1.12.15 Example \(y^{\prime }=3-2\frac {y}{x}\)

This is homogeneous ODE of Class A of form \(y^{\prime }=F\left ( \frac {y}{x}\right ) \), hence from the lookup table

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

The first step is to verify that \(\bar {x}=\epsilon x,\bar {y}=\epsilon y\) leaves the ode invariant.

\[ \frac {d\bar {y}}{d\bar {x}}=\frac {\bar {y}_{x}+\bar {y}_{y}y^{\prime }}{\bar {x}_{x}+\bar {x}_{y}y^{\prime }}=\frac {\epsilon y^{\prime }}{\epsilon }=y^{\prime }\]

Hence the ode becomes

\begin{align*} \frac {d\bar {y}}{d\bar {x}} & =3-2\frac {\bar {y}}{\bar {x}}\\ y^{\prime } & =3-2\frac {\epsilon y}{\epsilon x}\\ & =3-2\frac {y}{x}\end{align*}

Verified. Now the ode is solved. The tangent curves are computed directly from the Lie group symmetry given above

\begin{align*} \xi & =\left . \frac {\partial \bar {x}}{\partial \epsilon }\right \vert _{\epsilon =0}=x\\ \eta & =\left . \frac {\partial \bar {y}}{\partial \epsilon }\right \vert _{\epsilon =0}=y \end{align*}

The canonical coordinates \(\left ( X,Y\right ) \) are now found. Using

\begin{align} \frac {dx}{\xi } & =\frac {dy}{\eta }=dY\nonumber \\ \frac {dx}{x} & =\frac {dy}{y}=dY \tag {1}\end{align}

The first pair gives

\begin{align*} \frac {dy}{dx} & =\frac {y}{x}\\ \ln y & =\ln x+c_{1}\\ y & =cx \end{align*}

Hence

\begin{align*} X & =c\\ & =\frac {y}{x}\end{align*}

Now we find \(Y\) from the last pair of equations

\begin{align*} \frac {dy}{y} & =dY\\ Y & =\ln y \end{align*}

What is left is to find \(\frac {dY}{dX}\). This is given by

\[ \frac {dY}{dX}=f\left ( X\right ) \]

To find \(f\left ( X\right ) \), we use  \(dY=Y_{x}dx+Y_{y}dy=\frac {1}{y}dy\) and \(dX=X_{x}dx+X_{y}dy=-\frac {y}{x^{2}}dx+\frac {1}{x}dy\). Hence

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

But \(\frac {dy}{dx}=3-2\frac {y}{x}=3-2X\), hence

\begin{align*} \frac {dY}{dX} & =\frac {3-2X}{-X^{2}+X\left ( 3-2X\right ) }\\ & =\frac {3-2X}{3\left ( X-X^{2}\right ) }\end{align*}

Which is a quadrature. In Lie method, for first order ode, we always obtain \(\frac {dY}{dX}=f\left ( X\right ) \). Integrating the above gives

\begin{align*} \int dY & =\int \frac {3-2X}{3\left ( X-X^{2}\right ) }dX\\ Y & =\ln X-\frac {1}{3}\ln \left ( X-1\right ) +c_{1}\end{align*}

Final step is to replace \(X,Y\) back with \(x,y\) which gives

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