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## Study Notes On Using Lie Symmetry For Solving Diﬀerential Equations

January 30, 2024   Compiled on January 30, 2024 at 3:54am

### 1 First order ODE

#### 1.1 Terminology used and high level introduction

1. $$x,y$$ are the natural coordinates used in the input ode $$\frac {dy}{dx}=\omega \left ( x,y\right )$$.
2. $$\bar {x},\bar {y}$$ are called the Lie group (local) transformation coordinates. The ode remains invariant (same shape) when written in $$\bar {x},\bar {y}$$. The coordinates $$R,S$$ (some books use lower case $$r,s)$$ are called the canonical coordinates in which the input ode becomes a quadrature and therefore easily solved by just integration.
3. $$\xi ,\eta$$ are called the Lie inﬁnitesimals. $$\xi \left ( x,y\right ) ,\eta \left ( x,y\right )$$ can be calculated knowing $$\bar {x},\bar {y}$$. Also $$\bar {x},\bar {y}$$ can be calculated given $$\xi ,\eta$$. It is $$\xi ,\eta$$ which are the most important quantities that need to be determined in order to ﬁnd the canonical coordinates $$R,S$$. These quantities are called the tangent vectors. These specify how the orbit moves. The orbit is the path the point $$\left ( x,y\right )$$ point travels on as it move toward $$\bar {x},\bar {y}$$. The tangent vectors $$\xi ,\eta$$ are calculated at $$\epsilon =0$$. The point $$\bar {x}=x+\xi \epsilon$$ and the point $$\bar {y}=y+\eta \epsilon$$.
4. The ultimate goal is write $$\frac {dy}{dx}=\omega \left ( x,y\right )$$ in $$R,S$$ coordinates where it is solved by integration only as it will have the form $$\frac {dS}{dR}=F\left ( R\right )$$. The right hand side should always be a function of $$R$$ only in canonical coordinates.
5. $$\bar {x},\bar {y}$$ can be calculated knowing the canonical coordinates $$R,S$$.
6. The ideal transformation has the form $$\left ( \bar {x},\bar {y}\right ) \rightarrow \left ( x,y+\epsilon \right )$$ because with this transformation the ode becomes quadrature in the transformed coordinates. But because not all ode’s have this transformation available, the ode is transformed to canonical coordinates $$\left ( R,S\right )$$ where the transformation $$\left ( \bar {R},\bar {S}\right ) \rightarrow \left ( R,S+\epsilon \right )$$ can be used.
7. The main goal of Lie symmetry method is to determine $$S,R$$. To be able to do this, the quantities $$\xi ,\eta$$ must be determined ﬁrst.
8. The remarkable thing about this method, is that regardless of how complicated the original ode $$\frac {dy}{dx}=\omega \left ( x,y\right )$$ is, if the similarity condition PDE can be solved for $$\xi ,\eta$$, then $$R,S$$ are found and the ode becomes quadrature $$\frac {dS}{dR}=F\left ( R\right )$$. The ode is then solved in canonical coordinates and the solution transformed back to $$x,y$$.
9. The quantity $$\epsilon$$ is called the Lie parameter. This is a real quantity which as it goes to zero, gives the identity transformation. In other words, when $$\epsilon =0$$ then $$\left ( x,y\right ) =\left ( \bar {x},\bar {y}\right )$$.
10. But there is no free lunch, even in Mathematics. The problem comes down to ﬁnding $$\xi ,\eta$$. This requires solving a PDE. This is done using ansatz and trial and error. This reason possibly explains why the Lie symmetry method have not become standard in textbooks for solving ODE’s as the algebra and computation needed to ﬁnd $$\xi ,\eta$$ from the PDE becomes very complex to do by hand.
11. Total derivative operator: Given $$f\left ( x,y\right )$$ then $$\frac {df}{dx}=\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}\frac {dy}{dx}$$ where it is assumed that $$y\left ( x\right )$$ depends on $$x$$. Total derivative operator will be used extensively in all the derivatiations below, so good to practice this. It is written as $$D_{x}=\partial _{x}+\partial _{y}y^{\prime }$$ for ﬁrst order ode, and as $$D_{x}=\partial _{x}+\partial _{y}y^{\prime }+\partial _{y^{\prime }}y^{\prime \prime }$$ for second order ode and as $$D_{x}=\partial _{x}+\partial _{y}y^{\prime }+\partial _{y^{\prime }}y^{\prime \prime }+\partial _{y^{\prime \prime }}y^{\prime \prime \prime }$$ for third order ode and so on.
12. The notation $$f_{x}$$ means partial derivative. Hence $$\frac {\partial f}{\partial x}$$ is written as $$f_{x}$$. Total derivative will always be written as $$\frac {df}{dx}$$. It is important to distinguish between these two as the algebra will get messy with Lie symmetry. Sometimes we write $$f^{\prime }$$ to mean $$\frac {df}{dx}$$ but it is better to avoid $$f^{\prime }$$ and just write $$\frac {df}{dx}$$ when $$f$$ is function of more than one variable.
13. Given ﬁrst ode $$\frac {dy}{dx}=\omega \left ( x,y\right )$$, where $$\bar {y}\equiv \bar {y}\left ( x,y\right )$$ and $$\bar {x}\equiv \bar {x}\left ( x,y\right )$$ then then$$\frac {d\bar {y}}{d\bar {x}}$$ is given by the following (using the total derivative operator)\begin {align*} \frac {d\bar {y}}{d\bar {x}} & =\frac {D_{x}\bar {y}}{D_{x}\bar {x}}\\ & =\frac {\bar {y}_{x}+\bar {y}_{y}y^{\prime }}{\bar {x}_{x}+\bar {x}_{y}y^{\prime }}\\ & =\frac {\bar {y}_{x}+\bar {y}_{y}\omega }{\bar {x}_{x}+\bar {x}_{y}\omega } \end {align*}
14. Given second order ode $$\frac {d^{2}y}{dx^{2}}=\omega \left ( x,y,y^{\prime }\right )$$ where $$\bar {y}\equiv \bar {y}\left ( x,y,y^{\prime }\right )$$ and $$\bar {x}\equiv \bar {x}\left ( x,y,y^{\prime }\right )$$ then $$\frac {d^{2}\bar {y}}{d\bar {x}^{2}}$$ is given by\begin {align*} \frac {d^{2}\bar {y}}{d\bar {x}^{2}} & =\frac {D_{x}\frac {d\bar {y}}{d\bar {x}}}{D_{x}\bar {x}}\\ & =\frac {\bar {y}_{x}^{\prime }+\bar {y}_{y}^{\prime }y^{\prime }+\bar {y}_{y^{\prime }}^{\prime }y^{\prime \prime }}{\bar {x}_{x}^{\prime }+\bar {x}_{y}^{\prime }y^{\prime }} \end {align*}

To simplify notation we used $$\bar {y}^{\prime }$$ for $$\frac {d\bar {y}}{d\bar {x}}$$ above. The above simpliﬁes to$\frac {d^{2}\bar {y}}{d\bar {x}^{2}}=\frac {\bar {y}_{x}^{\prime }+\bar {y}_{y}^{\prime }y^{\prime }+\bar {y}_{y^{\prime }}^{\prime }\omega }{\bar {x}_{x}^{\prime }+\bar {x}_{y}^{\prime }y^{\prime }}$ Keeping in mind that $$\left ( \circ \right ) _{x}$$ or $$\left ( \circ \right ) _{y}$$ mean partial derivative.

15. Given third order ode $$\frac {d^{3}y}{dx^{3}}=\omega \left ( x,y,y^{\prime },y^{\prime \prime }\right )$$ where $$\bar {y}\equiv \bar {y}\left ( x,y,y^{\prime },y^{\prime \prime }\right )$$ and $$\bar {x}\equiv \bar {x}\left ( x,y,y^{\prime },y^{\prime }\right )$$ then $$\frac {d^{3}\bar {y}}{d\bar {x}^{3}}$$ is given by \begin {align*} \frac {d^{3}\bar {y}}{d\bar {x}^{3}} & =\frac {D_{x}\frac {d^{2}\bar {y}}{d\bar {x}^{2}}}{D_{x}\bar {x}}\\ & =\frac {\bar {y}_{x}^{^{\prime \prime }}+\bar {y}_{y}^{\prime \prime }y^{\prime }+\bar {y}_{y^{\prime }}^{\prime \prime }y^{\prime \prime }+\bar {y}_{y^{\prime \prime }}^{\prime \prime }y^{\prime \prime \prime }}{\bar {x}_{x}^{\prime }+\bar {x}_{y}^{\prime }y^{\prime }}\\ & =\frac {\bar {y}_{x}^{^{\prime \prime }}+\bar {y}_{y}^{\prime \prime }y^{\prime }+\bar {y}_{y^{\prime }}^{\prime \prime }y^{\prime \prime }+\bar {y}_{y^{\prime \prime }}^{\prime \prime }\omega }{\bar {x}_{x}^{\prime }+\bar {x}_{y}^{\prime }y^{\prime }} \end {align*}

To simplify notation we used $$\bar {y}^{\prime \prime }$$ for $$\frac {d^{2}\bar {y}}{d\bar {x}^{2}}$$ above. And so on for higher order ode’s.

#### 1.2 Introduction

Given any ﬁrst order ODE \begin {equation} \frac {dy}{dx}=\omega \left ( x,y\right ) \tag {A} \end {equation} The ﬁrst goal is to ﬁnd a one parameter invariant Lie group transformation that keeps the ode invariant. The Lie parameter the transformation depends on is called $$\epsilon$$. This means ﬁnding transformation of $$\left ( x,y\right )$$ to new coordinates $$\left ( \bar {x},\bar {y}\right )$$ that keeps the ode the same form when written using $$\bar {x},\bar {y}$$.

This view looks at the transformation on the ode itself. Another view is to look at the family of the solution curves of the ode instead. Looking at solution curves transformation is geometrical in nature and can lead to more insight.

What does the transformation mean when looking at solution curves instead of the ODE itself? It is the mapping of a point $$\left ( x,y\right )$$ on one solution curve to another point $$\left ( \bar {x},\bar {y}\right )$$ on another solution curve. If the mapping sends point $$\left ( x,y\right )$$ to another point $$\left ( \bar {x},\bar {y}\right )$$ on the same solution curve, then it is called a trivial mapping or trivial transformation.

As an example, given the ode $$y^{\prime }=0$$, this has solutions $$y=c_{1}$$. For any constant $$c_{1}$$ there is a solution curve. There are inﬁnite number of solution curves. All solution curves are horizontal lines. The mapping $$\left ( x,y\right ) \rightarrow \left ( x+\epsilon ,y\right )$$ is trivial transformation as it moves the point $$\left ( x,y\right )$$ to another point $$\left ( \bar {x},\bar {y}\right )$$ on the same solution curve.

The transformation $$\left ( x,y\right ) \rightarrow \left ( x,y+\epsilon \right )$$ however is non trivial as it moves the point $$\left ( x,y\right )$$ to point $$\left ( \bar {x},\bar {y}\right )$$ on another solution curve. Here $$\bar {x}=x$$ and $$\bar {y}=y+\epsilon$$. This can also be written $$\left ( x,y\right ) \rightarrow \left ( x,e^{\epsilon }y\right )$$ which is the preferred way.

The transformation $$\left ( x,y\right ) \rightarrow \left ( x+\epsilon ,y+\epsilon \right )$$ is non trivial for this ode. The simplest non trivial transformation that map all points on one solution curve to another solution curve is selected. In canonical coordinates the transformation used has the form $$\left ( R,S\right ) \rightarrow \left ( R,S+\epsilon \right )$$.

Another example is $$y^{\prime }=y$$. This has solution curves given by $$y=ce^{x}$$. This is a plot showing two such curves for diﬀerent $$c$$ values.

The above shows that a non trivial transformation is given by $$\bar {x}=x+\epsilon ,\bar {y}=y$$. This can be found analytically by solving the symmetry condition as will be illustrated below using examples. For this case, the tangent vectors are $$\xi =\left . \frac {\partial \bar {x}}{\partial \epsilon }\right \vert _{\epsilon =0}=1$$ and $$\eta =\left . \frac {\partial \bar {y}}{\partial \epsilon }\right \vert _{\epsilon =0}=0$$. In Maple this is found using

ode:=diff(y(x),x)=y(x);
DEtools:-symgen(ode)
[_xi = 1, _eta = 0]



But the following transformation $$\bar {x}=x,\bar {y}=y+\epsilon$$ does not work

This is because it does not leave the ode invariant because $$\frac {d\bar {y}}{d\bar {x}}=\bar {y}$$ becomes $$\frac {\bar {y}_{x}+\bar {y}_{y}y^{\prime }}{\bar {x}_{x}+\bar {x}_{y}y^{\prime }}=\bar {y}$$, where now $$\bar {y}_{x}=0,\bar {y}_{y}=1,\bar {x}_{x}=1,\bar {x}_{y}=0,\bar {y}=y+\epsilon$$, and hence $$\frac {\bar {y}_{x}+\bar {y}_{y}y^{\prime }}{\bar {x}_{x}+\bar {x}_{y}y^{\prime }}=\bar {y}$$ simpliﬁes to $$y^{\prime }=y+\epsilon$$ which is not the same ode. This shows that $$\bar {x}=x,\bar {y}=y+\epsilon$$ is not valid Lie point symmetry.

