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1 First order ODE

Considering first order ODE\begin {equation} \frac {dy}{dx}=\omega \left ( x,y\right ) \tag {A} \end {equation} Let the infinitesimal pointwise transformation from point \(\left ( x,y\right ) \) to another point in same plane \(\left ( X,Y\right ) \) be\begin {align*} X & =x+\epsilon \xi \left ( x,y\right ) +O\left ( \epsilon ^{2}\right ) \\ Y & =y+\epsilon \eta \left ( x,y\right ) +O\left ( \epsilon ^{2}\right ) \end {align*}

Or ignoring higher order terms\begin {align} X\left ( x,y\right ) & =x+\epsilon \xi \left ( x,y\right ) \tag {1}\\ Y\left ( x,y\right ) & =y+\epsilon \eta \left ( x,y\right ) \tag {2} \end {align}

In the above \(\epsilon \) is the one parameter in the Lie symmetry group. The symmetry condition for (A) is that \[ \frac {dY}{dX}=\omega \left ( X,Y\right ) \] Whenever\[ \frac {dy}{dx}=\omega \left ( x,y\right ) \] Symmetry of an ODE means the ODE in \(\left ( x,y\right ) \) remain the same form (but using new variables \(\left ( X,Y\right ) \)) after applying the (non-trivial) transformation (1,2). Nontrivial transformation means \(\epsilon \neq 0\). The first goal is to find the functions \(\xi \left ( x,y\right ) ,\eta \left ( x,y\right ) \) which satisfy the symmetry condition above. This turns out to be the hardest step in this method.

The symmetry condition is written as\begin {equation} \frac {dY}{dX}=\frac {\frac {dY}{dx}}{\frac {dX}{dx}}=\omega \left ( X,Y\right ) \tag {3} \end {equation} Where \(\frac {dY}{dx}\) is the total derivative with respect to the \(x\) variable. Similarly for \(\frac {dX}{dx}\). But\begin {align} \frac {dY}{dx} & =Y_{x}+Y_{y}\frac {dy}{dx}\nonumber \\ & =Y_{x}+Y_{y}\omega \left ( x,y\right ) \tag {4} \end {align}

And\begin {align} \frac {dX}{dx} & =X_{x}+X_{y}\frac {dy}{dx}\nonumber \\ & =X_{x}+X_{y}\omega \left ( x,y\right ) \tag {5} \end {align}

Substituting (4,5) into (3) gives the symmetry condition as

\begin {equation} \frac {Y_{x}+\omega \left ( x,y\right ) Y_{y}}{X_{x}+\omega \left ( x,y\right ) X_{y}}=\omega \left ( X,Y\right ) \tag {6} \end {equation} But\begin {equation} X_{x}=1+\varepsilon \xi _{x} \tag {7} \end {equation} And similarly \begin {equation} X_{y}=\varepsilon \xi _{y} \tag {8} \end {equation} And\begin {equation} Y_{x}=\varepsilon \eta _{x} \tag {9} \end {equation} And\begin {equation} Y_{y}=1+\varepsilon \eta _{y} \tag {10} \end {equation} Substituting (7,8,9,10) back into the symmetry condition (6) gives\begin {align} \frac {\varepsilon \eta _{x}+\omega \left ( 1+\varepsilon \eta _{y}\right ) }{\left ( 1+\varepsilon \xi _{x}\right ) +\omega \varepsilon \xi _{y}} & =\omega \left ( x+\varepsilon \xi ,y+\varepsilon \eta \right ) \nonumber \\ \frac {\varepsilon \eta _{x}+\omega +\omega \varepsilon \eta _{y}}{1+\varepsilon \xi _{x}+\omega \varepsilon \xi _{y}} & =\omega \left ( x+\varepsilon \xi ,y+\varepsilon \eta \right ) \nonumber \\ \frac {\omega +\varepsilon \left ( \eta _{x}+\omega \eta _{y}\right ) }{1+\varepsilon \left ( \xi _{x}+\omega \xi _{y}\right ) } & =\omega \left ( x+\varepsilon \xi ,y+\varepsilon \eta \right ) \tag {11} \end {align}

The above is what will be used to determine \(\xi \left ( x,y\right ) ,\eta \left ( x,y\right ) \). But the above is a PDE which is too complicated to use as is. It is linearized, and the linearized version is what is used to solve for \(\xi ,\eta \).

Eq (11) is linearized by expanding the LHS and the RHS using Taylor series, by expanding around \(\varepsilon =0\). Starting with the LHS first, let \(\frac {\omega +\varepsilon \left ( \eta _{x}+\omega \eta _{y}\right ) }{1+\varepsilon \left ( \xi _{x}+\omega \xi _{y}\right ) }=\Delta \). Expanding this using Taylor series around\(\ \varepsilon =0\) gives\begin {equation} \Delta =\Delta _{\varepsilon =0}+\varepsilon \frac {d}{d\varepsilon }\left ( \Delta \right ) _{\varepsilon =0}+h.o.t.\tag {11A} \end {equation}

But \(\Delta _{\varepsilon =0}=\omega \) and \begin {align*} \frac {d}{d\varepsilon }\left ( \Delta \right ) & =\frac {\frac {d}{d\varepsilon }\left [ \omega +\varepsilon \left ( \eta _{x}+\omega \eta _{y}\right ) \right ] \left ( 1+\varepsilon \left ( \xi _{x}+\omega \xi _{y}\right ) \right ) -\left ( \omega +\varepsilon \left ( \eta _{x}+\omega \eta _{y}\right ) \right ) \frac {d}{d\varepsilon }\left [ 1+\varepsilon \left ( \xi _{x}+\omega \xi _{y}\right ) \right ] }{\left ( 1+\varepsilon \left ( \xi _{x}+\omega \xi _{y}\right ) \right ) ^{2}}\\ & =\frac {\left ( \eta _{x}+\omega \eta _{y}\right ) \left ( 1+\varepsilon \left ( \xi _{x}+\omega \xi _{y}\right ) \right ) -\left ( \omega +\varepsilon \left ( \eta _{x}+\omega \eta _{y}\right ) \right ) \left ( \xi _{x}+\omega \xi _{y}\right ) }{\left ( 1+\varepsilon \left ( \xi _{x}+\omega \xi _{y}\right ) \right ) ^{2}} \end {align*}

