1.3.6.2 Example when G is zero \(y^{\prime }=5-x-\frac {2}{x}y-x^{2}y^{2}\)

Comparing to \(y^{\prime }=f_{0}+f_{1}y+f_{2}y^{2}\)

\begin{align*} f_{0} & =5-x\\ f_{1} & =-\frac {2}{x}\\ f_{2} & =-x^{2}\end{align*}

Hence

\begin{align*} F & =f_{0}f_{2}=\left ( 5-x\right ) \left ( -x^{2}\right ) \\ & =-5x^{2}+x^{3}\\ G & =f_{1}+\frac {f_{2}^{\prime }}{f_{2}}=-\frac {2}{x}+\frac {-2x}{-x^{2}}=-\frac {2}{x}+\frac {2}{x}\\ & =0 \end{align*}

The new ode is (after substituting \(y=\frac {u}{f_{2}}\)) becomes

\begin{align*} u^{\prime } & =F\left ( x\right ) +G\left ( x\right ) u+u^{2}\\ & =\left ( -5x^{2}+x^{3}\right ) +u^{2}\end{align*}

Now we look at the degree of \(F\left ( x\right ) \). Since degree is odd, then no polynomial solution exist for the above Riccati ode in \(u\).