1.3.4.4.3 Example \(y^{\prime }=x+xy-2xy^{2}\) Comparing to \(y^{\prime }=f_{0}+f_{1}y+f_{2}y^{2}\) shows that

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

Hence

\begin{align*} a^{2}f_{0}+abf_{1}+b^{2}f_{2} & =0\\ a^{2}x+xab-2b^{2}x & =0\\ a^{2}+ab-2b^{2} & =0 \end{align*}

A solution is \(a=2,b=-1\). Since \(a\neq 0\) then a particular solution is

\begin{align*} y_{1} & =\frac {b}{a}\\ & =\frac {-1}{2}\end{align*}

Now that we know a particular solution, finding the general solution to the Riccati ode is easy. See the section below on how this is done.