Example 4 Solve\[ y^{\prime }=-\frac {1}{2}\frac {y\left ( 1+\sqrt {x^{2}y^{4}+1}\right ) }{x}\] 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 {y\left ( 1+\sqrt {x^{2}y^{4}+1}\right ) }{x}\right ) \\ & =-\frac {1}{2}\left ( 1+\sqrt {x^{2}y^{4}+1}\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}\\ & =-\frac {1}{2}\frac {x^{2}y^{4}}{\sqrt {x^{2}y^{4}+1}} \end {align*}

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

Solving the integral gives\[ \ln x-c_{1}-\operatorname {arctanh}\left ( \frac {1}{\sqrt {x^{2}y^{4}+1}}\right ) =0 \]