_2} p}\, {\mathrm e}^{2 c_2} p -{\mathrm e}^{3 c_2}}&= 0 \end{align*}

for \(p \left (x \right )\). This is because of dividing by the above earlier. This gives the following singular solution(s), which also has to satisfy the given ODE.

\begin{align*} p \left (x \right ) = 0\\ p \left (x \right ) = -\left (-{\mathrm e}^{2 c_2}\right )^{{2}/{3}} {\mathrm e}^{-c_2} \end{align*}

Substituing the above solution for \(p\) in (2A) gives

\begin{align*} y &= -\frac {{\left (-\operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C15} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x -4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{6} {\mathrm e}^{-3 c_2}+{\mathrm e}^{c_2}\right )}^{2}}{-9 \operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C15} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x -4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{6} {\mathrm e}^{-3 c_2}-18 \sqrt {\operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C15} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x -4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{2}}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C15} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x -4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{2} {\mathrm e}^{-c_2}-9 \,{\mathrm e}^{c_2}} \\ y &= \frac {{\mathrm e}^{c_2}}{9} \\ \end{align*}

Solving ode 5A

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 &= 0\\ g &= -\frac {\left (p^{3}+{\mathrm e}^{c_2}\right )^{2}}{9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p -9 \,{\mathrm e}^{c_2}} \end{align*}

Hence (2) becomes

\begin{equation} \tag{2A} p = \left (-\frac {6 p^{5}}{9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p -9 \,{\mathrm e}^{c_2}}-\frac {6 p^{2} {\mathrm e}^{c_2}}{9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p -9 \,{\mathrm e}^{c_2}}+\frac {27 p^{8}}{\left (9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p -9 \,{\mathrm e}^{c_2}\right )^{2}}-\frac {9 p^{7} {\mathrm e}^{c_2}}{\left (9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p -9 \,{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p}}+\frac {18 p^{6} \sqrt {-{\mathrm e}^{c_2} p}}{\left (9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p -9 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {54 \,{\mathrm e}^{c_2} p^{5}}{\left (9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p -9 \,{\mathrm e}^{c_2}\right )^{2}}-\frac {18 \,{\mathrm e}^{2 c_2} p^{4}}{\left (9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p -9 \,{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p}}+\frac {36 \,{\mathrm e}^{c_2} p^{3} \sqrt {-{\mathrm e}^{c_2} p}}{\left (9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p -9 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {27 \,{\mathrm e}^{2 c_2} p^{2}}{\left (9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p -9 \,{\mathrm e}^{c_2}\right )^{2}}-\frac {9 \,{\mathrm e}^{3 c_2} p}{\left (9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p -9 \,{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p}}+\frac {18 \,{\mathrm e}^{2 c_2} \sqrt {-{\mathrm e}^{c_2} p}}{\left (9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p -9 \,{\mathrm e}^{c_2}\right )^{2}}\right ) p^{\prime }\left (x \right ) \end{equation}

The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives

\begin{align*} p = 0 \end{align*}

Solving the above for \(p\) results in

\begin{align*} p_{1} &=0 \end{align*}

Substituting these in (1A) and keeping singular solution that verifies the ode gives

