1.7.2.2 Example 2 \(y^{\prime }=x+y+y^{3}\)

\[ y^{\prime }=x+y+y^{3}\]

Comparing to

\[ y^{\prime }=f_{0}+f_{1}y+f_{2}y^{2}+f_{3}y^{3}\]

Shows that

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

Applying transformation \(y=\frac {1}{u}\) gives

\[ u^{\prime }u=k_{0}+k_{1}u+k_{2}u^{2}+k_{3}u^{3}\]

Where

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

Hence (1) becomes

\[ u^{\prime }u=-1-u^{2}-xu^{3}\]

Which is second kind Abel.