2.22.2.4 Example 4 \(y^{\prime }=-\frac {1}{3}x+\frac {2}{3x}y+\frac {1}{3x}y^{3}\)

\[ y^{\prime }=-\frac {1}{3}x+\frac {2}{3x}y+\frac {1}{3x}y^{3}\]

Comparing to

\[ y^{\prime }=h+gy+fy^{n}\]

Shows that

\begin{align*} h & =-\frac {1}{3}x\\ g & =\frac {2}{3x}\\ f & =\frac {1}{3x}\\ n & =3 \end{align*}

Hence

\begin{align*} \Delta & =f^{-n-1}h^{-2n+1}\left ( fh^{\prime }-f^{\prime }h-ngfh\right ) ^{n}n^{-n}\\ & =\left ( \frac {1}{3x}\right ) ^{-3-1}\left ( -\frac {1}{3}x\right ) ^{-6+1}\left ( \left ( \frac {1}{3x}\right ) \left ( -\frac {1}{3}x\right ) ^{\prime }-\left ( \frac {1}{3x}\right ) ^{\prime }\left ( -\frac {1}{3}x\right ) -3\left ( \frac {2}{3x}\right ) \left ( \frac {1}{3x}\right ) \left ( -\frac {1}{3}x\right ) \right ) ^{3}3^{-3}\\ & =\left ( \frac {1}{3x}\right ) ^{-4}\left ( -\frac {1}{3}x\right ) ^{-5}\left ( \left ( \frac {1}{3x}\right ) \left ( \frac {-1}{3}\right ) -\left ( -\frac {1}{3x^{2}}\right ) \left ( -\frac {1}{3}x\right ) -3\left ( \frac {2}{3x}\right ) \left ( \frac {1}{3x}\right ) \left ( -\frac {1}{3}x\right ) \right ) ^{3}\left ( 3^{-3}\right ) \\ & =0 \end{align*}

Hence the invariant is constant.

\begin{equation} z^{\prime }-gz=\alpha h \tag {4}\end{equation}

Where \(z=\left ( \frac {h}{f}\right ) ^{\frac {1}{n}}=\left ( \frac {-\frac {1}{3}x}{\frac {1}{3x}}\right ) ^{\frac {1}{3}}=\left ( -x^{2}\right ) ^{\frac {1}{3}}\). Hence \(z^{\prime }=\frac {d}{dx}\left ( \left ( -x^{2}\right ) ^{\frac {1}{3}}\right ) =-\frac {2}{3}\frac {x}{\left ( -x^{2}\right ) ^{\frac {2}{3}}}\). Therefore (4) becomes

\begin{align*} -\frac {2}{3}\frac {x}{\left ( -x^{2}\right ) ^{\frac {2}{3}}}-\frac {2}{3x}\left ( \left ( -x^{2}\right ) ^{\frac {1}{3}}\right ) & =\alpha \left ( -\frac {1}{3}x\right ) \\ \alpha & =0 \end{align*}

Since \(\Delta =0\) then we use the EQ. (6) in introduction

\begin{align*} \int ^{\left ( \frac {h}{f}\right ) ^{\frac {-1}{n}}y\left ( x\right ) }\frac {1}{u^{n}-\alpha u+1}du-\int \left ( \frac {h}{f}\right ) ^{\frac {-1}{n}}hdx+c_{1} & =0\\ \int ^{\allowbreak \left ( -x^{2}\right ) ^{\frac {-1}{3}}y\left ( x\right ) }\frac {1}{u^{3}+1}du-\int \left ( -x^{2}\right ) ^{\frac {-1}{3}}\left ( -\frac {1}{3}x\right ) dx+c_{1} & =0\\ \int ^{\allowbreak \left ( -x^{2}\right ) ^{\frac {-1}{3}}y\left ( x\right ) }\frac {1}{u^{3}+1}du+\frac {1}{3}\int \frac {x}{\sqrt [3]{-x^{2}}}dx+c_{1} & =0\\ \int ^{\allowbreak \left ( -x^{2}\right ) ^{\frac {-1}{3}}y\left ( x\right ) }\frac {1}{u^{3}+1}du+\frac {1}{3}\left ( \frac {3}{4}\frac {x^{2}}{\left ( -x^{2}\right ) ^{\frac {1}{3}}}\right ) +c_{1} & =0\\ \int ^{\allowbreak \left ( -x^{2}\right ) ^{\frac {-1}{3}}y\left ( x\right ) }\frac {1}{u^{3}+1}du+\frac {1}{4}\frac {x^{2}}{\left ( -x^{2}\right ) ^{\frac {1}{3}}}+c_{1} & =0 \end{align*}