Problem 54

2.1.54 Problem 54

Internal problem ID [10312]
Book : First order enumerated odes
Section : section 1
Problem number : 54
Date solved : Wednesday, February 11, 2026 at 06:29:38 AM
CAS classiﬁcation : [[_homogeneous, `class G`], _rational]

2.1.54.1 Solved using ﬁrst_order_ode_homog_type_G

5.198 (sec)

Entering ﬁrst order ode homog type G solver

\begin{align*} {y^{\prime }}^{3}&=\frac {y^{2}}{x} \\ \end{align*}

Multiplying the right side of the ode, which is \(\frac {\left (x^{2} y^{2}\right )^{{1}/{3}}}{x}\) by \(\frac {x}{y}\) gives

\begin{align*} y^{\prime } &= \left (\frac {x}{y}\right ) \frac {\left (x^{2} y^{2}\right )^{{1}/{3}}}{x}\\ &= \frac {\left (x^{2} y^{2}\right )^{{1}/{3}}}{y}\\ &= F(x,y) \end{align*}

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

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

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

Let

\begin{align*} y&=\frac {z}{x^ \alpha }\\ &=\frac {z}{\frac {1}{x^{2}}} \end{align*}

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

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

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

Therefore, the implicit solution is given by

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

Which gives

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

The value of the above is

\[ \ln \left (x \right )-c_1 +\frac {3 \ln \left (-\frac {-x^{2} \left (\frac {y^{2}}{x^{4}}\right )^{{1}/{3}}+2 y}{x^{2} \left (\frac {y^{2}}{x^{4}}\right )^{{1}/{3}}}\right )}{2} = 0 \]

Multiplying the right side of the ode, which is \(-\frac {\left (x^{2} y^{2}\right )^{{1}/{3}}}{2 x}+\frac {i \sqrt {3}\, \left (x^{2} y^{2}\right )^{{1}/{3}}}{2 x}\) by \(\frac {x}{y}\) gives

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

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

\begin{align*} f_x&= x \left (\frac {\partial }{\partial x}F \left (x , y\right )\right )\\ &= \frac {x^{2} \left (i \sqrt {3}-1\right ) y}{3 \left (x^{2} y^{2}\right )^{{2}/{3}}}\\ f_y&= y \left (\frac {\partial }{\partial y}F \left (x , y\right )\right )\\ &= -\frac {x^{2} \left (i \sqrt {3}-1\right ) y}{6 \left (x^{2} y^{2}\right )^{{2}/{3}}}\\ \alpha &= \frac {f_x}{f_y} \\ &=-2 \end{align*}

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

Let

\begin{align*} y&=\frac {z}{x^ \alpha }\\ &=\frac {z}{\frac {1}{x^{2}}} \end{align*}

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

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

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

Therefore, the implicit solution is given by

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

Which gives

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

The value of the above is

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

Multiplying the right side of the ode, which is \(-\frac {\left (x^{2} y^{2}\right )^{{1}/{3}}}{2 x}-\frac {i \sqrt {3}\, \left (x^{2} y^{2}\right )^{{1}/{3}}}{2 x}\) by \(\frac {x}{y}\) gives

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

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

\begin{align*} f_x&= x \left (\frac {\partial }{\partial x}F \left (x , y\right )\right )\\ &= -\frac {x^{2} \left (1+i \sqrt {3}\right ) y}{3 \left (x^{2} y^{2}\right )^{{2}/{3}}}\\ f_y&= y \left (\frac {\partial }{\partial y}F \left (x , y\right )\right )\\ &= \frac {x^{2} \left (1+i \sqrt {3}\right ) y}{6 \left (x^{2} y^{2}\right )^{{2}/{3}}}\\ \alpha &= \frac {f_x}{f_y} \\ &=-2 \end{align*}

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

Let

\begin{align*} y&=\frac {z}{x^ \alpha }\\ &=\frac {z}{\frac {1}{x^{2}}} \end{align*}

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

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

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

Therefore, the implicit solution is given by

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

Which gives

\[ \ln \left (x \right )-c_3 +\int _{}^{\frac {y}{x^{2}}}\frac {1}{z \left (2+\frac {\left (z^{2}\right )^{{1}/{3}} \left (1+i \sqrt {3}\right )}{2 z}\right )}d z = 0 \]

The value of the above is

\[ \ln \left (x \right )-c_3 +\int _{}^{\frac {y}{x^{2}}}\frac {1}{z \left (2+\frac {\left (z^{2}\right )^{{1}/{3}} \left (1+i \sqrt {3}\right )}{2 z}\right )}d z = 0 \]

Simplifying the above gives

\begin{align*} \ln \left (x \right )-c_1 +\frac {3 \ln \left (\frac {x^{2} \left (\frac {y^{2}}{x^{4}}\right )^{{1}/{3}}-2 y}{x^{2} \left (\frac {y^{2}}{x^{4}}\right )^{{1}/{3}}}\right )}{2} &= 0 \\ \ln \left (x \right )-c_2 -2 \int _{}^{\frac {y}{x^{2}}}\frac {1}{i \left (z^{2}\right )^{{1}/{3}} \sqrt {3}-\left (z^{2}\right )^{{1}/{3}}-4 z}d z &= 0 \\ \ln \left (x \right )-c_3 +2 \int _{}^{\frac {y}{x^{2}}}\frac {1}{i \sqrt {3}\, \left (z^{2}\right )^{{1}/{3}}+\left (z^{2}\right )^{{1}/{3}}+4 z}d z &= 0 \\ \end{align*}

Summary of solutions found

2.1.54.2 Solved using ﬁrst_order_nonlinear_p_but_separable

3.566 (sec)

Entering ﬁrst order nonlinear \(p\) but separable solver

\begin{align*} {y^{\prime }}^{3}&=\frac {y^{2}}{x} \\ \end{align*}

The ode has the form

\begin{align*} (y')^{\frac {n}{m}} &= f(x) g(y)\tag {1} \end{align*}

Where \(n=3, m=1, f=\frac {1}{x} , g=y^{2}\). Hence the ode is

\begin{align*} (y')^{3} &= \frac {y^{2}}{x} \end{align*}

Solving for \(y^{\prime }\) from (1) gives

\begin{align*} y^{\prime } &=\left (f g \right )^{{1}/{3}}\\ y^{\prime } &=-\frac {\left (f g \right )^{{1}/{3}}}{2}+\frac {i \sqrt {3}\, \left (f g \right )^{{1}/{3}}}{2}\\ y^{\prime } &=-\frac {\left (f g \right )^{{1}/{3}}}{2}-\frac {i \sqrt {3}\, \left (f g \right )^{{1}/{3}}}{2} \end{align*}

To be able to solve as separable ode, we have to now assume that \(f>0,g>0\).

\begin{align*} \frac {1}{x} &> 0\\ y^{2} &> 0 \end{align*}

Under the above assumption the diﬀerential equations become separable and can be written as

\begin{align*} y^{\prime } &=f^{{1}/{3}} g^{{1}/{3}}\\ y^{\prime } &=\frac {f^{{1}/{3}} g^{{1}/{3}} \left (i \sqrt {3}-1\right )}{2}\\ y^{\prime } &=-\frac {f^{{1}/{3}} g^{{1}/{3}} \left (1+i \sqrt {3}\right )}{2} \end{align*}

Therefore

\begin{align*} \frac {1}{g^{{1}/{3}}} \, dy &= \left (f^{{1}/{3}}\right )\,dx\\ \frac {2}{g^{{1}/{3}} \left (i \sqrt {3}-1\right )} \, dy &= \left (f^{{1}/{3}}\right )\,dx\\ -\frac {2}{g^{{1}/{3}} \left (1+i \sqrt {3}\right )} \, dy &= \left (f^{{1}/{3}}\right )\,dx \end{align*}

Replacing \(f(x),g(y)\) by their values gives

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

Integrating now gives the following solutions

\begin{align*} \int \frac {1}{\left (y^{2}\right )^{{1}/{3}}}d y &= \int \left (\frac {1}{x}\right )^{{1}/{3}}d x +c_1\\ \frac {3 y}{\left (y^{2}\right )^{{1}/{3}}} &= \frac {3 x \left (\frac {1}{x}\right )^{{1}/{3}}}{2}\\ \int \frac {2}{\left (y^{2}\right )^{{1}/{3}} \left (i \sqrt {3}-1\right )}d y &= \int \left (\frac {1}{x}\right )^{{1}/{3}}d x +c_1\\ \frac {6 y}{\left (y^{2}\right )^{{1}/{3}} \left (i \sqrt {3}-1\right )} &= \frac {3 x \left (\frac {1}{x}\right )^{{1}/{3}}}{2}\\ \int -\frac {2}{\left (y^{2}\right )^{{1}/{3}} \left (1+i \sqrt {3}\right )}d y &= \int \left (\frac {1}{x}\right )^{{1}/{3}}d x +c_1\\ -\frac {6 y}{\left (y^{2}\right )^{{1}/{3}} \left (1+i \sqrt {3}\right )} &= \frac {3 x \left (\frac {1}{x}\right )^{{1}/{3}}}{2} \end{align*}

Therefore

\begin{align*} y &= \frac {x^{2}}{8}+\frac {x^{2} \left (\frac {1}{x}\right )^{{2}/{3}} c_1}{4}+\frac {x \left (\frac {1}{x}\right )^{{1}/{3}} c_1^{2}}{6}+\frac {c_1^{3}}{27} \\ y &= \frac {x^{2}}{8}+\frac {x^{2} \left (\frac {1}{x}\right )^{{2}/{3}} c_1}{4}+\frac {x \left (\frac {1}{x}\right )^{{1}/{3}} c_1^{2}}{6}+\frac {c_1^{3}}{27} \\ y &= \frac {x^{2}}{8}+\frac {x^{2} \left (\frac {1}{x}\right )^{{2}/{3}} c_1}{4}+\frac {x \left (\frac {1}{x}\right )^{{1}/{3}} c_1^{2}}{6}+\frac {c_1^{3}}{27} \\ \end{align*}

Summary of solutions found

\begin{align*} y &= \frac {x^{2}}{8}+\frac {x^{2} \left (\frac {1}{x}\right )^{{2}/{3}} c_1}{4}+\frac {x \left (\frac {1}{x}\right )^{{1}/{3}} c_1^{2}}{6}+\frac {c_1^{3}}{27} \\ \end{align*}

2.1.54.3 Solved using ﬁrst_order_ode_parametric method

154.809 (sec)

Entering ﬁrst order ode parametric solver

\begin{align*} {y^{\prime }}^{3}&=\frac {y^{2}}{x} \\ \end{align*}

Let \(y^{\prime }\) be a parameter \(\lambda \). The ode becomes

\begin{align*} \lambda ^{3}-\frac {y^{2}}{x} = 0 \end{align*}

Isolating \(x\) gives

\begin{align*} x = \frac {y^{2}}{\lambda ^{3}}\\ x = F \left (y , \lambda \right ) \end{align*}

Now we generate an ode in \(y \left (\lambda \right )\) using

\begin{align*} \frac {d}{d \lambda }y \left (\lambda \right ) &= \frac { \lambda \frac {\partial F}{\partial \lambda }} { 1- \frac {\partial F}{\partial y} } \\ &= -\frac {3 y^{2}}{\lambda ^{3} \left (1-\frac {2 y}{\lambda ^{2}}\right )}\\ &= -\frac {3 y \left (\lambda \right )^{2}}{\lambda \left (\lambda ^{2}-2 y \left (\lambda \right )\right )} \end{align*}

Which is now solved for \(y\).

