Find what is called the abel invariant and check if constant.\[ \Delta =-\frac {\left ( -f_{0}^{\prime }f_{3}+f_{0}f_{3}^{\prime }+3f_{0}f_{3}f_{1}\right ) ^{3}}{27f_{3}^{4}f_{0}^{5}}\] The substitution \(y=\frac {1}{u}\) is now applied. Therefore \(y^{\prime }=-\frac {1}{u^{2}}u^{\prime }\). Substituting this in (1) gives\begin {align} -\frac {1}{u^{2}}u^{\prime } & =f_{0}(x)+f_{1}(x)\frac {1}{u}+f_{2}(x)\frac {1}{u^{2}}+f_{3}(x)\frac {1}{u^{3}}\nonumber \\ -uu^{\prime } & =u^{3}f_{0}(x)+u^{2}f_{1}(x)+uf_{2}(x)+f_{3}(x)\nonumber \\ uu^{\prime } & =-u^{3}f_{0}(x)-u^{2}f_{1}(x)-uf_{2}(x)-f_{3}(x) \tag {2} \end {align}
Using the substitution \(u=\frac {1}{E}\left ( y+\frac {f_{2}}{3f_{3}}\right ) \) where \(E=\exp \left ( \int f_{1}-\frac {f_{2}^{2}}{3f_{3}}dx\right ) \) in the above gives\[ \frac {1}{E}\left ( y+\frac {f_{2}}{3f_{3}}\right ) u^{\prime }=-u^{3}f_{0}(x)-u^{2}f_{1}(x)-uf_{2}(x)-f_{3}(x) \] Hence\begin {align*} u^{\prime } & =\frac {1}{E^{2}}\frac {dE}{dx}\left ( y+\frac {f_{2}}{3f_{3}}\right ) +\frac {1}{E}\left ( y^{\prime }+\frac {1}{3}\frac {f_{2}^{\prime }f_{3}-f_{2}f_{3}^{\prime }}{f_{3}^{2}}\right ) \\ & =\frac {1}{E^{2}}\frac {dE}{dx}\left ( \frac {1}{u}+\frac {f_{2}}{3f_{3}}\right ) +\frac {1}{E}\left ( -\frac {1}{u^{2}}u^{\prime }+\frac {1}{3}\frac {f_{2}^{\prime }f_{3}-f_{2}f_{3}^{\prime }}{f_{3}^{2}}\right ) \\ u^{\prime }+\frac {u^{\prime }}{Eu^{2}} & =\frac {1}{E^{2}}\frac {dE}{dx}\left ( \frac {1}{u}+\frac {f_{2}}{3f_{3}}\right ) +\frac {1}{3E}\frac {f_{2}^{\prime }f_{3}-f_{2}f_{3}^{\prime }}{f_{3}^{2}}\\ u^{\prime }\left ( 1+\frac {1}{Eu^{2}}\right ) & =\frac {1}{E^{2}}\frac {dE}{dx}\left ( \frac {1}{u}+\frac {f_{2}}{3f_{3}}\right ) +\frac {1}{3E}\frac {f_{2}^{\prime }f_{3}-f_{2}f_{3}^{\prime }}{f_{3}^{2}}\\ u^{\prime } & =\frac {Eu^{2}}{1+Eu^{2}}\left ( \frac {1}{E^{2}}\frac {dE}{dx}\left ( \frac {1}{u}+\frac {f_{2}}{3f_{3}}\right ) +\frac {1}{3E}\frac {f_{2}^{\prime }f_{3}-f_{2}f_{3}^{\prime }}{f_{3}^{2}}\right ) \\ u^{\prime } & =\frac {u^{2}}{1+Eu^{2}}\left ( \frac {1}{E}\frac {dE}{dx}\left ( \frac {1}{u}+\frac {f_{2}}{3f_{3}}\right ) +\frac {1}{3}\frac {f_{2}^{\prime }f_{3}-f_{2}f_{3}^{\prime }}{f_{3}^{2}}\right ) \end {align*}
Substituting the above into (2) gives\[ u\frac {u^{2}}{1+Eu^{2}}\left ( \frac {1}{E}\frac {dE}{dx}\left ( \frac {1}{u}+\frac {f_{2}}{3f_{3}}\right ) +\frac {1}{3}\frac {f_{2}^{\prime }f_{3}-f_{2}f_{3}^{\prime }}{f_{3}^{2}}\right ) =-u^{3}f_{0}-u^{2}f_{1}-uf_{2}-f_{3}\] Therefore\begin {align*} E & =\exp \left ( \int f_{1}\left ( x\right ) -\frac {f_{2}^{2}\left ( x\right ) }{3f_{3}\left ( x\right ) }dx\right ) \\ \xi & =\int f_{3}\left ( x\right ) E^{2}dx\\ u & =\frac {1}{E}\left ( y+\frac {f_{2}\left ( x\right ) }{3f_{3}\left ( x\right ) }\right ) \end {align*}
The above are used to convert the first kind Abel ode to canonical form. (To finish).