2.20.3.4.4 Examples for section 3-1-3 (b) Relations between the
coefficients, subcase (i)
2.20.3.4.4.1 Algorithm
Given \(y^{\prime }=f_{0}+f_{1}y+f_{2}y^{2}\), we look for \(a,b\) constants not both \(a,b\) zero such that
\[ a^{2}f_{0}+abf_{1}+b^{2}f_{2}=0 \]
If we can find such
\(a,b\), then there are
three cases to consider.
- \(a\neq 0\). In this case a particular solution is \(y_{1}=\frac {b}{a}\). Hence now we can solve the Riccati ode
since a particular solution is known.
- \(a=1,b=1\), then this means \(f_{0}+f_{1}+f_{2}=0\) where now a particular solution is \(y_{1}=1\), and now we can solve
the Riccati ode since a particular solution is known
- case \(a=0\) and \(b\neq 0\) can not show up, since this implies \(f_{2}=0\) and hence the ode is not Riccati
to start with.
2.20.3.4.4.2 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.
2.20.3.4.4.3 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.