2.1.54.4 Solved using ﬁrst_order_ode_isobaric

3.274 (sec)

Entering ﬁrst order ode isobaric solverSolving for \(y'\) gives

\[ y' = -\frac {3 y \left (\lambda \right )^{2}}{\lambda \left (\lambda ^{2}-2 y \left (\lambda \right )\right )} \]

An ode \(\frac {d}{d \lambda }y \left (\lambda \right )=f(\lambda ,y)\) is isobaric if

\[ f(t \lambda , t^m y) = t^{m-1} f(\lambda ,y)\tag {1} \]

Where here

\[ f(\lambda ,y) = -\frac {3 y \left (\lambda \right )^{2}}{\lambda \left (\lambda ^{2}-2 y \left (\lambda \right )\right )}\tag {2} \]

\(m\) is the order of isobaric. Substituting (2) into (1) and solving for \(m\) gives

\[ m = 2 \]

Since the ode is isobaric of order \(m=2\), then the substitution

\begin{align*} y&=u \lambda ^m \\ &=u \,\lambda ^{2} \end{align*}

Converts the ODE to a separable in \(u \left (\lambda \right )\). Performing this substitution gives

\[ 2 \lambda u \left (\lambda \right )+\lambda ^{2} \left (\frac {d}{d \lambda }u \left (\lambda \right )\right ) = -\frac {3 \lambda ^{3} u \left (\lambda \right )^{2}}{\lambda ^{2}-2 \lambda ^{2} u \left (\lambda \right )} \]

Entering ﬁrst order ode separable solverThe ode

\begin{equation} \frac {d}{d \lambda }u \left (\lambda \right ) = -\frac {u \left (\lambda \right ) \left (u \left (\lambda \right )-2\right )}{\lambda \left (2 u \left (\lambda \right )-1\right )} \end{equation}

is separable as it can be written as

\begin{align*} \frac {d}{d \lambda }u \left (\lambda \right )&= -\frac {u \left (\lambda \right ) \left (u \left (\lambda \right )-2\right )}{\lambda \left (2 u \left (\lambda \right )-1\right )}\\ &= f(\lambda ) g(u) \end{align*}

Where

\begin{align*} f(\lambda ) &= -\frac {1}{\lambda }\\ g(u) &= \frac {u \left (u -2\right )}{2 u -1} \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(\lambda ) \,d\lambda } \\ \int { \frac {2 u -1}{u \left (u -2\right )}\,du} &= \int { -\frac {1}{\lambda } \,d\lambda } \\ \end{align*}

\[ \frac {\ln \left (u \left (\lambda \right )\right )}{2}+\frac {3 \ln \left (u \left (\lambda \right )-2\right )}{2}=\ln \left (\frac {1}{\lambda }\right )+c_1 \]

We now need to ﬁnd the singular solutions, these are found by ﬁnding for what values \(g(u)\) is zero, since we had to divide by this above. Solving \(g(u)=0\) or

\[ \frac {u \left (u -2\right )}{2 u -1}=0 \]

for \(u \left (\lambda \right )\) gives

\begin{align*} u \left (\lambda \right )&=0\\ u \left (\lambda \right )&=2 \end{align*}

Now we go over each such singular solution and check if it veriﬁes the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} \frac {\ln \left (u \left (\lambda \right )\right )}{2}+\frac {3 \ln \left (u \left (\lambda \right )-2\right )}{2} &= \ln \left (\frac {1}{\lambda }\right )+c_1 \\ u \left (\lambda \right ) &= 0 \\ u \left (\lambda \right ) &= 2 \\ \end{align*}

Converting \(\frac {\ln \left (u \left (\lambda \right )\right )}{2}+\frac {3 \ln \left (u \left (\lambda \right )-2\right )}{2} = \ln \left (\frac {1}{\lambda }\right )+c_1\) back to \(y \left (\lambda \right )\) gives

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

Converting \(u \left (\lambda \right ) = 0\) back to \(y \left (\lambda \right )\) gives

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

Converting \(u \left (\lambda \right ) = 2\) back to \(y \left (\lambda \right )\) gives

\begin{align*} \frac {y \left (\lambda \right )}{\lambda ^{2}} = 2 \end{align*}

Solving for \(y \left (\lambda \right )\) gives

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

Summary of solutions found

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

2.1.54.5 Solved using ﬁrst_order_ode_homog_type_G

0.461 (sec)

Entering ﬁrst order ode homog type G solver

\begin{align*} \frac {d}{d \lambda }y \left (\lambda \right )&=-\frac {3 y \left (\lambda \right )^{2}}{\lambda \left (\lambda ^{2}-2 y \left (\lambda \right )\right )} \\ \end{align*}

Multiplying the right side of the ode, which is \(-\frac {3 y^{2}}{\lambda \left (\lambda ^{2}-2 y \right )}\) by \(\frac {\lambda }{y}\) gives

\begin{align*} \frac {d}{d \lambda }y \left (\lambda \right ) &= \left (\frac {\lambda }{y}\right ) -\frac {3 y^{2}}{\lambda \left (\lambda ^{2}-2 y \right )}\\ &= -\frac {3 y}{\lambda ^{2}-2 y}\\ &= F(\lambda ,y) \end{align*}

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

\begin{align*} f_\lambda &= \lambda \left (\frac {\partial }{\partial \lambda }F \left (\lambda , y\right )\right )\\ &= \frac {6 \lambda ^{2} y}{\left (\lambda ^{2}-2 y \right )^{2}}\\ f_y&= y \left (\frac {\partial }{\partial y}F \left (\lambda , y\right )\right )\\ &= -\frac {3 \lambda ^{2} y}{\left (\lambda ^{2}-2 y \right )^{2}}\\ \alpha &= \frac {f_\lambda }{f_y} \\ &=-2 \end{align*}

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

Let

\begin{align*} y&=\frac {z}{\lambda ^ \alpha }\\ &=\frac {z}{\frac {1}{\lambda ^{2}}} \end{align*}

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

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

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

Therefore, the implicit solution is given by

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

Which gives

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

The value of the above is

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

Summary of solutions found

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

Entering ﬁrst order ode abel second kind solver

\begin{align*} \frac {d}{d \lambda }y \left (\lambda \right )&=-\frac {3 y \left (\lambda \right )^{2}}{\lambda \left (\lambda ^{2}-2 y \left (\lambda \right )\right )} \\ \end{align*}

2.1.54.6 Solved using ﬁrst_order_ode_abel_second_kind_case_3

1.053 (sec)

Applying transformation

\begin{align*} y \left (\lambda \right )&=\frac {1}{u(\lambda )}-g \end{align*}

Results in the new ode which is Abel ﬁrst kind

\begin{align*} 2 \lambda u \left (\lambda \right )+\lambda ^{2} \left (\frac {d}{d \lambda }u \left (\lambda \right )\right ) = -\frac {3 \lambda ^{3} u \left (\lambda \right )^{2}}{\lambda ^{2}-2 \lambda ^{2} u \left (\lambda \right )} \end{align*}

Which is now solved. Entering ﬁrst order ode separable solverThe ode

is separable as it can be written as

Where

\begin{align*} f(\lambda ) &= -\frac {1}{\lambda }\\ g(u) &= \frac {\left (u -2\right ) u}{2 u -1} \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(\lambda ) \,d\lambda } \\ \int { \frac {2 u -1}{\left (u -2\right ) u}\,du} &= \int { -\frac {1}{\lambda } \,d\lambda } \\ \end{align*}

\[ \frac {\ln \left (u \left (\lambda \right )\right )}{2}+\frac {3 \ln \left (u \left (\lambda \right )-2\right )}{2}=\ln \left (\frac {1}{\lambda }\right )+c_1 \]

We now need to ﬁnd the singular solutions, these are found by ﬁnding for what values \(g(u)\) is zero, since we had to divide by this above. Solving \(g(u)=0\) or

\[ \frac {\left (u -2\right ) u}{2 u -1}=0 \]

for \(u \left (\lambda \right )\) gives

\begin{align*} u \left (\lambda \right )&=0\\ u \left (\lambda \right )&=2 \end{align*}

Now we go over each such singular solution and check if it veriﬁes the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

Substituting \(u \left (\lambda \right )=\frac {1}{y \left (\lambda \right )-\frac {\lambda ^{2}}{2}}\) in the above solution gives

Now we transform the solution \(u \left (\lambda \right ) = 2\) to \(y \left (\lambda \right )\) using \(u \left (\lambda \right )=\frac {1}{y \left (\lambda \right )-\frac {\lambda ^{2}}{2}}\) which gives

\begin{align*} y \left (\lambda \right ) &= \frac {1}{u \left (\lambda \right )} - -\frac {\lambda ^{2}}{2}\\ &= \frac {1}{2}+\frac {\lambda ^{2}}{2} \end{align*}

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

Summary of solutions found

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

2.1.54.7 Solved using ﬁrst_order_ode_abel_second_kind_solved_by_converting_to_ﬁrst_kind

117.632 (sec)

This is Abel second kind ODE, it has the form

\[ \left (y \left (\lambda \right )+g\right )\frac {d}{d \lambda }y \left (\lambda \right )= f_0(\lambda )+f_1(\lambda ) y \left (\lambda \right ) +f_2(\lambda )y \left (\lambda \right )^{2}+f_3(\lambda )y \left (\lambda \right )^{3} \]

Comparing the above to given ODE which is

\begin{align*}\frac {d}{d \lambda }y \left (\lambda \right ) = -\frac {3 y \left (\lambda \right )^{2}}{\lambda \left (\lambda ^{2}-2 y \left (\lambda \right )\right )}\tag {1} \end{align*}

Shows that

\begin{align*} g &= -\frac {\lambda ^{2}}{2}\\ f_0 &= 0\\ f_1 &= 0\\ f_2 &= \frac {3}{2 \lambda }\\ f_3 &= 0 \end{align*}

Applying transformation

\begin{align*} y \left (\lambda \right )&=\frac {1}{u(\lambda )}-g \end{align*}

Results in the new ode which is Abel ﬁrst kind

\begin{align*} \frac {d}{d \lambda }u \left (\lambda \right ) = -\frac {3 \lambda ^{3} u \left (\lambda \right )^{3}}{8}-\frac {u \left (\lambda \right )^{2} \lambda }{2}-\frac {3 u \left (\lambda \right )}{2 \lambda } \end{align*}

Which is now solved.

