This is homogeneous class D \(y^{\prime }=\frac {y}{x}+g\left ( x\right ) F\left ( \frac {y}{x}\right ) \). Hence from lookup table\begin {align*} \xi & =x^{2}\\ \eta & =xy \end {align*}
Now we just need to find canonical coordinates \(\left ( R,S\right ) \) since \(\xi ,\eta \) are known. Using\begin {align} \frac {dx}{\xi } & =\frac {dy}{\eta }=dS\nonumber \\ \frac {dx}{x^{2}} & =\frac {dy}{xy}=dS \tag {1} \end {align}
The first pair gives\begin {align*} \frac {dy}{dx} & =\frac {y}{x}\\ \ln y & =\ln x+c_{1}\\ y & =cx \end {align*}
Hence \begin {align*} R & =c\\ & =\frac {y}{x} \end {align*}
Now we find \(S\) from the last pair of equations (we could also use the first and last equations in (1)).\begin {align*} \frac {dy}{xy} & =dS\\ S & =\frac {1}{x}\ln y \end {align*}
What is left is to find \(\frac {dS}{dR}\). This is given by\begin {align*} \frac {dS}{dR} & =G\left ( R\right ) \\ & =\frac {S_{x}+S_{y}y^{\prime }}{R_{x}+R_{y}y^{\prime }} \end {align*}
To find \(G\left ( R\right ) \), we use \(S_{x}=\frac {-1}{x^{2}}\ln y,S_{y}=\frac {1}{xy}\) and \(R_{x}=-\frac {y}{x^{2}},R_{y}=\frac {1}{x}\). Hence\begin {align*} \frac {dS}{dR} & =\frac {\frac {-1}{x^{2}}\ln y+\frac {1}{xy}y^{\prime }}{-\frac {y}{x^{2}}+\frac {1}{x}y^{\prime }}\\ & =\frac {-\ln y-\frac {x}{y}y^{\prime }}{y+xy^{\prime }}\\ & =\frac {-\ln y-\frac {1}{R}y^{\prime }}{y+xy^{\prime }} \end {align*}
But \(y^{\prime }=\frac {y}{x}+\frac {1}{x}F\left ( \frac {y}{x}\right ) =R+\frac {1}{x}F\left ( R\right ) \). The above becomes\begin {align*} \frac {dS}{dR} & =\frac {-\ln y-\frac {1}{R}\left ( R+\frac {1}{x}F\left ( R\right ) \right ) }{y+x\left ( R+\frac {1}{x}F\left ( R\right ) \right ) }\\ & =\frac {-\ln y-1-\frac {1}{xR}F\left ( R\right ) }{y+xR+F\left ( R\right ) }\\ & =\frac {-\ln y-1-\frac {1}{x\frac {y}{x}}F\left ( R\right ) }{y+x\frac {y}{x}+F\left ( R\right ) }\\ & =\frac {-\ln y-1-\frac {1}{y}F\left ( R\right ) }{2y+F\left ( R\right ) } \end {align*}
Something is wrong. \(\frac {dS}{dR}\) should only be a function of \(R\). Need to find out why. Let me try the other pair of equations from (1) to solve for \(S\) and see what happens.\begin {align*} \frac {dx}{x^{2}} & =dS\\ S & =-\frac {1}{x} \end {align*}
What is left is to find \(\frac {dS}{dR}\). This is given by\begin {align*} \frac {dS}{dR} & =G\left ( R\right ) \\ & =\frac {S_{x}+S_{y}y^{\prime }}{R_{x}+R_{y}y^{\prime }} \end {align*}
To find \(G\left ( R\right ) \), we use \(S_{x}=\frac {1}{x^{2}},S_{y}=0\) and \(R_{x}=-\frac {y}{x^{2}},R_{y}=\frac {1}{x}\). Hence\begin {align*} \frac {dS}{dR} & =\frac {\frac {1}{x^{2}}}{-\frac {y}{x^{2}}+\frac {1}{x}y^{\prime }}\\ & =\frac {1}{-y+xy^{\prime }} \end {align*}
But \(y^{\prime }=\frac {y}{x}+\frac {1}{x}F\left ( \frac {y}{x}\right ) =R+\frac {1}{x}F\left ( R\right ) \). The above becomes\begin {align*} \frac {dS}{dR} & =\frac {1}{-y+x\left ( R+\frac {1}{x}F\left ( R\right ) \right ) }\\ & =\frac {1}{-y+xR+F\left ( R\right ) }\\ & =\frac {1}{-y+x\frac {y}{x}+F\left ( R\right ) }\\ & =\frac {1}{F\left ( R\right ) } \end {align*}
This worked. But why the first choice did not work? OK, let me continue now. Integrating the above gives\[ S=\int \frac {1}{F\left ( R\right ) }dR+c \] But \(S=-\frac {1}{x}\), hence \begin {align*} -\frac {1}{x} & =\int ^{\frac {y}{x}}\frac {1}{F\left ( r\right ) }dr+c\\ 0 & =\int ^{\frac {y}{x}}\frac {1}{F\left ( r\right ) }dr+c+\frac {1}{x} \end {align*}
This example shows that when solving for \(S\) from\[ \frac {dx}{x^{2}}=\frac {dy}{xy}=dS \] There are two choice. One is \(dS=\frac {dy}{xy}\) and the other \(dS=\frac {dx}{x^{2}}\). Using the first choice did not work here (unless I made a mistake, but do not see it)., Only the second choice worked because we must end up with \(\frac {dS}{dR}=G\left ( R\right ) \) where RHS is function of \(R\) only. I need to look more into this. In theory, any choice should have worked.