However $$\bar {x}=x+\epsilon ,\bar {y}=y$$ leaves the ODE invariant. In this case $$\bar {y}_{x}=0,\bar {y}_{y}=1,\bar {x}_{x}=1,\bar {x}_{y}=0,\bar {y}=y$$ and hence $$\frac {\bar {y}_{x}+\bar {y}_{y}y^{\prime }}{\bar {x}_{x}+\bar {x}_{y}y^{\prime }}=\bar {y}$$ becomes $$y^{\prime }=y$$ which is the same ode.

The transformation must keep the ode invariant as this is the main deﬁnition of symmetry transformation.

In the above, the path the point $$\left ( x,y\right )$$ travels over as it moves to $$\left ( \bar {x},\bar {y}\right )$$ as $$\epsilon$$ changes is called the orbit. Each point $$\left ( x,y\right )$$ travels on its orbit during transformation.

In all such transformations, there is a parameter $$\epsilon$$ that the transformation depends on. This is why this is called the Lie one parameter symmetry transformation group. There are inﬁnite number of such transformations.

Lie symmetry is called point symmetry, because of the above. It transforms points from an ODE solution curves to points on another solution curves for the same ODE. The identity transformation is when $$\epsilon =0$$, since then the point is transformed to itself.

An example using an ODE. The Clairaut ode of the form $$y=xf\left ( p\right ) +g\left ( p\right )$$ where $$p\equiv y^{\prime }$$. \begin {align} x\left ( y^{\prime }\right ) ^{2}-yy^{\prime }+m & =0\tag {1}\\ y & =x\frac {\left ( y^{\prime }\right ) ^{2}}{m}+y\frac {y^{\prime }}{m}\nonumber \end {align}

Where $$f\left ( p\right ) =\frac {\left ( y^{\prime }\right ) ^{2}}{m}$$ and $$g\left ( p\right ) =\frac {y^{\prime }}{m}$$. Using the dilation transformation Lie group\begin {align} \bar {x} & \equiv \bar {x}\left ( x,y;\epsilon \right ) =e^{2\epsilon }x\tag {2}\\ \bar {y} & \equiv \bar {y}\left ( x,y;\epsilon \right ) =e^{\epsilon }y \tag {3} \end {align}

Eq. (1) is now expressed in the new coordinates $$\bar {x},\bar {y}\,$$. If this results in same same ode form but written in $$\bar {x},\bar {y}\,$$ then the transformation is invariant. But how to ﬁnd $$\frac {d\bar {y}}{d\bar {x}}$$ ? This is done as follows\begin {align*} \frac {d\bar {y}}{d\bar {x}} & =\frac {\frac {d\bar {y}}{dx}}{\frac {d\bar {x}}{dx}}\\ & =\frac {\bar {y}_{x}+\bar {y}_{y}\frac {dy}{dx}}{\bar {x}_{x}+\bar {x}_{y}\frac {dy}{dx}} \end {align*}

In this example $$\bar {y}_{x}=0,\bar {y}_{y}=e^{\epsilon },\bar {x}_{x}=e^{2\epsilon },\bar {x}_{y}=0.$$ The above now becomes\begin {align*} \frac {d\bar {y}}{d\bar {x}} & =\frac {e^{\epsilon }\frac {dy}{dx}}{e^{2\epsilon }}\\ & =e^{-\epsilon }\frac {dy}{dx} \end {align*}

Writing (1) in terms of $$\bar {x},\bar {y}$$ now gives\begin {align} \bar {x}\left ( \frac {d\bar {y}}{d\bar {x}}\right ) ^{2}-\bar {y}\frac {d\bar {y}}{d\bar {x}}+m & =0\tag {4}\\ \left ( e^{2\epsilon }x\right ) \left ( e^{-\epsilon }\frac {dy}{dx}\right ) ^{2}-\left ( e^{\epsilon }y\right ) e^{-\epsilon }\frac {dy}{dx}+m & =0\nonumber \\ x\left ( \frac {dy}{dx}\right ) ^{2}-y\frac {dy}{dx}+m & =0 \tag {5} \end {align}

Which gives the same ode. The above method starts by replacing the given ode by $$\bar {x},\bar {y},\frac {d\bar {y}}{d\bar {x}}$$ and ﬁnds if the result gives back the original ode in $$x,y,\frac {dy}{dx}$$. This is simpler than having to transform the original ode to $$\bar {x},\bar {y},\frac {d\bar {y}}{d\bar {x}}$$. This transformation can be veriﬁed in Maple as follows

ode:=x*diff(y(x),x)^2-y(x)*diff(y(x),x)+m=0;
the_tr:={x=X*exp(-2*s),y(x)=Y(X)*exp(-s)};
newode:=PDEtools:-dchange(the_tr,ode,{Y(X),X},'known'={y(x)},'uknown'={Y(X)});
diff(Y(X), X)^2*X - Y(X)*diff(Y(X), X) + m = 0



Comparing (4) to (5) shows that the ode form did not change, only the letters changed from $$x$$ to $$\bar {x}$$ and $$y$$ to $$\bar {y}$$. The resulting ode must never have the parameter $$\epsilon$$ show or remain in it.

The above shows how to verify that a transformation is invariant or not. In Lie group transformation there is only one parameter $$\epsilon$$ and the transformation is obtained by evaluating the group as $$\epsilon$$ goes to zero.

But how does this help in solving the ode? If the ode in $$x,y$$ is hard to solve, then the ode written with $$\bar {x},\bar {y}\,$$ will also be hard to solve since it is the same. But Eq. (4) is not what is used to solve the ode, but the above is just to verify the transformation is invariant. Similarity transformation is used to determine tangent vectors $$\xi ,\eta$$ only. Then the ode in canonical coordinates is used instead. In the canonical coordinates $$\left ( R,S\right )$$ the ode becomes quadrature and solved by integration. The transformation found above is only one step toward ﬁnding $$\left ( R,S\right )$$ and it is these canonical coordinates that are the goal and not $$\bar {x},\bar {y}$$.

#### 1.3 Outline of the steps in solving a diﬀerential equation using Lie symmetry method

These are the steps in solving an ODE using Lie symmetry method.

1. Given an ode $$y^{\prime }=\omega \left ( x,y\right )$$ to solve in natural coordinates.
2. Now the tangent vector $$\xi \left ( x,y\right ) ,\eta \left ( x,y\right )$$ are found. There are two options.

1. If Lie group coordinates $$\left ( \bar {x},\bar {y}\right )$$ are given, then it is easy to determine $$\xi \left ( x,y\right ) ,\eta \left ( x,y\right )$$ using\begin {align*} \xi \left ( x,y\right ) & =\left . \frac {\partial \bar {x}}{\partial \epsilon }\right \vert _{\epsilon =0}\\ \eta \left ( x,y\right ) & =\left . \frac {\partial \bar {y}}{\partial \epsilon }\right \vert _{\epsilon =0} \end {align*}

Lie group coordinates $$\left ( \bar {x},\bar {y}\right )$$ must also satisfy $\bar {x}_{x}\bar {y}_{y}-\bar {x}_{y}\bar {y}_{x}\neq 0$

2. In practice Lie group coordinates $$\left ( \bar {x},\bar {y}\right )$$ are not given and are not known. In this case $$\xi \left ( x,y\right ) ,\eta \left ( x,y\right )$$ are found by solving the similarity condition which results in a PDE (derivation is given below). The PDE is $\eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0$
3. $$\xi ,\eta$$ are now used to determine the canonical coordinates $$\left ( R,S\right )$$. In the canonical coordinates, only $$S$$ translation is needed to make the ode quadrature. The transformation is $$\left ( R,S\right ) \rightarrow \left ( R,S+\varepsilon \right )$$. This transforms the original ode $$y^{\prime }=\omega \left ( x,y\right )$$ to $$\frac {dS}{dR}=F\left ( R\right )$$ which is then solved by only integration. This is the main advantage of moving to canonical coordinates $$\left ( R,S\right )$$.
4. The ODE is solved in $$\left ( R,S\right )$$ space where $$R\equiv R\left ( x,y\right ) ,S\equiv S\left ( x,y\right )$$. The transformation from $$\left ( x,y\right )$$ to $$\left ( R,S\right )$$ is found by solving two set of PDEs using the characteristic method. After ﬁnding $$R\left ( x,y\right ) ,S\left ( x,y\right )$$ the ode will then be given by $$\frac {dS}{dR}=\frac {S_{x}+S_{y}\frac {dy}{dx}}{R_{x}+R_{y}\frac {dy}{dx}}$$ which will be quadrature. If this ode does not come out as $$\frac {dS}{dR}=F\left ( R\right )$$ then something went wrong in the process. This ode is now solved for $$S\left ( R\right ) .$$ It is the symmetry of the form $$\left ( R,S\right ) \rightarrow \left ( R,S+\varepsilon \right )$$ which is of the most interest in the Lie method. This is called a translation transformation along the $$y$$ axis (or the $$S$$ axis). This is because this transformation leads to an ode which is solved by just integration.
5. Transform the solution from $$S\left ( R\right )$$ to $$y\left ( x\right )$$.
6. An alternative to steps (3) to (5) is to use $$\xi ,\eta$$ to determine an integrating factor $$\mu \left ( x,y\right )$$ which is given by $$\mu \left ( x,y\right ) =\frac {1}{\eta -\xi \omega }$$ then the general solution to $$y^{\prime }=\omega \left ( x,y\right )$$ can be written directly as $$\int \mu \left ( x,y\right ) \left ( dy-\omega dx\right ) =c_{1}$$ or $$\int \frac {dy-\omega dx}{\eta -\xi \omega }=c_{1}$$ but this requires ﬁnding a function $$F\left ( x,y\right )$$ whose diﬀerential is $$dF=\frac {dy-\omega dx}{\eta -\xi \omega }$$ and now the solution becomes $$\int dF=c_{1}\,$$ or $$F=c_{1}$$. If we can integrate this using $$\int \mu dy-\int \mu \omega dx=c_{1}$$ then this is the solution to the ode. It is implicit in $$y\left ( x\right )$$. Currently my program does not implement Lie symmetry to ﬁnd an integrating factor due to diﬃculty of ﬁnding $$dF$$ that satisﬁes$$\ dF=\frac {dy-\omega dx}{\eta -\xi \omega }$$ or in carrying out the integration in all general cases but I hope to add this soon as a backup algorithm if the main one fails.
7. An important property, at least for ﬁrst order ode’s (I do not know now if this carries to higher order) is that given $$\xi =f\left ( x,y\right ) ,\eta =g\left ( x,y\right )$$, then we can always shift and use $$\xi \equiv 0,\eta =g-\omega f$$  where $$y^{\prime }=\omega \left ( x,y\right )$$. This means we can always base everything on $$\xi \equiv 0$$ after this shift is done to $$\eta$$. This can simplify some parts of the computation. Ofcourse if $$\xi$$ was found to be zero initially, i.e. just after solving the linearized similarity PDE, then there is nothing more to do.

The most diﬃcult step in all of the above is 2(b) which requires ﬁnding $$\xi \left ( x,y\right ) ,\eta \left ( x,y\right )$$. In practice Lie group $$\bar {x},\bar {y}$$ transformation is not given. Lie inﬁnitesimal $$\xi \left ( x,y\right ) ,\eta \left ( x,y\right )$$ have to be found directly from the linearized symmetry condition PDE using ansatz and by trial and error. The following diagram illustrates the above steps.

The following diagram illustrates the above steps when we carry the shifting step in order to force $$\xi =0$$. We see that It simpliﬁes the algorithm as now we can just assume $$\xi =0$$ and we do not have to check for diﬀerent cases as before.

#### 1.4 Finding $$\xi \left ( x,y\right ) ,\eta \left ( x,y\right )$$ knowing the ﬁrst order ode type. Table lookup method.

There is a short cut to obtaining $$\xi \left ( x,y\right ) ,\eta \left ( x,y\right )$$ if the ﬁrst order ode type is known or can be determined. (of course, if we know the ode type, then a direct method for solving the ode can be used, since the type is known and there is no need to use Lie symmetry), but still Lie symmetry can be useful in this case, and also it allows us to ﬁnd the integrating factor quickly, which provides one more method to solve the ode. An example of a ﬁrst order ode which does not have known type is $\left ( x\cos y-e^{-\sin y}\right ) y^{\prime }+1=0$

The above can be solved using Lie symmetry but with functional form of anstaz $$\xi =f\left ( x\right ) g\left ( y\right ) ,\eta =0$$. which gives $$\xi =e^{-\sin y},\eta =0$$.

