1.6.2.2 Example 2 \(y^{\prime }=3+2xy+y^{2}+xy^{3}\)
This is already an Abel first kind
\[ y^{\prime }=f_{0}+f_{1}y+f_{2}y^{2}+f_{3}y^{3}\]
Where
\begin{align*} f_{0} & =3\\ f_{1} & =2x\\ f_{2} & =1\\ f_{3} & =x \end{align*}
The first step is to remove \(f_{2}\) using
\begin{align*} y\left ( x\right ) & =u\left ( x\right ) -\frac {f_{2}}{3f_{3}}\\ & =u\left ( x\right ) -\frac {1}{3x}\end{align*}
Which results in
\begin{align} u^{\prime } & =\frac {7}{3}-\frac {7}{27x^{2}}+\left ( 2x-\frac {1}{3x}\right ) u+xu^{3}\tag {A}\\ & =f_{0}+f_{1}u+f_{3}u^{3}\nonumber \end{align}
Where
\begin{align*} f_{0} & =\frac {7}{3}-\frac {7}{27x^{2}}\\ f_{1} & =2x-\frac {1}{3x}\\ f_{2} & =0\\ f_{3} & =x \end{align*}
We see the ode is missing the quadratic term. Now we check the invariant
\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 {11664\left ( 27x^{4}-3x^{2}-1\right ) ^{2}x\left ( 9x^{4}-4x^{2}+2\right ) }{49\left ( 9x^{2}-1\right ) ^{6}}\end{align*}
Which is not constant. Hence no direct solution exist. Let now apply the transformation to convert the ode (A) 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 \left ( \int f_{1}\left ( x\right ) dx\right ) \\ & =\exp \int 2x-\frac {1}{3x}dx\\ & =\exp \left ( x^{2}-\frac {1}{3}\ln x\right ) \\ & =\frac {e^{x^{2}}}{x^{\frac {1}{3}}}\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}{x\left ( \frac {e^{x^{2}}}{x^{\frac {1}{3}}}\right ) ^{3}}\left ( \frac {7}{3}-\frac {7}{27x^{2}}\right ) \\ & =\frac {\frac {7}{3}-\frac {7}{27x^{2}}}{x\left ( \frac {e^{x^{2}}}{x^{\frac {1}{3}}}\right ) ^{3}}\\ & =-\frac {1}{27x^{2}}\left ( 7e^{-3x^{2}}-63x^{2}e^{-3x^{2}}\right ) \end{align*}
And where
\begin{align*} \xi & =\int f_{3}U^{2}dx\\ & =\int x\left ( \frac {e^{x^{2}}}{x^{\frac {1}{3}}}\right ) ^{2}dx \end{align*}
But can’t inverse this also. So stop here. Need to find how this transformation is done. This transformation is not practical to use as it can only be done parametrically. Hence will not implement. The above examples are left here just for reference