3.4.10.10 Example \(y^{\prime }=\frac {-1-2yx}{x^{2}+2y}\)

Solve

\begin{align*} y^{\prime } & =\frac {-1-2yx}{x^{2}+2y}\\ 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}

Let anstaz be

\begin{align*} \xi & =0\\ \eta & =f\left ( x\right ) g\left ( y\right ) \end{align*}

Substituting this into (1) gives

\[ g\frac {df}{dx}+\omega f\frac {dg}{dy}-\omega _{y}fg=0 \]

But \(\omega =\frac {-1-2yx}{x^{2}+2y},\omega _{y}=\frac {d}{dy}\left ( \frac {-1-2yx}{x^{2}+2y}\right ) =\allowbreak \frac {2-2x^{3}}{\left ( x^{2}+2y\right ) ^{2}}\). The above becomes

\[ g\frac {df}{dx}+\left ( \frac {-1-2yx}{x^{2}+2y}\right ) f\frac {dg}{dy}-\left ( \frac {2-2x^{3}}{\left ( x^{2}+2y\right ) ^{2}}\right ) fg=0 \]

The numerator of the normal form is

\begin{align} g\frac {df}{dx}\left ( x^{2}+2y\right ) ^{2}+\left ( x^{2}+2y\right ) \left ( -1-2yx\right ) f\frac {dg}{dy}-\left ( 2-2x^{3}\right ) fg & =0\nonumber \\ g\frac {df}{dx}\left ( x^{4}+4x^{2}y+4y^{2}\right ) +\left ( -2x^{3}y-x^{2}-4xy^{2}-2y\right ) f\frac {dg}{dy}-\left ( 2-2x^{3}\right ) fg & =0 \tag {2}\end{align}

To solve this for \(f\left ( x\right ) ,g\left ( y\right ) \) we start by collecting on either \(x\) or \(y\). Let us start by collecting on \(y\). This gives

\begin{equation} \left [ 4\frac {df}{dx}\right ] \left ( gy^{2}\right ) +\left [ 4\frac {df}{dx}x^{2}\right ] \left ( yg\right ) +\left [ \frac {df}{dx}x^{4}-\left ( -2x^{3}+2\right ) f\right ] g+\left [ \left ( -2x^{3}-4x-2\right ) f\right ] \left ( \frac {dg}{dy}\right ) -\left [ x^{2}f\right ] \frac {dg}{dy}=0 \tag {3}\end{equation}

The other option was to collect on \(x\) terms. This would give

\begin{equation} \left [ -2y\frac {dg}{dy}+2g\right ] \left ( x^{3}f\right ) -\left [ x^{2}f\right ] \left ( \frac {dg}{dy}\right ) -\left [ 4xf\right ] \left ( y\frac {dg}{dy}\right ) +\left [ -2\frac {dg}{dy}y-2g\right ] \left ( f\right ) +\left [ g\right ] \left ( x^{4}\frac {df}{dx}\right ) +\left [ yg\right ] \left ( 4\frac {df}{dx}x^{2}\right ) +\left [ y^{2}g\right ] \left ( 4\frac {df}{dx}\right ) =0 \tag {4}\end{equation}

We start from (3), and if this yields no solutions for \(f\left ( x\right ) ,g\left ( y\right ) \) then we come back and try (4). In either form, the terms inside the \(\left [ \cdot \right ] \) must all be zero to satisfy the ode. From (3) this gives

\begin{align*} 4\frac {df}{dx} & =0\\ 4\frac {df}{dx}x^{2} & =0\\ \frac {df}{dx}x^{4}-\left ( -2x^{3}+2\right ) f & =0\\ \left ( -2x^{3}-4x-2\right ) f & =0\\ x^{2}f & =0 \end{align*}

If one of these results in \(f\left ( x\right ) \) which is function of \(x\). Then we try it to solve for \(g\left ( y\right ) \). If the solutions end up verifying the pde, then we are done. From the above, we start with the first one. This gives \(f=c_{1}\). Which is not function of \(x\). The second give same result. The this option which is \(\frac {df}{dx}x^{4}-\left ( -2x^{3}+2\right ) f=0\) gives

\[ f\left ( x\right ) =c_{1}\frac {e^{-\frac {2}{3x^{3}}}}{x^{2}}\]

Which is function of \(x\). We now use this to find \(g\left ( y\right ) \). It turns out this does not work. The whole anstaz will fail. So need to try different anstaz.