At \(\varepsilon =0\) the above reduces to\begin {align} \frac {d}{d\varepsilon }\left ( \Delta \right ) _{\varepsilon =0} & =\left ( \eta _{x}+\omega \eta _{y}\right ) -\omega \left ( \xi _{x}+\omega \xi _{y}\right ) \nonumber \\ & =\eta _{x}+\omega \eta _{y}-\omega \xi _{x}-\omega ^{2}\xi _{y}\nonumber \\ & =\eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}\tag {12} \end {align}

Therefore the LHS, EQ (11A), after expansion becomes\begin {equation} \Delta _{LHS}=\omega +\varepsilon \left ( \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}\right ) \tag {11B} \end {equation} Now RHS of EQ (11) is linearized. Let \(\omega \left ( x+\varepsilon \xi ,y+\varepsilon \eta \right ) =\Delta _{RHS}\). Expansion around \(\varepsilon =0\) gives\[ \Delta _{RHS}=\Delta _{\varepsilon =0}+\varepsilon \left ( \frac {d}{d\varepsilon }\Delta \right ) _{\varepsilon =0}+h.o.t. \] But \(\Delta _{\varepsilon =0}=\omega \left ( x,y\right ) \) and \[ \frac {d}{d\varepsilon }\Delta _{RHS}=\omega _{x}\xi +\omega _{y}\eta \] Hence the linearized RHS of (11) becomes \begin {equation} \Delta _{RHS}=\omega \left ( x,y\right ) +\varepsilon \left ( \omega _{x}\xi +\omega _{y}\eta \right ) \tag {13} \end {equation} Substituting (11B,13) back into (11), gives the linearized version of (11) as\begin {align} \Delta _{LHS} & =\Delta _{RHS}\nonumber \\ \omega +\varepsilon \left ( \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}\right ) & =\omega +\varepsilon \left ( \omega _{x}\xi +\omega _{y}\eta \right ) \nonumber \\ \varepsilon \left ( \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}\right ) & =\varepsilon \left ( \omega _{x}\xi +\omega _{y}\eta \right ) \nonumber \\ \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y} & =\omega _{x}\xi +\omega _{y}\eta \nonumber \\ \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta & =0\tag {14} \end {align}

The above equation is what is used to determine \(\xi ,\eta \). It is the linearized symmetry condition.

Now that we determined \(\xi ,\eta \), what to do with them and how to use them to solve the original ODE?

We need to find what is called the canonical coordinates \(\left ( R,S\right ) \). In these canonical coordinates the ODE becomes a quadrature and easily solved. Once solved, the solution is transformed back to \(\left ( x,y\right ) \). The canonical coordinates \(\left ( R,S\right ) \) are found by setting up the following\begin {equation} \frac {dx}{\xi }=\frac {dy}{\eta }=dS \tag {15} \end {equation} Where in the above \(\xi ,\eta \) are now known (These we found by solving  the symmetry condition given in EQ. 14).  

From EQ. (15), the first pair are used to set up the ode \(\frac {dy}{dx}=\frac {\eta }{\xi }\) and this is solved to obtain \(y\left ( x\right ) =f\left ( x\right ) +R\), which gives \(R\left ( x,y\right ) \). Now that \(R\) is found, then \(S\left ( R\right ) \) is found by solving the pair of equations\[ dS=\frac {dx}{\xi }\] Which gives \(S\left ( x,y\right ) \). Next the following ODE is set up\[ \frac {dS}{dR}=\frac {\frac {dS}{dx}+\omega \left ( x,y\right ) \frac {dS}{dy}}{\frac {dR}{dx}+\omega \left ( x,y\right ) \frac {dR}{dy}}=\Omega \left ( R\right ) \] Which is always a quadrature. Its solution is \[ S\left ( R\right ) =F\left ( R\right ) \] The final step is to replace \(S,R\) in the above in terms of \(x,y\,\) (since \(S,R\) were found earlier as functions of \(x,y\)).

This gives the solution to the original ODE \(y\left ( x\right ) \).

The only way to understand this method well, is to workout some problems showing the method on some ODE’s.

To learn more about the theory of Lie transformation itself and why it works, there are many links in my links page on the subject.

1.1 Examples

1.1.1 Example 1

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

For \(x\neq 0\). The first step is to find \(\xi \) and \(\eta \). The end of this problem shows how these are found.

Let \(X=x+\varepsilon x\) which implies that \(\xi =x\). And let \(Y=y-2\varepsilon y\) which implies \(\eta =-2y\). Therefore\begin {align*} \xi \left ( x,y\right ) & =x\\ \eta \left ( x,y\right ) & =-2y \end {align*}

The next step is to determine what is called the canonical coordinates \(R,S\). Where \(R\) is the independent variable and \(S\) is the dependent variable. So we are looking for \(S\left ( R\right ) \) function. This is done by using the standard characteristic equation by writing\begin {align} \frac {dx}{\xi } & =\frac {dy}{\eta }=dS\nonumber \\ \frac {dx}{x} & =\frac {dy}{-2y}=dS \tag {1} \end {align}

The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x}+\eta \frac {\partial }{\partial y}\right ) S\left ( x,y\right ) =1\). Which is a first order PDE.  This is solved for \(S\), which gives (1) using the method of characteristic to solve first order PDE which is standard method. Starting with the first pair of ODE gives\[ \frac {dy}{dx}=-\frac {2y}{x}\] Integrating gives \(yx^{2}=c\) where \(c\) is constant of integration. In this method \(R\) is always \(c\). Hence\[ R\left ( x,y\right ) =yx^{2}\] \(S\left ( x,y\right ) \) is now found from the first equation in (1) and the last equation which gives\begin {align*} dS & =\frac {dx}{\xi }\\ S & =\int \frac {dx}{x}\\ S & =\ln \left \vert x\right \vert \end {align*}

Now that \(R\left ( x,y\right ) ,S\left ( x,y\right ) \) are found, the ODE \(\frac {dS}{dR}=\Omega \left ( R\right ) \) is setup. The ODE comes out to be function of \(R\) only, so it is quadrature. This is the main idea of this method. By solving for \(R\) we go back to \(x,y\) and solve for \(y\left ( x\right ) \). How to find \(\frac {dS}{dR}\)? There is an equation to determine this given by\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*}

Everything on the RHS is known. Substituting gives\begin {align*} \frac {dS}{dR} & =\frac {\frac {1}{x}+\left ( xy^{2}-\frac {2y}{x}-\frac {1}{x^{3}}\right ) \left ( 0\right ) }{2xy+\left ( xy^{2}-\frac {2y}{x}-\frac {1}{x^{3}}\right ) x^{2}}\\ & =\frac {\frac {1}{x}}{2xy+\left ( xy^{2}-\frac {2y}{x}-\frac {1}{x^{3}}\right ) x^{2}}\\ & =\frac {1}{x^{4}y^{2}-1} \end {align*}

