1.11.1 Example 1  \(y^{\prime }=xy^{2}-\frac {2y}{x}-\frac {1}{x^{3}}\)

\begin{align*} y^{\prime } & =xy^{2}-\frac {2y}{x}-\frac {1}{x^{3}}\\ & =\omega \left ( x,y\right ) \end{align*}

This has Lie tangents

\begin{align*} \xi & =x\\ \eta & =-2y \end{align*}

Let solve it first without shifting. To find \(X\) we solve

\begin{align*} \frac {dy}{dx} & =\frac {\eta }{\xi }\\ & =\frac {-2y}{x}\end{align*}

Hence

\[ y=\frac {c_{1}}{x^{2}}\]

Therefore \(X=c_{1}\) or

\[ X=yx^{2}\]

To find \(Y\) we evaluate

\begin{align*} Y & =\int \frac {dx}{\xi }\\ & =\int \frac {dx}{x}\\ & =\ln x \end{align*}

Hence the canonical coordinates are  

\begin{align*} X & =yx^{2}\\ Y & =\ln x \end{align*}

Now we need to find the ODE in the canonical coordinates space. This is found using

\[ \frac {dY}{dX}=\frac {Y_{x}+\omega Y_{y}}{X_{x}+\omega X_{y}}\]

but \(Y_{x}=\frac {1}{x},Y_{y}=0,X_{x}=2yx,X_{y}=x^{2}\). The above simplifies to

\begin{align*} \frac {dY}{dX} & =\frac {1}{x^{4}y^{2}-1}\\ & =\frac {1}{X^{2}-1}\end{align*}

We see as expected that the ODE in canonical coordinates is always quadrature. Solving this gives

\[ Y\left ( X\right ) =-\operatorname {arctanh}\left ( X\right ) +c_{1}\]

Converting back to natural coordinates \(x,y\) gives

\begin{align} \ln x & =-\operatorname {arctanh}\left ( yx^{2}\right ) +c_{1}\nonumber \\ -\ln x+c_{1} & =\operatorname {arctanh}\left ( yx^{2}\right ) \nonumber \\ yx^{2} & =\tanh \left ( -\ln x+c_{1}\right ) \nonumber \\ y & =\frac {\tanh \left ( -\ln x+c_{1}\right ) }{x^{2}} \tag {1}\end{align}

Now we will solve the same ode using shifting. Hence, since \(\xi =x,\eta =-2y\) then

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

And

\[ \xi =0 \]

Since \(\xi =0\) then

\begin{align*} X & =x\\ Y & =\int \frac {dy}{\eta }\\ & =-x^{2}\int \frac {dy}{\left ( x^{4}y^{2}-1\right ) }\\ & =-x^{2}\left ( \frac {\ln \left ( x^{2}y-1\right ) -\ln \left ( x^{2}y+1\right ) }{2x^{2}}\right ) \\ & =-\frac {1}{2}\left ( \ln \left ( x^{2}y-1\right ) -\ln \left ( x^{2}y+1\right ) \right ) \end{align*}

Now we need to find the ODE in the canonical coordinates space. This is found using

\[ \frac {dY}{dX}=\frac {Y_{x}+\omega Y_{y}}{X_{x}+\omega X_{y}}\]

Where now

\begin{align*} Y_{x} & =-\frac {2xy}{x^{4}y^{2}-1}\\ Y_{y} & =-\frac {x^{2}}{x^{4}y^{2}-1}\\ X_{x} & =1\\ X_{y} & =0 \end{align*}

Hence the ode in canonical coordinates becomes

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

Solving gives

\[ S=-\ln R+c_{1}\]

Moving back to natural coordinates gives

\begin{align} -\frac {1}{2}\left ( \ln \left ( x^{2}y-1\right ) -\ln \left ( x^{2}y+1\right ) \right ) & =-\ln x+c_{1}\nonumber \\ \ln \left ( x^{2}y-1\right ) -\ln \left ( x^{2}y+1\right ) & =2\ln x+c_{2}\nonumber \\ \ln \left ( \frac {x^{2}y-1}{x^{2}y+1}\right ) & =\ln x^{2}+c_{2}\nonumber \\ y & =-\frac {1+c_{3}x^{2}}{x^{2}\left ( c_{3}x^{2}-1\right ) }\nonumber \\ & =\frac {c_{4}+x^{2}}{x^{2}\left ( c_{4}-x^{2}\right ) } \tag {2}\end{align}

In summary we have solution for \(y^{\prime }=xy^{2}-\frac {2y}{x}-\frac {1}{x^{3}}\) as

solution using no shift	\(y=\frac {\tanh \left ( -\ln x+c_{1}\right ) }{x^{2}}\)
solution using shift	\(y=\frac {c_{4}+x^{2}}{x^{2}\left ( c_{4}-x^{2}\right ) }\)

Maple 2025 gives the first solution as default. But when asked to use Lie method, it does give the second solution, which means it used \(\xi =0\).

Both solutions are verified correct. The solution using shifting is simpler since it involves no functions at all.