2.1.54.8 Solved using ﬁrst_order_ode_isobaric

19.257 (sec)

Entering ﬁrst order ode isobaric solverSolving for \(u'\) gives

\[ u' = -\frac {3 \lambda ^{3} u \left (\lambda \right )^{3}}{8}-\frac {u \left (\lambda \right )^{2} \lambda }{2}-\frac {3 u \left (\lambda \right )}{2 \lambda } \]

An ode \(\frac {d}{d \lambda }u \left (\lambda \right )=f(\lambda ,u)\) is isobaric if

\[ f(t \lambda , t^m u) = t^{m-1} f(\lambda ,u)\tag {1} \]

Where here

\[ f(\lambda ,u) = -\frac {3 \lambda ^{3} u \left (\lambda \right )^{3}}{8}-\frac {u \left (\lambda \right )^{2} \lambda }{2}-\frac {3 u \left (\lambda \right )}{2 \lambda }\tag {2} \]

\(m\) is the order of isobaric. Substituting (2) into (1) and solving for \(m\) gives

\[ m = -2 \]

Since the ode is isobaric of order \(m=-2\), then the substitution

\begin{align*} u&=u \lambda ^m \\ &=\frac {u}{\lambda ^{2}} \end{align*}

Converts the ODE to a separable in \(u \left (\lambda \right )\). Performing this substitution gives

\[ -\frac {2 u \left (\lambda \right )}{\lambda ^{3}}+\frac {\frac {d}{d \lambda }u \left (\lambda \right )}{\lambda ^{2}} = -\frac {3 u \left (\lambda \right )^{3}}{8 \lambda ^{3}}-\frac {u \left (\lambda \right )^{2}}{2 \lambda ^{3}}-\frac {3 u \left (\lambda \right )}{2 \lambda ^{3}} \]

Entering ﬁrst order ode separable solverThe ode

\begin{equation} \frac {d}{d \lambda }u \left (\lambda \right ) = -\frac {u \left (\lambda \right ) \left (u \left (\lambda \right )+2\right ) \left (3 u \left (\lambda \right )-2\right )}{8 \lambda } \end{equation}

is separable as it can be written as

\begin{align*} \frac {d}{d \lambda }u \left (\lambda \right )&= -\frac {u \left (\lambda \right ) \left (u \left (\lambda \right )+2\right ) \left (3 u \left (\lambda \right )-2\right )}{8 \lambda }\\ &= f(\lambda ) g(u) \end{align*}

Where

\begin{align*} f(\lambda ) &= -\frac {1}{8 \lambda }\\ g(u) &= u \left (u +2\right ) \left (3 u -2\right ) \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(\lambda ) \,d\lambda } \\ \int { \frac {1}{u \left (u +2\right ) \left (3 u -2\right )}\,du} &= \int { -\frac {1}{8 \lambda } \,d\lambda } \\ \end{align*}

\[ \frac {\ln \left (u \left (\lambda \right )+2\right )}{16}-\frac {\ln \left (u \left (\lambda \right )\right )}{4}+\frac {3 \ln \left (3 u \left (\lambda \right )-2\right )}{16}=\ln \left (\frac {1}{\lambda ^{{1}/{8}}}\right )+c_2 \]

We now need to ﬁnd the singular solutions, these are found by ﬁnding for what values \(g(u)\) is zero, since we had to divide by this above. Solving \(g(u)=0\) or

\[ u \left (u +2\right ) \left (3 u -2\right )=0 \]

for \(u \left (\lambda \right )\) gives

\begin{align*} u \left (\lambda \right )&=-2\\ u \left (\lambda \right )&=0 \end{align*}

Now we go over each such singular solution and check if it veriﬁes the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} \frac {\ln \left (u \left (\lambda \right )+2\right )}{16}-\frac {\ln \left (u \left (\lambda \right )\right )}{4}+\frac {3 \ln \left (3 u \left (\lambda \right )-2\right )}{16} &= \ln \left (\frac {1}{\lambda ^{{1}/{8}}}\right )+c_2 \\ u \left (\lambda \right ) &= -2 \\ u \left (\lambda \right ) &= 0 \\ \end{align*}

Converting \(\frac {\ln \left (u \left (\lambda \right )+2\right )}{16}-\frac {\ln \left (u \left (\lambda \right )\right )}{4}+\frac {3 \ln \left (3 u \left (\lambda \right )-2\right )}{16} = -\frac {\ln \left (\lambda \right )}{8}+c_2\) back to \(u \left (\lambda \right )\) gives

\begin{align*} \frac {\ln \left (u \left (\lambda \right ) \lambda ^{2}+2\right )}{16}-\frac {\ln \left (u \left (\lambda \right ) \lambda ^{2}\right )}{4}+\frac {3 \ln \left (3 u \left (\lambda \right ) \lambda ^{2}-2\right )}{16} = -\frac {\ln \left (\lambda \right )}{8}+c_2 \end{align*}

Converting \(u \left (\lambda \right ) = -2\) back to \(u \left (\lambda \right )\) gives

\begin{align*} u \left (\lambda \right ) \lambda ^{2} = -2 \end{align*}

Converting \(u \left (\lambda \right ) = 0\) back to \(u \left (\lambda \right )\) gives

\begin{align*} u \left (\lambda \right ) \lambda ^{2} = 0 \end{align*}

Solving for \(u \left (\lambda \right )\) gives

\begin{align*} \frac {\ln \left (u \left (\lambda \right ) \lambda ^{2}+2\right )}{16}-\frac {\ln \left (u \left (\lambda \right ) \lambda ^{2}\right )}{4}+\frac {3 \ln \left (3 u \left (\lambda \right ) \lambda ^{2}-2\right )}{16} &= -\frac {\ln \left (\lambda \right )}{8}+c_2 \\ u \left (\lambda \right ) &= 0 \\ u \left (\lambda \right ) &= -\frac {2}{\lambda ^{2}} \\ \end{align*}

Summary of solutions found

\begin{align*} \frac {\ln \left (u \left (\lambda \right ) \lambda ^{2}+2\right )}{16}-\frac {\ln \left (u \left (\lambda \right ) \lambda ^{2}\right )}{4}+\frac {3 \ln \left (3 u \left (\lambda \right ) \lambda ^{2}-2\right )}{16} &= -\frac {\ln \left (\lambda \right )}{8}+c_2 \\ u \left (\lambda \right ) &= 0 \\ u \left (\lambda \right ) &= -\frac {2}{\lambda ^{2}} \\ \end{align*}

2.1.54.9 Solved using ﬁrst_order_ode_homog_type_G

6.742 (sec)

Entering ﬁrst order ode homog type G solver

\begin{align*} \frac {d}{d \lambda }u \left (\lambda \right )&=-\frac {3 \lambda ^{3} u \left (\lambda \right )^{3}}{8}-\frac {u \left (\lambda \right )^{2} \lambda }{2}-\frac {3 u \left (\lambda \right )}{2 \lambda } \\ \end{align*}

Multiplying the right side of the ode, which is \(-\frac {3 \lambda ^{3} u^{3}}{8}-\frac {u^{2} \lambda }{2}-\frac {3 u}{2 \lambda }\) by \(\frac {\lambda }{u}\) gives

\begin{align*} \frac {d}{d \lambda }u \left (\lambda \right ) &= \left (\frac {\lambda }{u}\right ) -\frac {3 \lambda ^{3} u^{3}}{8}-\frac {u^{2} \lambda }{2}-\frac {3 u}{2 \lambda }\\ &= -\frac {3}{8} \lambda ^{4} u^{2}-\frac {1}{2} u \,\lambda ^{2}-\frac {3}{2}\\ &= F(\lambda ,u) \end{align*}

Since \(F \left (\lambda , u\right )\) has \(u\), then let

\begin{align*} f_\lambda &= \lambda \left (\frac {\partial }{\partial \lambda }F \left (\lambda , u\right )\right )\\ &= -\frac {u \,\lambda ^{2} \left (3 u \,\lambda ^{2}+2\right )}{2}\\ f_u&= u \left (\frac {\partial }{\partial u}F \left (\lambda , u\right )\right )\\ &= -\frac {u \,\lambda ^{2} \left (3 u \,\lambda ^{2}+2\right )}{4}\\ \alpha &= \frac {f_\lambda }{f_u} \\ &=2 \end{align*}

Since \(\alpha \) is independent of \(\lambda ,u\) then this is Homogeneous type G.

Let

\begin{align*} u&=\frac {z}{\lambda ^ \alpha }\\ &=\frac {z}{\lambda ^{2}} \end{align*}

Substituting the above back into \(F(\lambda ,u)\) gives

\begin{align*} F \left (z \right ) &=-\frac {3}{8} z^{2}-\frac {1}{2} z -\frac {3}{2} \end{align*}

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

Therefore, the implicit solution is given by

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

Which gives

\[ \ln \left (\lambda \right )-c_1 +\int _{}^{u \left (\lambda \right ) \lambda ^{2}}\frac {1}{z \left (-\frac {1}{2}+\frac {3}{8} z^{2}+\frac {1}{2} z \right )}d z = 0 \]

The value of the above is

\[ \ln \left (\lambda \right )-c_1 +\frac {\ln \left (u \left (\lambda \right ) \lambda ^{2}+2\right )}{2}-2 \ln \left (u \left (\lambda \right ) \lambda ^{2}\right )+\frac {3 \ln \left (3 u \left (\lambda \right ) \lambda ^{2}-2\right )}{2} = 0 \]

Summary of solutions found

\begin{align*} \ln \left (\lambda \right )-c_1 +\frac {\ln \left (u \left (\lambda \right ) \lambda ^{2}+2\right )}{2}-2 \ln \left (u \left (\lambda \right ) \lambda ^{2}\right )+\frac {3 \ln \left (3 u \left (\lambda \right ) \lambda ^{2}-2\right )}{2} &= 0 \\ \end{align*}

2.1.54.10 Solved using ﬁrst_order_ode_abel_ﬁrst_kind

71.959 (sec)

Entering ﬁrst order ode abel ﬁrst kind solver

This is Abel ﬁrst kind ODE, it has the form

\[ \frac {d}{d \lambda }u \left (\lambda \right )= f_0(\lambda )+f_1(\lambda ) u \left (\lambda \right ) +f_2(\lambda )u \left (\lambda \right )^{2}+f_3(\lambda )u \left (\lambda \right )^{3} \]

Comparing the above to given ODE which is

\begin{align*}\frac {d}{d \lambda }u \left (\lambda \right )&=-\frac {3 \lambda ^{3} u \left (\lambda \right )^{3}}{8}-\frac {u \left (\lambda \right )^{2} \lambda }{2}-\frac {3 u \left (\lambda \right )}{2 \lambda }\tag {1} \end{align*}

Therefore

\begin{align*} f_0 &= 0\\ f_1 &= -\frac {3}{2 \lambda }\\ f_2 &= -\frac {\lambda }{2}\\ f_3 &= -\frac {3 \lambda ^{3}}{8} \end{align*}

Hence

\begin{align*} f'_{0} &= 0\\ f'_{3} &= -\frac {9 \lambda ^{2}}{8} \end{align*}

Since \(f_2(\lambda )=-\frac {\lambda }{2}\) is not zero, then the followingtransformation is used to remove \(f_2\). Let \(u \left (\lambda \right ) = u(\lambda ) - \frac {f_2}{3 f_3}\) or

\begin{align*} u \left (\lambda \right ) &= u(\lambda ) - \left ( \frac {-\frac {\lambda }{2}}{-\frac {9 \lambda ^{3}}{8}} \right ) \\ &= u \left (\lambda \right )-\frac {4}{9 \lambda ^{2}} \end{align*}

The above transformation applied to (1) gives a new ODE as

\begin{align*} \frac {d}{d \lambda }u \left (\lambda \right ) = -\frac {3 \lambda ^{3} u \left (\lambda \right )^{3}}{8}-\frac {23 u \left (\lambda \right )}{18 \lambda }-\frac {70}{243 \lambda ^{3}}\tag {2} \end{align*}

The above ODE (2) can now be solved.