I am in the process of building table for ready to use inﬁnitesimal based on the ﬁrst ode type. The following small list is the current ones determined. For some ﬁrst order ode such as linear $$y^{\prime }=f\left ( x\right ) y\left ( x\right ) +g\left ( x\right )$$ or separable $$y^{\prime }=f\left ( x\right ) g\left ( y\right )$$ the inﬁnitesimals can be written directly (but again, for these simple ode’s Lie method is not really needed but it provides good illustration on how to use it. Lie method is meant to be used for ode’s which have no known type or diﬃcult to solve otherwise). For an ode type not given in this list, an anstaz have to be used to solve the similarity PDE.

 ode type form $$\xi$$ $$\eta$$ notes linear ode $$y'=f(x) y(x) +g(x)$$ $$0$$ $$e^{\int fdx}$$ Notice that $$g\left ( x\right )$$ does not aﬀect the result separable ode $$y^{\prime }=f\left ( x\right ) g\left ( y\right )$$ $$\frac {1}{f}$$ $$0$$ This works for any $$g$$ function that depends on $$y$$ only quadrature ode $$y^{\prime }=f\left ( x\right )$$ $$0$$ $$1$$ of course for quadrature we do not need Lie symmetry as ode is already quadrature quadrature ode $$y^{\prime }=g\left ( y\right )$$ $$1$$ $$0$$ For example $$y^{\prime }=\frac {x+y}{-x+y}$$ or $$y^{\prime }=\frac {y+2\sqrt {yx}}{x}$$ homogeneous ODEs of Class A $$y^{\prime }=f\left ( \frac {y}{x}\right )$$ $$x$$ $$y$$ homogeneous ODEs of Class C $$y^{\prime }=\left ( a+bx+cy\right ) ^{\frac {n}{m}}$$ $$1$$ $$-\frac {b}{c}$$ Also $$\xi =0,\eta =c\left ( bx+cy+a\right ) ^{\frac {n}{m}}$$ are possible. For example, for $$y^{\prime }=\left ( 1+2x+3y\right ) ^{\frac {1}{2}}$$ then use the ﬁrst option as simpler which is $$\xi =1,\eta =-\frac {2}{3}$$. Notice that $$\xi =1,\eta =-\frac {b}{c}$$ does not depend on $$a$$ and not on $$n,m$$. Hence these odes $$y^{\prime }=\left ( 1+x+y\right ) ^{\frac {1}{3}}$$,$$y^{\prime }=\left ( 10+x+y\right ) ^{\frac {1}{3}}$$ and $$y^{\prime }=\left ( 10+x+y\right ) ^{\frac {2}{3}}$$ all have the same inﬁnitesimals $$\xi =1,\eta =-\frac {b}{c}=-1$$ homogeneous class D $$y^{\prime }=\frac {y}{x}+g\left ( x\right ) F\left (\frac {y}{x}\right )$$ $$x^{2}$$ $$xy$$ example $$y^{\prime }=\frac {y}{x}+\frac {1}{x}e^{-\frac {y}{x}}$$. Where here $$g\left ( x\right ) =\frac {1}{x},F\left ( \frac {y}{x}\right ) =e^{-\frac {y}{x}}$$. First order special form ID 1 $$y^{\prime }=g\left ( x\right ) e^{h\left (x\right ) +by}+f\left ( x\right )$$ $$\frac {e^{-\int bf\left ( x\right )dx-h\left ( x\right ) }}{g\left ( x\right ) }$$ $$\frac {f\left ( x\right )e^{-\int bf\left ( x\right ) dx-h\left ( x\right ) }}{g\left ( x\right ) }$$ For an example, for the ode $$y^{\prime }=5e^{x^{2}+20y}+\sin x$$, here $$g\left ( x\right ) =5,h\left ( x\right ) =x^{2},b=20,f\left ( x\right ) =\sin x$$, hence $$\xi =\frac {e^{-\int 20\sin dx-x^{2}}}{5},\eta =\frac {\sin xe^{-\int 20\sin xdx-x^{2}}}{5}$$ or $$\xi =\frac {1}{5}\sin x\left ( e^{20\cos \left ( x\right ) -x^{2}}\right ) ,\eta =\frac {\sin \left ( x\right ) }{5}\left ( e^{20\cos \left ( x\right ) -x^{2}}\right )$$. In this form, $$b$$ must be constant. polynomial type ode $$y^{\prime }=\frac {a_{1}x+b_{1}y+c_{1}}{a_{2}x+b_{2}y+c_{2}}$$ $$\frac {a_{1}b_{2}x-a_{2}b_{1}x-b_{1}c_{2}+b_{2}c_{1}}{a_{1}b_{2}-a_{2}b_{1}}$$ $$\frac {a_{1}b_{2}y-a_{2}b_{1}y-a_{1}c_{2}-a_{2}c_{1}}{a_{1}b_{2}-a_{2}b_{1}}$$ For example for $$y^{\prime }=\frac {x+y+3}{2x+y}$$ then $$a_{1}=1,b_{1}=1,c_{1}=3,a_{2}=2,b_{2}=1,c_{2}=0$$. Hence $$\xi =x-3,\eta =y+6$$. Bernoulli ode $$y' = f(x) y+ g(x)y^{n}$$ $$0$$ $$y^n e^{\int (1-n) f(x) \,dx}$$ $$n$$ is integer $$n\neq 1,n\neq 0$$. For example, for $$y^{\prime }=-\sin \left ( x\right ) y+x^{2}y^{2}$$ then $$f\left ( x\right ) =-\sin x,g\left ( x\right ) =x^{2},n=2$$ and $$\xi =0,\eta =e^{\int \sin xdx}y^{2}$$ or $$\xi =0,\eta =e^{-\cos x}y^{2}$$. Notice that $$g\left ( x\right )$$ does not show up in the inﬁnitesimals Another example is $$y^{\prime }=2\frac {y}{x}+\frac {y^{3}}{x^{2}}$$ where here $$f\left ( x\right ) =\frac {2}{x}$$. Hence $$\xi =0,\eta =e^{-\int \left ( 3-1\right ) \frac {2}{x}dx}y^{3}$$ or $$\xi =0,\eta =\eta =\frac {y^{3}}{x^{4}}$$ Reduced Riccati $$y^{\prime }=f_{1}\left ( x\right ) y+f_{2}\left ( x\right )y^{2}$$ $$0$$ $$e^{-\int f_{1}dx}$$ For example, for $$y^{\prime }=xy+\sin \left ( x\right ) y^{2}$$ then $$f_{1}=x,f_{2}=\sin x$$ and hence $$\xi =0,\eta =e^{-\int xdx}$$ or $$\xi =0,\eta =e^{\frac {1}{2}x^{2}}$$. Notice that $$f_{2}\left ( x\right )$$ does not show up in the inﬁnitesimals. I could not ﬁnd inﬁnitesimals for the full Riccati ode $$y^{\prime }=f_{0}\left ( x\right ) +f_{1}\left ( x\right ) y+f_{2}\left ( x\right ) y^{2}$$. Notice that $$f_{1},f_{2}$$ can not be both constants, else this becomes separable Abel ﬁrst kind $$y^{\prime }=f_{0}\left ( x\right ) +f_{1}\left ( x\right ) y+f_{2}\left ( x\right ) y^{2}+f_{3}\left ( x\right ) y^{3}$$ No inﬁnitesimals found

Currently the above are the ones I am able to determine for known ﬁrst order ode’s. If I ﬁnd more, will add them. The table lookup is much faster to use than having to solve the similarity PDE each time using anstaz in order to ﬁnd $$\xi ,\eta$$.

#### 1.5 Finding $$\xi \left ( x,y\right ) ,\eta \left ( x,y\right )$$ from linearized symmetry condition

Given any ﬁrst order ODE  \begin {equation} \frac {dy}{dx}=\omega \left ( x,y\right ) \tag {A} \end {equation} $$\xi \left ( x,y\right ) ,\eta \left ( x,y\right )$$ are called the inﬁnitesimals of the transformation. Maple has function called symgen in the DEtools package to determine these using 16 diﬀerent algorithms. Starting with the Lie point transformation group \begin {align*} \bar {x} & \equiv \bar {x}\left ( x,y;\epsilon \right ) \\ \bar {y} & \equiv \bar {y}\left ( x,y;\epsilon \right ) \end {align*}

Expanding using Taylor series near$$\ \epsilon =0$$ gives\begin {align*} \bar {x} & =x+\left . \frac {\partial \bar {x}}{\partial \epsilon }\right \vert _{\epsilon =0}\epsilon +O\left ( \epsilon ^{2}\right ) \\ & =x+\epsilon \xi \left ( x,y\right ) +O\left ( \epsilon ^{2}\right ) \\ \bar {y} & =y+\left . \frac {\partial \bar {y}}{\partial \epsilon }\right \vert _{\epsilon =0}\epsilon +O\left ( \epsilon ^{2}\right ) \\ & =y+\epsilon \eta \left ( x,y\right ) +O\left ( \epsilon ^{2}\right ) \end {align*}

Ignoring higher order terms gives\begin {align} \bar {x}\left ( x,y\right ) & =x+\epsilon \xi \left ( x,y\right ) \tag {1}\\ \bar {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 $\frac {d\bar {y}}{d\bar {x}}=\omega \left ( \bar {x},\bar {y}\right )$ 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 $$\left ( \bar {x},\bar {y}\right )$$) 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.

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

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

Substituting (4,5) into (3) gives the symmetry condition as\begin {equation} \frac {\bar {y}_{x}+\omega \left ( x,y\right ) \bar {y}_{y}}{\bar {x}_{x}+\omega \left ( x,y\right ) \bar {x}_{y}}=\omega \left ( \bar {x},\bar {y}\right ) \tag {6} \end {equation} But\begin {equation} \bar {x}_{x}=1+\epsilon \xi _{x} \tag {7} \end {equation} And similarly \begin {equation} \bar {x}_{y}=\epsilon \xi _{y} \tag {8} \end {equation} And\begin {equation} \bar {y}_{x}=\epsilon \eta _{x} \tag {9} \end {equation} And\begin {equation} \bar {y}_{y}=1+\epsilon \eta _{y} \tag {10} \end {equation} Substituting (7,8,9,10) back into the symmetry condition (6) gives \begin {align} \frac {\epsilon \eta _{x}+\omega \left ( 1+\epsilon \eta _{y}\right ) }{\left ( 1+\epsilon \xi _{x}\right ) +\omega \epsilon \xi _{y}} & =\omega \left ( x+\epsilon \xi ,y+\epsilon \eta \right ) \nonumber \\ \frac {\epsilon \eta _{x}+\omega +\omega s\eta _{y}}{1+\epsilon \xi _{x}+\omega \epsilon \xi _{y}} & =\omega \left ( x+\epsilon \xi ,y+\epsilon \eta \right ) \nonumber \\ \frac {\omega +s\left ( \eta _{x}+\omega \eta _{y}\right ) }{1+\epsilon \left ( \xi _{x}+\omega \xi _{y}\right ) } & =\omega \left ( x+\epsilon \xi ,y+\epsilon \eta \right ) \tag {11} \end {align}

The above is used to determine $$\xi \left ( x,y\right ) ,\eta \left ( x,y\right )$$. The above PDE is too complicated to use as is. It is linearized, and the linearized version is used to solve for $$\xi ,\eta$$ near small $$\epsilon$$.

Eq. (11) is linearized by expanding the LHS and the RHS using Taylor series around $$\epsilon =0$$ . Starting with the LHS ﬁrst, let $$\frac {\omega +\epsilon \left ( \eta _{x}+\omega \eta _{y}\right ) }{1+\epsilon \left ( \xi _{x}+\omega \xi _{y}\right ) }=\Delta _{LHS}$$. Expanding this using Taylor series around$$\ \epsilon =0$$ gives\begin {equation} \Delta _{LHS}=\Delta _{\epsilon =0}+\epsilon \frac {d}{d\epsilon }\ \left ( \Delta \right ) _{\epsilon =0}+h.o.t. \tag {11A} \end {equation} But $$\Delta _{\epsilon =0}=\omega$$ and \begin {align*} \frac {d}{d\epsilon }\ \left ( \Delta _{LHS}\right ) & =\frac {\frac {d}{d\epsilon }\ \left [ \omega +\epsilon \left ( \eta _{x}+\omega \eta _{y}\right ) \right ] \left ( 1+\epsilon \left ( \xi _{x}+\omega \xi _{y}\right ) \right ) -\left ( \omega +\epsilon \left ( \eta _{x}+\omega \eta _{y}\right ) \right ) \frac {d}{d\epsilon }\ \left [ 1+\epsilon \left ( \xi _{x}+\omega \xi _{y}\right ) \right ] }{\left ( 1+\epsilon \left ( \xi _{x}+\omega \xi _{y}\right ) \right ) ^{2}}\\ & =\frac {\left ( \eta _{x}+\omega \eta _{y}\right ) \left ( 1+\epsilon \left ( \xi _{x}+\omega \xi _{y}\right ) \right ) -\left ( \omega +\epsilon \left ( \eta _{x}+\omega \eta _{y}\right ) \right ) \left ( \xi _{x}+\omega \xi _{y}\right ) }{\left ( 1+\epsilon \left ( \xi _{x}+\omega \xi _{y}\right ) \right ) ^{2}} \end {align*}

At $$\epsilon =0$$ the above reduces to\begin {align} \frac {d}{d\epsilon }\ \left ( \Delta _{LHS}\right ) _{\epsilon =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 of Eq. (11A) becomes\begin {equation} \Delta _{LHS}=\omega +\epsilon \left ( \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}\right ) \tag {11B} \end {equation} Now the RHS of Eq. (11) is linearized. Let $$\omega \left ( x+s\xi ,y+s\eta \right ) =\Delta _{RHS}$$. Expansion around $$\epsilon =0$$ gives$\Delta _{RHS}=\Delta _{\epsilon =0}+\epsilon \left ( \frac {d}{d\epsilon }\ \Delta \right ) _{\epsilon =0}+h.o.t.$ But $$\Delta _{\epsilon =0}=\omega \left ( x,y\right )$$ and $\frac {d}{d\epsilon }\ \Delta _{RHS}=\omega _{x}\xi +\omega _{y}\eta$ Hence the linearized RHS of (11) becomes \begin {equation} \Delta _{RHS}=\omega \left ( x,y\right ) +\epsilon \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 +\epsilon \left ( \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}\right ) & =\omega +\epsilon \left ( \omega _{x}\xi +\omega _{y}\eta \right ) \nonumber \\ \epsilon \left ( \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}\right ) & =\epsilon \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 \end {align}

Hence\begin {equation} \fbox {$\eta _x+\omega \left ( \eta _y-\xi _x\right ) -\omega ^2\xi _y-\omega _x\xi -\omega _y\eta =0$} \tag {14} \end {equation} The above equation (14) is what is used to determine $$\xi ,\eta$$. It is the linearized symmetry condition. There is an additional constraint not mentioned above which is $\bar {x}_{x}\bar {y}_{y}\neq \bar {x}_{y}\bar {y}_{x}$ The restricted form of (14) is$\chi _{x}+\chi _{y}\omega -\chi \omega _{y}=0$ An important property is the following. Given any$\xi =A,\eta =B$ Then we can always write the above as$\xi =0,\eta =B-\omega A$ So that $$\xi =0$$ can always be used if needed to simplify some things.