\begin{align*} y = \frac {{\mathrm e}^{c_2}}{9} \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{equation} \tag{3} p^{\prime }\left (x \right ) = \frac {p \left (x \right )}{-\frac {6 p \left (x \right )^{5}}{9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )-9 \,{\mathrm e}^{c_2}}-\frac {6 p \left (x \right )^{2} {\mathrm e}^{c_2}}{9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )-9 \,{\mathrm e}^{c_2}}+\frac {27 p \left (x \right )^{8}}{\left (9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )-9 \,{\mathrm e}^{c_2}\right )^{2}}-\frac {9 p \left (x \right )^{7} {\mathrm e}^{c_2}}{\left (9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )-9 \,{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}}+\frac {18 p \left (x \right )^{6} \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}}{\left (9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )-9 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {54 \,{\mathrm e}^{c_2} p \left (x \right )^{5}}{\left (9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )-9 \,{\mathrm e}^{c_2}\right )^{2}}-\frac {18 \,{\mathrm e}^{2 c_2} p \left (x \right )^{4}}{\left (9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )-9 \,{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}}+\frac {36 \,{\mathrm e}^{c_2} p \left (x \right )^{3} \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}}{\left (9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )-9 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {27 \,{\mathrm e}^{2 c_2} p \left (x \right )^{2}}{\left (9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )-9 \,{\mathrm e}^{c_2}\right )^{2}}-\frac {9 \,{\mathrm e}^{3 c_2} p \left (x \right )}{\left (9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )-9 \,{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}}+\frac {18 \,{\mathrm e}^{2 c_2} \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}}{\left (9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )-9 \,{\mathrm e}^{c_2}\right )^{2}}} \end{equation}

This ODE is now solved for \(p \left (x \right )\). No inversion is needed.

Integrating gives

\begin{align*} \int \frac {-\sqrt {-{\mathrm e}^{c_2} p}\, p^{7}+3 \,{\mathrm e}^{c_2} p^{6}+2 \sqrt {-{\mathrm e}^{c_2} p}\, {\mathrm e}^{c_2} p^{4}+2 \,{\mathrm e}^{2 c_2} p^{3}+3 \sqrt {-{\mathrm e}^{c_2} p}\, {\mathrm e}^{2 c_2} p -{\mathrm e}^{3 c_2}}{3 \left (-p^{3}-2 \sqrt {-{\mathrm e}^{c_2} p}\, p +{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p}}d p &= dx\\ \frac {2 \,{\mathrm e}^{-2 c_2} \left (-\frac {{\mathrm e}^{2 c_2} p^{2}}{4}+{\mathrm e}^{2 c_2} \sqrt {-{\mathrm e}^{c_2} p}\right )}{3}&= x +\textit {\_C16} \end{align*}

Singular solutions are found by solving

\begin{align*} \frac {3 \left (-p^{3}-2 \sqrt {-{\mathrm e}^{c_2} p}\, p +{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p}}{-\sqrt {-{\mathrm e}^{c_2} p}\, p^{7}+3 \,{\mathrm e}^{c_2} p^{6}+2 \sqrt {-{\mathrm e}^{c_2} p}\, {\mathrm e}^{c_2} p^{4}+2 \,{\mathrm e}^{2 c_2} p^{3}+3 \sqrt {-{\mathrm e}^{c_2} p}\, {\mathrm e}^{2 c_2} p -{\mathrm e}^{3 c_2}}&= 0 \end{align*}

for \(p \left (x \right )\). This is because of dividing by the above earlier. This gives the following singular solution(s), which also has to satisfy the given ODE.

\begin{align*} p \left (x \right ) = 0\\ p \left (x \right ) = -\left (-{\mathrm e}^{2 c_2}\right )^{{2}/{3}} {\mathrm e}^{-c_2} \end{align*}

Substituing the above solution for \(p\) in (2A) gives

\begin{align*} y &= -\frac {{\left (-\operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C16} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x -4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{6} {\mathrm e}^{-3 c_2}+{\mathrm e}^{c_2}\right )}^{2}}{-9 \operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C16} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x -4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{6} {\mathrm e}^{-3 c_2}-18 \sqrt {\operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C16} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x -4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{2}}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C16} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x -4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{2} {\mathrm e}^{-c_2}-9 \,{\mathrm e}^{c_2}} \\ y &= \frac {{\mathrm e}^{c_2}}{9} \\ \end{align*}

Solving ode 6A

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 &= 0\\ g &= -\frac {\left (p^{3}+{\mathrm e}^{c_2}\right )^{2}}{9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p -9 \,{\mathrm e}^{c_2}} \end{align*}

