Given one Abel ode \(y^{\prime }\left ( x\right ) =f_{0}(x)+f_{1}(x)y+f_{2}(x)y^{2}+f_{3}(x)y^{3}\), it is called equivalent to another Abel ode \(u^{\prime }\left ( t\right ) =g_{0}(t)+g_{1}(t)u+g_{2}(t)u^{2}+g_{3}(t)u^{3}\) if there is transformation which converts one to the other. This transformation is given by
\begin {align} x & =F\left ( t\right ) \tag {1}\\ y\left ( x\right ) & =P\left ( t\right ) u\left ( t\right ) +Q\left ( t\right ) \nonumber \end {align}
Where \(F^{\prime }\neq 0,P\neq 0\). If such transformation can be found, then if given the solution of one of these ode’s, the solution to the other ode can directly be fond using this transformation. In this case, we also call these two ode as belonging to same Abel equivalence class. In other words, an Abel equivalence class is the set of all Abel ode’s that can be transformed to each others using the same transformation given in (1).
There are many disjoint Abel equivalence classes, each class will have all the ode that can be transformed to each others using some specific transformation (1). Here is one example below taken from paper by A.D.Roch and E.S.Cheb-Terrab called "Abel ODEs: Equivalence and integrable classes".
Given one Abel ode
\begin {equation} y^{\prime }\left ( x\right ) =\frac {1}{2x+8}y^{2}+\frac {x}{2x+8}y^{3} \tag {2} \end {equation} Which is known to have solution\begin {equation} c_{1}+\frac {\sqrt {y^{2}x-4y-1}}{y}+2\arctan \left ( \frac {1+2y}{\sqrt {y^{2}x-4y-1}}\right ) =0 \tag {3} \end {equation} And now we are given a second Abel ode\begin {equation} u^{\prime }\left ( t\right ) =\frac {1}{t}u+\frac {f^{\prime }t-f}{2\left ( f+3t\right ) }u^{2}+\frac {\left ( f^{\prime }t-f\right ) \left ( t-f\right ) }{2\left ( f+3t\right ) }u^{3} \tag {4} \end {equation} And asked to find its solution. If we can determine if (4) is equivalent to (2) then the solution of (4) can be obtained directly. It can be found that\begin {align*} F\left ( t\right ) & =\frac {f\left ( t\right ) }{t}-1\\ Q\left ( t\right ) & =0\\ P\left ( t\right ) & =t \end {align*}
Where see that \(F^{\prime }\left ( t\right ) \neq 0\) and \(P\left ( t\right ) \neq 0\). Hence (1) becomes\begin {align} x & =\frac {f\left ( t\right ) }{t}-1\tag {5}\\ y\left ( x\right ) & =tu\left ( t\right ) \nonumber \end {align}
Applying the transformation (5) on the solution (3) results in the solution of (4) as\begin {align} A & =\sqrt {\left ( \frac {f}{t}-1\right ) t^{2}u^{2}-4tu-1}\nonumber \\ c_{1}+\frac {A}{tu}+2\arctan \left ( \frac {1+2tu}{A}\right ) & =0 \tag {6} \end {align}
Equation (6) above is the implicit solution to (4) obtained from the solution to (2) by using equivalence transformation as the two ode’s are found to be equivalent. Finding the transformation (5) requires more calculation and not trivial. See the above paper for more information.