But \(R=\) \(yx^{2}\), hence the above becomes\[ \frac {dS}{dR}=\frac {1}{R^{2}-1}\] This is just quadrature. Integrating gives \[ S=-\operatorname {arctanh}\left ( R\right ) +c_{1}\] This solution is converted back to \(x,y\). Since \(S=\ln \left \vert x\right \vert ,R=yx^{2}\), the above becomes\[ \ln \left \vert x\right \vert =-\operatorname {arctanh}\left ( yx^{2}\right ) +c_{1}\] Or\begin {align*} -\ln \left \vert x\right \vert +c_{1} & =\operatorname {arctanh}\left ( yx^{2}\right ) \\ yx^{2} & =\tanh \left ( -\ln \left \vert x\right \vert +c_{1}\right ) \\ y\left ( x\right ) & =\frac {\tanh \left ( -\ln \left \vert x\right \vert +c_{1}\right ) }{x^{2}} \end {align*}

Which is the solution to the original ODE.

The above shows the basic steps in this method. Let us solve more ODE’s to practice this method more.

Finding Lie symmetries for this example

The condition of symmetry is given above in equation (14) as\begin {equation} \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0 \tag {14} \end {equation} We now need to solve the above for \(\xi ,\eta \,\) given a specific \(\omega \left ( x,y\right ) \) for the ODE at hand. This PDE can not be solved as is for \(\xi ,\eta \) without an ansatz. One common ansatz is to use \(\xi =\alpha \left ( x\right ) \) and \(\eta =\beta \left ( x\right ) y+\gamma \left ( x\right ) \) and plugging these into the above and then compare coefficients to solve for \(\alpha \left ( x\right ) ,\beta \left ( x\right ) ,\gamma \left ( x\right ) \).

Another ansatz is to use a polynomials for \(\xi ,\eta \). And this is what we will start with.

Using polynomial as ansatz

Let\begin {align} \xi & =c_{1}x+c_{2}y+c_{3}\tag {1}\\ \eta & =c_{4}x+c_{5}y+c_{6} \tag {2} \end {align}

One of these ansatz will generate a solution if one exists. If not, then it means there is no non-trivial symmetry. But instead of doing the above here, we will start with \(\xi =c_{1}x+c_{2}y+c_{3},\eta =c_{4}x+c_{5}y+c_{6}\) and use the computer to solve the set of the the generates \(Ac=b\) system.

Eq (14) becomes\begin {align*} \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta & =0\\ c_{4}+\omega \left ( c_{5}-c_{1}\right ) -\omega ^{2}c_{2}-\omega _{x}\left ( c_{1}x+c_{2}y+c_{3}\right ) -\omega _{y}\left ( c_{4}x+c_{5}y+c_{6}\right ) & =0 \end {align*}

But in this ODE \(\omega =xy^{2}-\frac {2y}{x}-\frac {1}{x^{3}}\), hence \(\omega _{x}=y^{2}+\frac {2y}{x^{2}}+\frac {3}{x^{4}}\) and \(\omega _{y}=2yx-\frac {2}{x}\). The above becomes

\begin {align*} c_{4}+\left ( xy^{2}-\frac {2y}{x}-\frac {1}{x^{3}}\right ) \left ( c_{5}-c_{1}\right ) -\left ( xy^{2}-\frac {2y}{x}-\frac {3}{x^{3}}\right ) ^{2}c_{2}-\left ( y^{2}+\frac {2y}{x^{2}}+\frac {1}{x^{4}}\right ) \left ( c_{1}x+c_{2}y+c_{3}\right ) -\left ( 2yx-\frac {2}{x}\right ) \left ( c_{4}x+c_{5}y+c_{6}\right ) & =0\\ 3c_{4}+\frac {2}{x}c_{6}-\frac {1}{x^{4}}c_{3}-y^{2}c_{3}+3y^{3}c_{2}-\frac {1}{x^{3}}c_{5}-\frac {9}{x^{6}}\allowbreak c_{2}-x^{2}y^{4}c_{2}-2xyc_{6}-2xy^{2}c_{1}-\frac {2}{x^{2}}yc_{3}-2x^{2}\allowbreak yc_{4}-\frac {13}{x^{4}}yc_{2}-xy^{2}c_{5} & =0 \end {align*}

Each coefficient to each monomial must be zero. Hence

We can set up \(Ac=b\) and solve for \(c_{i}\). But no need here. We see by inspection that \(c_{2}=c_{3}=c_{4}=c_{6}=0\) and \(-2c_{1}-c_{5}=0\). This means \[ c_{1}=-\frac {c_{5}}{2}\] So if we let \(c_{1}=1\) then \(c_{5}=-2\). Therefore, since\begin {align} \xi & =c_{1}x+c_{2}y+c_{3}\tag {10}\\ \eta & =c_{4}x+c_{5}y+c_{6} \tag {11} \end {align}

Then\begin {align} \xi & =x\tag {10}\\ \eta & =-2y \tag {11} \end {align}

Which is what we wanted to show for this ODE. These are the values we used earlier to solve the ODE using symmetry method.

Using functions as ansatz

Now \(\xi ,\eta \) are found using \(\xi =\alpha \left ( x\right ) \) and \(\eta =\beta \left ( x\right ) y+\gamma \left ( x\right ) \) as ansatz. EQ (14) is\begin {equation} \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0 \tag {14} \end {equation} But \[ \eta _{x}=\beta ^{\prime }\left ( x\right ) y+\gamma ^{\prime }\left ( x\right ) \] And\[ \eta _{y}=\beta \left ( x\right ) \] And\begin {align*} \xi _{y} & =0\\ \xi _{x} & =\alpha ^{\prime }\left ( x\right ) \end {align*}