Hence (2) becomes

\begin{equation} \tag{2A} p = \left (-\frac {6 p^{5}}{9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p -9 \,{\mathrm e}^{c_2}}-\frac {6 p^{2} {\mathrm e}^{c_2}}{9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p -9 \,{\mathrm e}^{c_2}}+\frac {27 p^{8}}{\left (9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p -9 \,{\mathrm e}^{c_2}\right )^{2}}-\frac {9 p^{7} {\mathrm e}^{c_2}}{\left (9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p -9 \,{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p}}+\frac {18 p^{6} \sqrt {-{\mathrm e}^{c_2} p}}{\left (9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p -9 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {54 \,{\mathrm e}^{c_2} p^{5}}{\left (9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p -9 \,{\mathrm e}^{c_2}\right )^{2}}-\frac {18 \,{\mathrm e}^{2 c_2} p^{4}}{\left (9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p -9 \,{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p}}+\frac {36 \,{\mathrm e}^{c_2} p^{3} \sqrt {-{\mathrm e}^{c_2} p}}{\left (9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p -9 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {27 \,{\mathrm e}^{2 c_2} p^{2}}{\left (9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p -9 \,{\mathrm e}^{c_2}\right )^{2}}-\frac {9 \,{\mathrm e}^{3 c_2} p}{\left (9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p -9 \,{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p}}+\frac {18 \,{\mathrm e}^{2 c_2} \sqrt {-{\mathrm e}^{c_2} p}}{\left (9 p^{3}+18 \sqrt {-{\mathrm e}^{c_2} p}\, p -9 \,{\mathrm e}^{c_2}\right )^{2}}\right ) p^{\prime }\left (x \right ) \end{equation}

The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives

\begin{align*} p = 0 \end{align*}

Solving the above for \(p\) results in

\begin{align*} p_{1} &=0 \end{align*}

Substituting these in (1A) and keeping singular solution that verifies the ode gives

\begin{align*} y = \frac {{\mathrm e}^{c_2}}{9} \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{equation} \tag{3} p^{\prime }\left (x \right ) = \frac {p \left (x \right )}{-\frac {6 p \left (x \right )^{5}}{9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )-9 \,{\mathrm e}^{c_2}}-\frac {6 p \left (x \right )^{2} {\mathrm e}^{c_2}}{9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )-9 \,{\mathrm e}^{c_2}}+\frac {27 p \left (x \right )^{8}}{\left (9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )-9 \,{\mathrm e}^{c_2}\right )^{2}}-\frac {9 p \left (x \right )^{7} {\mathrm e}^{c_2}}{\left (9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )-9 \,{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}}+\frac {18 p \left (x \right )^{6} \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}}{\left (9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )-9 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {54 \,{\mathrm e}^{c_2} p \left (x \right )^{5}}{\left (9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )-9 \,{\mathrm e}^{c_2}\right )^{2}}-\frac {18 \,{\mathrm e}^{2 c_2} p \left (x \right )^{4}}{\left (9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )-9 \,{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}}+\frac {36 \,{\mathrm e}^{c_2} p \left (x \right )^{3} \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}}{\left (9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )-9 \,{\mathrm e}^{c_2}\right )^{2}}+\frac {27 \,{\mathrm e}^{2 c_2} p \left (x \right )^{2}}{\left (9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )-9 \,{\mathrm e}^{c_2}\right )^{2}}-\frac {9 \,{\mathrm e}^{3 c_2} p \left (x \right )}{\left (9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )-9 \,{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}}+\frac {18 \,{\mathrm e}^{2 c_2} \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}}{\left (9 p \left (x \right )^{3}+18 \sqrt {-{\mathrm e}^{c_2} p \left (x \right )}\, p \left (x \right )-9 \,{\mathrm e}^{c_2}\right )^{2}}} \end{equation}

This ODE is now solved for \(p \left (x \right )\). No inversion is needed.