After ﬁnding $$\xi ,\eta$$ from (14), the question now becomes is how to use them to solve the original ODE?

#### 1.6 Moving to canonical coordinates $$R,S$$

The next step is to determine what is called the canonical coordinates $$\left ( R,S\right )$$. In these canonical coordinates the ODE becomes a quadrature and solved by integration. Once solved, the solution is transformed back to $$\left ( x,y\right )$$. The canonical coordinates $$\left ( R,S\right )$$ are found as follows. Selecting the transformation to be \begin {align} \bar {R} & =R\tag {15}\\ \bar {S} & =S+\epsilon \tag {16} \end {align}

Eq. (15) becomes$\left . \frac {\partial \bar {R}}{\partial \epsilon }\right \vert _{\epsilon =0}=\left . \left ( \frac {\partial \bar {R}}{\partial x}\frac {dx}{d\epsilon }\right ) \right \vert _{\epsilon =0}+\left . \left ( \frac {\partial \bar {R}}{\partial y}\frac {dy}{d\epsilon }\right ) \right \vert _{\epsilon =0}$ But $$\left . \frac {\partial \bar {R}}{\partial x}\right \vert _{\epsilon =0}=\frac {\partial R}{\partial x}$$ and $$\left . \frac {dx}{d\epsilon }\right \vert _{\epsilon =0}=\xi \left ( x,y\right )$$ and similarly $$\left . \frac {\partial \bar {R}}{\partial y}\right \vert _{\epsilon =0}=\frac {\partial R}{\partial y}$$ and $$\left . \frac {dy}{d\epsilon }\right \vert _{\epsilon =0}=\eta \left ( x,y\right )$$. The above becomes$\left . \frac {\partial \bar {R}}{\partial \epsilon }\right \vert _{\epsilon =0}=\frac {\partial R}{\partial x}\xi +\frac {\partial R}{\partial y}\eta$ But $$\left . \frac {\partial \bar {R}}{\partial \epsilon }\right \vert _{\epsilon =0}=0$$ since $$\bar {R}=R$$. The above reduces to $0=\frac {\partial R}{\partial x}\xi +\frac {\partial R}{\partial y}\eta$ This PDE have solution using symmetry method given by\begin {align} \frac {dR}{dt} & =0\tag {15A}\\ \frac {dx}{dt} & =\xi \tag {15B}\\ \frac {dy}{dt} & =\eta \tag {15C} \end {align}

The same procedure is applied to Eq. (16) which gives$\left . \frac {\partial \bar {S}}{\partial \epsilon }\right \vert _{\epsilon =0}=\left . \left ( \frac {\partial \bar {S}}{\partial x}\frac {dx}{d\epsilon }\right ) \right \vert _{\epsilon =0}+\left . \left ( \frac {\partial \bar {S}}{\partial y}\frac {dy}{d\epsilon }\right ) \right \vert _{\epsilon =0}$ But $$\left . \frac {\partial \bar {S}}{\partial x}\right \vert _{\epsilon =0}=\frac {\partial S}{\partial x}$$ and $$\left . \frac {dx}{d\epsilon }\right \vert _{\epsilon =0}=\xi \left ( x,y\right )$$ and similarly $$\left . \frac {\partial \bar {S}}{\partial y}\right \vert _{\epsilon =0}=\frac {\partial S}{\partial y}$$ and $$\left . \frac {dy}{d\epsilon }\right \vert _{\epsilon =0}=\eta \left ( x,y\right ) \,$$. The above becomes$\left . \frac {\partial \bar {S}}{\partial \epsilon }\right \vert _{\epsilon =0}=\frac {\partial R}{\partial x}\xi +\frac {\partial R}{\partial y}\eta$ But $$\left . \frac {\partial \bar {S}}{\partial \epsilon }\right \vert _{\epsilon =0}=1$$ since $$\bar {S}=S+\epsilon$$. The above reduces to $1=\frac {\partial S}{\partial x}\xi +\frac {\partial S}{\partial y}\eta$ This PDE have solution using symmetry method given by\begin {align} \frac {dS}{dt} & =1\tag {16A}\\ \frac {dx}{dt} & =\xi \tag {16B}\\ \frac {dy}{dt} & =\eta \tag {16C} \end {align}

Equations (15A,B,C) are used to solve for $$R\left ( x,y\right )$$ and equations (16A,B,C) are used to solve for $$S\left ( x,y\right )$$. Starting with $$R$$. In the case when $$\xi =0$$ the equations become\begin {align*} \frac {dR}{dt} & =0\\ \frac {dx}{dt} & =0\\ \frac {dy}{dt} & =\eta \end {align*}

First equation above gives $$R=c_{1}$$. Second equation gives $$x=c_{2}$$.  Letting $$c_{1}=c_{2}$$ then $R=x$ If $$\xi \neq 0$$ then combining Eqs. (15B,15C) gives  \begin {align*} \frac {dy}{dx} & =\frac {\eta }{\xi }\\ R & =c_{1} \end {align*}

The ODE $$\frac {dy}{dx}=\frac {\eta }{\xi }$$ is solved ﬁrst and the constant of integration is replaced by $$R$$. Hence $$R$$ is now found. $$S\left ( x,y\right )$$ is found similarly using Eqs. (16A,B,C).  If $$\xi =0$$ then \begin {align*} \frac {dS}{dt} & =1\\ \frac {dx}{dt} & =0\\ \frac {dy}{dt} & =\eta \end {align*}

The ﬁrst and third equations give\begin {align*} \frac {dS}{dy} & =\frac {1}{\eta }\\ S & =\int \frac {1}{\eta }dy \end {align*}

If $$\xi \neq 0$$ then using the second and third equation gives\begin {align*} \frac {dS}{dx} & =\frac {1}{\xi }\\ S & =\int \frac {1}{\xi }dx \end {align*}

Now that $$R,S$$ are found and the problem is solved. The ode in $$\left ( R,S\right )$$ space is set up using \begin {equation} \frac {dS}{dR}=\frac {S_{x}+S_{y}\frac {dy}{dx}}{R_{x}+R_{y}\frac {dy}{dx}} \tag {16} \end {equation} Where $$\frac {dy}{dx}=\omega \left ( x,y\right )$$ which is given. The solution $$S\left ( R\right )$$ is next converted back to $$y\left ( x\right )$$.

Examples below illustrate how this done on a number of ODE’s. Eq. (16) is solved by quadrature. This is the whole point of Lie symmetry method, is that the original ode is solved in canonical coordinates where it is much easier to solve and the solution is transformed back to natural coordinates.

The only way to understand this method well, is to workout some problems. 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.

#### 1.7 Deﬁnitions and various notes

1. inﬁnitesimal generator operator. $$\boldsymbol {\Gamma }=\xi \left ( x,y\right ) \frac {\partial }{\partial x}+\eta \left ( x,y\right ) \frac {\partial }{\partial y}$$. Any ﬁrst order ode has such generator. For instance, for the ode $$y^{\prime }=\omega \left ( x,y\right )$$ then $$\boldsymbol {\Gamma }\omega =\xi \frac {\partial \omega }{\partial x}+\eta \frac {\partial \omega }{\partial y}$$. The ode $$y^{\prime }=\omega \left ( x,y\right ) =\frac {y}{x}+x$$ has solution $$y=x^{2}+xc_{1}$$, therefore the solution family is $$\phi \left ( x,y\right ) =\frac {y-x^{2}}{x}=c$$. Using $$\xi =0,\eta =x$$ then $$\boldsymbol {\Gamma }\phi =x\frac {\partial \left ( \frac {y-x^{2}}{x}\right ) }{\partial y}=1$$. This is another example: using $$\xi =x,\eta =2y\,$$, hence $$\boldsymbol {\Gamma }\phi =x\frac {\partial \left ( \frac {y-x^{2}}{x}\right ) }{\partial x}+2y\frac {\partial \left ( \frac {y-x^{2}}{x}\right ) }{\partial y}=x\left ( -\frac {y}{x^{2}}-1\right ) +2y\left ( \frac {1}{x}\right ) =-\frac {y}{x}-1+2\frac {y}{x}=\frac {y}{x}-1\neq 1$$. I must be not applying the symmetry generator correct as the result supposed to be $$1$$. Need to visit this again. See book Bluman and Anco, page 109. Maybe some of the assumptions for using this generator are not satisﬁed for this ode.
2. $$\omega \left ( x,y\right )$$ is invariant iﬀ $$\boldsymbol {\Gamma }\omega =\xi \left ( x,y\right ) \frac {\partial }{\partial x}+\eta \left ( x,y\right ) \frac {\partial }{\partial y}=0$$.
3. The linearized PDE from the symmetry condition is $$\omega \xi _{x}+\omega ^{2}\xi _{y}+\omega _{x}\xi =\omega _{y}\eta +\eta _{x}+\omega \eta _{y}$$. This is used to determine tangent vector $$\left ( \xi \left ( x,y\right ) ,\eta \left ( x,y\right ) \right )$$ which is one of the core parts of the algorithm to solve the ode using symmetry methods. There are inﬁnite number of solutions and only one is needed.
4. Symmetries and ﬁrst integrals are the two most important structures of diﬀerential equations. First integral is quantity that depends on $$x,y$$ and when integrated over any solution curve is constant.
5. Lie symmetry allows one to reduce the order of an ode by one. So if we have third order ode and we know the symmetry for it, we can change the ode to second order ode. Then if apply the symmetry for this second order ode, its order is reduced to one now.
6. If $$\xi ,\eta$$ are known then the canonical coordinates $$R,S$$ can now be found as functions of $$x,y$$. We just $$\xi ,\eta$$ to ﬁnd $$R,S$$. Once $$R,S$$ are known then $$\frac {dS}{dR}=f\left ( R\right )$$ can be formulated. This ode is solved for $$S$$ by quadrature. Final solution is found by replacing $$R,S$$ back by $$x,y$$. I have functions and a solver now written and complete to do all of this but just for ﬁrst order ode’s only. I need to start on second order ode’s after that. The main and most diﬃcult step is in ﬁnding $$\xi ,\eta$$. Currently I only use multivariable polynomial ansatz up to second order for $$\xi$$ and multivariable polynomial ansatz up to third order for $$\eta$$ and then try all possible combinations. This is not very eﬃcient. But works for now. I need to add better and more eﬃcient methods to ﬁnding $$\xi ,\eta$$ but need to do more research on this.
7. When using polynomial ansatz to ﬁnd $$\xi ,\eta$$ do not mix $$x,y$$ in both ansatz. For example if we use $$\xi =p\left ( x\right )$$ then can use $$\eta =q\left ( x\right )$$ or $$\eta =q\left ( x,y\right )$$ polynomial ansatz to ﬁnd $$\eta$$. But do not try $$\xi =p\left ( x,y\right )$$ ansatz with $$\eta =q\left ( x,y\right )$$ ansatz. In other words, if one ansatz polynomial is multivariable, then the other should be single variable. Otherwise results will be complicated and this defeats the whole ides of using Lie symmetry as the ode generated will be as complicated or more than the original ode we are trying to solve. I found this the hard way. I was generating all permutations of $$\xi ,\eta$$ ansatz’s but with both as multivariable polynomials. This did not work well.
8. Symmetries on the ode itself, is same as talking about symmetries on solution curves. i.e. given an ode $$y^{\prime }=\omega \left ( x,y\right )$$ with solution $$y=f\left ( x\right )$$, then when we look for symmetry on the ode which leaves the ode looking the same but using the new variables $$\bar {x},\bar {y}$$. This is the same as when we look for symmetry which maps any point $$\left ( x,y\right )$$ on solution curve $$y=f\left ( x\right )$$ to another solution curve.  In other words, the symmetry will map all solution curves of $$y^{\prime }=\omega \left ( x,y\right )$$ to the same solution curves. i.e. a speciﬁc solution curve $$y=f\left ( x,c_{1}\right )$$ will be mapped to $$y=f\left ( x,c_{2}\right )$$. All solution curves of $$y^{\prime }=\omega \left ( x,y\right )$$ will be mapped to the same of solution curves. But each curve maps to another curve within the same set. If the same curve maps to itself, then this is called invariant curve.
9. An orbit is the name given to the path the transformation moves the point $$\left ( x,y\right )$$ from one solution curve to another point on another solution curve due to the symmetry transformation.
10. A solution curve of $$y^{\prime }=\omega \left ( x,y\right )$$ that maps to itself under the symmetry transformation is called an invariant curve.
11. Not every ﬁrst order ode has symmetry. At least according to Maple. For example $$y^{\prime }+y^{3}+xy^{2}=0$$ which is Abel ode type, it found no symmetries using way=all. May be with special hint it can ﬁnd symmetry?
12. After trying polynomials ansatz, I ﬁnd it is limited. Since it will only ﬁnd symmetries that has polynomials form. A more powerful ansatz is the functional form. But these are much harder to work with but they are more general at same time and can ﬁnd symmetries that can’t be found with just polynomials. So I have to learn how to use functional ansatz’s. Currently I only use Polynomials.
13. $$\xi ,\eta$$ are called Lie inﬁnitesimal and $$\bar {x},\bar {y}$$ are called the Lie group.
14. If we given the $$\xi ,\eta$$ then we can ﬁnd Lie group $$\left ( \bar {x},\bar {y}\right )$$. See example below.
15. If we are given Lie group $$\left ( \bar {x},\bar {y}\right )$$ then we can ﬁnd the inﬁnitesimal using $$\xi \left ( x,y\right ) =\left . \frac {\partial }{\partial \epsilon }\bar {x}\right \vert _{\epsilon =0}$$ and $$\eta \left ( x,y\right ) =\left . \frac {\partial }{\partial \epsilon }\bar {y}\right \vert _{\epsilon =0}$$.
16. First order ode have inﬁnite number of symmetries. Talking about symmetry of an ode is the same as talking about symmetry between solution curves of the ode itself. i.e. symmetry then becomes ﬁnding mapping that maps each solution curve to another one in the same family of solutions of the ode.
17. $$\xi ,\eta$$ can also be used to ﬁnd the integrating factor for the ﬁrst order ode. This is given by $$\mu \left ( x,y\right ) =\frac {1}{\eta -\xi \omega }$$ where the ode is $$y^{\prime }\left ( x\right ) =\omega \left ( x,y\right ) \,$$. This gives an alternative approach to solve the ode. I still need to add examples using $$\mu \left ( x,y\right )$$.
18. For ﬁrst order ode, to ﬁnd Lie inﬁnitesmilas, we have to solve ﬁrst order PDE in 2 variables. For second order ode, to ﬁnd Lie inﬁnitesmilas, we have to solve second order PDE in 3 variables. For third order ode, to ﬁnd Lie inﬁnitesmilas, we have to solve third order PDE in 4 variables and so on. Hence in general, for $$n^{th}$$ order ode, we have to solve $$n^{th}$$ order PDE in $$n+1$$ variables to ﬁnd the required Lie inﬁnitesmilas. For ﬁrst order, these variables are $$\xi ,\eta$$ and the PDE is $$\eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0$$. Currently my program only handles ﬁrst order odes. Once I am more familar with Lie method for second order ode, will update these notes. See at the end a section on just second order ode that I started working on.

