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

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

Hence

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

A solution is \(a=1,b=1\). This is case 2. Hence a particular solution is

\[ y_{1}=1 \]

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.