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## Notes on using Lie symmetry group for solving ODE’s

October 20, 2020   Compiled on October 20, 2020 at 4:30pm

1.1 Examples
1.1.1 Example 1
1.1.2 Example 2
1.1.3 Example 3

### 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 (1) is linearized by expanding the LHS and the RHS using Taylor series, by expanding around $$\varepsilon =0$$. Starting wit 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 linearlized 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.

#### 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

 monomial equation $$x^{0}y^{0}$$ $$3c_{4}=0$$ $$\frac{1}{x}$$ $$2c_{6}=0$$ $$\frac{1}{x^{4}}$$ $$-c_{3}=0$$ $$y^{2}$$ $$-c_{3}=0$$ $$y^{3}$$ $$3c_{2}=0$$ $$\frac{1}{x^{3}}$$ $$-c_{5}=0$$ $$\frac{1}{x^{6}}$$ $$-9c_{2}=0$$ $$\frac{y^{2}}{x^{2}}$$ $$-4c_{2}=0$$ $$x^{2}y^{4}$$ $$-c_{2}=0$$ $$xy$$ $$-2c_{6}=0$$ $$xy^{2}$$ $$-2c_{1}-c_{5}=0$$ $$\frac{y}{x^{2}}$$ $$-2c_{3}=0$$ $$x^{2}y$$ $$-2c_{4}=0$$ $$\frac{y}{x^{4}}$$ $$-13c_{2}=0$$ $$xy^{2}$$ $$-c_{5}=0$$

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

 monomial equation $$\frac{1}{x^{2}}$$ $$c_{3}-c_{2}=0$$ $$\frac{1}{x}$$ $$c_{5}-c_{6}=0$$ $$\frac{y^{2}}{x^{3}}$$ $$2c_{1}-c_{5}=0$$ $$\frac{y^{2}}{x^{4}}$$ $$-2c_{2}+3c_{3}=0$$ $$\frac{y^{3}}{x^{4}}$$ $$c_{2}=0$$ $$\frac{y^{2}}{x^{3}}$$ $$-c_{5}=0$$ $$\frac{y^{4}}{x^{6}}$$ $$-c_{2}=0$$ $$\frac{y}{x^{2}}$$ $$-c_{2}+c_{3}-2c_{4}=0$$ $$\frac{y}{x^{3}}$$ $$-2c_{6}=0$$

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

 monomial equation $$x^{0}y^{0}$$ $$c_{9}-c_{5}=0$$ $$y$$ $$-c_{5}+c_{9}=0$$ $$\frac{1}{x^{2}}$$ $$c_{1}-c_{3}=0$$ $$\frac{1}{x}$$ $$-c_{4}-c_{6}+c_{8}=0$$ $$\frac{y^{2}}{x}$$ $$-c_{4}+c_{10}=0$$ $$\frac{y^{2}}{x^{3}}$$ $$2c_{2}-2c_{4}-c_{8}=0$$ $$\frac{y^{2}}{x^{4}}$$ $$3c_{1}-2c_{3}=0$$ $$\frac{y^{2}}{x^{2}}$$ $$c_{5}-c_{9}=0$$ $$\frac{y^{3}}{x^{4}}$$ $$c_{3}=0$$ $$\frac{y^{4}}{x^{5}}$$ $$-c_{4}=0$$ $$\frac{y^{4}}{x^{6}}$$ $$-c_{3}=0$$ $$\frac{y}{x^{2}}$$ $$c_{1}-c_{3}-2c_{7}=0$$ $$\frac{y}{x^{3}}$$ $$2c_{6}=0$$ $$\frac{y}{x}$$ $$-2c_{4}+2c_{10}=0$$

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 $$c_{i}$$

 equation $$-256 c_{2} -64 c_{4}=0$$ $$128 c_{1} -128 c_{5}=0$$ $$-64 c_{6}=0$$ $$-64 c_{2} -16 c_{4}=0$$ $$64 c_{3}=0$$ $$48 c_{1} -16 c_{5}=0$$ $$64 c_{1} -64 c_{5}=0$$ $$-32 c_{6}=0$$ $$64 c_{2} +16 c_{4}=0$$ $$32 c_{3}=0$$ $$16 c_{2} +4 c_{4}=0$$ $$32 c_{3}=0$$ $$8 c_{1} +8 c_{5}=0$$ $$8 c_{6}=0$$ $$8 c_{1} -8 c_{5}=0$$ $$-4 c_{6}=0$$ $$16 c_{2} +4 c_{4}=0$$ $$8 c_{3}=0$$ $$-c_{6}=0$$ $$4 c_{2} +c_{4}=0$$ $$4 c_{3}=0$$ $$-c_{1} +3 c_{5}=0$$ $$2 c_{6}=0$$ $$c_{3}=0$$

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

 equation $$-256c_{2}-64c_{4}=0$$ $$128c_{1}-128c_{5}=0$$ $$-64c_{2}-16c_{4}=0$$ $$48c_{1}-16c_{5}=0$$ $$64c_{1}-64c_{5}=0$$ $$64c_{2}+16c_{4}=0$$ $$16c_{2}+4c_{4}=0$$ $$8c_{1}+8c_{5}=0$$ $$8c_{1}-8c_{5}=0$$ $$16c_{2}+4c_{4}=0$$ $$4c_{2}+c_{4}=0$$ $$-c_{1}+3c_{5}=0$$

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.