1.5.3.3 Example 3  \(\left ( xy-2x\right ) y^{\prime }=y-y^{2}+3x^{2}y^{3}\)

Given

\[ \left ( xy-2x\right ) y^{\prime }=y-y^{2}+3x^{2}y^{3}\]

We start by normalizing this to form \(\left ( y+g\right ) y^{\prime }=f_{0}+f_{1}y+f_{2}y^{2}+f_{3}y^{3}\) by dividing by \(x\neq 0\) which gives

\[ \left ( y-2\right ) y^{\prime }=\frac {1}{x}y-\frac {1}{x}y^{2}+3xy^{3}\]

Comparing this to (1) \(\left ( y+g\right ) y^{\prime }=f_{0}+f_{1}y+f_{2}y^{2}+f_{3}y^{3}\) shows that

\begin{align*} g & =-2\\ f_{0} & =0\\ f_{1} & =\frac {1}{x}\\ f_{2} & =-\frac {1}{x}\\ f_{3} & =3x \end{align*}

Applying the transformation in (2) gives

\begin{align*} y & =\frac {1}{u}-g\\ y & =\frac {1}{u}+2 \end{align*}

Results in the ode (4)

\begin{equation} u^{\prime }=k_{0}+k_{1}u+k_{2}u^{2}+k_{3}u^{3} \tag {3}\end{equation}

Where, using (5) gives

\begin{align*} k_{0} & =-f_{3}\\ & =-3x\\ k_{1} & =3gf_{3}-f_{2}\\ & =3\left ( -2\right ) \left ( 3x\right ) +\frac {1}{x}\\ & =-18x+\frac {1}{x}\\ k_{2} & =-3g^{2}f_{3}+2gf_{2}-g^{\prime }-f_{1}\\ & =-3\left ( -2\right ) ^{2}\left ( 3x\right ) +2\left ( -2\right ) \left ( -\frac {1}{x}\right ) -\frac {1}{x}\\ & =-\frac {3}{x}\left ( 12x^{2}-1\right ) \\ k_{3} & =g^{3}f_{3}-g^{2}f_{2}+f_{1}g-f_{0}\\ & =\left ( -8\right ) \left ( 3x\right ) -\left ( -2\right ) ^{2}\left ( -\frac {1}{x}\right ) +\frac {1}{x}\left ( -2\right ) \\ & =-\frac {2}{x}\left ( 12x^{2}-1\right ) \end{align*}

Hence (3) becomes

\begin{align*} u^{\prime } & =k_{0}+k_{1}u+k_{2}u^{2}+k_{3}u^{3}\\ & =-3x+\left ( -18x+\frac {1}{x}\right ) u+\left ( -\frac {3}{x}\left ( 12x^{2}-1\right ) \right ) u^{2}+\left ( -\frac {2}{x}\left ( 12x^{2}-1\right ) \right ) u^{3}\end{align*}

Which is Abel first kind.