#### 1.8 Closer look at orbits and tangent vectors

This section takes a closer look at orbits and tangent vectors $$\xi ,\eta$$ which are the core of Lie symmetry method. By deﬁnition\begin {align} \xi \left ( x,y\right ) & =\left . \frac {d\bar {x}}{d\epsilon }\right \vert _{\epsilon =0}\tag {1}\\ \eta \left ( x,y\right ) & =\left . \frac {d\bar {y}}{d\epsilon }\right \vert _{\epsilon =0}\nonumber \end {align}

Hence $$\xi \left ( x,y\right )$$ shows how $$\bar {x}$$ changes as function of $$\left ( x,y\right )$$. And $$\eta \left ( x,y\right )$$ shows how $$\bar {y}$$ changes as function of $$\left ( x,y\right )$$. This is because\begin {align} \bar {x} & =x+\xi \epsilon \tag {2}\\ \bar {y} & =y+\eta \epsilon \nonumber \end {align}

Comparing (2) to equation of motion where $$\bar {x}$$ represents ﬁnal position and $$x$$ is initial position, then $$\xi$$ is the speed and $$\epsilon$$ is the time. When time is zero, initial and ﬁnal position is the same. As time increases ﬁnal position changes depending on the speed as time (here represented as $$\epsilon$$) increases. So it helps to think of $$\xi ,\eta$$ as the rate at which $$\bar {x},\bar {y}$$ change location depending on the value $$\epsilon$$. $$\xi ,\eta$$ are calculated when $$\epsilon$$ is very small in the limit as it reaches zero.

As $$\epsilon$$ increases the point $$\left ( x,y\right )$$ moves closer to the ﬁnal destination point $$\left ( \bar {x},\bar {y}\right )$$. So these quantities $$\xi ,\eta$$ specify the orbit shape. The orbit is the path taken by point transformation from $$\left ( x,y\right )$$ to $$\left ( \bar {x},\bar {y}\right )$$ and depends on $$\epsilon$$ such that the ode remain invariant in $$\bar {x},\bar {y}$$ and points on solution curves are mapped to points on other solution curves.

Diﬀerent $$\xi ,\eta$$ give diﬀerent orbits between two solution curves. The following example shows this. Given the ode $y^{\prime }=\frac {x-y}{x+y}$ This is Abel type ode. Also Homogeneous class A$$.$$

It has two solutions. One solution is given by Mathematica as $$y=-x-\sqrt {c_{1}+2x^{2}}$$. A small program was now written that plots the orbit for 4 solutions $$\xi ,\eta$$ found for the similarity conditions. The similarity solution were found by Maple’s symgen command.

The program starts from the same $$\left ( x,y\right )$$ point from one solution curve and determines $$\left ( \bar {x},\bar {y}\right )$$ location on anther solution curve using each pair of $$\xi ,\eta$$ found. The same solution curves are used in order to compare the orbits. The following plot was generated showing the result

The source code used to generate the above plot is

<<MaTeX
ode=y'[x]==(x-y[x])/(x+y[x]);
ysol=DSolve[ode,y[x],x]
ysol=-x-Sqrt[C[1]+2 x^2];

x1 = 1.5;
y1 = ysol /. {C[1] -> 1, x -> x1};

ysol2=ysol/.C[1]->1.1

getSolutions[inf_List, titles_List, x_Symbol, ysol1_, ysol2_, x1_,
y1_, from_, to_] :=
Module[{xbar, ybar, eps, eq, soleps, p, data, n, xi, eta, texStyle},
data = Table[0, {n, Length@inf}];
texStyle = {FontFamily -> "Latin Modern Roman", FontSize -> 12};

Do[
xi = First[inf[[n]]];
eta = Last[inf[[n]]];
xbar = x1 + eps*xi ;
ybar = y1 + eps*eta;
eq = ybar == ysol2 /. x -> xbar;
soleps = SolveValues[eq, eps];
soleps = First@SortBy[soleps, Abs];
ybar = ybar /. eps -> soleps;
xbar = xbar /. eps -> soleps;
p = Plot[{ysol1, ysol2}, {x, from, to},
PlotLabel -> MaTeX[titles[[n]], Magnification -> 1.5],
BaseStyle -> texStyle,
Epilog -> {{Arrowheads[.02], Arrow[{{x1, y1}, {xbar, ybar}}]},
Text[MaTeX["\\left( x,y \\right)"], {x1, y1}, {-1, -1}],
Text[
MaTeX["\\left( \\bar{x},\\bar{y}\\right)"], {xbar, ybar}, {1,
1}]},
ImageSize -> 400];
data[[n]] = p
,
{n, 1, Length@inf}
];

data

];

inf = {{1/x1, -1/x1},
{0, 1/(x1 + y1)},
{-(x1^2 - 2*x1*y1 - y1^2)/(x1 - y1), 0},
{2*x1 + y1, x1}
};
titles = {"\\xi=\\frac{1}{x},\\eta=-\\frac{1}{x}",
"\\xi=0,\\eta=\\frac{1}{x+y}",
"\\xi=\\frac{-(x^2-2 x y-y^2)}{x=y},\\eta=0", "\\xi=2 x+y,\\eta=x"};
data = getSolutions[inf, titles, x, ysol /. C[1] -> 1, ysol2, x1, y1,
1.45, 1.51];
p = Grid[Partition[data, 2], Frame -> All, Spacings -> {1, 1}]



#### 1.9 Selection of ansatz to try

The following are selection of ansatz to try for solving the linearized PDE above generated from the symmetry condition in order to solve for $$\xi \left ( x,y\right ) ,\eta \left ( x,y\right )$$. These use the functional form. As a general rule, the simpler that ansatz that works, the better it is.  Functional form of ansatz is better than explicit polynomials but much harder to use and implement. Maple’s symgen has 16 diﬀerent algorithms include HINT option to support functional forms. The following are possible cases to use.

1. $$\xi =0,\eta =f\left ( x\right )$$
2. $$\xi =0,\eta =f\left ( y\right )$$
3. $$\xi =f\left ( x\right ) ,\eta =0$$
4. $$\xi =f\left ( y\right ) ,\eta =0$$
5. $$\xi =f\left ( x\right ) ,\eta =xg\left ( y\right )$$. An example: applied to $$y^{\prime }=\frac {x+\cos \left ( e^{y}+\left ( 1+x\right ) e^{-x}\right ) }{e^{y+x}}$$ should give $$\xi =e^{x},\eta =xe^{-y}$$ which leads to solution $$y=\ln \left ( 2\arctan \left ( \frac {e^{-\left ( c_{1}+e^{-x}\right ) -1}}{e^{-\left ( c_{1}+e^{-x}\right ) +1}}\right ) -\left ( 1+x\right ) e^{-x}\right )$$.
6. $$\xi =f\left ( x\right ) ,\eta =g\left ( y\right )$$
7. $$\xi =0,\eta =f\left ( x\right ) g\left ( y\right )$$. For example, applied to $$y^{\prime }=\frac {x\sqrt {1+y}+\sqrt {1+y}+1+y}{1+x}$$ should give $$f\left ( x\right ) =\sqrt {1+x},g\left ( y\right ) =\sqrt {1+y}$$.
8. $$\xi =f\left ( x\right ) g\left ( y\right ) ,\eta =0$$

#### 1.10 Examples

##### 1.10.1 Example 1 on how to ﬁnd Lie group $$\left ( \bar {x},\bar {y}\right )$$ given Lie inﬁnitesimal $$\xi ,\eta$$

Given$$\ \xi =1,\eta =2x$$ ﬁnd Lie group $$\bar {x},\bar {y}$$.  Since$\xi \left ( x,y\right ) =\left . \frac {\partial \bar {x}}{\partial \epsilon }\right \vert _{\epsilon =0}$ Then \begin {align} \frac {d\bar {x}}{d\epsilon } & =\xi \left ( \bar {x},\bar {y}\right ) \nonumber \\ & =1 \tag {1} \end {align}

Similarly, since $\eta \left ( x,y\right ) =\left . \frac {\partial \bar {y}}{\partial \epsilon }\right \vert _{\epsilon =0}$ Then \begin {align} \frac {d\bar {y}}{d\epsilon } & =\eta \left ( \bar {x},\bar {y}\right ) \nonumber \\ & =2\bar {y} \tag {2} \end {align}

Where in both odes (1,2) we have the condition that at $$\epsilon =0$$ then $$\bar {x}=x,\bar {y}=y$$. Starting with (1), solving it gives$\bar {x}=\epsilon +c_{1}\left ( x,y\right )$ Where $$c_{1}\left ( x,y\right )$$ is arbitrary function which acts like constant of integration since $$\bar {x}\left ( x,y\right )$$ is function of two variables. At $$\epsilon =0$$ then $$c_{1}\left ( x,y\right ) =x$$. Hence the above is\begin {equation} \bar {x}=\epsilon +x \tag {3} \end {equation} And from (2), solving give$\bar {y}=2\bar {x}\epsilon +c_{2}\left ( x,y\right )$ But at $$\epsilon =0$$ $$,\bar {y}=y,\bar {x}=x$$ then the above gives $$c_{2}=y$$. Hence the above becomes$\bar {y}=2\bar {x}\epsilon +y$ But $$\bar {x}=\epsilon +x$$ from (3), hence  the above becomes\begin {align*} \bar {y} & =2\left ( \epsilon +x\right ) \epsilon +y\\ & =2\epsilon ^{2}+2\epsilon x+y \end {align*}

Therefore Lie group is\begin {align*} \bar {x} & =\epsilon +x\\ \bar {y} & =2\epsilon ^{2}+2\epsilon x+y \end {align*}

##### 1.10.2 Example how to ﬁnd Lie group $$\left ( \bar {x},\bar {y}\right )$$ given canonical coordinates $$R,S$$

Given$$\ R=x,S=\frac {y}{x}$$ ﬁnd Lie group $$\bar {x},\bar {y}$$.  Solving for $$x,y$$ from $$R,S$$ gives\begin {align*} x & =R\\ y & =SR \end {align*}

Hence\begin {align*} \bar {x} & =\bar {R}\\ \bar {y} & =\bar {S}\bar {R} \end {align*}

But $$\bar {S}=S+\epsilon$$ by deﬁnition of canonical coordinates and $$\bar {R}=R$$ by deﬁnition of canonical coordinates. Hence the above becomes\begin {align*} \bar {x} & =R\\ \bar {y} & =\left ( S+\epsilon \right ) R \end {align*}

