Example when G not zero y′ = 1 + x2 − 2xy + y2

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

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

Hence

\begin{align*} F & =f_{0}f_{2}\\ & =1+x^{2}\\ G & =f_{1}+\frac {f_{2}^{\prime }}{f_{2}}\\ & =-2x \end{align*}

Since \(F,G\) are polynomials, and \(G\neq 0\) then let \(y=\frac {u}{f_{2}}\,\) which gives

\begin{align} u^{\prime } & =F\left ( x\right ) +G\left ( x\right ) u+u^{2}\tag {1}\\ & =\left ( 1+x^{2}\right ) -2xu+u^{2}\nonumber \end{align}

Since \(G\neq 0\) then let

\begin{align} u & =w-\frac {G}{2}\nonumber \\ & =w+x \tag {2}\end{align}

When \(G\neq 0\), book says to calculate \(Q\) given by

\begin{align*} Q & =G^{2}-4F-2G^{\prime }\\ & =\left ( -2x\right ) ^{2}-4\left ( 1+x^{2}\right ) -2\left ( -2\right ) \\ & =0 \end{align*}

There are 3 cases, either \(Q\) is polynomial of even or odd degree, or \(Q\) is constant as in this case.

Substituting (2) into (1) gives

\[ w^{\prime }=w^{2}\]

We see this is now separable. And we can solve it. This gives

\[ w=\frac {1}{c_{1}-x}\]

Hence

\begin{align*} u & =w+x\\ & =\frac {1}{c_{1}-x}+x \end{align*}

And since \(y=\frac {u}{f_{2}}\) then

\begin{align*} y & =\frac {\frac {1}{c_{1}-x}+x}{1}\\ & =\frac {x^{2}+c_{2}x-1}{x+c_{2}}\end{align*}