1.11.2 Example 2  \(y^{\prime }=\sqrt {x-y}\)

\begin{align*} y^{\prime } & =\sqrt {x-y}\\ & =\omega \left ( x,y\right ) \end{align*}

This has Lie tangents

\begin{align*} \xi & =1\\ \eta & =1 \end{align*}

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

\begin{align*} \frac {dy}{dx} & =\frac {\eta }{\xi }\\ & =1 \end{align*}

Hence

\[ y=x+c_{1}\]

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

\[ X=y-x \]

To find \(Y\) we evaluate

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

Hence the canonical coordinates are

\begin{align*} X & =y-x\\ Y & =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 \(Y_{x}=1,Y_{y}=0,X_{x}=-1,X_{y}=1\). The above simplifies to

\begin{align*} \frac {dY}{dX} & =\frac {1}{-1+\sqrt {x-y}}\\ & =\frac {1}{-1+\sqrt {-\left ( y-x\right ) }}\\ & =\frac {1}{-1+\sqrt {-X}}\end{align*}

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

\[ Y=-\ln \left ( -X-1\right ) -2\sqrt {-X}-\ln \left ( -1+\sqrt {-X}\right ) +\ln \left ( 1+\sqrt {-X}\right ) +c_{1}\]

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

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

Now we will solve the same ode using shifting. Hence, since \(\xi =1,\eta =1\) then

\begin{align*} \eta & =\eta -\omega \xi \\ & =1-\sqrt {x-y}\end{align*}

And

\[ \xi =0 \]

Since \(\xi =0\) then

\begin{align*} X & =x\\ Y & =\int \frac {dy}{\eta }\\ & =\int \frac {dy}{1-\sqrt {x-y}}\\ & =2\sqrt {x-y}+2\ln \left ( -1+\sqrt {x-y}\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 {1}{\sqrt {x-y}}+\frac {1}{\left ( -1+\sqrt {x-y}\right ) \sqrt {x-y}}\\ Y_{y} & =-\frac {1}{\sqrt {x-y}}-\frac {1}{\left ( -1+\sqrt {x-y}\right ) \sqrt {x-y}}\\ X_{x} & =1\\ X_{y} & =0 \end{align*}

Hence the ode in canonical coordinates becomes

\begin{align*} \frac {dY}{dX} & =\frac {Y_{x}+\omega Y_{y}}{X_{x}+\omega X_{y}}\\ & =\left ( \frac {1}{\sqrt {x-y}}+\frac {1}{\left ( -1+\sqrt {x-y}\right ) \sqrt {x-y}}\right ) +\sqrt {x-y}\left ( -\frac {1}{\sqrt {x-y}}-\frac {1}{\left ( -1+\sqrt {x-y}\right ) \sqrt {x-y}}\right ) \\ & =-1 \end{align*}

Solving gives

\[ Y=-X+c_{1}\]

Moving back to natural coordinates gives

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

In summary we have solution for \(y^{\prime }=\sqrt {x-y}\) as

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

Both are valid solutions. We notice that when using shifting the algebra can become more complicated since \(Y=\int \frac {dy}{\eta }\) where \(\eta \) in this case becomes more complicated. But both methods will produce valid solution.