Substituting the above into EQ. (14) gives\[ \beta ^{\prime }\left ( x\right ) y+\gamma ^{\prime }\left ( x\right ) +\omega \left ( \beta \left ( x\right ) -\alpha ^{\prime }\left ( x\right ) \right ) -\omega _{x}\alpha \left ( x\right ) -\omega _{y}\left ( \beta \left ( x\right ) y+\gamma \left ( x\right ) \right ) =0 \] But in this ODE \(\omega =xy^{2}-\frac {2y}{x}-\frac {1}{x^{3}}\), hence \(\omega _{x}=y^{2}+\frac {2y}{x^{2}}+\frac {3}{x^{4}}\) and \(\omega _{y}=2yx-\frac {2}{x}\). The above becomes\[ \beta ^{\prime }y+\gamma ^{\prime }+\left ( xy^{2}-\frac {2y}{x}-\frac {1}{x^{3}}\right ) \left ( \beta -\alpha ^{\prime }\right ) -\left ( y^{2}+\frac {2y}{x^{2}}+\frac {3}{x^{4}}\right ) \alpha -\left ( 2yx-\frac {2}{x}\right ) \left ( \beta y+\gamma \right ) =0 \] Or\[ \gamma ^{\prime }+y\beta ^{\prime }+\frac {2}{x}\gamma -\frac {1}{x^{3}}\beta -\frac {3}{x^{4}}\alpha -y^{2}\alpha +\frac {1}{x^{3}}\alpha ^{\prime }-2xy\gamma -\frac {2}{x^{2}}y\alpha -xy^{2}\beta +\frac {2}{x}y\alpha ^{\prime }-xy^{2}\alpha ^{\prime }=0 \] Collecting on \(y\) gives\[ y^{0}\left ( \gamma ^{\prime }+\frac {2}{x}\gamma -\frac {1}{x^{3}}\beta -\frac {3}{x^{4}}\alpha +\frac {1}{x^{3}}\alpha ^{\prime }\right ) +y\left ( \beta ^{\prime }-2xy\gamma -\frac {2}{x^{2}}\alpha +\frac {2}{x}\alpha ^{\prime }\right ) +y^{2}\left ( -\alpha -x\beta -x\alpha ^{\prime }\right ) =0 \] Each term above is zero. This gives the following equations\begin {align*} \gamma ^{\prime }\left ( x\right ) +\frac {2}{x}\gamma \left ( x\right ) -\frac {1}{x^{3}}\beta \left ( x\right ) -\frac {3}{x^{4}}\alpha \left ( x\right ) +\frac {1}{x^{3}}\alpha ^{\prime }\left ( x\right ) & =0\\ \beta ^{\prime }\left ( x\right ) -2xy\gamma \left ( x\right ) -\frac {2}{x^{2}}\alpha \left ( x\right ) +\frac {2}{x}\alpha ^{\prime }\left ( x\right ) & =0\\ -\alpha \left ( x\right ) -x\beta \left ( x\right ) -x\alpha ^{\prime }\left ( x\right ) & =0 \end {align*}

Solving these coupled ODE on the computer gives\begin {align*} \alpha \left ( x\right ) & =\frac {1}{x}\left ( c_{3}x^{4}+c_{1}x^{2}+c_{2}\right ) \\ \beta \left ( x\right ) & =-4c_{3}x^{2}-2c_{1}\\ \gamma \left ( x\right ) & =-2c_{3}-2\frac {c_{2}}{x^{4}} \end {align*}

Where the \(c_{1},c_{2},c_{3}\) above are constant of integration. Let \(c_{2}=c_{3}=0\). Hence\begin {align*} \alpha \left ( x\right ) & =\frac {1}{x}\left ( c_{3}x^{4}+c_{1}x^{2}\right ) \\ \beta \left ( x\right ) & =-4c_{3}x^{2}-2c_{1}\\ \gamma \left ( x\right ) & =0 \end {align*}

Let \(c_{3}=0\). Hence\begin {align*} \alpha \left ( x\right ) & =\frac {1}{x}c_{1}x^{2}\\ \beta \left ( x\right ) & =-2c_{1}\\ \gamma \left ( x\right ) & =0 \end {align*}

Let \(c_{1}=1\), hence\begin {align*} \alpha \left ( x\right ) & =x\\ \beta \left ( x\right ) & =-2\\ \gamma \left ( x\right ) & =0 \end {align*}

Therefore, since \(\xi =\alpha \left ( x\right ) \) and \(\eta =\beta \left ( x\right ) y+\gamma \left ( x\right ) \) then \(\xi =x,\eta =-2y\) which is the same as the earlier method. After working using this ansatz, I find using the polynomial ansatz better. First of all, I had to set constants above to values in order to obtain the same result as earlier. Setting these constants other values will give different result.  For example, the following are another set of possible solutions obtained from Maple for this ODE\begin {align*} & \left \{ \alpha \left ( x\right ) =\frac {1}{x},\beta \left ( x\right ) =0,\gamma \left ( x\right ) =-\frac {2}{x^{4}}\right \} \\ & \left \{ \alpha \left ( x\right ) =-\frac {x}{2},\beta \left ( x\right ) =1,\gamma \left ( x\right ) =0\right \} \\ & \left \{ \alpha \left ( x\right ) =-\frac {x^{3}}{4},\beta \left ( x\right ) =x^{2},\gamma \left ( x\right ) =\frac {1}{2}\right \} \end {align*}

Which gives\begin {align*} & \left \{ \xi =\frac {1}{x},\eta =-\frac {2}{x^{4}}\right \} \\ & \left \{ \xi =-\frac {x}{2},\eta =y\right \} \\ & \left \{ \xi =\frac {-x^{3}}{4},\eta =x^{2}y+\frac {1}{2}\right \} \end {align*}

1.1.2 Example 2

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

The first step is to find \(\xi \) and \(\eta \). This is shown at the end of this problem below. Let \(x_{1}=x+\varepsilon x^{2}\) which implies \(\xi =x^{2}\). And let \(y_{1}=y+\varepsilon xy\) which implies \(\eta =xy.\)\begin {align*} \xi & =x^{2}\\ \eta & =xy \end {align*}

The next step is to determine what is called the canonical coordinates \(R,S\). Where \(R\) is the independent variable and \(S\) is the dependent variable. So we are looking for \(S\left ( R\right ) \) function. This is done by using the standard characteristic equation by writing\begin {align} \frac {dx}{\xi } & =\frac {dy}{\eta }=dS\nonumber \\ \frac {dx}{x^{2}} & =\frac {dy}{xy}=dS \tag {1} \end {align}