2.1.54.11 Solved using ﬁrst_order_ode_isobaric

12.308 (sec)

Entering ﬁrst order ode isobaric solverSolving for \(u'\) gives

\[ u' = -\frac {3 \lambda ^{3} u \left (\lambda \right )^{3}}{8}-\frac {23 u \left (\lambda \right )}{18 \lambda }-\frac {70}{243 \lambda ^{3}} \]

An ode \(\frac {d}{d \lambda }u \left (\lambda \right )=f(\lambda ,u)\) is isobaric if

\[ f(t \lambda , t^m u) = t^{m-1} f(\lambda ,u)\tag {1} \]

Where here

\[ f(\lambda ,u) = -\frac {3 \lambda ^{3} u \left (\lambda \right )^{3}}{8}-\frac {23 u \left (\lambda \right )}{18 \lambda }-\frac {70}{243 \lambda ^{3}}\tag {2} \]

\(m\) is the order of isobaric. Substituting (2) into (1) and solving for \(m\) gives

\[ m = -2 \]

Since the ode is isobaric of order \(m=-2\), then the substitution

\begin{align*} u&=u \lambda ^m \\ &=\frac {u}{\lambda ^{2}} \end{align*}

Converts the ODE to a separable in \(u \left (\lambda \right )\). Performing this substitution gives

\[ -\frac {2 u \left (\lambda \right )}{\lambda ^{3}}+\frac {\frac {d}{d \lambda }u \left (\lambda \right )}{\lambda ^{2}} = -\frac {3 u \left (\lambda \right )^{3}}{8 \lambda ^{3}}-\frac {23 u \left (\lambda \right )}{18 \lambda ^{3}}-\frac {70}{243 \lambda ^{3}} \]

Entering ﬁrst order ode separable solverThe ode

\begin{equation} \frac {d}{d \lambda }u \left (\lambda \right ) = -\frac {\left (9 u \left (\lambda \right )-10\right ) \left (9 u \left (\lambda \right )-4\right ) \left (9 u \left (\lambda \right )+14\right )}{1944 \lambda } \end{equation}

is separable as it can be written as

\begin{align*} \frac {d}{d \lambda }u \left (\lambda \right )&= -\frac {\left (9 u \left (\lambda \right )-10\right ) \left (9 u \left (\lambda \right )-4\right ) \left (9 u \left (\lambda \right )+14\right )}{1944 \lambda }\\ &= f(\lambda ) g(u) \end{align*}

Where

\begin{align*} f(\lambda ) &= -\frac {1}{1944 \lambda }\\ g(u) &= \left (9 u -10\right ) \left (9 u -4\right ) \left (9 u +14\right ) \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(\lambda ) \,d\lambda } \\ \int { \frac {1}{\left (9 u -10\right ) \left (9 u -4\right ) \left (9 u +14\right )}\,du} &= \int { -\frac {1}{1944 \lambda } \,d\lambda } \\ \end{align*}

\[ -\frac {\ln \left (9 u \left (\lambda \right )-4\right )}{972}+\frac {\ln \left (9 u \left (\lambda \right )+14\right )}{3888}+\frac {\ln \left (9 u \left (\lambda \right )-10\right )}{1296}=\ln \left (\frac {1}{\lambda ^{\frac {1}{1944}}}\right )+c_1 \]

We now need to ﬁnd the singular solutions, these are found by ﬁnding for what values \(g(u)\) is zero, since we had to divide by this above. Solving \(g(u)=0\) or

\[ \left (9 u -10\right ) \left (9 u -4\right ) \left (9 u +14\right )=0 \]

for \(u \left (\lambda \right )\) gives

\begin{align*} u \left (\lambda \right )&=-{\frac {14}{9}}\\ u \left (\lambda \right )&={\frac {4}{9}} \end{align*}

Now we go over each such singular solution and check if it veriﬁes the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} -\frac {\ln \left (9 u \left (\lambda \right )-4\right )}{972}+\frac {\ln \left (9 u \left (\lambda \right )+14\right )}{3888}+\frac {\ln \left (9 u \left (\lambda \right )-10\right )}{1296} &= \ln \left (\frac {1}{\lambda ^{\frac {1}{1944}}}\right )+c_1 \\ u \left (\lambda \right ) &= -{\frac {14}{9}} \\ u \left (\lambda \right ) &= {\frac {4}{9}} \\ \end{align*}

Converting \(-\frac {\ln \left (9 u \left (\lambda \right )-4\right )}{972}+\frac {\ln \left (9 u \left (\lambda \right )+14\right )}{3888}+\frac {\ln \left (9 u \left (\lambda \right )-10\right )}{1296} = -\frac {\ln \left (\lambda \right )}{1944}+c_1\) back to \(u \left (\lambda \right )\) gives

\begin{align*} -\frac {\ln \left (9 u \left (\lambda \right ) \lambda ^{2}-4\right )}{972}+\frac {\ln \left (9 u \left (\lambda \right ) \lambda ^{2}+14\right )}{3888}+\frac {\ln \left (9 u \left (\lambda \right ) \lambda ^{2}-10\right )}{1296} = -\frac {\ln \left (\lambda \right )}{1944}+c_1 \end{align*}

Converting \(u \left (\lambda \right ) = -{\frac {14}{9}}\) back to \(u \left (\lambda \right )\) gives

\begin{align*} u \left (\lambda \right ) \lambda ^{2} = -{\frac {14}{9}} \end{align*}

Converting \(u \left (\lambda \right ) = {\frac {4}{9}}\) back to \(u \left (\lambda \right )\) gives

\begin{align*} u \left (\lambda \right ) \lambda ^{2} = {\frac {4}{9}} \end{align*}

Solving for \(u \left (\lambda \right )\) gives

\begin{align*} -\frac {\ln \left (9 u \left (\lambda \right ) \lambda ^{2}-4\right )}{972}+\frac {\ln \left (9 u \left (\lambda \right ) \lambda ^{2}+14\right )}{3888}+\frac {\ln \left (9 u \left (\lambda \right ) \lambda ^{2}-10\right )}{1296} &= -\frac {\ln \left (\lambda \right )}{1944}+c_1 \\ u \left (\lambda \right ) &= -\frac {14}{9 \lambda ^{2}} \\ u \left (\lambda \right ) &= \frac {4}{9 \lambda ^{2}} \\ \end{align*}

Summary of solutions found

\begin{align*} -\frac {\ln \left (9 u \left (\lambda \right ) \lambda ^{2}-4\right )}{972}+\frac {\ln \left (9 u \left (\lambda \right ) \lambda ^{2}+14\right )}{3888}+\frac {\ln \left (9 u \left (\lambda \right ) \lambda ^{2}-10\right )}{1296} &= -\frac {\ln \left (\lambda \right )}{1944}+c_1 \\ u \left (\lambda \right ) &= -\frac {14}{9 \lambda ^{2}} \\ u \left (\lambda \right ) &= \frac {4}{9 \lambda ^{2}} \\ \end{align*}

2.1.54.12 Solved using ﬁrst_order_ode_homog_type_G

7.330 (sec)

Entering ﬁrst order ode homog type G solver

\begin{align*} \frac {d}{d \lambda }u \left (\lambda \right )&=-\frac {3 \lambda ^{3} u \left (\lambda \right )^{3}}{8}-\frac {23 u \left (\lambda \right )}{18 \lambda }-\frac {70}{243 \lambda ^{3}} \\ \end{align*}

Multiplying the right side of the ode, which is \(-\frac {3 \lambda ^{3} u^{3}}{8}-\frac {23 u}{18 \lambda }-\frac {70}{243 \lambda ^{3}}\) by \(\frac {\lambda }{u}\) gives

\begin{align*} \frac {d}{d \lambda }u \left (\lambda \right ) &= \left (\frac {\lambda }{u}\right ) -\frac {3 \lambda ^{3} u^{3}}{8}-\frac {23 u}{18 \lambda }-\frac {70}{243 \lambda ^{3}}\\ &= -\frac {729 \lambda ^{6} u^{3}+2484 u \,\lambda ^{2}+560}{1944 \lambda ^{2} u}\\ &= F(\lambda ,u) \end{align*}

Since \(F \left (\lambda , u\right )\) has \(u\), then let

\begin{align*} f_\lambda &= \lambda \left (\frac {\partial }{\partial \lambda }F \left (\lambda , u\right )\right )\\ &= -\frac {729 \lambda ^{6} u^{3}-280}{486 \lambda ^{2} u}\\ f_u&= u \left (\frac {\partial }{\partial u}F \left (\lambda , u\right )\right )\\ &= -\frac {729 \lambda ^{6} u^{3}-280}{972 \lambda ^{2} u}\\ \alpha &= \frac {f_\lambda }{f_u} \\ &=2 \end{align*}

Since \(\alpha \) is independent of \(\lambda ,u\) then this is Homogeneous type G.

Let

\begin{align*} u&=\frac {z}{\lambda ^ \alpha }\\ &=\frac {z}{\lambda ^{2}} \end{align*}

Substituting the above back into \(F(\lambda ,u)\) gives

\begin{align*} F \left (z \right ) &=-\frac {729 z^{3}+2484 z +560}{1944 z} \end{align*}

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

Therefore, the implicit solution is given by

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

Which gives

\[ \ln \left (\lambda \right )-c_1 +\int _{}^{u \left (\lambda \right ) \lambda ^{2}}\frac {1}{z \left (-2+\frac {729 z^{3}+2484 z +560}{1944 z}\right )}d z = 0 \]

The value of the above is

\[ \ln \left (\lambda \right )-c_1 -2 \ln \left (9 u \left (\lambda \right ) \lambda ^{2}-4\right )+\frac {\ln \left (9 u \left (\lambda \right ) \lambda ^{2}+14\right )}{2}+\frac {3 \ln \left (9 u \left (\lambda \right ) \lambda ^{2}-10\right )}{2} = 0 \]

Summary of solutions found

\begin{align*} \ln \left (\lambda \right )-c_1 -2 \ln \left (9 u \left (\lambda \right ) \lambda ^{2}-4\right )+\frac {\ln \left (9 u \left (\lambda \right ) \lambda ^{2}+14\right )}{2}+\frac {3 \ln \left (9 u \left (\lambda \right ) \lambda ^{2}-10\right )}{2} &= 0 \\ \end{align*}

2.1.54.13 Solved using ﬁrst_order_ode_chini

0.105 (sec)

Entering ﬁrst order ode chini solver

The solution to this Chini ode is (more steps will be added showing how soon).

\[ \int _{}^{\frac {351 u \left (\lambda \right ) \lambda ^{2}}{140}}\frac {1}{\frac {4900}{59319} u^{3}-u +1}d u +\frac {13 \,560^{{1}/{3}} 729^{{2}/{3}} \ln \left (\lambda \right ) 70^{{2}/{3}}}{204120}+c_1 = 0 \]

Summary of solutions found

\begin{align*} \int _{}^{\frac {351 u \left (\lambda \right ) \lambda ^{2}}{140}}\frac {1}{\frac {4900}{59319} u^{3}-u +1}d u +\frac {13 \,560^{{1}/{3}} 729^{{2}/{3}} \ln \left (\lambda \right ) 70^{{2}/{3}}}{204120}+c_1 &= 0 \\ \end{align*}

2.1.54.14 Solved using ﬁrst_order_ode_LIE

36.260 (sec)

Entering ﬁrst order ode LIE solver

Writing the ode as

\begin{align*} \frac {d}{d \lambda }u \left (\lambda \right )&=-\frac {729 \lambda ^{6} u^{3}+2484 u \,\lambda ^{2}+560}{1944 \lambda ^{3}}\\ \frac {d}{d \lambda }u \left (\lambda \right )&= \omega \left ( \lambda ,u \left (\lambda \right )\right ) \end{align*}