Integrating gives

\begin{align*} \int \frac {-\sqrt {-{\mathrm e}^{c_2} p}\, p^{7}+3 \,{\mathrm e}^{c_2} p^{6}+2 \sqrt {-{\mathrm e}^{c_2} p}\, {\mathrm e}^{c_2} p^{4}+2 \,{\mathrm e}^{2 c_2} p^{3}+3 \sqrt {-{\mathrm e}^{c_2} p}\, {\mathrm e}^{2 c_2} p -{\mathrm e}^{3 c_2}}{3 \left (-p^{3}-2 \sqrt {-{\mathrm e}^{c_2} p}\, p +{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p}}d p &= dx\\ \frac {2 \,{\mathrm e}^{-2 c_2} \left (-\frac {{\mathrm e}^{2 c_2} p^{2}}{4}+{\mathrm e}^{2 c_2} \sqrt {-{\mathrm e}^{c_2} p}\right )}{3}&= x +\textit {\_C17} \end{align*}

Singular solutions are found by solving

\begin{align*} \frac {3 \left (-p^{3}-2 \sqrt {-{\mathrm e}^{c_2} p}\, p +{\mathrm e}^{c_2}\right )^{2} \sqrt {-{\mathrm e}^{c_2} p}}{-\sqrt {-{\mathrm e}^{c_2} p}\, p^{7}+3 \,{\mathrm e}^{c_2} p^{6}+2 \sqrt {-{\mathrm e}^{c_2} p}\, {\mathrm e}^{c_2} p^{4}+2 \,{\mathrm e}^{2 c_2} p^{3}+3 \sqrt {-{\mathrm e}^{c_2} p}\, {\mathrm e}^{2 c_2} p -{\mathrm e}^{3 c_2}}&= 0 \end{align*}

for \(p \left (x \right )\). This is because of dividing by the above earlier. This gives the following singular solution(s), which also has to satisfy the given ODE.

\begin{align*} p \left (x \right ) = 0\\ p \left (x \right ) = -\left (-{\mathrm e}^{2 c_2}\right )^{{2}/{3}} {\mathrm e}^{-c_2} \end{align*}

Substituing the above solution for \(p\) in (2A) gives

\begin{align*} y &= -\frac {{\left (-\operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C17} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x -4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{6} {\mathrm e}^{-3 c_2}+{\mathrm e}^{c_2}\right )}^{2}}{-9 \operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C17} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x -4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{6} {\mathrm e}^{-3 c_2}-18 \sqrt {\operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C17} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x -4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{2}}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C17} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x -4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{2} {\mathrm e}^{-c_2}-9 \,{\mathrm e}^{c_2}} \\ y &= \frac {{\mathrm e}^{c_2}}{9} \\ \end{align*}

The solution \(y = \frac {{\left (-\operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C12} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x +4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{6} {\mathrm e}^{-3 c_2}+{\mathrm e}^{c_2}\right )}^{2}}{9 \operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C12} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x +4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{6} {\mathrm e}^{-3 c_2}-18 \sqrt {\operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C12} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x +4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{2}}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C12} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x +4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{2} {\mathrm e}^{-c_2}+9 \,{\mathrm e}^{c_2}}\) simplifies to

\begin{align*} \text {Expression too large to display} \\ \text {Expression too large to display} \\ \text {Expression too large to display} \\ \text {Expression too large to display} \\ \end{align*}

The solution \(y = \frac {{\left (-\operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C13} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x +4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{6} {\mathrm e}^{-3 c_2}+{\mathrm e}^{c_2}\right )}^{2}}{9 \operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C13} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x +4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{6} {\mathrm e}^{-3 c_2}-18 \sqrt {\operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C13} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x +4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{2}}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C13} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x +4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{2} {\mathrm e}^{-c_2}+9 \,{\mathrm e}^{c_2}}\) simplifies to

\begin{align*} \text {Expression too large to display} \\ \text {Expression too large to display} \\ \text {Expression too large to display} \\ \text {Expression too large to display} \\ \end{align*}

