3.4.10.14 Example \(y^{\prime }=\frac {1}{3}\frac {2y+y^{3}-x^{2}}{x}\)

Solve

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

Using Maple the infinitesimals are

\begin{align*} \xi & =\frac {3}{2x^{\frac {1}{3}}}\\ \eta & =\frac {y}{x^{\frac {4}{3}}}\end{align*}

(Will need to show how to obtain these). Lets solve this using the integration factor method first. The integrating factor is given by

\begin{align*} \mu \left ( x,y\right ) & =\frac {1}{\eta -\xi \omega }\\ & =\frac {1}{\frac {y}{x^{\frac {4}{3}}}-\frac {3}{2x^{\frac {1}{3}}}\left ( \frac {1}{3}\frac {2y+y^{3}-x^{2}}{x}\right ) }\\ & =2\frac {x^{\frac {4}{3}}}{x^{2}-y^{3}}\end{align*}

Then the general solution is

\begin{align*} \int \mu \left ( x,y\right ) \left ( dy-\omega dx\right ) & =c_{1}\\ \int 2\frac {x^{\frac {4}{3}}}{x^{2}-y^{3}}\left ( dy-\left ( \frac {1}{3}\frac {2y+y^{3}-x^{2}}{x}\right ) dx\right ) & =c_{1}\\ \int \left ( 2\frac {x^{\frac {4}{3}}}{x^{2}-y^{3}}dy-\left ( 2\frac {x^{\frac {4}{3}}}{x^{2}-y^{3}}\right ) \left ( \frac {1}{3}\frac {2y+y^{3}-x^{2}}{x}\right ) dx\right ) & =c_{1}\\ \int \left ( 2\frac {x^{\frac {4}{3}}}{x^{2}-y^{3}}dy-\left ( \frac {2}{3}\frac {x^{\frac {1}{3}}}{x^{2}-y^{3}}\right ) \left ( 2y+y^{3}-x^{2}\right ) dx\right ) & =c_{1}\end{align*}

Hence we need to find \(F\left ( x,y\right ) \) s.t. \(dF=\left ( 2\frac {x^{\frac {4}{3}}}{x^{2}-y^{3}}dy-\left ( \frac {2}{3}\frac {x^{\frac {1}{3}}}{x^{2}-y^{3}}\right ) \left ( 2y+y^{3}-x^{2}\right ) dx\right ) \) which will make the solution \(F=c\). Therefore

\begin{align*} dF & =\frac {\partial F}{\partial x}dx+\frac {\partial F}{\partial y}dy\\ & =2\frac {x^{\frac {4}{3}}}{x^{2}-y^{3}}dy-\left ( \frac {2}{3}\frac {x^{\frac {1}{3}}}{x^{2}-y^{3}}\right ) \left ( 2y+y^{3}-x^{2}\right ) dx \end{align*}

Hence

\begin{align} \frac {\partial F}{\partial x} & =-\frac {2}{3}\frac {x^{\frac {1}{3}}\left ( 2y+y^{3}-x^{2}\right ) }{x^{2}-y^{3}}\tag {1}\\ \frac {\partial F}{\partial y} & =2\frac {x^{\frac {4}{3}}}{x^{2}-y^{3}} \tag {2}\end{align}

Integrating (1) gives

\begin{align} F & =\left ( \int -\frac {2}{3}\frac {x^{\frac {1}{3}}\left ( 2y+y^{3}-x^{2}\right ) }{x^{2}-y^{3}}dx\right ) +g\left ( y\right ) \nonumber \\ & =\frac {1}{2}x^{\frac {4}{3}}+\frac {1}{3}\ln \left ( x^{\frac {4}{3}}+x^{\frac {2}{3}}y+y^{2}\right ) -\frac {2}{3}\sqrt {3}\arctan \left ( \frac {1}{3}\frac {\left ( 2x^{\frac {2}{3}}+y\right ) \sqrt {3}}{y}\right ) -\frac {2}{3}\ln \left ( x^{\frac {2}{3}}-y\right ) +g\left ( y\right ) \tag {3}\end{align}

Where \(g\left ( y\right ) \) acts as the integration constant but \(F\) depends on \(x,y\) it becomes an arbitrary function. Taking derivative of the above w.r.t. \(y\) gives

\begin{equation} \frac {\partial F}{\partial y}=2\frac {x^{\frac {4}{3}}}{x^{2}-y^{3}}+g^{\prime }\left ( y\right ) \tag {4}\end{equation}

Equating (4,2) gives

\begin{align*} 2\frac {x^{\frac {4}{3}}}{x^{2}-y^{3}} & =2\frac {x^{\frac {4}{3}}}{x^{2}-y^{3}}+g^{\prime }\left ( y\right ) \\ 0 & =g^{\prime }\left ( y\right ) \\ g\left ( y\right ) & =c_{1}\end{align*}

Hence (3) becomes

\[ F=\frac {1}{2}x^{\frac {4}{3}}+\frac {1}{3}\ln \left ( x^{\frac {4}{3}}+x^{\frac {2}{3}}y+y^{2}\right ) -\frac {2}{3}\sqrt {3}\arctan \left ( \frac {1}{3}\frac {\left ( 2x^{\frac {2}{3}}+y\right ) \sqrt {3}}{y}\right ) -\frac {2}{3}\ln \left ( x^{\frac {2}{3}}-y\right ) +c_{1}\]

