1.5.3.1 Example 1  \(\left ( y-2x\right ) y^{\prime }=x-y+y^{2}\)

Given

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

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

Applying the transformation in (2B) gives

\begin{align*} y & =\frac {1}{u}-g\\ y & =\frac {1}{u}+2x \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}\\ & =0\\ k_{1} & =3gf_{3}-f_{2}\\ & =-1\\ k_{2} & =-3g^{2}f_{3}+2gf_{2}-g^{\prime }-f_{1}\\ & =2\left ( -2x\right ) \left ( 1\right ) +2+1\\ & =-4x+3\\ k_{3} & =g^{3}f_{3}-g^{2}f_{2}+f_{1}g-f_{0}\\ & =-\left ( -2x\right ) ^{2}-1\left ( -2x\right ) -x\\ & =x-4x^{2}\end{align*}

Hence (3) becomes

\begin{align*} u^{\prime } & =k_{0}+k_{1}u+k_{2}u^{2}+k_{3}u^{3}\\ & =-u+\left ( -4x+3\right ) u^{2}+\left ( -4x^{2}+x\right ) u^{3}\end{align*}

Which is Abel ode of first kind. But this can not be solved. (not all Abel odes can be solved)