The solution \(y = \frac {{\left (-\operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C14} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x +4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{6} {\mathrm e}^{-3 c_2}+{\mathrm e}^{c_2}\right )}^{2}}{9 \operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C14} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x +4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{6} {\mathrm e}^{-3 c_2}-18 \sqrt {\operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C14} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x +4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{2}}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C14} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x +4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{2} {\mathrm e}^{-c_2}+9 \,{\mathrm e}^{c_2}}\) simplifies to

\begin{align*} \text {Expression too large to display} \\ \text {Expression too large to display} \\ \text {Expression too large to display} \\ \text {Expression too large to display} \\ \end{align*}

The solution \(y = -\frac {{\left (-\operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C15} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x -4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{6} {\mathrm e}^{-3 c_2}+{\mathrm e}^{c_2}\right )}^{2}}{-9 \operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C15} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x -4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{6} {\mathrm e}^{-3 c_2}-18 \sqrt {\operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C15} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x -4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{2}}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C15} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x -4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{2} {\mathrm e}^{-c_2}-9 \,{\mathrm e}^{c_2}}\) simplifies to

\begin{align*} \text {Expression too large to display} \\ \text {Expression too large to display} \\ \text {Expression too large to display} \\ \text {Expression too large to display} \\ \end{align*}

The solution \(y = -\frac {{\left (-\operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C16} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x -4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{6} {\mathrm e}^{-3 c_2}+{\mathrm e}^{c_2}\right )}^{2}}{-9 \operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C16} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x -4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{6} {\mathrm e}^{-3 c_2}-18 \sqrt {\operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C16} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x -4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{2}}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C16} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x -4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{2} {\mathrm e}^{-c_2}-9 \,{\mathrm e}^{c_2}}\) simplifies to

\begin{align*} \text {Expression too large to display} \\ \text {Expression too large to display} \\ \text {Expression too large to display} \\ \text {Expression too large to display} \\ \end{align*}

The solution \(y = -\frac {{\left (-\operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C17} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x -4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{6} {\mathrm e}^{-3 c_2}+{\mathrm e}^{c_2}\right )}^{2}}{-9 \operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C17} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x -4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{6} {\mathrm e}^{-3 c_2}-18 \sqrt {\operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C17} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x -4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{2}}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}+6 \textit {\_C17} \,{\mathrm e}^{2 c_2}+6 \,{\mathrm e}^{2 c_2} x -4 \textit {\_Z} \,{\mathrm e}^{2 c_2}\right )^{2} {\mathrm e}^{-c_2}-9 \,{\mathrm e}^{c_2}}\) simplifies to

\begin{align*} \text {Expression too large to display} \\ \text {Expression too large to display} \\ \text {Expression too large to display} \\ \text {Expression too large to display} \\ \end{align*}

The above solution was found not to satisfy the ode or the IC. Hence it is removed.

The above solution was found not to satisfy the ode or the IC. Hence it is removed.

The above solution was found not to satisfy the ode or the IC. Hence it is removed.

The above solution was found not to satisfy the ode or the IC. Hence it is removed.

The above solution was found not to satisfy the ode or the IC. Hence it is removed.

The above solution was found not to satisfy the ode or the IC. Hence it is removed.

The above solution was found not to satisfy the ode or the IC. Hence it is removed.

The above solution was found not to satisfy the ode or the IC. Hence it is removed.

The above solution was found not to satisfy the ode or the IC. Hence it is removed.

The above solution was found not to satisfy the ode or the IC. Hence it is removed.

The above solution was found not to satisfy the ode or the IC. Hence it is removed.

The above solution was found not to satisfy the ode or the IC. Hence it is removed.

The above solution was found not to satisfy the ode or the IC. Hence it is removed.

The above solution was found not to satisfy the ode or the IC. Hence it is removed.

The above solution was found not to satisfy the ode or the IC. Hence it is removed.