Therefore the solution is

\begin{align*} F & =c\\ \frac {1}{2}x^{\frac {4}{3}}+\frac {1}{3}\ln \left ( x^{\frac {4}{3}}+x^{\frac {2}{3}}y+y^{2}\right ) -\frac {2}{3}\sqrt {3}\arctan \left ( \frac {1}{3}\frac {\left ( 2x^{\frac {2}{3}}+y\right ) \sqrt {3}}{y}\right ) -\frac {2}{3}\ln \left ( x^{\frac {2}{3}}-y\right ) & =c_{2}\end{align*}

Where constants \(c_{1},c\,\) were combined into \(c_{2}\). Now this ode will be solved using direct symmetry by converting to canonical coordinates.  This is done by using the standard characteristic equation by writing

\begin{align*} \frac {dx}{\xi } & =\frac {dy}{\eta }=dS\\ \frac {dx}{\frac {3}{2x^{\frac {1}{3}}}} & =\frac {dy}{\frac {y}{x^{\frac {4}{3}}}}=dS \end{align*}

First pair of ode’s give

\[ \frac {dy}{dx}=\frac {\frac {y}{x^{\frac {4}{3}}}}{\frac {3}{2x^{\frac {1}{3}}}}=\frac {2}{3x}y \]

Hence

\[ y=c_{1}x^{\frac {2}{3}}\]

Therefore

\[ R=yx^{-\frac {2}{3}}\]

And

\[ dS=\frac {dx}{\xi }=\frac {2}{3}x^{\frac {1}{3}}dx \]

Integrating gives

\begin{align*} S & =\int \frac {2}{3}x^{\frac {1}{3}}dx\\ & =\frac {1}{2}x^{\frac {4}{3}}+c_{1}\\ & =\frac {1}{2}x^{\frac {4}{3}}\end{align*}

By choosing \(c_{1}=0\). Now the ODE \(\frac {dS}{dR}=F\left ( R\right ) \) is found from

\begin{align*} \frac {dS}{dR} & =\frac {\frac {dS}{dx}+\omega \left ( x,y\right ) \frac {dS}{dy}}{\frac {dR}{dx}+\omega \left ( x,y\right ) \frac {dR}{dy}}\\ & =\frac {S_{x}+\omega \left ( x,y\right ) S_{y}}{R_{x}+\omega \left ( x,y\right ) R_{y}}\end{align*}

But \(S_{x}=\frac {2}{3}x^{\frac {1}{3}},R_{x}=-\frac {2}{3}yx^{-\frac {5}{3}},S_{y}=0,R_{y}=x^{-\frac {2}{3}}\). Substituting these into the above and simplifying gives

\begin{align*} \frac {dS}{dR} & =\frac {\frac {2}{3}x^{\frac {1}{3}}}{-\frac {2}{3}yx^{-\frac {5}{3}}+\omega \left ( x,y\right ) x^{-\frac {2}{3}}}\\ & =\frac {\frac {2}{3}x^{\frac {1}{3}}}{-\frac {2}{3}yx^{-\frac {5}{3}}+\left ( \frac {1}{3}\frac {2y+y^{3}-x^{2}}{x}\right ) x^{-\frac {2}{3}}}\\ & =-2\frac {x^{2}}{x^{2}-y^{3}}\end{align*}

But \(R=yx^{-\frac {2}{3}}\) or \(y=Rx^{\frac {2}{3}}\). The above becomes

\begin{align*} \frac {dS}{dR} & =-2\frac {x^{2}}{x^{2}-R^{3}x^{2}}\\ & =\frac {-2}{1-R^{3}}\end{align*}

Which is a quadrature. Solving gives

\begin{align*} \int dS & =\int \frac {-2}{1-R^{3}}dR\\ S & =-\frac {1}{3}\ln \left ( R^{2}+x+1\right ) -\frac {2}{3}\sqrt {3}\arctan \left ( \frac {1}{3}\left ( 1+2R\right ) \sqrt {3}\right ) +\frac {2}{3}\ln \left ( R-1\right ) +c_{1}\end{align*}

Converting back to \(x,y\) gives

\begin{align*} \frac {1}{2}x^{\frac {4}{3}} & =-\frac {1}{3}\ln \left ( \left ( yx^{-\frac {2}{3}}\right ) ^{2}+x+1\right ) -\frac {2}{3}\sqrt {3}\arctan \left ( \frac {1}{3}\left ( 1+2\left ( yx^{-\frac {2}{3}}\right ) \right ) \sqrt {3}\right ) +\frac {2}{3}\ln \left ( \left ( yx^{-\frac {2}{3}}\right ) -1\right ) +c_{1}\\ \frac {1}{2}x^{\frac {4}{3}} & =-\frac {1}{3}\ln \left ( y^{2}x^{-\frac {4}{3}}+x+1\right ) -\frac {2}{3}\sqrt {3}\arctan \left ( \frac {1}{3}\left ( 1+2yx^{-\frac {2}{3}}\right ) \sqrt {3}\right ) +\frac {2}{3}\ln \left ( yx^{-\frac {2}{3}}-1\right ) +c_{1}\end{align*}