The condition of Lie symmetry is the linearized PDE given by

\begin{align*} \eta _{\lambda }+\omega \left ( \eta _{u \left (\lambda \right )}-\xi _{\lambda }\right ) -\omega ^{2}\xi _{u \left (\lambda \right )}-\omega _{\lambda }\xi -\omega _{u \left (\lambda \right )}\eta =0\tag {A} \end{align*}

To determine \(\xi ,\eta \) then (A) is solved using ansatz. Making bivariate polynomials of degree 1 to use as anstaz gives

\begin{align*} \tag{1E} \xi &= \lambda a_{2}+u a_{3}+a_{1} \\ \tag{2E} \eta &= \lambda b_{2}+u 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 (729 \lambda ^{6} u^{3}+2484 u \,\lambda ^{2}+560\right ) \left (b_{3}-a_{2}\right )}{1944 \lambda ^{3}}-\frac {\left (729 \lambda ^{6} u^{3}+2484 u \,\lambda ^{2}+560\right )^{2} a_{3}}{3779136 \lambda ^{6}}-\left (-\frac {4374 u^{3} \lambda ^{5}+4968 u \lambda }{1944 \lambda ^{3}}+\frac {729 \lambda ^{6} u^{3}+2484 u \,\lambda ^{2}+560}{648 \lambda ^{4}}\right ) \left (\lambda a_{2}+u a_{3}+a_{1}\right )+\frac {\left (2187 u^{2} \lambda ^{6}+2484 \lambda ^{2}\right ) \left (\lambda b_{2}+u b_{3}+b_{1}\right )}{1944 \lambda ^{3}} = 0 \end{equation}

Putting the above in normal form gives

\[ -\frac {531441 \lambda ^{12} u^{6} a_{3}-4251528 \lambda ^{10} u^{2} b_{2}-5668704 \lambda ^{9} u^{3} a_{2}-2834352 \lambda ^{9} u^{3} b_{3}-629856 \lambda ^{8} u^{4} a_{3}-4251528 \lambda ^{9} u^{2} b_{1}-4251528 \lambda ^{8} u^{3} a_{1}+816480 \lambda ^{6} u^{3} a_{3}-8608032 b_{2} \lambda ^{6}+10999152 \lambda ^{4} u^{2} a_{3}-4828896 \lambda ^{5} b_{1}+4828896 \lambda ^{4} u a_{1}+2177280 \lambda ^{3} a_{2}+1088640 \lambda ^{3} b_{3}+6048000 \lambda ^{2} u a_{3}+3265920 \lambda ^{2} a_{1}+313600 a_{3}}{3779136 \lambda ^{6}} = 0 \]

Setting the numerator to zero gives

\begin{equation} \tag{6E} -531441 \lambda ^{12} u^{6} a_{3}+4251528 \lambda ^{10} u^{2} b_{2}+5668704 \lambda ^{9} u^{3} a_{2}+2834352 \lambda ^{9} u^{3} b_{3}+629856 \lambda ^{8} u^{4} a_{3}+4251528 \lambda ^{9} u^{2} b_{1}+4251528 \lambda ^{8} u^{3} a_{1}-816480 \lambda ^{6} u^{3} a_{3}+8608032 b_{2} \lambda ^{6}-10999152 \lambda ^{4} u^{2} a_{3}+4828896 \lambda ^{5} b_{1}-4828896 \lambda ^{4} u a_{1}-2177280 \lambda ^{3} a_{2}-1088640 \lambda ^{3} b_{3}-6048000 \lambda ^{2} u a_{3}-3265920 \lambda ^{2} a_{1}-313600 a_{3} = 0 \end{equation}

Looking at the above PDE shows the following are all the terms with \(\{\lambda , u\}\) in them.

\[ \{\lambda , u\} \]

The following substitution is now made to be able to collect on all terms with \(\{\lambda , u\}\) in them

\[ \{\lambda = v_{1}, u = v_{2}\} \]

The above PDE (6E) now becomes

\begin{equation} \tag{7E} -531441 a_{3} v_{1}^{12} v_{2}^{6}+5668704 a_{2} v_{1}^{9} v_{2}^{3}+629856 a_{3} v_{1}^{8} v_{2}^{4}+4251528 b_{2} v_{1}^{10} v_{2}^{2}+2834352 b_{3} v_{1}^{9} v_{2}^{3}+4251528 a_{1} v_{1}^{8} v_{2}^{3}+4251528 b_{1} v_{1}^{9} v_{2}^{2}-816480 a_{3} v_{1}^{6} v_{2}^{3}-10999152 a_{3} v_{1}^{4} v_{2}^{2}+8608032 b_{2} v_{1}^{6}-4828896 a_{1} v_{1}^{4} v_{2}+4828896 b_{1} v_{1}^{5}-2177280 a_{2} v_{1}^{3}-6048000 a_{3} v_{1}^{2} v_{2}-1088640 b_{3} v_{1}^{3}-3265920 a_{1} v_{1}^{2}-313600 a_{3} = 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} -531441 a_{3} v_{1}^{12} v_{2}^{6}+4251528 b_{2} v_{1}^{10} v_{2}^{2}+\left (5668704 a_{2}+2834352 b_{3}\right ) v_{1}^{9} v_{2}^{3}+4251528 b_{1} v_{1}^{9} v_{2}^{2}+629856 a_{3} v_{1}^{8} v_{2}^{4}+4251528 a_{1} v_{1}^{8} v_{2}^{3}-816480 a_{3} v_{1}^{6} v_{2}^{3}+8608032 b_{2} v_{1}^{6}+4828896 b_{1} v_{1}^{5}-10999152 a_{3} v_{1}^{4} v_{2}^{2}-4828896 a_{1} v_{1}^{4} v_{2}+\left (-2177280 a_{2}-1088640 b_{3}\right ) v_{1}^{3}-6048000 a_{3} v_{1}^{2} v_{2}-3265920 a_{1} v_{1}^{2}-313600 a_{3} = 0 \end{equation}

Setting each coeﬃcients in (8E) to zero gives the following equations to solve

\begin{align*} -4828896 a_{1}&=0\\ -3265920 a_{1}&=0\\ 4251528 a_{1}&=0\\ -10999152 a_{3}&=0\\ -6048000 a_{3}&=0\\ -816480 a_{3}&=0\\ -531441 a_{3}&=0\\ -313600 a_{3}&=0\\ 629856 a_{3}&=0\\ 4251528 b_{1}&=0\\ 4828896 b_{1}&=0\\ 4251528 b_{2}&=0\\ 8608032 b_{2}&=0\\ -2177280 a_{2}-1088640 b_{3}&=0\\ 5668704 a_{2}+2834352 b_{3}&=0 \end{align*}

Solving the above equations for the unknowns gives

\begin{align*} a_{1}&=0\\ a_{2}&=a_{2}\\ a_{3}&=0\\ b_{1}&=0\\ b_{2}&=0\\ b_{3}&=-2 a_{2} \end{align*}

Substituting the above solution in the anstaz (1E,2E) (using \(1\) as arbitrary value for any unknown in the RHS) gives

\begin{align*} \xi &= \lambda \\ \eta &= -2 u \\ \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 (\lambda ,u\right ) \xi \\ &= -2 u - \left (-\frac {729 \lambda ^{6} u^{3}+2484 u \,\lambda ^{2}+560}{1944 \lambda ^{3}}\right ) \left (\lambda \right ) \\ &= \frac {729 \lambda ^{6} u^{3}-1404 u \,\lambda ^{2}+560}{1944 \lambda ^{2}}\\ \xi &= 0 \end{align*}

The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( \lambda ,u\right ) \to \left ( R,S \right )\) 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 \lambda }{\xi } &= \frac {d u}{\eta } = dS \tag {1} \end{align*}

The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial \lambda } + \eta \frac {\partial }{\partial u}\right ) S(\lambda ,u) = 1\). 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 = \lambda \end{align*}

\(S\) is found from

\begin{align*} S &= \int { \frac {1}{\eta }} dy\\ &= \int { \frac {1}{\frac {729 \lambda ^{6} u^{3}-1404 u \,\lambda ^{2}+560}{1944 \lambda ^{2}}}} dy \end{align*}

Which results in

\begin{align*} S&= -2 \ln \left (9 u \,\lambda ^{2}-4\right )+\frac {3 \ln \left (9 u \,\lambda ^{2}-10\right )}{2}+\frac {\ln \left (9 u \,\lambda ^{2}+14\right )}{2} \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_{\lambda } + \omega (\lambda ,u) S_{u} }{ R_{\lambda } + \omega (\lambda ,u) R_{u} }\tag {2} \end{align*}

Where in the above \(R_{\lambda },R_{u},S_{\lambda },S_{u}\) are all partial derivatives and \(\omega (\lambda ,u)\) is the right hand side of the original ode given by

\begin{align*} \omega (\lambda ,u) &= -\frac {729 \lambda ^{6} u^{3}+2484 u \,\lambda ^{2}+560}{1944 \lambda ^{3}} \end{align*}

Evaluating all the partial derivatives gives

\begin{align*} R_{\lambda } &= 1\\ R_{u} &= 0\\ S_{\lambda } &= \frac {3888 u \lambda }{729 \lambda ^{6} u^{3}-1404 u \,\lambda ^{2}+560}\\ S_{u} &= \frac {1944 \lambda ^{2}}{729 \lambda ^{6} u^{3}-1404 u \,\lambda ^{2}+560} \end{align*}

Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.

\begin{align*} \frac {dS}{dR} &= -\frac {1}{\lambda }\tag {2A} \end{align*}

We now need to express the RHS as function of \(R\) only. This is done by solving for \(\lambda ,u\) 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 \(R,S\).

Since the ode has the form \(\frac {d}{d R}S \left (R \right )=f(R)\), then we only need to integrate \(f(R)\).

\begin{align*} \int {dS} &= \int {-\frac {1}{R}\, dR}\\ S \left (R \right ) &= -\ln \left (R \right ) + c_2 \end{align*}

\begin{align*} S \left (R \right )&= -\ln \left (R \right )+c_2 \end{align*}

To complete the solution, we just need to transform the above back to \(\lambda ,u\) coordinates. This results in

\begin{align*} -2 \ln \left (9 u \left (\lambda \right ) \lambda ^{2}-4\right )+\frac {3 \ln \left (9 u \left (\lambda \right ) \lambda ^{2}-10\right )}{2}+\frac {\ln \left (9 u \left (\lambda \right ) \lambda ^{2}+14\right )}{2} = -\ln \left (\lambda \right )+c_2 \end{align*}

Summary of solutions found

Substituting \(u=u \left (\lambda \right )+\frac {4}{9 \lambda ^{2}}\) in the above solution gives

\begin{align*} -\frac {\ln \left (9 \left (u \left (\lambda \right )+\frac {4}{9 \lambda ^{2}}\right ) \lambda ^{2}-4\right )}{972}+\frac {\ln \left (9 \left (u \left (\lambda \right )+\frac {4}{9 \lambda ^{2}}\right ) \lambda ^{2}+14\right )}{3888}+\frac {\ln \left (9 \left (u \left (\lambda \right )+\frac {4}{9 \lambda ^{2}}\right ) \lambda ^{2}-10\right )}{1296} = -\frac {\ln \left (\lambda \right )}{1944}+c_1 \end{align*}