The above solution was found not to satisfy the ode or the IC. Hence it is removed.

The above solution was found not to satisfy the ode or the IC. Hence it is removed.

The above solution was found not to satisfy the ode or the IC. Hence it is removed.

The above solution was found not to satisfy the ode or the IC. Hence it is removed.

The above solution was found not to satisfy the ode or the IC. Hence it is removed.

The above solution was found not to satisfy the ode or the IC. Hence it is removed.

The above solution was found not to satisfy the ode or the IC. Hence it is removed.

The above solution was found not to satisfy the ode or the IC. Hence it is removed.

The above solution was found not to satisfy the ode or the IC. Hence it is removed.

For solution (3) found earlier, since \(p=y^{\prime }\) then the new first order ode to solve is

\begin{align*} y^{\prime } = 0 \end{align*}

Entering first order ode quadrature solverSince the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).

\begin{align*} \int {dy} &= \int {0\, dx} + \textit {\_C18} \\ y &= \textit {\_C18} \end{align*}

Summary of solutions found

\begin{align*} \int _{}^{y}-\frac {4}{\left ({\mathrm e}^{-\frac {\ln \left (\tau \right )}{2}+\frac {c_1}{2}}+3\right )^{{2}/{3}} \left (1+i \sqrt {3}\right )^{2} \tau ^{{1}/{3}}}d \tau &= x +c_9 \\ y &= \textit {\_C18} \\ y &= \frac {{\mathrm e}^{c_2}}{9} \\ \end{align*}

2.1.45.3 ✓ Maple. Time used: 0.039 (sec). Leaf size: 263

ode:=y(x)*diff(diff(y(x),x),x)^2+diff(y(x),x) = 0; 
dsolve(ode,y(x), singsol=all);
 

\begin{align*} y &= c_1 \\ y &= 0 \\ -\int _{}^{y}\frac {\textit {\_a}}{\left (\textit {\_a}^{{3}/{2}} \left (c_1 -3 \sqrt {\textit {\_a}}\right )\right )^{{2}/{3}}}d \textit {\_a} -x -c_2 &= 0 \\ -\int _{}^{y}\frac {\textit {\_a}}{\left (\textit {\_a}^{{3}/{2}} \left (c_1 +3 \sqrt {\textit {\_a}}\right )\right )^{{2}/{3}}}d \textit {\_a} -x -c_2 &= 0 \\ \frac {-4 \int _{}^{y}\frac {\textit {\_a}}{\left (\textit {\_a}^{{3}/{2}} \left (c_1 -3 \sqrt {\textit {\_a}}\right )\right )^{{2}/{3}}}d \textit {\_a} -2 i \left (c_2 +x \right ) \sqrt {3}+2 x +2 c_2}{\left (-1-i \sqrt {3}\right )^{2}} &= 0 \\ \frac {-4 \int _{}^{y}\frac {\textit {\_a}}{\left (\textit {\_a}^{{3}/{2}} \left (c_1 -3 \sqrt {\textit {\_a}}\right )\right )^{{2}/{3}}}d \textit {\_a} +2 i \left (c_2 +x \right ) \sqrt {3}+2 x +2 c_2}{\left (1-i \sqrt {3}\right )^{2}} &= 0 \\ \frac {-4 \int _{}^{y}\frac {\textit {\_a}}{\left (\textit {\_a}^{{3}/{2}} \left (c_1 +3 \sqrt {\textit {\_a}}\right )\right )^{{2}/{3}}}d \textit {\_a} -2 i \left (c_2 +x \right ) \sqrt {3}+2 x +2 c_2}{\left (-1-i \sqrt {3}\right )^{2}} &= 0 \\ \frac {-4 \int _{}^{y}\frac {\textit {\_a}}{\left (\textit {\_a}^{{3}/{2}} \left (c_1 +3 \sqrt {\textit {\_a}}\right )\right )^{{2}/{3}}}d \textit {\_a} +2 i \left (c_2 +x \right ) \sqrt {3}+2 x +2 c_2}{\left (1-i \sqrt {3}\right )^{2}} &= 0 \\ \end{align*}