Using the values given for $$R,S$$ in terms of $$x,y$$ the above becomes\begin {align*} \bar {x} & =x\\ \bar {y} & =\left ( \frac {y}{x}+\epsilon \right ) x\\ & =y+\epsilon x \end {align*}

##### 1.10.3 Example $$y^{\prime }=\frac {y}{x}+x$$

This is linear ﬁrst order which can be easily solved using integrating factor. But this is just to illustrate Lie symmetry method.\begin {align} y^{\prime } & =\frac {y}{x}+x\tag {1}\\ y^{\prime } & =\omega \left ( x,y\right ) \nonumber \end {align}

The ﬁrst step is to ﬁnd $$\xi$$ and $$\eta$$. Using lookup method, since this is linear ode of form $$y^{\prime }=f\left ( x\right ) y+g\left ( x\right )$$ then

\begin {align*} \xi & =0\\ \eta & =e^{\int fdx}=e^{\int \frac {1}{x}dx}=x \end {align*}

The end of this problem shows also how to ﬁnd these from the symmetry conditions. Therefore we write\begin {align} \bar {x} & =x+\xi \epsilon \nonumber \\ & =x\nonumber \\ \bar {y} & =y+\eta \epsilon \nonumber \\ & =y+\eta x \tag {2} \end {align}

The integrating factor is therefore \begin {align*} \mu \left ( x,y\right ) & =\frac {1}{\eta -\xi \omega }\\ & =\frac {1}{x} \end {align*}

Before solving this, let us ﬁrst verify that transformation (2) is invariant which means it leaves the ode in same form but using $$\bar {x},\bar {y}$$. We do the same as in the above introduction.\begin {align*} \frac {d\bar {y}}{d\bar {x}} & =\frac {\frac {d\bar {y}}{dx}}{\frac {d\bar {x}}{dx}}\\ & =\frac {\bar {y}_{x}+\bar {y}_{y}\frac {dy}{dx}}{\bar {x}_{x}+\bar {x}_{y}\frac {dy}{dx}} \end {align*}

But $$\bar {y}_{x}=s,\bar {y}_{y}=1,\bar {x}_{x}=1,\bar {x}_{y}=0$$ and the above becomes\begin {align*} \frac {d\bar {y}}{d\bar {x}} & =\frac {\epsilon +\frac {dy}{dx}}{1}\\ & =\epsilon +\frac {dy}{dx} \end {align*}

Substituting $$\bar {x},\bar {y},\frac {d\bar {y}}{d\bar {x}}$$ in the original ode gives\begin {align*} \frac {d\bar {y}}{d\bar {x}} & =\frac {\bar {y}}{\bar {x}}+\bar {x}\\ \epsilon +\frac {dy}{dx} & =\frac {y+\epsilon x}{x}+x\\ \epsilon +\frac {dy}{dx} & =\frac {y}{x}+\epsilon +x\\ \frac {dy}{dx} & =\frac {y}{x}+x \end {align*}

Which is the original ODE. Therefore (2) are indeed an invariant Lie group transformation as it leaves the ODE unchanged. 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}{0} & =\frac {dy}{x}=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.  In the special case when $$\xi =0$$ and $$\eta \neq 0$$ these give\begin {align*} R & =x\\ S & =\int \frac {1}{\eta }dy\\ & =\int \frac {1}{x}dy\\ & =\frac {y}{x}+c \end {align*}

We are free to set $$c=0$$, hence $$S=\frac {y}{x}$$. Therefore the transformation to canonical coordinates is $\left ( x,y\right ) \rightarrow \left ( R,S\right ) =\left ( x,\frac {y}{x}\right )$ The derivative in $$\left ( R,S\right )$$ is found same as with $$\frac {d\bar {y}}{d\bar {x}}$$ giving$\frac {dS}{dR}=\frac {S_{x}+S_{y}\frac {dy}{dx}}{R_{x}+R_{y}\frac {dy}{dx}}$ But $$S_{x}=-\frac {y}{x^{2}},S_{y}=\frac {1}{x},R_{x}=1,R_{y}=0$$ and the above becomes\begin {align*} \frac {dS}{dR} & =\frac {-\frac {y}{x^{2}}+\frac {1}{x}\frac {dy}{dx}}{1}\\ & =-\frac {y}{x^{2}}+\frac {1}{x}\frac {dy}{dx} \end {align*}

But $$\frac {dy}{dx}=\frac {y}{x}+x$$ hence the above becomes\begin {align*} \frac {dS}{dR} & =-\frac {y}{x^{2}}+\frac {1}{x}\left ( \frac {y}{x}+x\right ) \\ & =1 \end {align*}

Solving this gives $S=R+c_{1}$ But $$S=\frac {y}{x},R=x$$. Therefore the above becomes\begin {align*} \frac {y}{x} & =x+c_{1}\\ y & =x^{2}+c_{1}x \end {align*}

Which is the solution to the original ode. Of course this was just an example showing how to use Lie symmetry method. The original ode is linear and can be easily solved using an integrating factor\begin {align*} y^{\prime }-\frac {y}{x} & =x\\ I & =e^{-\int \frac {1}{x}dx}=e^{-\ln x}=\frac {1}{x} \end {align*}

Multiplying the ode by $$I$$ gives\begin {align*} \frac {d}{dx}\left ( yI\right ) & =Ix\\ \frac {y}{x} & =\int \frac {x}{x}dx\\ & =x+c_{1} \end {align*}

Hence$y=x^{2}+xc_{1}$ Which is same solution. But Lie symmetry method works the same way for any given ode. And this is where it powers are. It can solve much more complicated odes than this using the same procedure. The main diﬃculty is in ﬁnding the inﬁnitesimals for the group, which are $$\xi ,\eta$$ that leaves the ode invariant.

Finding Lie symmetries for this example

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

The condition of symmetry is a the linearized PDE 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 ﬁrst ﬁnd the determining equation before solving for $$\xi ,\eta$$. Since $$\omega =\frac {y}{x}+x$$ then $$\omega _{y}=\frac {1}{x},\omega _{x}=-\frac {y}{x^{2}}+1$$. Hence the above becomes\begin {align*} \eta _{x}+\left ( \frac {y}{x}+x\right ) \left ( \eta _{y}-\xi _{x}\right ) -\left ( \frac {y}{x}+x\right ) ^{2}\xi _{y}-\left ( -\frac {y}{x^{2}}+1\right ) \xi -\frac {1}{x}\eta & =0\\ \eta _{x}+\left ( \frac {y}{x}+x\right ) \left ( \eta _{y}-\xi _{x}\right ) -\left ( \frac {y^{2}}{x^{2}}+x^{2}+2y\right ) \xi _{y}-\left ( -\frac {y}{x^{2}}+1\right ) \xi -\frac {1}{x}\eta & =0\\ \eta _{x}+\left ( \frac {y}{x}+x\right ) \eta _{y}-\xi _{x}\left ( \frac {y}{x}+x\right ) -\left ( \frac {y^{2}}{x^{2}}+x^{2}+2y\right ) \xi _{y}-\left ( -\frac {y}{x^{2}}+1\right ) \xi -\frac {1}{x}\eta & =0 \end {align*}

Multiplying by $$x^{2}$$ to normalize gives\begin {equation} x^{2}\eta _{x}+\left ( yx+x^{3}\right ) \eta _{y}-\xi _{x}\left ( yx+x^{3}\right ) -\left ( y^{2}+x^{4}+2yx^{2}\right ) \xi _{y}-\left ( -y+x^{2}\right ) \xi -x\eta =0 \tag {A} \end {equation} Equation (A) is called the determining equation. Using diﬀerent ansatz can result in more solutions.

Trying ansatz \begin {align*} \xi & =0\\ \eta & =b_{0}x \end {align*}

Plugging these into (A) and comparing coeﬃcients to solve for the unknown gives\begin {align*} x^{2}\left ( b_{0}\right ) -x\eta & =0\\ b_{0}x^{2}-x\left ( b_{0}x\right ) & =0\\ b_{0}x^{2}-b_{0}x^{2} & =0\\ b_{0}\left ( 0\right ) & =0 \end {align*}

So any $$b_{0}$$ will work. Let $$b_{0}=1$$. Hence \begin {align*} \xi & =0\\ \eta & =x \end {align*}

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

Then $$\xi _{x}=a_{1},\xi _{y}=0,\eta _{x}=0,\eta _{y}=b_{1}$$ and the determining equation (A) becomes\begin {align*} \left ( b_{0}+b_{1}y\right ) x+\left ( a_{0}+a_{1}x\right ) \left ( x^{2}-y\right ) +b_{1}\left ( -yx-x^{3}\right ) +a_{1}\left ( yx+x^{3}\right ) & =0\\ \left ( b_{0}+b_{1}y\right ) x+\left ( a_{0}+a_{1}x\right ) \left ( x^{2}-y\right ) +\left ( b_{1}-a_{1}\right ) \left ( -yx-x^{3}\right ) & =0\\ xb_{0}-ya_{0}+x^{2}a_{0}+x^{3}\left ( 2a_{1}-b_{1}\right ) & =0 \end {align*}

Setting each coeﬃcient to zero gives\begin {align*} b_{0} & =0\\ a_{0} & =0\\ a_{0} & =0\\ 2a_{1}-b_{1} & =0 \end {align*}

Hence the solution is $$a_{0}=0,b_{0}=0,a_{1}=\frac {b_{1}}{2}$$. Using $$b_{1}=2$$ gives $$a_{1}=1$$ and therefore \begin {align*} \xi & =x\\ \eta & =2y \end {align*}

And Trying ansatz as \begin {align*} \xi & =a_{0}+a_{1}x+a_{2}y\\ \eta & =b_{0}+b_{1}y+b_{2}x \end {align*}

Hence $$\xi _{x}=a_{1},\xi _{y}=a_{2},\eta _{x}=b_{2},\eta _{y}=b_{1}$$ and the determining equation (A) becomes\begin {align*} \left ( b_{0}+b_{1}y+b_{2}x\right ) x+\left ( a_{0}+a_{1}x+a_{2}y\right ) \left ( x^{2}-y\right ) +b_{1}\left ( -yx-x^{3}\right ) +a_{2}\left ( y^{2}+x^{4}+2yx^{2}\right ) +b_{2}\left ( -x^{2}\right ) +a_{1}\left ( yx+x^{3}\right ) & =0\\ x^{4}\left ( -a_{2}\right ) +x^{3}\left ( -2a_{1}\right ) +x^{2}y\left ( -3a_{2}\right ) +x^{3}\left ( b_{1}\right ) +x^{2}\left ( -a_{0}\right ) +y\left ( a_{0}\right ) -x\left ( b_{0}\right ) & =0 \end {align*}

Setting each coeﬃcient to zero gives\begin {align*} b_{0} & =0\\ a_{0} & =0\\ a_{1} & =0\\ b_{1} & =0\\ a_{2} & =0\\ b_{2} & =0 \end {align*}

This shows there is no solution for this ansatz. There are more solutions depending on what ansatz we used. We just need one to obtain the ﬁnal solution. In Maple, these solutions can be found as follows

ode:=diff(y(x),x)= y(x)/x+x;
DEtools:-symgen(ode,y(x),way=all)
[_xi = 0, _eta = x],
[_xi = 0, _eta = x],
[_xi = 0, _eta = x^2 - y],
[_xi = x, _eta = 2*y],
[_xi = 1, _eta = y/x],
[_xi = x^2 + y, _eta = 4*y*x],
[_xi = x^2 - 3*y, _eta = -4*y^2/x]



Trying ansatz using functional form. Let $$\xi =0,\eta =f\left ( x\right )$$ then $$\xi _{x}=0,\xi _{y}=0,\eta _{x}=f^{\prime }\left ( x\right ) ,\eta _{y}=0$$ and the determining equation (A) becomes\begin {align*} x^{2}\eta _{x}+\left ( yx+x^{3}\right ) \eta _{y}-\xi _{x}\left ( yx+x^{3}\right ) -\left ( y^{2}+x^{4}+2yx^{2}\right ) \xi _{y}-\left ( -y+x^{2}\right ) \xi -x\eta & =0\\ x^{2}f^{\prime }\left ( x\right ) -xf\left ( x\right ) & =0\\ xf^{\prime }\left ( x\right ) -f\left ( x\right ) & =0 \end {align*}

This is easily solved to give $$f=cx$$. Hence $$\xi =0,\eta =x$$ by choosing $$c=1$$. We see that this choice of ansatz was the easiest in this case, as the ode generated was linear. Let us try another and see what happens.

Trying ansatz as $$\xi =0,\eta =f\left ( y\right )$$ then $$\xi _{x}=0,\xi _{y}=0,\eta _{x}=0,\eta _{y}=f^{\prime }\left ( y\right )$$ and the determining equation (A) becomes\begin {align*} \left ( yx+x^{3}\right ) f^{\prime }\left ( y\right ) -xf\left ( y\right ) & =0\\ \left ( y+x^{2}\right ) f^{\prime }\left ( y\right ) -f\left ( y\right ) & =0 \end {align*}

This is separable and its solution is $$f=c_{1}\left ( x^{2}+y\right )$$. Hence $$\xi =0,\eta =\left ( x^{2}+y\right )$$ by using $$c_{1}=1$$. But this is not function of $$y$$ only. So this choice did not work. Trying $$\left [ \xi =f\left ( x\right ) ,\eta =0\right ] ,\left [ \xi =f\left ( y\right ) ,\eta =0\right ]$$ shows these also do not work.

$$\xi ,\eta$$ can be checked for validity by substituting them in the PDE. Maple’s symtest command does this. These functional ansatz’s lead to an ode which have to be solved.

