2.1.54 Problem 54
Internal
problem
ID
[10312]
Book
:
First
order
enumerated
odes
Section
:
section
1
Problem
number
:
54
Date
solved
:
Thursday, November 27, 2025 at 10:33:00 AM
CAS
classification
:
[[_homogeneous, `class G`], _rational]
Solved using first_order_ode_homog_type_G
Time used: 0.280 (sec)
Solve
\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
\[
\ln \left (x \right )-c_1 +\int _{}^{\frac {y}{x^{2}}}\frac {1}{z \left (2-\frac {\left (z^{2}\right )^{{1}/{3}}}{z}\right )}d z = 0
\]
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
\[
\ln \left (x \right )-c_2 +\int _{}^{\frac {y}{x^{2}}}\frac {1}{z \left (2-\frac {\left (z^{2}\right )^{{1}/{3}} \left (i \sqrt {3}-1\right )}{2 z}\right )}d z = 0
\]
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
\[
\ln \left (x \right )-c_3 +\int _{}^{\frac {y}{x^{2}}}\frac {1}{z \left (2+\frac {\left (z^{2}\right )^{{1}/{3}} \left (1+i \sqrt {3}\right )}{2 z}\right )}d z = 0
\]
Summary of solutions found
\begin{align*}
\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 \\
\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 \\
\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 \\
\end{align*}
Solved using first_order_nonlinear_p_but_separable
Time used: 0.792 (sec)
Solve
\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*}
Solved using first_order_ode_parametric method
Time used: 4.335 (sec)
Solve
\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 \(y\).
Solve Solving for \(y'\) gives
\begin{align*}
\tag{1} y' &= \frac {3 y \left (\lambda \right )^{2}}{\lambda \left (-\lambda ^{2}+2 y \left (\lambda \right )\right )} \\
\end{align*}
Each of the above ode’s is now solved 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
\[ 2 \lambda u \left (\lambda \right )+\lambda ^{2} \left (\frac {d}{d \lambda }u \left (\lambda \right )\right ) = \frac {3 \lambda ^{3} u \left (\lambda \right )^{2}}{-\lambda ^{2}+2 \lambda ^{2} u \left (\lambda \right )} \]
The 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*}
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 \,{\mathrm e}^{2 c_1} y+{\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 y \left (\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 ) x \sqrt {\left ({\mathrm e}^{2 c_1}+8 y\right ) y \left (4 \,{\mathrm e}^{2 c_1} y+{\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 \,{\mathrm e}^{2 c_1} y^{2}-\frac {y \,{\mathrm e}^{4 c_1}}{4}\right )}{\left ({\mathrm e}^{2 c_1}+8 y\right )^{2}} &= 0 \\
\end{align*}
Summary of solutions found
\begin{align*}
\frac {4 y \left (\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 ) x \sqrt {\left ({\mathrm e}^{2 c_1}+8 y\right ) y \left (4 \,{\mathrm e}^{2 c_1} y+{\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 \,{\mathrm e}^{2 c_1} y^{2}-\frac {y \,{\mathrm e}^{4 c_1}}{4}\right )}{\left ({\mathrm e}^{2 c_1}+8 y\right )^{2}} &= 0 \\
\end{align*}
✓ 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
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y \left (x \right )\right )^{3}=\frac {y \left (x \right )^{2}}{x} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d x}y \left (x \right )=\frac {\left (y \left (x \right )^{2} x^{2}\right )^{{1}/{3}}}{x}, \frac {d}{d x}y \left (x \right )=-\frac {\left (y \left (x \right )^{2} x^{2}\right )^{{1}/{3}}}{2 x}-\frac {\mathrm {I} \sqrt {3}\, \left (y \left (x \right )^{2} x^{2}\right )^{{1}/{3}}}{2 x}, \frac {d}{d x}y \left (x \right )=-\frac {\left (y \left (x \right )^{2} x^{2}\right )^{{1}/{3}}}{2 x}+\frac {\mathrm {I} \sqrt {3}\, \left (y \left (x \right )^{2} x^{2}\right )^{{1}/{3}}}{2 x}\right ] \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}y \left (x \right )=\frac {\left (y \left (x \right )^{2} x^{2}\right )^{{1}/{3}}}{x} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}y \left (x \right )=-\frac {\left (y \left (x \right )^{2} x^{2}\right )^{{1}/{3}}}{2 x}-\frac {\mathrm {I} \sqrt {3}\, \left (y \left (x \right )^{2} x^{2}\right )^{{1}/{3}}}{2 x} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}y \left (x \right )=-\frac {\left (y \left (x \right )^{2} x^{2}\right )^{{1}/{3}}}{2 x}+\frac {\mathrm {I} \sqrt {3}\, \left (y \left (x \right )^{2} x^{2}\right )^{{1}/{3}}}{2 x} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\mathit {workingODE} , \mathit {workingODE} , \mathit {workingODE}\right \} \end {array} \]
✓ 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*}
✓ Sympy. Time used: 1.453 (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)
\[
\left [ - \frac {3 x \sqrt [3]{\frac {y^{2}{\left (x \right )}}{x}}}{2 y^{\frac {2}{3}}{\left (x \right )}} + 3 \sqrt [3]{y{\left (x \right )}} = C_{1}, \ \frac {3 x \sqrt [3]{\frac {y^{2}{\left (x \right )}}{x}} \left (1 + \sqrt {3} i\right )}{4 y^{\frac {2}{3}}{\left (x \right )}} + 3 \sqrt [3]{y{\left (x \right )}} = C_{1}, \ \frac {3 x \sqrt [3]{\frac {y^{2}{\left (x \right )}}{x}} \left (1 - \sqrt {3} i\right )}{4 y^{\frac {2}{3}}{\left (x \right )}} + 3 \sqrt [3]{y{\left (x \right )}} = C_{1}\right ]
\]