1.3.6.7 Example when G not zero \(y^{\prime }=-x+\frac {33}{16}+\left ( x^{2}-x-\frac {1}{2}\right ) y+y^{2}\)

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

\begin{align*} f_{0} & =-x+\frac {33}{16}\\ f_{1} & =\left ( x^{2}-x-\frac {1}{2}\right ) \\ f_{2} & =1 \end{align*}

Since \(f_{1}=1\) already, we can calculate \(Q\) directly now

\begin{align*} Q & =f_{1}^{2}-4f_{0}-2f_{1}^{\prime }\\ & =\left ( x^{2}-x-\frac {1}{2}\right ) ^{2}-4\left ( -x+\frac {33}{16}\right ) -2\left ( 2x-1\right ) \\ & =x^{4}-2x^{3}+x-6 \end{align*}

Since the degree is even, then now we expand \(\sqrt {Q}\) and keep rational integer part and call that \(X\)

\begin{align*} x^{4}-2x^{3}+x-6 & =\left ( ax^{2}+bx+c\right ) ^{2}\\ & =a^{2}x^{4}+\left ( bx+c\right ) ^{2}+2ax^{2}\left ( bx+c\right ) \\ & =a^{2}x^{4}+\left ( b^{2}x^{2}+c^{2}+2bcx\right ) +2abx^{3}+2acx^{2}\\ & =a^{2}x^{4}+x^{3}\left ( 2ab\right ) +x^{2}\left ( b^{2}+2ac\right ) +x\left ( 2bc\right ) +c^{2}\end{align*}

Comparing terms, then \(a=1,2ab=-2\) or \(b=-1\). And \(2bc=1\) or \(c=-\frac {1}{2}\). Hence

\begin{align*} \sqrt {Q} & =X\\ & =ax^{2}+bx+c\\ & =x^{2}-x-\frac {1}{2}\end{align*}

Since \(X\) is not constant, then there are two possible particular polynomial solutions of \(y^{\prime }=f_{0}+f_{1}y+f_{2}y^{2}\). These are

\begin{align*} y & =-\frac {1}{2}\left ( f_{1}\pm X\right ) \\ & =-\frac {1}{2}\left ( \left ( x^{2}-x-\frac {1}{2}\right ) \pm \left ( x^{2}-x-\frac {1}{2}\right ) \right ) \end{align*}

These are

\begin{align*} y_{1} & =-\frac {1}{2}\left ( \left ( x^{2}-x-\frac {1}{2}\right ) +\left ( x^{2}-x-\frac {1}{2}\right ) \right ) \\ & =-x^{2}+x+\frac {1}{2}\\ y_{2} & =-\frac {1}{2}\left ( \left ( x^{2}-x-\frac {1}{2}\right ) -\left ( x^{2}-x-\frac {1}{2}\right ) \right ) \\ & =0 \end{align*}

Testing each shows that neither satisfies the ode. Hence this approach did not work in this example.