Now we transform the solution \(u \left (\lambda \right ) = -\frac {14}{9 \lambda ^{2}}\) to \(u \left (\lambda \right )\) using \(u \left (\lambda \right ) = u(\lambda ) - \frac {f_2}{3 f_3}\), which gives

\begin{align*} u \left (\lambda \right ) &= -\frac {14}{9 \lambda ^{2}} - \left (\frac {4}{9 \lambda ^{2}}\right ) \\ &= -\frac {2}{\lambda ^{2}}\\ &= -\frac {2}{\lambda ^{2}} \end{align*}

Now we transform the solution \(u \left (\lambda \right ) = \frac {4}{9 \lambda ^{2}}\) to \(u \left (\lambda \right )\) using \(u \left (\lambda \right ) = u(\lambda ) - \frac {f_2}{3 f_3}\), which gives

\begin{align*} u \left (\lambda \right ) &= \frac {4}{9 \lambda ^{2}} - \left (\frac {4}{9 \lambda ^{2}}\right ) \\ &= 0\\ &= 0 \end{align*}

Simplifying the above gives

\begin{align*} -\frac {\ln \left (3\right )}{1296}-\frac {\ln \left (u \left (\lambda \right ) \lambda ^{2}\right )}{972}+\frac {\ln \left (u \left (\lambda \right ) \lambda ^{2}+2\right )}{3888}+\frac {\ln \left (3 u \left (\lambda \right ) \lambda ^{2}-2\right )}{1296} &= -\frac {\ln \left (\lambda \right )}{1944}+c_1 \\ u \left (\lambda \right ) &= -\frac {2}{\lambda ^{2}} \\ u \left (\lambda \right ) &= 0 \\ \end{align*}

Summary of solutions found

\begin{align*} -\frac {\ln \left (3\right )}{1296}-\frac {\ln \left (u \left (\lambda \right ) \lambda ^{2}\right )}{972}+\frac {\ln \left (u \left (\lambda \right ) \lambda ^{2}+2\right )}{3888}+\frac {\ln \left (3 u \left (\lambda \right ) \lambda ^{2}-2\right )}{1296} &= -\frac {\ln \left (\lambda \right )}{1944}+c_1 \\ u \left (\lambda \right ) &= 0 \\ u \left (\lambda \right ) &= -\frac {2}{\lambda ^{2}} \\ \end{align*}

2.1.54.15 Solved using ﬁrst_order_ode_LIE

4.682 (sec)

Entering ﬁrst order ode LIE solver

Writing the ode as

\begin{align*} \frac {d}{d \lambda }u \left (\lambda \right )&=-\frac {u \left (3 \lambda ^{4} u^{2}+4 u \,\lambda ^{2}+12\right )}{8 \lambda }\\ \frac {d}{d \lambda }u \left (\lambda \right )&= \omega \left ( \lambda ,u \left (\lambda \right )\right ) \end{align*}

The condition of Lie symmetry is the linearized PDE given by

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 &= \lambda a_{2}+u a_{3}+a_{1} \\ \tag{2E} \eta &= \lambda b_{2}+u 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 {u \left (3 \lambda ^{4} u^{2}+4 u \,\lambda ^{2}+12\right ) \left (b_{3}-a_{2}\right )}{8 \lambda }-\frac {u^{2} \left (3 \lambda ^{4} u^{2}+4 u \,\lambda ^{2}+12\right )^{2} a_{3}}{64 \lambda ^{2}}-\left (-\frac {u \left (12 u^{2} \lambda ^{3}+8 u \lambda \right )}{8 \lambda }+\frac {u \left (3 \lambda ^{4} u^{2}+4 u \,\lambda ^{2}+12\right )}{8 \lambda ^{2}}\right ) \left (\lambda a_{2}+u a_{3}+a_{1}\right )-\left (-\frac {3 \lambda ^{4} u^{2}+4 u \,\lambda ^{2}+12}{8 \lambda }-\frac {u \left (6 \lambda ^{4} u +4 \lambda ^{2}\right )}{8 \lambda }\right ) \left (\lambda b_{2}+u b_{3}+b_{1}\right ) = 0 \end{equation}

Putting the above in normal form gives

\[ -\frac {9 \lambda ^{8} u^{6} a_{3}+24 \lambda ^{6} u^{5} a_{3}-72 \lambda ^{6} u^{2} b_{2}-96 \lambda ^{5} u^{3} a_{2}-48 \lambda ^{5} u^{3} b_{3}+16 \lambda ^{4} u^{4} a_{3}-72 \lambda ^{5} u^{2} b_{1}-72 \lambda ^{4} u^{3} a_{1}-64 \lambda ^{4} u b_{2}-64 \lambda ^{3} u^{2} a_{2}-32 \lambda ^{3} u^{2} b_{3}+64 \lambda ^{2} u^{3} a_{3}-64 \lambda ^{3} u b_{1}-32 \lambda ^{2} u^{2} a_{1}-160 b_{2} \lambda ^{2}+240 u^{2} a_{3}-96 \lambda b_{1}+96 u a_{1}}{64 \lambda ^{2}} = 0 \]

Setting the numerator to zero gives

\begin{equation} \tag{6E} -9 \lambda ^{8} u^{6} a_{3}-24 \lambda ^{6} u^{5} a_{3}+72 \lambda ^{6} u^{2} b_{2}+96 \lambda ^{5} u^{3} a_{2}+48 \lambda ^{5} u^{3} b_{3}-16 \lambda ^{4} u^{4} a_{3}+72 \lambda ^{5} u^{2} b_{1}+72 \lambda ^{4} u^{3} a_{1}+64 \lambda ^{4} u b_{2}+64 \lambda ^{3} u^{2} a_{2}+32 \lambda ^{3} u^{2} b_{3}-64 \lambda ^{2} u^{3} a_{3}+64 \lambda ^{3} u b_{1}+32 \lambda ^{2} u^{2} a_{1}+160 b_{2} \lambda ^{2}-240 u^{2} a_{3}+96 \lambda b_{1}-96 u a_{1} = 0 \end{equation}

Looking at the above PDE shows the following are all the terms with \(\{\lambda , u\}\) in them.

\[ \{\lambda , u\} \]

The following substitution is now made to be able to collect on all terms with \(\{\lambda , u\}\) in them

\[ \{\lambda = v_{1}, u = v_{2}\} \]

The above PDE (6E) now becomes

\begin{equation} \tag{7E} -9 a_{3} v_{1}^{8} v_{2}^{6}-24 a_{3} v_{1}^{6} v_{2}^{5}+96 a_{2} v_{1}^{5} v_{2}^{3}-16 a_{3} v_{1}^{4} v_{2}^{4}+72 b_{2} v_{1}^{6} v_{2}^{2}+48 b_{3} v_{1}^{5} v_{2}^{3}+72 a_{1} v_{1}^{4} v_{2}^{3}+72 b_{1} v_{1}^{5} v_{2}^{2}+64 a_{2} v_{1}^{3} v_{2}^{2}-64 a_{3} v_{1}^{2} v_{2}^{3}+64 b_{2} v_{1}^{4} v_{2}+32 b_{3} v_{1}^{3} v_{2}^{2}+32 a_{1} v_{1}^{2} v_{2}^{2}+64 b_{1} v_{1}^{3} v_{2}-240 a_{3} v_{2}^{2}+160 b_{2} v_{1}^{2}-96 a_{1} v_{2}+96 b_{1} v_{1} = 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} -9 a_{3} v_{1}^{8} v_{2}^{6}-24 a_{3} v_{1}^{6} v_{2}^{5}+72 b_{2} v_{1}^{6} v_{2}^{2}+\left (96 a_{2}+48 b_{3}\right ) v_{1}^{5} v_{2}^{3}+72 b_{1} v_{1}^{5} v_{2}^{2}-16 a_{3} v_{1}^{4} v_{2}^{4}+72 a_{1} v_{1}^{4} v_{2}^{3}+64 b_{2} v_{1}^{4} v_{2}+\left (64 a_{2}+32 b_{3}\right ) v_{1}^{3} v_{2}^{2}+64 b_{1} v_{1}^{3} v_{2}-64 a_{3} v_{1}^{2} v_{2}^{3}+32 a_{1} v_{1}^{2} v_{2}^{2}+160 b_{2} v_{1}^{2}+96 b_{1} v_{1}-240 a_{3} v_{2}^{2}-96 a_{1} v_{2} = 0 \end{equation}

Setting each coeﬃcients in (8E) to zero gives the following equations to solve

\begin{align*} -96 a_{1}&=0\\ 32 a_{1}&=0\\ 72 a_{1}&=0\\ -240 a_{3}&=0\\ -64 a_{3}&=0\\ -24 a_{3}&=0\\ -16 a_{3}&=0\\ -9 a_{3}&=0\\ 64 b_{1}&=0\\ 72 b_{1}&=0\\ 96 b_{1}&=0\\ 64 b_{2}&=0\\ 72 b_{2}&=0\\ 160 b_{2}&=0\\ 64 a_{2}+32 b_{3}&=0\\ 96 a_{2}+48 b_{3}&=0 \end{align*}

Solving the above equations for the unknowns gives

\begin{align*} a_{1}&=0\\ a_{2}&=a_{2}\\ a_{3}&=0\\ b_{1}&=0\\ b_{2}&=0\\ b_{3}&=-2 a_{2} \end{align*}

Substituting the above solution in the anstaz (1E,2E) (using \(1\) as arbitrary value for any unknown in the RHS) gives

\begin{align*} \xi &= \lambda \\ \eta &= -2 u \\ \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 (\lambda ,u\right ) \xi \\ &= -2 u - \left (-\frac {u \left (3 \lambda ^{4} u^{2}+4 u \,\lambda ^{2}+12\right )}{8 \lambda }\right ) \left (\lambda \right ) \\ &= \frac {3}{8} u^{3} \lambda ^{4}+\frac {1}{2} u^{2} \lambda ^{2}-\frac {1}{2} u\\ \xi &= 0 \end{align*}

The characteristic pde which is used to ﬁnd the canonical coordinates is

\begin{align*} \frac {d \lambda }{\xi } &= \frac {d u}{\eta } = dS \tag {1} \end{align*}

\begin{align*} R = \lambda \end{align*}

\(S\) is found from

\begin{align*} S &= \int { \frac {1}{\eta }} dy\\ &= \int { \frac {1}{\frac {3}{8} u^{3} \lambda ^{4}+\frac {1}{2} u^{2} \lambda ^{2}-\frac {1}{2} u}} dy \end{align*}

Which results in

\begin{align*} S&= \frac {\ln \left (u \,\lambda ^{2}+2\right )}{2}+\frac {3 \ln \left (3 u \,\lambda ^{2}-2\right )}{2}-2 \ln \left (u \right ) \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_{\lambda } + \omega (\lambda ,u) S_{u} }{ R_{\lambda } + \omega (\lambda ,u) R_{u} }\tag {2} \end{align*}

Where in the above \(R_{\lambda },R_{u},S_{\lambda },S_{u}\) are all partial derivatives and \(\omega (\lambda ,u)\) is the right hand side of the original ode given by

\begin{align*} \omega (\lambda ,u) &= -\frac {u \left (3 \lambda ^{4} u^{2}+4 u \,\lambda ^{2}+12\right )}{8 \lambda } \end{align*}

Evaluating all the partial derivatives gives

\begin{align*} R_{\lambda } &= 1\\ R_{u} &= 0\\ S_{\lambda } &= \frac {12 u^{2} \lambda ^{3}+16 u \lambda }{3 \lambda ^{4} u^{2}+4 u \,\lambda ^{2}-4}\\ S_{u} &= \frac {8}{3 u^{3} \lambda ^{4}+4 u^{2} \lambda ^{2}-4 u} \end{align*}

Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.

\begin{align*} \frac {dS}{dR} &= \frac {3}{\lambda }\tag {2A} \end{align*}

We now need to express the RHS as function of \(R\) only. This is done by solving for \(\lambda ,u\) in terms of \(R,S\) from the result obtained earlier and simplifying. This gives

\begin{align*} \frac {dS}{dR} &= \frac {3}{R} \end{align*}

Since the ode has the form \(\frac {d}{d R}S \left (R \right )=f(R)\), then we only need to integrate \(f(R)\).

\begin{align*} \int {dS} &= \int {\frac {3}{R}\, dR}\\ S \left (R \right ) &= 3 \ln \left (R \right ) + c_2 \end{align*}

\begin{align*} S \left (R \right )&= 3 \ln \left (R \right )+c_2 \end{align*}

To complete the solution, we just need to transform the above back to \(\lambda ,u\) coordinates. This results in

\begin{align*} \frac {\ln \left (u \left (\lambda \right ) \lambda ^{2}+2\right )}{2}+\frac {3 \ln \left (3 u \left (\lambda \right ) \lambda ^{2}-2\right )}{2}-2 \ln \left (u \left (\lambda \right )\right ) = 3 \ln \left (\lambda \right )+c_2 \end{align*}

Summary of solutions found

Substituting \(u \left (\lambda \right )=\frac {1}{y \left (\lambda \right )-\frac {\lambda ^{2}}{2}}\) in the above solution gives

\[ \frac {\ln \left (\frac {\lambda ^{2}}{y \left (\lambda \right )-\frac {\lambda ^{2}}{2}}+2\right )}{16}-\frac {\ln \left (\frac {\lambda ^{2}}{y \left (\lambda \right )-\frac {\lambda ^{2}}{2}}\right )}{4}+\frac {3 \ln \left (\frac {3 \lambda ^{2}}{y \left (\lambda \right )-\frac {\lambda ^{2}}{2}}-2\right )}{16} = -\frac {\ln \left (\lambda \right )}{8}+c_2 \]

Now we transform the solution \(u \left (\lambda \right ) = -\frac {2}{\lambda ^{2}}\) to \(y \left (\lambda \right )\) using \(u \left (\lambda \right )=\frac {1}{y \left (\lambda \right )-\frac {\lambda ^{2}}{2}}\) which gives

\[ y \left (\lambda \right ) = 0 \]

Simplifying the above gives

\begin{align*} \frac {\ln \left (2\right )}{4}+\frac {\ln \left (-\frac {y \left (\lambda \right )}{\lambda ^{2}-2 y \left (\lambda \right )}\right )}{16}-\frac {\ln \left (-\frac {\lambda ^{2}}{\lambda ^{2}-2 y \left (\lambda \right )}\right )}{4}+\frac {3 \ln \left (\frac {-2 \lambda ^{2}+y \left (\lambda \right )}{\lambda ^{2}-2 y \left (\lambda \right )}\right )}{16} &= -\frac {\ln \left (\lambda \right )}{8}+c_2 \\ y \left (\lambda \right ) &= 0 \\ \end{align*}

Summary of solutions found

2.1.54.16 Solved using ﬁrst_order_ode_LIE

13.553 (sec)

Entering ﬁrst order ode LIE solver

\begin{align*} \frac {d}{d \lambda }y \left (\lambda \right )&=-\frac {3 y \left (\lambda \right )^{2}}{\lambda \left (\lambda ^{2}-2 y \left (\lambda \right )\right )} \\ \end{align*}

Writing the ode as

\begin{align*} \frac {d}{d \lambda }y \left (\lambda \right )&=\frac {3 y^{2}}{\lambda \left (-\lambda ^{2}+2 y \right )}\\ \frac {d}{d \lambda }y \left (\lambda \right )&= \omega \left ( \lambda ,y \left (\lambda \right )\right ) \end{align*}

The condition of Lie symmetry is the linearized PDE given by

\begin{align*} \eta _{\lambda }+\omega \left ( \eta _{y \left (\lambda \right )}-\xi _{\lambda }\right ) -\omega ^{2}\xi _{y \left (\lambda \right )}-\omega _{\lambda }\xi -\omega _{y \left (\lambda \right )}\eta =0\tag {A} \end{align*}

To determine \(\xi ,\eta \) then (A) is solved using ansatz. Making bivariate polynomials of degree 1 to use as anstaz gives

\begin{align*} \tag{1E} \xi &= \lambda a_{2}+y a_{3}+a_{1} \\ \tag{2E} \eta &= \lambda 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 {3 y^{2} \left (b_{3}-a_{2}\right )}{\lambda \left (-\lambda ^{2}+2 y \right )}-\frac {9 y^{4} a_{3}}{\lambda ^{2} \left (-\lambda ^{2}+2 y \right )^{2}}-\left (-\frac {3 y^{2}}{\lambda ^{2} \left (-\lambda ^{2}+2 y \right )}+\frac {6 y^{2}}{\left (-\lambda ^{2}+2 y \right )^{2}}\right ) \left (\lambda a_{2}+y a_{3}+a_{1}\right )-\left (\frac {6 y}{\lambda \left (-\lambda ^{2}+2 y \right )}-\frac {6 y^{2}}{\lambda \left (-\lambda ^{2}+2 y \right )^{2}}\right ) \left (\lambda b_{2}+y b_{3}+b_{1}\right ) = 0 \end{equation}

Putting the above in normal form gives

\[ \frac {\lambda ^{6} b_{2}+2 \lambda ^{4} y b_{2}-6 \lambda ^{3} y^{2} a_{2}+3 \lambda ^{3} y^{2} b_{3}-9 \lambda ^{2} y^{3} a_{3}+6 \lambda ^{3} y b_{1}-9 \lambda ^{2} y^{2} a_{1}-2 \lambda ^{2} y^{2} b_{2}-3 y^{4} a_{3}-6 \lambda \,y^{2} b_{1}+6 y^{3} a_{1}}{\lambda ^{2} \left (\lambda ^{2}-2 y \right )^{2}} = 0 \]

Setting the numerator to zero gives

\begin{equation} \tag{6E} \lambda ^{6} b_{2}+2 \lambda ^{4} y b_{2}-6 \lambda ^{3} y^{2} a_{2}+3 \lambda ^{3} y^{2} b_{3}-9 \lambda ^{2} y^{3} a_{3}+6 \lambda ^{3} y b_{1}-9 \lambda ^{2} y^{2} a_{1}-2 \lambda ^{2} y^{2} b_{2}-3 y^{4} a_{3}-6 \lambda \,y^{2} b_{1}+6 y^{3} a_{1} = 0 \end{equation}

Looking at the above PDE shows the following are all the terms with \(\{\lambda , y\}\) in them.

\[ \{\lambda , y\} \]

The following substitution is now made to be able to collect on all terms with \(\{\lambda , y\}\) in them

\[ \{\lambda = v_{1}, y = v_{2}\} \]

The above PDE (6E) now becomes

\begin{equation} \tag{7E} b_{2} v_{1}^{6}-6 a_{2} v_{1}^{3} v_{2}^{2}-9 a_{3} v_{1}^{2} v_{2}^{3}+2 b_{2} v_{1}^{4} v_{2}+3 b_{3} v_{1}^{3} v_{2}^{2}-9 a_{1} v_{1}^{2} v_{2}^{2}-3 a_{3} v_{2}^{4}+6 b_{1} v_{1}^{3} v_{2}-2 b_{2} v_{1}^{2} v_{2}^{2}+6 a_{1} v_{2}^{3}-6 b_{1} v_{1} v_{2}^{2} = 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} b_{2} v_{1}^{6}+2 b_{2} v_{1}^{4} v_{2}+\left (-6 a_{2}+3 b_{3}\right ) v_{1}^{3} v_{2}^{2}+6 b_{1} v_{1}^{3} v_{2}-9 a_{3} v_{1}^{2} v_{2}^{3}+\left (-9 a_{1}-2 b_{2}\right ) v_{1}^{2} v_{2}^{2}-6 b_{1} v_{1} v_{2}^{2}-3 a_{3} v_{2}^{4}+6 a_{1} v_{2}^{3} = 0 \end{equation}

Setting each coeﬃcients in (8E) to zero gives the following equations to solve

\begin{align*} b_{2}&=0\\ 6 a_{1}&=0\\ -9 a_{3}&=0\\ -3 a_{3}&=0\\ -6 b_{1}&=0\\ 6 b_{1}&=0\\ 2 b_{2}&=0\\ -9 a_{1}-2 b_{2}&=0\\ -6 a_{2}+3 b_{3}&=0 \end{align*}

Solving the above equations for the unknowns gives

\begin{align*} a_{1}&=0\\ a_{2}&=a_{2}\\ a_{3}&=0\\ b_{1}&=0\\ b_{2}&=0\\ b_{3}&=2 a_{2} \end{align*}

Substituting the above solution in the anstaz (1E,2E) (using \(1\) as arbitrary value for any unknown in the RHS) gives

\begin{align*} \xi &= \lambda \\ \eta &= 2 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 (\lambda ,y\right ) \xi \\ &= 2 y - \left (\frac {3 y^{2}}{\lambda \left (-\lambda ^{2}+2 y \right )}\right ) \left (\lambda \right ) \\ &= \frac {y \left (2 \lambda ^{2}-y \right )}{\lambda ^{2}-2 y}\\ \xi &= 0 \end{align*}

The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( \lambda ,y\right ) \to \left ( R,S \right )\) 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 \lambda }{\xi } &= \frac {d y}{\eta } = dS \tag {1} \end{align*}

The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial \lambda } + \eta \frac {\partial }{\partial y}\right ) S(\lambda ,y) = 1\). 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 = \lambda \end{align*}

\(S\) is found from

\begin{align*} S &= \int { \frac {1}{\eta }} dy\\ &= \int { \frac {1}{\frac {y \left (2 \lambda ^{2}-y \right )}{\lambda ^{2}-2 y}}} dy \end{align*}

Which results in

\begin{align*} S&= \frac {3 \ln \left (-2 \lambda ^{2}+y \right )}{2}+\frac {\ln \left (y \right )}{2} \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_{\lambda } + \omega (\lambda ,y) S_{y} }{ R_{\lambda } + \omega (\lambda ,y) R_{y} }\tag {2} \end{align*}

Where in the above \(R_{\lambda },R_{y},S_{\lambda },S_{y}\) are all partial derivatives and \(\omega (\lambda ,y)\) is the right hand side of the original ode given by

\begin{align*} \omega (\lambda ,y) &= \frac {3 y^{2}}{\lambda \left (-\lambda ^{2}+2 y \right )} \end{align*}

Evaluating all the partial derivatives gives

\begin{align*} R_{\lambda } &= 1\\ R_{y} &= 0\\ S_{\lambda } &= -\frac {6 \lambda }{-2 \lambda ^{2}+y}\\ S_{y} &= \frac {\lambda ^{2}-2 y}{y \left (2 \lambda ^{2}-y \right )} \end{align*}

Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.

\begin{align*} \frac {dS}{dR} &= \frac {3}{\lambda }\tag {2A} \end{align*}

We now need to express the RHS as function of \(R\) only. This is done by solving for \(\lambda ,y\) in terms of \(R,S\) from the result obtained earlier and simplifying. This gives

\begin{align*} \frac {dS}{dR} &= \frac {3}{R} \end{align*}

Since the ode has the form \(\frac {d}{d R}S \left (R \right )=f(R)\), then we only need to integrate \(f(R)\).

