2.12.1.2 Example 2 \(y^{\prime }=\frac {2x\left ( -x^{4}-2x^{2}y+y^{2}\right ) }{y^{2}+2x^{2}y-x^{4}}\)

Solve

\[ y^{\prime }=\frac {2x\left ( -x^{4}-2x^{2}y+y^{2}\right ) }{y^{2}+2x^{2}y-x^{4}}\]

The first step is to identify if this is class G and find \(F\). We start by multiplying the RHS by \(\frac {x}{y}\) (regardless of what is in the RHS) which gives

\begin{align*} y^{\prime } & =\frac {x}{y}\left ( \frac {2x\left ( -x^{4}-2x^{2}y+y^{2}\right ) }{y^{2}+2x^{2}y-x^{4}}\right ) \\ & =\frac {2x^{2}\left ( x^{4}+2x^{2}y-y^{2}\right ) }{y\left ( x^{4}-2x^{2}y-y^{2}\right ) }\\ & =F\left ( x,y\right ) \end{align*}

Next we check if \(F\left ( x,y\right ) \) has \(y\) or not in it. If so, then let the RHS above be \(F\left ( x,y\right ) \) and now do

\begin{align*} f_{x} & =x\frac {\partial F}{\partial x}\\ & =x\left ( \frac {4x\left ( x^{8}-4x^{6}y-6x^{4}y^{2}-4x^{2}y^{3}+y^{4}\right ) }{y\left ( x^{4}-2x^{2}y-y^{2}\right ) ^{2}}\right ) \\ & =\frac {4x^{2}\left ( x^{8}-4x^{6}y-6x^{4}y^{2}-4x^{2}y^{3}+y^{4}\right ) }{y\left ( x^{4}-2x^{2}y-y^{2}\right ) ^{2}}\end{align*}

And let

\begin{align*} f_{y} & =y\frac {\partial F}{\partial y}\\ & =y\left ( \frac {-2x^{2}\left ( x^{8}-4x^{6}y-6x^{4}y^{2}-4x^{2}y^{2}-4x^{2}y^{3}+y^{4}\right ) }{y^{2}\left ( x^{4}-2x^{2}y-y^{2}\right ) ^{2}}\right ) \\ & =\frac {-2x^{2}\left ( x^{8}-4x^{6}y-6x^{4}y^{2}-4x^{2}y^{2}-4x^{2}y^{3}+y^{4}\right ) }{y\left ( x^{4}-2x^{2}y-y^{2}\right ) ^{2}}\end{align*}

Now we check, if \(f_{y}=0\) then this is not Homogeneous type G. Else we now need to determine value of \(\alpha \). This is done as follows.

\begin{align*} \alpha & =\frac {fx}{f_{y}}\\ & =-2 \end{align*}

If \(\alpha \) comes out not to have in it \(x\) nor \(y\) as in this case, then we are done. This ode is Homogeneous type G and the ode can be written as

\[ y^{\prime }=\frac {y}{x}F\left ( \frac {y}{x^{\alpha }}\right ) \]

Hence the solution is

\begin{equation} \ln x-c_{1}+\int ^{yx^{\alpha }}\frac {1}{\tau \left ( -\alpha -F\left ( \tau \right ) \right ) }d\tau =0 \tag {1}\end{equation}

Now let \(y=\frac {\tau }{x^{\alpha }}\) and substitute this into \(F\left ( x,y\right ) \) which results in

\begin{align*} F\left ( \tau \right ) & =\frac {2x^{2}\left ( x^{4}+2x^{2}\frac {\tau }{x^{\alpha }}-\left ( \frac {\tau }{x^{\alpha }}\right ) ^{2}\right ) }{\frac {\tau }{x^{\alpha }}\left ( x^{4}-2x^{2}\frac {\tau }{x^{\alpha }}-\left ( \frac {\tau }{x^{\alpha }}\right ) ^{2}\right ) }\\ & =\frac {2x^{2}\left ( x^{4}+2x^{2}\frac {\tau }{x^{-2}}-\left ( \frac {\tau }{x^{-2}}\right ) ^{2}\right ) }{\frac {\tau }{x^{-2}}\left ( x^{4}-2x^{2}\frac {\tau }{x^{-2}}-\left ( \frac {\tau }{x^{-2}}\right ) ^{2}\right ) }\\ & =\frac {2x^{2}\left ( x^{4}+2x^{4}\tau -x^{4}\tau ^{2}\right ) }{\tau x^{2}\left ( x^{4}-2x^{4}\tau -\tau ^{2}x^{4}\right ) }\\ & =\frac {2\left ( x^{4}+2x^{4}\tau -x^{4}\tau ^{2}\right ) }{\tau \left ( x^{4}-2x^{4}\tau -\tau ^{2}x^{4}\right ) }\\ & =\frac {2}{\tau }\frac {\left ( 1+2\tau -\tau ^{2}\right ) }{\left ( 1-2\tau -\tau ^{2}\right ) }\\ & =\frac {2}{\tau }\frac {\left ( \tau ^{2}-2\tau -1\right ) }{\left ( \tau ^{2}+2\tau -1\right ) }\end{align*}

The solution(1) becomes

\begin{align*} \ln x-c_{1}+\int ^{yx^{\alpha }}\frac {1}{\tau \left ( -\alpha -F\left ( \alpha \right ) \right ) }d\tau & =0\\ \ln x-c_{1}+\int ^{\frac {y}{x^{2}}}\frac {1}{\tau \left ( 2-\left ( \frac {2}{\tau }\frac {\left ( \tau ^{2}-2\tau -1\right ) }{\left ( \tau ^{2}+2\tau -1\right ) }\right ) \right ) }d\tau & =0\\ \ln x-c_{1}+\int ^{\frac {y}{x^{2}}}\frac {1}{2}\frac {\tau ^{2}+2\tau -1}{\tau ^{3}+\tau ^{2}+\tau +1}d\tau & =0 \end{align*}

Solving the integral gives

\[ \ln x-c_{1}-\frac {1}{2}\ln \left ( \frac {x^{2}+y}{x^{2}}\right ) +\frac {1}{2}\ln \left ( \frac {x^{4}+y^{2}}{x^{4}}\right ) =0 \]