Maple trace

Methods for second order ODEs: 
   *** Sublevel 2 *** 
   Methods for second order ODEs: 
   Successful isolation of d^2y/dx^2: 2 solutions were found. Trying to solve e\ 
ach resulting ODE. 
      *** Sublevel 3 *** 
      Methods for second order ODEs: 
      --- Trying classification methods --- 
      trying 2nd order Liouville 
      trying 2nd order WeierstrassP 
      trying 2nd order JacobiSN 
      differential order: 2; trying a linearization to 3rd order 
      trying 2nd order ODE linearizable_by_differentiation 
      trying 2nd order, 2 integrating factors of the form mu(x,y) 
      trying differential order: 2; missing variables 
         -> Computing symmetries using: way = 3 
      -> Calling odsolve with the ODE, diff(_b(_a),_a)*_b(_a)-1/_a*(-_a*_b(_a)) 
^(1/2) = 0, _b(_a), HINT = [[_a, 1/3*_b]] 
         *** Sublevel 4 *** 
         symmetry methods on request 
         1st order, trying reduction of order with given symmetries: 
[_a, 1/3*_b] 
         1st order, trying the canonical coordinates of the invariance group 
            -> Calling odsolve with the ODE, diff(y(x),x) = 1/3*y(x)/x, y(x) 
               *** Sublevel 5 *** 
               Methods for first order ODEs: 
               --- Trying classification methods --- 
               trying a quadrature 
               trying 1st order linear 
               <- 1st order linear successful 
         <- 1st order, canonical coordinates successful 
      <- differential order: 2; canonical coordinates successful 
      <- differential order 2; missing variables successful 
   ------------------- 
   * Tackling next ODE. 
      *** Sublevel 3 *** 
      Methods for second order ODEs: 
      --- Trying classification methods --- 
      trying 2nd order Liouville 
      trying 2nd order WeierstrassP 
      trying 2nd order JacobiSN 
      differential order: 2; trying a linearization to 3rd order 
      trying 2nd order ODE linearizable_by_differentiation 
      trying 2nd order, 2 integrating factors of the form mu(x,y) 
      trying differential order: 2; missing variables 
         -> Computing symmetries using: way = 3 
      -> Calling odsolve with the ODE, diff(_b(_a),_a)*_b(_a)+1/_a*(-_a*_b(_a)) 
^(1/2) = 0, _b(_a), HINT = [[_a, 1/3*_b]] 
         *** Sublevel 4 *** 
         symmetry methods on request 
         1st order, trying reduction of order with given symmetries: 
[_a, 1/3*_b] 
         1st order, trying the canonical coordinates of the invariance group 
         <- 1st order, canonical coordinates successful 
      <- differential order: 2; canonical coordinates successful 
      <- differential order 2; missing variables successful 
-> Calling odsolve with the ODE, diff(y(x),x) = 0, y(x), singsol = none 
   *** Sublevel 2 *** 
   Methods for first order ODEs: 
   --- Trying classification methods --- 
   trying a quadrature 
   trying 1st order linear 
   <- 1st order linear successful
 

2.1.45.4 ✓ Mathematica. Time used: 60.637 (sec). Leaf size: 18473

ode=y[x]*D[y[x],{x,2}]^2+D[y[x],x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

Too large to display

2.1.45.5 ✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(y(x)*Derivative(y(x), (x, 2))**2 + Derivative(y(x), x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

NotImplementedError : solve: Cannot solve y(x)*Derivative(y(x), (x, 2))**2 + Derivative(y(x), x)
 

Python version: 3.12.3 (main, Aug 14 2025, 17:47:21) [GCC 13.3.0] 
Sympy version 1.14.0
 

classify_ode(ode,func=y(x)) 
 
()