The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x}+\eta \frac {\partial }{\partial y}\right ) S\left ( x,y\right ) =1\). Which is a first order PDE.  We need to solve this for \(S\), which gives (1) using method of characteristic to solve first order PDE which is standard method.  Starting with the first pair of ODE in (1) gives\[ \frac {dy}{dx}=\frac {xy}{x^{2}}=\frac {y}{x}\] Integrating gives \(\frac {y}{x}=c\) where \(c\) is constant of integration. In this method \(R\) is always \(c\). Hence\[ R\left ( x,y\right ) =\frac {y}{x}\] Now we find \(S\left ( x,y\right ) \) from the first equation in (1) and the last equation \begin {align*} dS & =\frac {dx}{\xi }\\ S & =\int \frac {dx}{x^{2}}\\ S & =\frac {-1}{x} \end {align*}

Now that we found \(R\) and \(S\), we determine the ODE \(\frac {dS}{dR}=\Omega \left ( R\right ) \).  The ODE comes out to be function of \(R\) only, so it is quadrature. This is the whole idea of this method. By solving for \(R\) we go back to \(x,y\) and solve for \(y\left ( x\right ) \). How to find \(\frac {dS}{dR}\)? There is an equation to determine this given by\[ \frac {dS}{dR}=\frac {S_{x}+\omega \left ( x,y\right ) S_{y}}{R_{x}+\omega \left ( x,y\right ) R_{y}}\] We know everything on the RHS. Substituting gives\begin {align*} \frac {dS}{dR} & =\frac {\frac {1}{x^{2}}+\left ( \frac {y+1}{x}+\frac {y^{2}}{x^{3}}\right ) \left ( 0\right ) }{-\frac {y}{x^{2}}+\left ( \frac {y+1}{x}+\frac {y^{2}}{x^{3}}\right ) \frac {1}{x}}\\ & =\frac {\frac {1}{x^{2}}}{-\frac {y}{x^{2}}+\left ( \frac {y+1}{x}+\frac {y^{2}}{x^{3}}\right ) \frac {1}{x}}\\ & =\frac {x^{2}}{x^{2}+y^{2}}\\ & =\frac {1}{1+\left ( \frac {y}{x}\right ) ^{2}} \end {align*}

But \(R=\) \(\frac {y}{x}\), hence the above becomes\[ \frac {dS}{dR}=\frac {1}{1+R^{2}}\] This is just quadrature. Integrating gives \[ S=\arctan \left ( R\right ) +c_{1}\] Now we go back to \(x,y\). Since \(S=-\frac {1}{x},R=\frac {y}{x}\), then the above becomes\begin {align*} -\frac {1}{x} & =\arctan \left ( \frac {y}{x}\right ) +c_{1}\\ \frac {-1}{x}+c_{2} & =\arctan \left ( \frac {y}{x}\right ) \\ \frac {y}{x} & =\tan \left ( \frac {-1}{x}+c_{2}\right ) \\ y\left ( x\right ) & =x\tan \left ( \frac {-1}{x}+c_{2}\right ) \end {align*}

And the above is the solution to original ODE.

Finding Lie symmetries for this example

The symmetry condition was derived earlier as\begin {equation} \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0 \tag {14} \end {equation} Let ansatz be \begin {align*} \xi & =c_{1}x+c_{2}y+c_{3}\\ \eta & =c_{4}x+c_{5}y+c_{6} \end {align*}

Eq 9 becomes\[ c_{4}+\omega \left ( c_{5}-c_{1}\right ) -\omega ^{2}c_{2}-\omega _{x}\left ( c_{1}x+c_{2}y+c_{3}\right ) -\omega _{y}\left ( c_{4}x+c_{5}y+c_{6}\right ) =0 \] But in this ODE \(\omega =\frac {y+1}{x}+\frac {y^{2}}{x^{3}}\), hence \(\omega _{x}=-\frac {y+1}{x^{2}}-3\frac {y^{2}}{x^{4}}\) and \(\omega _{y}=\frac {1}{x}+\frac {2y}{x^{3}}\). The above becomes

\begin {align*} c_{4}+\left ( \frac {y+1}{x}+\frac {y^{2}}{x^{3}}\right ) \left ( c_{5}-c_{1}\right ) -\left ( \frac {y+1}{x}+\frac {y^{2}}{x^{3}}\right ) ^{2}c_{2}-\left ( -\frac {y+1}{x^{2}}-3\frac {y^{2}}{x^{4}}\right ) \left ( c_{1}x+c_{2}y+c_{3}\right ) -\left ( \frac {1}{x}+\frac {2y}{x^{3}}\right ) \left ( c_{4}x+c_{5}y+c_{6}\right ) & =0\\ \frac {1}{x^{2}}c_{3}-\frac {1}{x^{2}}c_{2}+\frac {1}{x}c_{5}-\frac {1}{x}c_{6}+\frac {2}{x^{3}}y^{2}c_{1}-\frac {2}{x^{4}}y^{2}c_{2}+\allowbreak \frac {3}{x^{4}}y^{2}c_{3}+\frac {1}{x^{4}}y^{3}c_{2}-\frac {1}{x^{3}}y^{2}c_{5}-\frac {1}{x^{6}}y^{4}c_{2}-\frac {1}{x^{2}}y\allowbreak c_{2}+\frac {1}{x^{2}}yc_{3}-\frac {2}{x^{2}}yc_{4}-\frac {2}{x^{3}}yc_{6} & =0 \end {align*}

Each coefficient to each monomial must be zero. Hence

Hence \(c_{2}=0,c_{6}=0,c_{5}=0\) which implies \(c_{3}=0,c_{1}=0\). Therefore \(-c_{2}+c_{3}-2c_{4}=0\) gives \(c_{4}=0\) also. Hence only trivial solution. This means this ansatz did not work. Trying now\begin {align*} \xi & =c_{1}+c_{2}x+c_{3}y+c_{4}xy+c_{5}x^{2}\\ \eta & =c_{6}+c_{7}x+c_{8}y+c_{9}xy+c_{10}y^{2} \end {align*}