##### 1.10.4 Example $$y^{\prime }=xy^{2}-\frac {2y}{x}-\frac {1}{x^{3}}$$

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

For $$x\neq 0$$.  Given dilation transformation \begin {align} \bar {x} & =e^{\epsilon }x\tag {2}\\ \bar {y} & =e^{-2\epsilon }y\nonumber \end {align}

Hence\begin {align*} \xi \left ( x,y\right ) & =\left . \frac {d\bar {x}}{d\epsilon }\right \vert _{\epsilon =0}=x\\ \eta \left ( x,y\right ) & =\left . \frac {d\bar {y}}{d\epsilon }\right \vert _{\epsilon =0}=-2y \end {align*}

(At end shows how to obtain these). The integrating factor is therefore \begin {align*} \mu \left ( x,y\right ) & =\frac {1}{\eta -\xi \omega }\\ & =\frac {1}{-2y-x\left ( xy^{2}-\frac {2y}{x}-\frac {1}{x^{3}}\right ) }\\ & =-\frac {x^{2}}{x^{4}y^{2}-1} \end {align*}

Now\begin {align} \bar {x} & =x+\xi \epsilon =x+\epsilon x\tag {3}\\ \bar {y} & =y+\eta \epsilon =y-2y\epsilon \nonumber \end {align}

This transformation $$\bar {x}=e^{\epsilon }x,\bar {y}=e^{-2\epsilon }y$$ is now veriﬁed that it keeps the ode invariant. $\frac {d\bar {y}}{d\bar {x}}=\frac {\bar {y}_{x}+\bar {y}_{y}\frac {dy}{dx}}{\bar {x}_{x}+\bar {x}_{y}\frac {dy}{dx}}=\frac {e^{-2\epsilon }\frac {dy}{dx}}{e^{\epsilon }}=e^{-3\epsilon }\frac {dy}{dx}$ Substituting $$\bar {x},\bar {y},\frac {d\bar {y}}{d\bar {x}}$$ in the original ode gives\begin {align*} \frac {d\bar {y}}{d\bar {x}} & =\bar {x}\bar {y}^{2}-\frac {2\bar {y}}{\bar {x}}-\frac {1}{\bar {x}^{3}}\\ e^{-3\epsilon }\frac {dy}{dx} & =\left ( e^{\epsilon }x\right ) \left ( e^{-2\epsilon }y\right ) ^{2}-\frac {2\left ( e^{-2\epsilon }y\right ) }{\left ( e^{\epsilon }x\right ) }-\frac {1}{\left ( e^{\epsilon }x\right ) ^{3}}\\ e^{-3\epsilon }\frac {dy}{dx} & =e^{-3\epsilon }xy^{2}-\frac {2e^{-3\epsilon }y}{x}-\frac {e^{-3\epsilon }}{x^{3}}\\ \frac {dy}{dx} & =xy^{2}-\frac {2y}{x}-\frac {1}{x^{3}} \end {align*}

Which is the original ode. Hence the transformation (2) is invariant. It is important to use (2) and not (3) when doing the veriﬁcation.

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$R=yx^{2}$ $$S\left ( x,y\right )$$ 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 x \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. But \begin {align*} S_{x} & =\frac {1}{x}\\ S_{y} & =0\\ R_{x} & =2yx\\ R_{y} & =x^{2} \end {align*}

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 $$S=\ln x,R=yx^{2}$$, 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 & =\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

We start with order $$1$$ polynomials. Hence\begin {align} \xi & =a_{0}+a_{1}x\tag {1}\\ \eta & =b_{0}+b_{1}y \tag {2} \end {align}

If this does not generate solution, we will try higher order polynomials. 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\\ 0+\omega \left ( b_{1}-a_{1}\right ) -\omega ^{2}\left ( 0\right ) -\omega _{x}\left ( a_{0}+a_{1}x\right ) -\omega _{y}\left ( b_{0}+b_{1}y\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*} \left ( xy^{2}-\frac {2y}{x}-\frac {1}{x^{3}}\right ) \left ( b_{1}-a_{1}\right ) -\left ( y^{2}+\frac {2y}{x^{2}}+\frac {3}{x^{4}}\right ) \left ( a_{0}+a_{1}x\right ) -\left ( 2yx-\frac {2}{x}\right ) \left ( b_{0}+b_{1}y\right ) & =0\\ xy^{2}b_{1}-\frac {2y}{x}b_{1}-\frac {1}{x^{3}}b_{1}-xy^{2}a_{1}+\frac {2y}{x}a_{1}+\frac {1}{x^{3}}a_{1}-y^{2}a_{0}-\frac {2y}{x^{2}}a_{0}-\frac {3}{x^{4}}a_{0}-xy^{2}a_{1}-a_{1}\frac {2y}{x}-a_{1}\frac {3}{x^{3}}-2yxb_{0}+\frac {2}{x}b_{0}-2y^{2}xb_{1}+\frac {2y}{x}b_{1} & =0\\ xy^{2}\left ( b_{1}-a_{1}-a_{1}-2b_{1}\right ) +\frac {y}{x}\left ( -2b_{1}+2a_{1}-2a_{1}+2b_{1}\right ) +\frac {1}{x^{3}}\left ( -b_{1}+a_{1}-3a_{1}\right ) +y^{2}\left ( -a_{0}\right ) +\frac {y}{x^{2}}\left ( -2a_{0}\right ) +\frac {1}{x^{4}}\left ( -3a_{0}\right ) +yx\left ( -2b_{0}\right ) +\frac {1}{x}\left ( 2b_{0}\right ) & =0 \end {align*}

Each coeﬃcient to each monomial must be zero. Hence\begin {align*} -2a_{1}-b_{1} & =0\\ -b_{1}-2a_{1} & =0\\ -2a_{1}-2b_{1} & =0\\ a_{0} & =0\\ b_{0} & =0 \end {align*}

These are overdetermined equations. Solving gives $$a_{1}=-\frac {1}{2}b_{1}$$ and $$a_{0}=b_{0}=0$$. Choosing $$b_{1}=-2$$ gives $$a_{1}=1$$. Hence\begin {align*} \xi & =a_{0}+a_{1}x=x\\ \eta & =b_{0}+b_{1}y=-2y \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*}

##### 1.10.5 Example $$y^{\prime }=\frac {y+1}{x}+\frac {y^{2}}{x^{3}}$$

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

This can be written as

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

Hence this has the form $$y^{\prime }=\frac {y}{x}+g\left ( x\right ) F\left ( \frac {y}{x}\right )$$ where $$g\left ( x\right ) =\frac {1}{x}$$ and $$F=\left ( 1+\left ( \frac {y}{x}\right ) ^{2}\right )$$. Therefore this is homogeneous class D. Lookup table gives  \begin {align*} \xi & =x^{2}\\ \eta & =xy \end {align*}

Another way to ﬁnd $$\xi ,\eta$$ is by solving the symmetry condition PDE and this is shown at the end of this problem. Hence

\begin {align} \bar {x} & =x+\xi \epsilon \nonumber \\ & =x+x^{2}\epsilon \nonumber \\ \bar {y} & =y+\eta \epsilon \nonumber \\ & =y+xy\epsilon \tag {2} \end {align}

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

The ode is now veriﬁed that it remains invariant under (2) transformation.\begin {align*} \frac {d\bar {y}}{d\bar {x}} & =\frac {\frac {d\bar {y}}{dx}}{\frac {d\bar {x}}{dx}}\\ & =\frac {\bar {y}_{x}+\bar {y}_{y}\frac {dy}{dx}}{\bar {x}_{x}+\bar {x}_{y}\frac {dy}{dx}} \end {align*}

But from (2) $$\bar {y}_{x}=y\epsilon ,\bar {y}_{y}=1+x\epsilon ,\bar {x}_{x}=1+2x\epsilon ,\bar {x}_{y}=0$$ and the above becomes$\frac {d\bar {y}}{d\bar {x}}=\frac {1+\left ( 1+x\epsilon \right ) \frac {dy}{dx}}{1+2x\epsilon }$ Substituting $$\bar {x},\bar {y},\frac {d\bar {y}}{d\bar {x}}$$ in the original ode gives\begin {align*} \frac {d\bar {y}}{d\bar {x}} & =\frac {\bar {y}+1}{\bar {x}}+\frac {\bar {y}^{2}}{\bar {x}^{3}}\\ \frac {1+\left ( 1+x\epsilon \right ) \frac {dy}{dx}}{1+2x\epsilon } & =\frac {\left ( y+xy\epsilon \right ) +1}{x+x^{2}\epsilon }+\frac {\left ( y+xy\epsilon \right ) ^{2}}{\left ( x+x^{2}\epsilon \right ) ^{3}} \end {align*}

Which as $$\lim _{\epsilon \rightarrow 0}$$ gives $\frac {dy}{dx}=\frac {y+1}{x}+\frac {y^{2}}{x^{3}}$ The same original ode showing the transformation is valid symmetry.

Y:=y/(1-s*x):
X:=x/(1-s*x):
eq:=(diff(Y,x)+diff(Y,y)*Z)/(diff(X,x)+diff(X,y)*Z)=simplify((Y+1)/X+Y^2/X^3):
solve(simplify(eq),Z)
y/x + 1/x + y^2/x^3

`

Hence the transformation in (2) is invariant.

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

Each coeﬃcient to each monomial must be zero. Hence\begin {align*} c_{3}-c_{2} & =0\\ c_{5}-c_{6} & =0\\ 2c_{1}-c_{5} & =0\\ -2c_{2}+3c_{3} & =0\\ c_{2} & =0\\ -c_{2}+c_{3}-2c_{4} & =0\\ -2c_{6} & =0 \end {align*}

Which simpliﬁes to (since $$c_{2}=0,c_{6}=0$$)\begin {align*} c_{3} & =0\\ c_{5} & =0\\ c_{1}-c_{5} & =0\\ 3c_{3} & =0\\ c_{3}-2c_{4} & =0 \end {align*}

Which simpliﬁes to (since $$c_{3}=0,c_{5}=0$$)\begin {align*} c_{5} & =0\\ c_{1}-c_{5} & =0\\ c_{4} & =0 \end {align*}

Hence $$c_{5}=0,c_{1}=0,c_{4}=0$$. We see that all $$c_{i}=0$$, therefore there is no solution using this ansatz.

Trying ansatz\begin {align*} \xi & =a_{0}+a_{1}x+a_{2}y+a_{3}xy+a_{4}x^{2}\\ \eta & =b_{0}+b_{1}x+b_{2}y+b_{3}xy+b_{4}y^{2} \end {align*}

Eq 9 becomes$\eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0$ Substituting the ansatz and simplifying gives$-x^{2}y^{3}a_{2}+y^{4}a_{2}+x^{4}(-a_{0}+a_{2})+x^{2}y^{2}(-3a_{0}+2a_{2})+xy^{4}a_{3}+2x^{3}yb_{0}+x^{4}y(-a_{0}+a_{2}+2b_{1})+x^{5}(a_{3}+b_{0}-b_{2})+x^{3}y^{2}(-2a_{1}+2a_{3}+b_{2})+x^{6}(a_{4}-b_{3})+x^{6}y(a_{4}-b_{3})+x^{4}y^{2}(-a_{4}+b_{3})+x^{5}y(2a_{3}-2b_{4})+x^{5}y^{2}(a_{3}-b_{4})=0$ Each coeﬃcient to each monomial must be zero. Hence\begin {align*} a_{2} & =0\\ -a_{0}+a_{2} & =0\\ -3a_{0}+2a_{2} & =0\\ a_{3} & =0\\ b_{0} & =0\\ -a_{0}+a_{2}+2b_{1} & =0\\ a_{3}+b_{0}-b_{2} & =0\\ -2a_{1}+2a_{3}+b_{2} & =0\\ a_{4}-b_{3} & =0\\ 2a_{3}-2b_{4} & =0\\ a_{3}-b_{4} & =0 \end {align*}

Since $$a_{2}=a_{3}=b_{0}=0$$ the above simpliﬁes to \begin {align*} -a_{0} & =0\\ -3a_{0} & =0\\ -a_{0}+2b_{1} & =0\\ -b_{2} & =0\\ -2a_{1}+b_{2} & =0\\ a_{4}-b_{3} & =0\\ -2b_{4} & =0\\ -b_{4} & =0 \end {align*}

Since $$a_{0}=b_{2}=a_{4}=b_{4}=0$$, The above now simpliﬁes to$a_{4}-b_{3}=0$ Therefore, if we let $$a_{4}=1$$ then $$b_{3}=1$$ and the solution is \begin {align*} \xi & =a_{0}+a_{1}x+a_{2}y+a_{3}xy+a_{4}x^{2}\\ & =x^{2}\\ \eta & =b_{0}+b_{1}x+b_{2}y+b_{3}xy+b_{5}y^{2}\\ & =xy \end {align*}

Which is what we used above to solve the ode.

