1.5.2 Details

This ODE has the form

\begin{equation} \left ( g_{1}y+g_{0}\right ) y^{\prime }=f_{0}+f_{1}y+f_{2}y^{2}+f_{3}y^{3}\tag {1A}\end{equation}

In some places, this is only shows up to quadratic term

\begin{equation} \left ( g_{1}y+g_{0}\right ) y^{\prime }=f_{0}+f_{1}y+f_{2}y^{2}\tag {1B}\end{equation}

And the above second form seems more common. But it seems both can be used in everything I tried. This is transformed to first kind Abel ode in \(u\left ( x\right ) \) using the following transformation

\begin{equation} y=\frac {1}{ug_{1}}-\frac {g_{0}}{g_{1}}\tag {2A}\end{equation}

But a more common form is the following (used by handbook of exact solutions to ordinary differential equations) is this

\begin{equation} \left ( y+g\right ) y^{\prime }=f_{0}+f_{1}y+f_{2}y^{2}\tag {1C}\end{equation}

Which can be obtained from (1B) by dividing by \(g_{1}\) (where the \(f_{i}\) are now different). Using (1C)  form, then (2A) becomes

\begin{equation} y=\frac {1}{u}-g\tag {2B}\end{equation}

Which is simpler than (2A) and this is what will be used from now on. The canonical form for the second kind is the following

\begin{equation} yy^{\prime }=F\left ( x\right ) +y\tag {3}\end{equation}

Here \(F\) must be function of \(x\) and can not be scalar, else the ode becomes quadrature. When the second kind is in the canonical form above there are tabulated cases of known solutions for specific cases of \(f\left ( x\right ) \). See equation world and handbook of exact solutions.

Those tables are given for the ode in form \(yy^{\prime }=\left ( sx+Ax^{m}\right ) +y\). i.e. \(f=sx+Ax^{m}\) for different values of \(s,m\) where \(A\) is arbitrary parameter. There are tables for different forms also.

Applying transformation to first kind given in (2B) results in

\begin{equation} u^{\prime }=k_{0}+k_{1}u+k_{2}u^{2}+k_{3}u^{3} \tag {4}\end{equation}

Where

\begin{align} k_{0} & =-f_{3}\tag {5}\\ k_{1} & =3gf_{3}-f_{2}\nonumber \\ k_{2} & =-3g^{2}f_{3}+2gf_{2}-g^{\prime }-f_{1}\nonumber \\ k_{3} & =g^{3}f_{3}-g^{2}f_{2}+f_{1}g-f_{0}\nonumber \end{align}

Which is Abel of first kind.