Eq 9 becomes\[ \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0 \] Or\begin {multline*} \left ( c_{7}+c_{9}y\right ) +\omega \left ( c_{8}+c_{9}x+2yc_{10}-\left ( c_{2}+c_{4}y+2xc_{5}\right ) \right ) -\omega ^{2}\left ( c_{3}+c_{4}x\right ) -\\ \omega _{x}\left ( c_{1}+c_{2}x+c_{3}y+c_{4}xy+c_{5}x^{2}\right ) -\omega _{y}\left ( c_{6}+c_{7}x+c_{8}y+c_{9}xy+c_{10}y^{2}\right ) =0 \end {multline*} But in this ODE \(\omega =\frac {y+1}{x}+\frac {y^{2}}{x^{3}}\), hence \(\omega _{x}=-\frac {y+1}{x^{2}}-3\frac {y^{2}}{x^{4}}\) and \(\omega _{y}=\frac {1}{x}+\frac {2y}{x^{3}}\). The above becomes\begin {multline*} \left ( c_{7}+c_{9}y\right ) +\left ( \frac {y+1}{x}+\frac {y^{2}}{x^{3}}\right ) \left ( c_{8}+c_{9}x+2yc_{10}-\left ( c_{2}+c_{4}y+2xc_{5}\right ) \right ) -\left ( \frac {y+1}{x}+\frac {y^{2}}{x^{3}}\right ) ^{2}\left ( c_{3}+c_{4}x\right ) \\ -\left ( -\frac {y+1}{x^{2}}-3\frac {y^{2}}{x^{4}}\right ) \left ( c_{1}+c_{2}x+c_{3}y+c_{4}xy+c_{5}x^{2}\right ) -\left ( \frac {1}{x}+\frac {2y}{x^{3}}\right ) \left ( c_{6}+c_{7}x+c_{8}y+c_{9}xy+c_{10}y^{2}\right ) =0 \end {multline*} Expanding gives\begin {multline*} c_{9}-c_{5}-yc_{5}+yc_{9}+\frac {1}{x^{2}}c_{1}-\frac {1}{x}c_{4}-\frac {1}{x^{2}}c_{3}-\frac {1}{x}c_{6}+\frac {1}{x}c_{8}-\frac {1}{x}y^{2}c_{4}+\frac {2}{x^{3}}y^{2}c_{2}+\frac {3}{x^{4}}y^{2}c_{1}+\frac {1}{x^{2}}y^{2}c_{5}\\ -\frac {2}{x^{3}}y^{2}c_{4}-\frac {2}{x^{4}}y^{2}c_{3}+\frac {1}{x^{4}}y^{3}c_{3}-\frac {1}{x^{2}}y^{2}c_{9}-\frac {1}{x^{3}}y^{2}c_{8}-\frac {1}{x^{5}}y^{4}c_{4}-\frac {1}{x^{6}}y^{4}c_{3}+\frac {1}{x}y^{2}c_{10}+\frac {1}{x^{2}}yc_{1}-\frac {2}{x}yc_{4}-\frac {1}{x^{2}}yc_{3}-\frac {2}{x^{2}}yc_{7}-\frac {2}{x^{3}}yc_{6}+\frac {2}{x}yc_{10}=0 \end {multline*} Each coefficient to each monomial must be zero. Hence

Hence we see that \(c_{3}=0,c_{4}=0,c_{6}=0\). Therefore \(c_{1}=0,c_{8}=0,c_{7}=0,c_{10}=\). We have left the following equation \(c_{9}-c_{5}=0.\) Assuming \(c_{9}=1\) then \(c_{5}=1\). Therefore, since\begin {align*} \xi & =c_{1}+c_{2}x+c_{3}y+c_{4}xy+c_{5}x^{2}\\ \eta & =c_{6}+c_{7}x+c_{8}y+c_{9}xy+c_{10}y^{2} \end {align*}

Then\begin {align*} \xi & =x^{2}\\ \eta & =xy \end {align*}

Which is what we wanted to show for this ODE. These are the values used earlier to solve the ODE using symmetry method. This shows that this method requires large effort to do by hand.

1.1.3 Example 3

Solve \begin {align*} y^{\prime } & =\frac {y-4xy^{2}-16x^{3}}{y^{3}+4x^{2}y+x}\\ y^{\prime } & =\omega \left ( x,y\right ) \end {align*}

The first step is to find \(\xi \) and \(\eta \). This is shown at the end of this problem below. Let \(x_{1}=x-\varepsilon y\) which implies \(\xi =-y\). And let \(y_{1}=y+4\varepsilon x\) which implies \(\eta =4x.\)\begin {align*} \xi & =-y\\ \eta & =4x \end {align*}

The next step is to determine what is called the canonical coordinates \(R,S\). Where \(R\) is the independent variable and \(S\) is the dependent variable. So we are looking for \(S\left ( R\right ) \) function. This is done by using the standard characteristic equation by writing\begin {align} \frac {dx}{\xi } & =\frac {dy}{\eta }=dS\nonumber \\ \frac {dx}{-y} & =\frac {dy}{4x}=dS \tag {1} \end {align}

The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x}+\eta \frac {\partial }{\partial y}\right ) S\left ( x,y\right ) =1\). Which is a first order PDE.  We need to solve this for \(S\), which gives (1) using method of characteristic to solve first order PDE which is standard method.  Starting with the first pair of ODE in (1) gives\[ \frac {dy}{dx}=-\frac {4x}{y}\] Integrating gives \(y=\sqrt {-4x^{2}+c^{2}}\) where \(c^{2}\) is constant of integration (For \(y>0\) only). In this method \(R\) is always \(c\). Hence\begin {align} y^{2} & =-4x^{2}+R^{2}\nonumber \\ R^{2} & =y^{2}+4x^{2}\nonumber \\ R & =\sqrt {y^{2}+4x^{2}} \tag {2} \end {align}

Now we find \(S\left ( x,y\right ) \) from the first equation in (1) and the last equation \begin {align*} dS & =\frac {dx}{\xi }\\ S & =-\int \frac {dx}{y} \end {align*}

But \(y=\sqrt {R^{2}-4x^{2}}\). The above becomes\begin {align*} S & =-\int \frac {dx}{\sqrt {R^{2}-4x^{2}}}\\ & =-\frac {1}{2}\arctan \left ( \frac {2x}{\sqrt {R^{2}-4x^{2}}}\right ) \\ & =-\frac {1}{2}\arctan \left ( \frac {2x}{\sqrt {\left ( y^{2}+4x^{2}\right ) +4x^{2}}}\right ) \\ & =-\frac {1}{2}\arctan \left ( \frac {2x}{y}\right ) \end {align*}

