2.21.6.2.1 Example 1 \(y^{\prime }=\left ( \frac {1}{x^{2}}-\frac {1}{x}\right ) y^{2}+\frac {1}{x^{2}}u^{3}\)

Which is Abel first kind.

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

Since \(f_{2}\) is not zero, it is removed using

\begin{align*} y\left ( x\right ) & =u\left ( x\right ) -\frac {f_{2}}{3f_{3}}\\ & =u\left ( x\right ) -\frac {\left ( \frac {1}{x^{2}}-\frac {1}{x}\right ) }{3\left ( \frac {1}{x^{2}}\right ) }\\ & =u\left ( t\right ) -\left ( \frac {1}{3}-\frac {1}{3}x\right ) \end{align*}

This results in

\begin{align} u^{\prime }\left ( x\right ) & =\left ( -\frac {2}{27}x-\frac {1}{9}-\frac {2}{9x}+\frac {2}{27x^{2}}\right ) +\left ( \frac {-1}{3}+\frac {2}{3x}-\frac {1}{3x^{2}}\right ) u+\frac {1}{x^{2}}u^{3}\tag {A}\\ & =f_{0}+f_{1}u+f_{3}u^{3}\nonumber \end{align}

Where now

\begin{align*} f_{0} & =\left ( -\frac {2}{27}x-\frac {1}{9}-\frac {2}{9x}+\frac {2}{27x^{2}}\right ) \\ f_{1} & =\left ( \frac {-1}{3}+\frac {2}{3x}-\frac {1}{3x^{2}}\right ) \\ f_{2} & =0\\ f_{3} & =\frac {1}{x^{2}}\end{align*}

We see the ode is missing the \(u^{2}\) term. The above is Abel first kind. Now we check if the invariant is constant or not. (Recall, this check is only done when the quadratic term is missing). This gives

\begin{align*} \Delta & =-\frac {\left ( -f_{0}^{\prime }f_{3}+f_{0}f_{3}^{\prime }+3f_{0}f_{3}f_{1}\right ) ^{3}}{27f_{3}^{4}f_{0}^{5}}\\ & =\frac {1458\left ( 2x^{5}+5x^{4}+8x^{3}-5x^{2}+10x-2\right ) ^{2}x^{2}\left ( x^{3}+4x^{2}-2x-6\right ) }{\left ( 2x^{3}+3x^{2}+6x-2\right ) ^{6}}\end{align*}

Which is clearly not constant. Hence no direct solution exist. Let now apply the transformation to convert the ode in \(u\left ( x\right ) \) to canonical form

\[ \eta ^{\prime }\left ( \xi \right ) =\Phi \left ( \xi \right ) +\eta ^{3}\left ( \xi \right ) \]

We start by finding

\begin{align*} U & =\exp \left ( \int f_{1}\left ( x\right ) -\frac {f_{2}^{2}\left ( x\right ) }{3f_{3}\left ( x\right ) }dx\right ) \\ & =\exp \int f_{1}dx\\ & =\exp \int \frac {-1}{3}+\frac {2}{3x}-\frac {1}{3x^{2}}dx\\ & =\exp \left ( -\frac {x}{3}+\frac {2}{3}\ln x+\frac {1}{3x}\right ) \\ & =x^{\frac {2}{3}}e^{-\frac {x}{3}}e^{\frac {1}{3x}}\end{align*}

Next we find

\begin{align*} \Phi \left ( \xi \right ) & =\frac {1}{f_{3}U^{3}}\left ( f_{0}-\frac {f_{1}f_{2}}{3f_{3}}+\frac {2f_{2}^{3}}{27f_{3}^{2}}-\frac {1}{3}\frac {d}{dx}\left ( \frac {f_{2}}{f_{3}}\right ) \right ) \\ & =\frac {1}{\left ( \frac {1}{x^{2}}\right ) \left ( x^{\frac {2}{3}}e^{-\frac {x}{3}}e^{\frac {1}{3x}}\right ) ^{3}}\left ( -\frac {2}{27}x-\frac {1}{9}-\frac {2}{9x}+\frac {2}{27x^{2}}\right ) \\ & =\frac {-\frac {2}{27}x-\frac {1}{9}-\frac {2}{9x}+\frac {2}{27x^{2}}}{\left ( \frac {1}{x^{2}}\right ) \left ( x^{\frac {2}{3}}e^{-\frac {x}{3}}e^{\frac {1}{3x}}\right ) ^{3}}\end{align*}

Where

\begin{align*} \xi & =\int f_{3}U^{2}dx\\ & =\int \frac {1}{x^{2}}\left ( x^{\frac {2}{3}}e^{-\frac {x}{3}}e^{\frac {1}{3x}}\right ) ^{2}dx \end{align*}

But need to find \(x\) as function of \(\xi \) in order to make RHS of \(\Phi \left ( \xi \right ) \) all \(\xi \). Since this integral have no closed form, can not do it. I picked not a good example or do not understand this method. Unable to find one single worked example showing how this transformation works.