3.4.10.11 Example \(y^{\prime }=3\sqrt {yx}\)

Solve

\begin{align*} y^{\prime } & =3\sqrt {yx}\\ y^{\prime } & =\omega \left ( x,y\right ) \end{align*}

The symmetry condition results in the pde

\begin{equation} \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0 \tag {1}\end{equation}

Trying polynomial anstaz

\begin{align*} \xi & =a_{0}+a_{1}x\\ \eta & =b_{0}+b_{1}y \end{align*}

And substituting these into (1) and simplifying gives

\[ \left ( -9a_{1}+3b_{1}\right ) yx-3xb_{0}-3ya_{0}=0 \]

Setting all coefficients to zero gives

\begin{align*} -9a_{1}+3b_{1} & =0\\ b_{0} & =0\\ a_{0} & =0 \end{align*}

Hence \(a_{1}=\frac {1}{3}b_{1}\). Letting \(b_{1}=1\) then \(a_{1}=\frac {1}{3}\) and the infinitesimals are

\begin{align*} \xi & =\frac {1}{3}x\\ \eta & =y \end{align*}

The integrating factor is therefore

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

The next step is to determine the canonical coordinates \(R,S\). This is done by using the standard characteristic equation by writing

\begin{equation} \frac {dx}{\xi }=\frac {dy}{\eta }=dS\nonumber \end{equation}

The first pair of equations gives

\[ \frac {dy}{dx}=\frac {\eta }{\xi }=\frac {3y}{x}\]

Solving gives

\[ y=c_{1}x^{3}\]

Hence

\begin{equation} R=c_{1}=\frac {y}{x^{3}} \tag {2}\end{equation}

And \(S\) is found from

\[ dS=\frac {dx}{\xi }=3\frac {dx}{x}\]

Integrating gives

\begin{align*} S & =3\ln x+c_{1}\\ & =3\ln x \end{align*}

By choosing \(c_{1}=0\). Now that \(R\left ( x,y\right ) ,S\left ( x,y\right ) \) are found, the ODE \(\frac {dS}{dR}=F\left ( R\right ) \) is determined. This is determined from

\begin{align*} \frac {dS}{dR} & =\frac {\frac {dS}{dx}+\omega \left ( x,y\right ) \frac {dS}{dy}}{\frac {dR}{dx}+\omega \left ( x,y\right ) \frac {dR}{dy}}\\ & =\frac {S_{x}+\omega \left ( x,y\right ) S_{y}}{R_{x}+\omega \left ( x,y\right ) R_{y}}\end{align*}

But \(S_{x}=\frac {3}{x},R_{x}=-3\frac {y}{x^{4}},S_{y}=0,R_{y}=\frac {1}{x^{3}}\). Substituting these into the above gives

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

But \(\omega \left ( x,y\right ) =3\sqrt {yx}\). The above becomes

\begin{align} \frac {dS}{dR} & =\frac {3x^{3}}{-3y+3x\sqrt {yx}}\nonumber \\ & =\frac {x^{3}}{x\sqrt {yx}-y}\nonumber \\ & =\frac {-1}{\sqrt {\frac {y}{x^{3}}}-\frac {y}{x^{3}}} \tag {3}\end{align}

But \(R=\frac {y}{x^{3}}\) and the above becomes

\[ \frac {dS}{dR}=\frac {-1}{R-\sqrt {R}}\]

Which is a quadrature. Solving gives

\begin{align*} \int dS & =\int \frac {-1}{R-\sqrt {R}}dR\\ S & =-2\ln \left ( \sqrt {R}-1\right ) +c_{1}\end{align*}

Converting back to \(x,y\) gives

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

Or

\begin{align*} y_{1}\left ( x\right ) & =2x\left ( x^{2}+x\sqrt {xc_{1}}\right ) -x^{3}+c_{1}\\ y_{2}\left ( x\right ) & =-2x\left ( -x^{2}+x\sqrt {xc_{1}}\right ) -x^{3}+c_{1}\end{align*}