For \(y>0\). Now that we found \(R\) and \(S\), we determine the ODE \(\frac {dS}{dR}=\Omega \left ( R\right ) \).  The ODE comes out to be function of \(R\) only, so it is quadrature. This is the whole idea of this method. By solving for \(R\) we go back to \(x,y\) and solve for \(y\left ( x\right ) \). How to find \(\frac {dS}{dR}\)? There is an equation to determine this given by\[ \frac {dS}{dR}=\frac {S_{x}+\omega \left ( x,y\right ) S_{y}}{R_{x}+\omega \left ( x,y\right ) R_{y}}\] We know everything on the RHS. Substituting gives\begin {align*} \frac {dS}{dR} & =\frac {\frac {d}{dx}\left ( -\frac {1}{2}\arctan \left ( \frac {2x}{y}\right ) \right ) +\left ( \frac {y-4xy^{2}-16x^{3}}{y^{3}+4x^{2}y+x}\right ) \frac {d}{dy}\left ( -\frac {1}{2}\arctan \left ( \frac {2x}{y}\right ) \right ) }{\frac {d}{dx}\sqrt {y^{2}+4x^{2}}+\left ( \frac {y-4xy^{2}-16x^{3}}{y^{3}+4x^{2}y+x}\right ) \frac {d}{dy}\sqrt {y^{2}+4x^{2}}}\\ & =\frac {\frac {-1}{y\left ( \frac {4x^{2}}{y^{2}}+1\right ) }+\left ( \frac {y-4xy^{2}-16x^{3}}{y^{3}+4x^{2}y+x}\right ) \frac {x}{y^{2}\left ( \frac {4x^{2}}{y^{2}}+1\right ) }}{\frac {4x}{\sqrt {y^{2}+4x^{2}}}+\left ( \frac {y-4xy^{2}-16x^{3}}{y^{3}+4x^{2}y+x}\right ) \frac {y}{\sqrt {y^{2}+4x^{2}}}}\\ & =-\sqrt {4x^{2}+y^{2}}\\ & =-R \end {align*}

Hence\[ \frac {dS}{dR}=-R \] This is just quadrature. Integrating gives \[ S=-\frac {R^{2}}{2}+c \] Now we go back to \(x,y\). Since \(S=-\frac {1}{2}\arctan \left ( \frac {2x}{y}\right ) ,R=\sqrt {y^{2}+4x^{2}}\), then the above becomes\begin {align*} -\frac {1}{2}\arctan \left ( \frac {2x}{y}\right ) & =-\left ( \frac {y^{2}+4x^{2}}{2}\right ) +c\\ \frac {y^{2}}{2}-\frac {1}{2}\arctan \left ( \frac {2x}{y}\right ) +2x^{2}-c & =0\qquad y>0 \end {align*}

And the above is the solution to original ODE.

Finding Lie symmetries for this example

The symmetry condition was derived earlier as\begin {equation} \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0 \tag {14} \end {equation} Let ansatz be \begin {align*} \xi & =c_{1}x+c_{2}y+c_{3}\\ \eta & =c_{4}x+c_{5}y+c_{6} \end {align*}

Eq 9 becomes\[ c_{4}+\omega \left ( c_{5}-c_{1}\right ) -\omega ^{2}c_{2}-\omega _{x}\left ( c_{1}x+c_{2}y+c_{3}\right ) -\omega _{y}\left ( c_{4}x+c_{5}y+c_{6}\right ) =0 \] But in this ODE \(\omega =\frac {y-4xy^{2}-16x^{3}}{y^{3}+4x^{2}y+x}\), hence \(\omega _{x}=\frac {-4y^{5}-32x^{2}y^{3}-8xy^{2}+\left ( -64x^{4}-1\right ) y-32x^{3}}{\left ( 4x^{2}y+y^{3}+x\right ) ^{2}}\) and \(\omega _{y}=\frac {64x^{5}+32x^{3}y^{2}+4xy^{4}-8x^{2}y-2y^{3}+x}{\left ( 4x^{2}y+y^{3}+x\right ) ^{2}}\). Above becomes

\begin {multline*} c_{4}+\left ( \frac {y-4xy^{2}-16x^{3}}{y^{3}+4x^{2}y+x}\right ) \left ( c_{5}-c_{1}\right ) -\left ( \frac {y-4xy^{2}-16x^{3}}{y^{3}+4x^{2}y+x}\right ) ^{2}c_{2}-\left ( \frac {-4y^{5}-32x^{2}y^{3}-8xy^{2}+\left ( -64x^{4}-1\right ) y-32x^{3}}{\left ( 4x^{2}y+y^{3}+x\right ) ^{2}}\right ) \left ( c_{1}x+c_{2}y+c_{3}\right ) \\ -\left ( \frac {64x^{5}+32x^{3}y^{2}+4xy^{4}-8x^{2}y-2y^{3}+x}{\left ( 4x^{2}y+y^{3}+x\right ) ^{2}}\right ) \left ( c_{4}x+c_{5}y+c_{6}\right ) =0 \end {multline*}

