2.12.1.3 Example 3 \(y^{\prime }=-\frac {1}{2}\frac {3y^{2}-x}{y\left ( y^{2}-3x\right ) }\)

Solve

\[ y^{\prime }=-\frac {1}{2}\frac {3y^{2}-x}{y\left ( y^{2}-3x\right ) }\]

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 {1}{2}\frac {3y^{2}-x}{y\left ( y^{2}-3x\right ) }\right ) \\ & =-\frac {1}{2}\frac {3xy^{2}-x^{2}}{y^{4}-3xy^{2}}\\ & =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}\\ & =\frac {1}{2}\frac {x\left ( -3y^{4}+2xy^{2}-3x^{2}\right ) }{y^{2}\left ( -y^{2}+3x\right ) ^{2}}\end{align*}

And let

\begin{align*} f_{y} & =y\frac {\partial F}{\partial y}\\ & =\frac {3xy^{4}-3x^{2}y^{2}+3x^{3}}{y^{2}\left ( -y^{2}+3x\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}}\\ & =-\frac {1}{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 {1}{2}\frac {3xy^{2}-x^{2}}{y^{4}-3xy^{2}}\\ & =-\frac {1}{2}\frac {3x\left ( \frac {\tau }{x^{\alpha }}\right ) ^{2}-x^{2}}{\left ( \frac {\tau }{x^{\alpha }}\right ) ^{4}-3x\left ( \frac {\tau }{x^{\alpha }}\right ) ^{2}}\\ & =-\frac {1}{2}\frac {3x\left ( \frac {\tau }{x^{-\frac {1}{2}}}\right ) ^{2}-x^{2}}{\left ( \frac {\tau }{x^{-\frac {1}{2}}}\right ) ^{4}-3x\left ( \frac {\tau }{x^{-\frac {1}{2}}}\right ) ^{2}}\\ & =-\frac {1}{2}\frac {3x^{2}\tau ^{2}-x^{2}}{\tau ^{4}x^{2}-3x\tau ^{2}x}\\ & =-\frac {1}{2}\frac {3\tau ^{2}-1}{\tau ^{4}-3\tau ^{2}}\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}{\sqrt {x}}}\frac {1}{\tau \left ( \frac {1}{2}-\left ( -\frac {1}{2}\frac {3\tau ^{2}-1}{\tau ^{4}-3\tau ^{2}}\right ) \right ) }d\tau & =0\\ \ln x-c_{1}+\int ^{\frac {y}{\sqrt {x}}}2\tau \frac {\tau ^{2}-3}{\tau ^{4}-1}d\tau & =0 \end{align*}

Solving the integral gives

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