\begin{align*} \int {dS} &= \int {\frac {3}{R}\, dR}\\ S \left (R \right ) &= 3 \ln \left (R \right ) + c_2 \end{align*}

\begin{align*} S \left (R \right )&= 3 \ln \left (R \right )+c_2 \end{align*}

To complete the solution, we just need to transform the above back to \(\lambda ,y\) coordinates. This results in

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

Summary of solutions found

\begin{align*} \frac {3 \ln \left (-2 \lambda ^{2}+y \left (\lambda \right )\right )}{2}+\frac {\ln \left (y \left (\lambda \right )\right )}{2} &= 3 \ln \left (\lambda \right )+c_2 \\ \end{align*}

Now that we have found solution \(y\), we have two equations with parameter \(\lambda \). They are

\begin{align*} \frac {\ln \left (\frac {y}{\lambda ^{2}}\right )}{2}+\frac {3 \ln \left (\frac {y}{\lambda ^{2}}-2\right )}{2} &= \ln \left (\frac {1}{\lambda }\right )+c_1 \\ x &= \frac {y^{2}}{\lambda ^{3}} \\ \end{align*}

Eliminating \(\lambda \) gives the solution for \(y\).

\[ \ln \left (\frac {y}{\operatorname {RootOf}\left (x \,\textit {\_Z}^{3}-y^{2}\right )^{2}}\right )+3 \ln \left (-\frac {2 \operatorname {RootOf}\left (x \,\textit {\_Z}^{3}-y^{2}\right )^{2}-y}{\operatorname {RootOf}\left (x \,\textit {\_Z}^{3}-y^{2}\right )^{2}}\right )-2 \ln \left (\frac {1}{\operatorname {RootOf}\left (x \,\textit {\_Z}^{3}-y^{2}\right )}\right )-2 c_1 \]

Which can be written as

\begin{align*} \frac {x \,{\mathrm e}^{-3 c_1} {\left (\left ({\mathrm e}^{2 c_1}+8 y\right ) y \left (4 y \,{\mathrm e}^{2 c_1}+{\mathrm e}^{2 c_1} \left (y \,{\mathrm e}^{4 c_1}\right )^{{1}/{3}}-2 \left (y \,{\mathrm e}^{4 c_1}\right )^{{2}/{3}}\right )\right )}^{{3}/{2}}}{\left ({\mathrm e}^{2 c_1}+8 y\right )^{3}}-y^{2} &= 0 \\ \end{align*}

Simplifying the above gives

\begin{align*} \frac {4 \left (x \left (-\frac {{\mathrm e}^{-3 c_1} \left (y \,{\mathrm e}^{4 c_1}\right )^{{2}/{3}}}{2}+{\mathrm e}^{-c_1} \left (y+\frac {\left (y \,{\mathrm e}^{4 c_1}\right )^{{1}/{3}}}{4}\right )\right ) \sqrt {\left ({\mathrm e}^{2 c_1}+8 y\right ) y \left (4 y \,{\mathrm e}^{2 c_1}+{\mathrm e}^{2 c_1} \left (y \,{\mathrm e}^{4 c_1}\right )^{{1}/{3}}-2 \left (y \,{\mathrm e}^{4 c_1}\right )^{{2}/{3}}\right )}-16 y^{3}-4 y^{2} {\mathrm e}^{2 c_1}-\frac {y \,{\mathrm e}^{4 c_1}}{4}\right ) y}{\left ({\mathrm e}^{2 c_1}+8 y\right )^{2}} &= 0 \\ \end{align*}

Summary of solutions found

2.1.54.17 ✓ Maple. Time used: 0.482 (sec). Leaf size: 353

ode:=diff(y(x),x)^3 = y(x)^2/x; 
dsolve(ode,y(x), singsol=all);

\begin{align*} y &= 0 \\ y &= -\frac {3 x^{{4}/{3}} c_1}{8}+\frac {3 x^{{2}/{3}} c_1^{2}}{8}-\frac {c_1^{3}}{8}+\frac {x^{2}}{8} \\ y &= \frac {3 \left (-1-i \sqrt {3}\right ) c_1^{2} x^{{2}/{3}}}{16}+\frac {3 c_1 \left (1-i \sqrt {3}\right ) x^{{4}/{3}}}{16}-\frac {c_1^{3}}{8}+\frac {x^{2}}{8} \\ y &= \frac {3 c_1^{2} \left (i \sqrt {3}-1\right ) x^{{2}/{3}}}{16}+\frac {3 c_1 \left (1+i \sqrt {3}\right ) x^{{4}/{3}}}{16}-\frac {c_1^{3}}{8}+\frac {x^{2}}{8} \\ y &= \frac {3 x^{{4}/{3}} c_1}{16}+\frac {3 x^{{2}/{3}} c_1^{2}}{32}+\frac {c_1^{3}}{64}+\frac {x^{2}}{8} \\ y &= \frac {3 \left (-1-i \sqrt {3}\right ) c_1^{2} x^{{2}/{3}}}{64}+\frac {3 c_1 \left (i \sqrt {3}-1\right ) x^{{4}/{3}}}{32}+\frac {c_1^{3}}{64}+\frac {x^{2}}{8} \\ y &= \frac {3 c_1^{2} \left (i \sqrt {3}-1\right ) x^{{2}/{3}}}{64}-\frac {3 c_1 \left (1+i \sqrt {3}\right ) x^{{4}/{3}}}{32}+\frac {c_1^{3}}{64}+\frac {x^{2}}{8} \\ y &= -\frac {3 x^{{4}/{3}} c_1}{16}+\frac {3 x^{{2}/{3}} c_1^{2}}{32}-\frac {c_1^{3}}{64}+\frac {x^{2}}{8} \\ y &= \frac {3 \left (-1-i \sqrt {3}\right ) c_1^{2} x^{{2}/{3}}}{64}+\frac {3 c_1 \left (1-i \sqrt {3}\right ) x^{{4}/{3}}}{32}-\frac {c_1^{3}}{64}+\frac {x^{2}}{8} \\ y &= \frac {3 c_1^{2} \left (i \sqrt {3}-1\right ) x^{{2}/{3}}}{64}+\frac {3 c_1 \left (1+i \sqrt {3}\right ) x^{{4}/{3}}}{32}-\frac {c_1^{3}}{64}+\frac {x^{2}}{8} \\ \end{align*}

Maple trace

Methods for first order ODEs: 
   *** Sublevel 2 *** 
   Methods for first order ODEs: 
   -> Solving 1st order ODE of high degree, 1st attempt 
   trying 1st order WeierstrassP solution for high degree ODE 
   trying 1st order WeierstrassPPrime solution for high degree ODE 
   trying 1st order JacobiSN solution for high degree ODE 
   trying 1st order ODE linearizable_by_differentiation 
   trying differential order: 1; missing variables 
   trying simple symmetries for implicit equations 
   Successful isolation of dy/dx: 3 solutions were found. Trying to solve each \ 
resulting ODE. 
      *** Sublevel 3 *** 
      Methods for first order ODEs: 
      --- Trying classification methods --- 
      trying homogeneous types: 
      trying homogeneous G 
      trying an integrating factor from the invariance group 
      <- integrating factor successful 
      <- homogeneous successful 
   ------------------- 
   * Tackling next ODE. 
      *** Sublevel 3 *** 
      Methods for first order ODEs: 
      --- Trying classification methods --- 
      trying homogeneous types: 
      trying homogeneous G 
      trying an integrating factor from the invariance group 
      <- integrating factor successful 
      <- homogeneous successful 
   ------------------- 
   * Tackling next ODE. 
      *** Sublevel 3 *** 
      Methods for first order ODEs: 
      --- Trying classification methods --- 
      trying homogeneous types: 
      trying homogeneous G 
      trying an integrating factor from the invariance group 
      <- integrating factor successful 
      <- homogeneous successful

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y \left (x \right )\right )^{3}=\frac {y \left (x \right )^{2}}{x} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d x}y \left (x \right )=\frac {\left (y \left (x \right )^{2} x^{2}\right )^{{1}/{3}}}{x}, \frac {d}{d x}y \left (x \right )=-\frac {\left (y \left (x \right )^{2} x^{2}\right )^{{1}/{3}}}{2 x}-\frac {\mathrm {I} \sqrt {3}\, \left (y \left (x \right )^{2} x^{2}\right )^{{1}/{3}}}{2 x}, \frac {d}{d x}y \left (x \right )=-\frac {\left (y \left (x \right )^{2} x^{2}\right )^{{1}/{3}}}{2 x}+\frac {\mathrm {I} \sqrt {3}\, \left (y \left (x \right )^{2} x^{2}\right )^{{1}/{3}}}{2 x}\right ] \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}y \left (x \right )=\frac {\left (y \left (x \right )^{2} x^{2}\right )^{{1}/{3}}}{x} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}y \left (x \right )=-\frac {\left (y \left (x \right )^{2} x^{2}\right )^{{1}/{3}}}{2 x}-\frac {\mathrm {I} \sqrt {3}\, \left (y \left (x \right )^{2} x^{2}\right )^{{1}/{3}}}{2 x} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}y \left (x \right )=-\frac {\left (y \left (x \right )^{2} x^{2}\right )^{{1}/{3}}}{2 x}+\frac {\mathrm {I} \sqrt {3}\, \left (y \left (x \right )^{2} x^{2}\right )^{{1}/{3}}}{2 x} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\mathit {workingODE} , \mathit {workingODE} , \mathit {workingODE}\right \} \end {array} \]

2.1.54.18 ✓ Mathematica. Time used: 0.05 (sec). Leaf size: 152

ode=(D[y[x],x])^3==y[x]^2/x; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]

\begin{align*} y(x)&\to \frac {1}{216} \left (3 x^{2/3}+2 c_1\right ){}^3\\ y(x)&\to \frac {1}{216} \left (18 i \left (\sqrt {3}+i\right ) c_1{}^2 x^{2/3}-27 i \left (\sqrt {3}-i\right ) c_1 x^{4/3}+27 x^2+8 c_1{}^3\right )\\ y(x)&\to \frac {1}{216} \left (-18 i \left (\sqrt {3}-i\right ) c_1{}^2 x^{2/3}+27 i \left (\sqrt {3}+i\right ) c_1 x^{4/3}+27 x^2+8 c_1{}^3\right )\\ y(x)&\to 0 \end{align*}

2.1.54.19 ✓ Sympy. Time used: 1.604 (sec). Leaf size: 109

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(Derivative(y(x), x)**3 - y(x)**2/x,0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)

\[ \left [ - \frac {3 x \sqrt [3]{\frac {y^{2}{\left (x \right )}}{x}}}{2 y^{\frac {2}{3}}{\left (x \right )}} + 3 \sqrt [3]{y{\left (x \right )}} = C_{1}, \ \frac {3 x \sqrt [3]{\frac {y^{2}{\left (x \right )}}{x}} \left (1 + \sqrt {3} i\right )}{4 y^{\frac {2}{3}}{\left (x \right )}} + 3 \sqrt [3]{y{\left (x \right )}} = C_{1}, \ \frac {3 x \sqrt [3]{\frac {y^{2}{\left (x \right )}}{x}} \left (1 - \sqrt {3} i\right )}{4 y^{\frac {2}{3}}{\left (x \right )}} + 3 \sqrt [3]{y{\left (x \right )}} = C_{1}\right ] \]

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)) 
 
('factorable', 'lie_group')

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