Which expands to

\begin {multline*} \frac {8c_{1}xy^{2}}{4x^{2}y+y^{3}+x}+\frac {4c_{5}xy^{2}}{4x^{2}y+y^{3}+x}-\frac {256c_{2}x^{4}y^{2}}{\left ( 4x^{2}y+y^{3}+x\right ) ^{2}}-\frac {48c_{2}x^{2}y^{4}}{\left ( 4x^{2}y+y^{3}+x\right ) ^{2}}+\frac {16c_{2}x^{3}y}{\left ( 4x^{2}y+y^{3}+x\right ) ^{2}}+\frac {12c_{2}xy^{3}}{\left ( 4x^{2}y+y^{3}+x\right ) ^{2}}\\ +\frac {48x^{2}c_{2}y}{4x^{2}y+y^{3}+x}-\frac {128x^{5}yc_{1}}{\left ( 4x^{2}y+y^{3}+x\right ) ^{2}}-\frac {128x^{4}yc_{3}}{\left ( 4x^{2}y+y^{3}+x\right ) ^{2}}-\frac {32x^{3}y^{3}c_{1}}{\left ( 4x^{2}y+y^{3}+x\right ) ^{2}}-\frac {32x^{2}y^{3}c_{3}}{\left ( 4x^{2}y+y^{3}+x\right ) ^{2}}\\ +\frac {4x^{2}y^{2}c_{1}}{\left ( 4x^{2}y+y^{3}+x\right ) ^{2}}+\frac {4xy^{2}c_{3}}{\left ( 4x^{2}y+y^{3}+x\right ) ^{2}}+\frac {yc_{1}x}{\left ( 4x^{2}y+y^{3}+x\right ) ^{2}}+\frac {8x^{2}yc_{4}}{4x^{2}y+y^{3}+x}+\frac {8xyc_{6}}{4x^{2}y+y^{3}+x}-\\ \frac {64x^{5}c_{5}y}{\left ( 4x^{2}y+y^{3}+x\right ) ^{2}}-\frac {64x^{4}y^{2}c_{4}}{\left ( 4x^{2}y+y^{3}+x\right ) ^{2}}-\frac {64x^{3}y^{3}c_{5}}{\left ( 4x^{2}y+y^{3}+x\right ) ^{2}}-\frac {64x^{3}y^{2}c_{6}}{\left ( 4x^{2}y+y^{3}+x\right ) ^{2}}-\frac {12x^{2}y^{4}c_{4}}{\left ( 4x^{2}y+y^{3}+x\right ) ^{2}}-\frac {16c_{5}x^{3}}{4x^{2}y+y^{3}+x}\\ -\frac {256c_{2}x^{6}}{\left ( 4x^{2}y+y^{3}+x\right ) ^{2}}+\frac {64c_{1}x^{3}}{4x^{2}y+y^{3}+x}-\frac {c_{1}y}{4x^{2}y+y^{3}+x}+\frac {48x^{2}c_{3}}{4x^{2}y+y^{3}+x}+\frac {4y^{3}c_{2}}{4x^{2}y+y^{3}+x}+\frac {4y^{2}c_{3}}{4x^{2}y+y^{3}+x}\\ -\frac {16x^{4}c_{1}}{\left ( 4x^{2}y+y^{3}+x\right ) ^{2}}-\frac {16x^{3}c_{3}}{\left ( 4x^{2}y+y^{3}+x\right ) ^{2}}+\frac {yc_{3}}{\left ( 4x^{2}y+y^{3}+x\right ) ^{2}}-\frac {c_{4}x}{4x^{2}y+y^{3}+x}-\frac {64x^{6}c_{4}}{\left ( 4x^{2}y+y^{3}+x\right ) ^{2}}-\frac {64x^{5}c_{6}}{\left ( 4x^{2}y+y^{3}+x\right ) ^{2}}\\ +\frac {3y^{4}c_{5}}{\left ( 4x^{2}y+y^{3}+x\right ) ^{2}}+\frac {3y^{3}c_{6}}{\left ( 4x^{2}y+y^{3}+x\right ) ^{2}}-\frac {c_{6}}{4x^{2}y+y^{3}+x}-\frac {12xy^{5}c_{5}}{\left ( 4x^{2}y+y^{3}+x\right ) ^{2}}-\frac {12xy^{4}c_{6}}{\left ( 4x^{2}y+y^{3}+x\right ) ^{2}}+\frac {4x^{3}yc_{4}}{\left ( 4x^{2}y+y^{3}+x\right ) ^{2}}\\ +\frac {4x^{2}y^{2}c_{5}}{\left ( 4x^{2}y+y^{3}+x\right ) ^{2}}+\frac {4x^{2}yc_{6}}{\left ( 4x^{2}y+y^{3}+x\right ) ^{2}}+\frac {3y^{3}c_{4}x}{\left ( 4x^{2}y+y^{3}+x\right ) ^{2}}+c_{4}=0 \end {multline*}

Multiplying each term by \(\left ( 4x^{2}y+y^{3}+x\right ) ^{2}\ \) and expanding gives the multivariable polynomial

\begin {multline*} 128x^{5}yc_{1}+64x^{3}y^{3}c_{1}+8c_{1}xy^{5}-256c_{2}x^{6}-64c_{2}x^{4}y^{2}+16c_{2}x^{2}y^{4}+4c_{2}y^{6}-64x^{6}c_{4}-16x^{4}y^{2}c_{4}+4x^{2}y^{4}c_{4}+c_{4}y^{6}\\ -128x^{5}c_{5}y-64x^{3}y^{3}c_{5}-8xy^{5}c_{5}+64x^{4}yc_{3}+32x^{2}y^{3}c_{3}+4c_{3}y^{5}-64x^{5}c_{6}-32x^{3}y^{2}c_{6}-4xy^{4}c_{6}+48x^{4}c_{1}+\\ 8x^{2}y^{2}c_{1}-c_{1}y^{4}+64c_{2}x^{3}y+16c_{2}xy^{3}+16x^{3}yc_{4}+4y^{3}c_{4}x-16c_{5}x^{4}+8x^{2}y^{2}c_{5}+3y^{4}c_{5}+32x^{3}c_{3}+8xy^{2}c_{3}+8x^{2}yc_{6}+2y^{3}c_{6}+yc_{3}-c_{6}x=0 \end {multline*}

Each monomial coefficient must be zero. This gives the following equations to solve for \(c_{i}\)

Hence we see that \(c_{6}=0,c_{3}=0\). The above reduces to

Hence \(Ac=b\) gives\[\begin {pmatrix} 0 & -256 & -64 & 0\\ 128 & 0 & 0 & -128\\ 0 & -64 & -16 & 0\\ 48 & 0 & 0 & -16\\ 64 & 0 & 0 & -64\\ 0 & 64 & 16 & 0\\ 0 & 16 & 4 & 0\\ 8 & 0 & 0 & -8\\ 0 & 16 & 4 & 0\\ 0 & 4 & 1 & 0\\ -1 & 0 & 0 & 3 \end {pmatrix}\begin {pmatrix} c_{1}\\ c_{2}\\ c_{4}\\ c_{5}\end {pmatrix} =\begin {pmatrix} 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0 \end {pmatrix} \] The rank of \(A\) is \(3\) and the number of columns is \(4\). Hence non-trivial solution exist. Solving the above gives \(c_{4}=-4\) and \(c_{2}=1\) and all other coefficients are zero. this means that , since\begin {align*} \xi & =c_{1}x+c_{2}y+c_{3}\\ \eta & =c_{4}x+c_{5}y+c_{6} \end {align*}

Then\begin {align*} \xi & =y\\ \eta & =-4x \end {align*}

Which is what we wanted to show for this ODE.

1.2 Notes, things to find out

1. Given an ODE \(y^{\prime }\left ( x\right ) =\omega \left ( x,y\right ) \) then we want to find nontrivial Lie symmetry. The condition for this is that \[ \eta \left ( x,y\right ) \neq \xi \left ( x,y\right ) \omega \left ( x,y\right ) \] so any values for \(\eta ,\xi \) must satisfies the above.
2. Can we always find \(\xi ,\eta \) for non-trivial symmetry for first order ODE?  When I tried some in Maple, it could not find symmetries for some first order ODE’s. How does one check if nontrivial symmetry exist before trying to find one?

2 References

1. Symmetry Methods for Differential Equations. By PETER E. HYDON. CAMBRIDGE UNIVERSITY PRESS. 2000.
2. Many other references, listed in my links page under computer algebra section.