##### 1.10.6 Example $$y^{\prime }=\frac {y-4xy^{2}-16x^{3}}{y^{3}+4x^{2}y+x}$$

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

The integrating factor is therefore \begin {align*} \mu \left ( x,y\right ) & =\frac {1}{\eta -\xi \omega }\\ & =\frac {1}{4x+y\left ( \frac {y-4xy^{2}-16x^{3}}{y^{3}+4x^{2}y+x}\right ) }\\ & =\frac {x^{2}y+x+y^{3}}{4x^{2}+y^{2}} \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. 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 ﬁrst pair of ode’s in (1) gives$\frac {dy}{dx}=-\frac {4x}{y}$ Solving gives $y=\sqrt {-4x^{2}+c}$ Where $$c$$ is constant of integration (For $$y>0$$ only). In this method $$R$$ is always $$c$$. Hence\begin {align} y^{2} & =-4x^{2}+c\nonumber \\ R & =y^{2}+4x^{2} \tag {2} \end {align}

The ﬁrst equation in (1) and the last equation gives\begin {align*} dS & =\frac {dx}{\xi }\\ S & =-\int \frac {dx}{y} \end {align*}

But $$y=\sqrt {-4x^{2}+c}$$. The above becomes \begin {align*} S & =-\int \frac {dx}{\sqrt {-4x^{2}+c}}\\ & =-\frac {1}{2}\arctan \left ( \frac {2x}{\sqrt {-4x^{2}+c}}\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 14 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$c_{4}+\left ( \frac {y-4xy^{2}-16x^{3}}{y^{3}+4x^{2}y+x}\right ) \left ( c_{5}-c_{1}\right ) -\left ( \frac {y-4xy^{2}-16x^{3}}{y^{3}+4x^{2}y+x}\right ) ^{2}c_{2}-\left ( \frac {-4y^{5}-32x^{2}y^{3}-8xy^{2}+\left ( -64x^{4}-1\right ) y-32x^{3}}{\left ( 4x^{2}y+y^{3}+x\right ) ^{2}}\right ) \left ( c_{1}x+c_{2}y+c_{3}\right ) -\left ( \frac {64x^{5}+32x^{3}y^{2}+4xy^{4}-8x^{2}y-2y^{3}+x}{\left ( 4x^{2}y+y^{3}+x\right ) ^{2}}\right ) \left ( c_{4}x+c_{5}y+c_{6}\right ) =0$ 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.

##### 1.10.7 Example $$y^{\prime }=\frac {-y^{2}}{e^{x}-y}$$

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

The symmetry condition results in the PDE$\eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0$ End of the problem shows how this is solved for $$\xi ,\eta$$ which results in\begin {align*} \xi \left ( x,y\right ) & =1\\ \eta \left ( x,y\right ) & =y \end {align*}

The integrating factor is therefore \begin {align*} \mu \left ( x,y\right ) & =\frac {1}{\eta -\xi \omega }\\ & =\frac {1}{y-\left ( \frac {-y^{2}}{e^{x}-y}\right ) }\\ & =\frac {1-ye^{-x}}{y} \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}{1} & =\frac {dy}{y}=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$\frac {dy}{dx}=y$ Integrating gives $$\ln \left \vert y\right \vert =x+c$$ or $$y=ce^{x}$$ where $$c$$ is constant of integration. In this method $$R$$ is always $$c$$. Hence$R\left ( x,y\right ) =ye^{-x}$ $$S\left ( x,y\right )$$ is now found from the ﬁrst equation in (1) and the last equation which gives\begin {align*} dS & =\frac {dx}{\xi }\\ dS & =\frac {dx}{1}\\ dS & =dx\\ S & =x \end {align*}

Hence\begin {align*} R & =ye^{-x}\\ S & =x \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. $$S_{x}=1,R_{x}=-ye^{-x},S_{y}=0,R_{y}=e^{-x}$$. Substituting gives\begin {align*} \frac {dS}{dR} & =\frac {1}{-ye^{-x}+\frac {-y^{2}}{e^{x}-y}e^{-x}}\\ & =\frac {ye^{-x}-1}{ye^{-x}} \end {align*}

But $$R=ye^{-x}$$, hence the above becomes$\frac {dS}{dR}=\frac {R-1}{R}$ This is just quadrature. Integrating gives \begin {align*} S & =\int \frac {R-1}{R}dR\\ & =R-\ln R+c_{1} \end {align*}

This solution is converted back to $$x,y$$. Since $$S=x,R=ye^{-x}$$, the above becomes$x=ye^{-x}-\ln \left ( ye^{-x}\right ) +c_{1}$ Which is the solution to the original ODE.

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

Hence $$\xi _{x}=c_{1},\xi _{y}=c_{2},\eta _{x}=c_{4},\eta _{y}=c_{5}$$  and (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 $$\omega =\frac {-y^{2}}{e^{x}-y},\omega _{x}=\frac {y^{2}e^{x}}{\left ( e^{x}-y\right ) ^{2}},\omega _{y}=\left ( -\frac {2y}{e^{x}-y}-\frac {y^{2}}{\left ( e^{x}-y\right ) ^{2}}\right )$$ and the above becomes$c_{4}+\frac {-y^{2}}{e^{x}-y}\left ( c_{5}-c_{1}\right ) -\left ( \frac {-y^{2}}{e^{x}-y}\right ) ^{2}c_{2}-\frac {y^{2}e^{x}}{\left ( e^{x}-y\right ) ^{2}}\left ( c_{1}x+c_{2}y+c_{3}\right ) -\left ( -\frac {2y}{e^{x}-y}-\frac {y^{2}}{\left ( e^{x}-y\right ) ^{2}}\right ) \left ( c_{4}x+c_{5}y+c_{6}\right ) =0$ Need to do this again. I should get $$c_{3}=1,c_{5}=1$$ and everything else zero.\begin {align*} \xi & =1\\ \eta & =y \end {align*}

##### 1.10.8 Example $$y^{\prime }=\frac {x\sqrt {1+y}+\sqrt {1+y}+1+y}{1+x}$$

Solve \begin {align*} y^{\prime } & =\frac {x\sqrt {1+y}+\sqrt {1+y}+1+y}{1+x}\\ 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 Ansatz be\begin {align*} \xi & =0\\ \eta & =f\left ( x\right ) g\left ( y\right ) \end {align*}

Hence (1) becomes$g\left ( y\right ) \frac {df}{dx}+\omega f\left ( x\right ) \frac {dg}{dy}-\omega _{y}f\left ( x\right ) g\left ( y\right ) =0$ But $$\omega _{x}=\frac {d}{dx}\left ( \frac {x\sqrt {1+y}+\sqrt {1+y}+1+y}{1+x}\right ) =-\frac {\left ( y+1\right ) }{\left ( x+1\right ) ^{2}}$$ and $$\omega _{y}=\frac {x+1+2\sqrt {1+y}}{\sqrt {1+y}\left ( 2+2x\right ) }$$. Hence the above becomes\begin {equation} g\left ( y\right ) \frac {df}{dx}+\left ( \frac {x\sqrt {1+y}+\sqrt {1+y}+1+y}{1+x}\right ) f\left ( x\right ) \frac {dg}{dy}-\frac {x+1+2\sqrt {1+y}}{\sqrt {1+y}\left ( 2+2x\right ) }\left ( f\left ( x\right ) g\left ( y\right ) \right ) =0 \tag {2} \end {equation} The numerator of the normal form of the above is\begin {equation} 2\frac {df}{dx}g\sqrt {1+y}x+2y\sqrt {1+y}f\frac {dg}{dy}+2f\frac {dg}{dy}xy+2\frac {df}{dx}g\sqrt {1+y}-2fg\sqrt {1+y}+2f\frac {dg}{dy}\sqrt {1+y}-fgx+2f\frac {dg}{dy}x+2fy\frac {dg}{dy}-fg+2f\frac {dg}{dy}=0 \tag {3} \end {equation} We can now either collect on $$y$$ or $$x$$ and try. Let us start with collecting on all terms with $$y$$. This gives\begin {equation} g\sqrt {1+y}\left ( 2x\frac {df}{dx}+2\frac {df}{dx}-2f\right ) +y\sqrt {1+y}\frac {dg}{dy}\left ( 2f\right ) +\frac {dg}{dy}\sqrt {1+y}\left ( 2f\right ) +g\left ( xf-f\right ) +y\frac {dg}{dy}\left ( 2xf+2f\right ) +\frac {dg}{dy}\left ( 2xf+2f\right ) =0 \tag {3A} \end {equation} The coeﬃcients of all terms with $$g\left ( y\right )$$ or $$y$$ in them are from the above are the following, which each must be zero\begin {align*} 2f & =0\\ xf-f & =0\\ 2xf+2f & =0\\ 2x\frac {df}{dx}+2\frac {df}{dx}-2f & =0 \end {align*}

Now we set each to zero and see if this produces $$f\left ( x\right )$$ which can be used. We have 4 choices to try above. Starting from the most simple one. The ﬁrst one above gives $$2f=0$$ or $$f=0$$. But this is not function of $$x$$. We try the next one $$xf-f=0$$. This gives $$f=0$$ or $$x=1$$. Hence this does not give $$f$$ as function of $$x.$$ Next we try $$2xf+2f.$$ This also does not give $$f$$ as function of $$x.$$ The last one is $$2x\frac {df}{dx}+2\frac {df}{dx}-2f=0$$ or $$\frac {df}{dx}=\frac {2f}{2x+2}$$. Solving this gives $$f=c_{1}\left ( x+1\right )$$. This is successful since $$f$$ is function of $$x$$. Hence\begin {align*} f\left ( x\right ) & =c_{1}\left ( x+1\right ) \\ \frac {df}{dx} & =c_{1} \end {align*}

Now we need to determine $$g\left ( y\right )$$. Substituting the above into (3) gives$2c_{1}g(y)\sqrt {1+y}x+2\sqrt {1+y}c_{1}(x+1)\frac {dg}{dy}y+2c_{1}(x+1)\frac {dg}{dy}xy+2c_{1}g\sqrt {1+y}-2c_{1}(x+1)g\sqrt {1+y}+2c_{1}(x+1)\frac {dg}{dy}\sqrt {1+y}-c_{1}(x+1)g(y)x+2c_{1}(x+1)\frac {dg}{dy}x+2c_{1}(x+1)\frac {dg}{dy}y-c_{1}(x+1)g(y)+2c_{1}(x+1)\frac {dg}{dy}=0$ Which simpliﬁes to\begin {equation} 2c_{1}\sqrt {1+y}\frac {dg}{dy}yx+2c_{1}\frac {dg}{dy}x^{2}y-c_{1}gx^{2}+2c_{1}\frac {dg}{dy}\sqrt {1+y}x+2\sqrt {1+y}c_{1}\frac {dg}{dy}y+2c_{1}\frac {dg}{dy}x^{2}+4c_{1}\frac {dg}{dy}xy-2c_{1}xg+2c_{1}\frac {dg}{dy}\sqrt {1+y}+4c_{1}\frac {dg}{dy}x+2c_{1}\frac {dg}{dy}y-c_{1}g(y)+2c_{1}\frac {dg}{dy}=0 \tag {4} \end {equation} Now factoring on all terms with $$x$$, and these are $$\left \{ x,x^{2}\right \}$$ gives\begin {equation} -c_{1}x^{2}\left ( -2\frac {dg}{dy}y+g-2\frac {dg}{dy}\right ) -c_{1}x\left ( -2\sqrt {1+y}\frac {dg}{dy}y-2\sqrt {1+y}\frac {dg}{dy}-2\frac {dg}{dy}y+g-2\frac {dg}{dy}\right ) +T=0 \tag {4A} \end {equation} Where $$T$$ are terms that depends on $$y$$ only.  Each factor of $$x,x^{2}$$ must be zero. Hence the ﬁrst above implies\begin {align*} -2\frac {dg}{dy}y+g-2\frac {dg}{dy} & =0\\ g^{\prime }\left ( y\right ) & =\frac {g}{2\left ( 1+y\right ) } \end {align*}

Solving gives\begin {equation} g=c_{2}\sqrt {1+y} \tag {5} \end {equation} Substituting (5) into (4) gives$c_{1}\left ( 1+x\right ) c_{2}\left ( 1+y\right ) =0$ Which is not zero. Hence this term does not work. Now we try the second term in (4A) which means\begin {align*} -2\sqrt {1+y}\frac {dg}{dy}y-2\sqrt {1+y}\frac {dg}{dy}-2\frac {dg}{dy}y+g-2\frac {dg}{dy} & =0\\ \frac {dg}{dy} & =\frac {-g}{-2\sqrt {1+y}y-2\sqrt {1+y}-2y-2} \end {align*}

Solving gives$g\left ( y\right ) =c_{2}\frac {\sqrt {1+y}}{1+\sqrt {1+y}}$ Again, substituting the above back in (4) gives$c_{1}\left ( 1+x\right ) c_{2}\frac {\left ( 1+y\right ) x}{\left ( 1+\sqrt {1+y}\right ) ^{2}}=0$ Which is not zero. Therefore starting with $$f\left ( x\right ) =c_{1}\left ( x+1\right )$$ has failed to produce a valid $$g\left ( y\right )$$ to satisfy the pde. This means we need to start all over again. Going back to (3) and now collecting on all terms with $$x$$ instead. Here is (3) again\begin {equation} 2\frac {df}{dx}g\sqrt {1+y}x+2y\sqrt {1+y}f\frac {dg}{dy}+2f\frac {dg}{dy}xy+2\frac {df}{dx}g\sqrt {1+y}-2fg\sqrt {1+y}+2f\frac {dg}{dy}\sqrt {1+y}-fgx+2f\frac {dg}{dy}x+2fy\frac {dg}{dy}-fg+2f\frac {dg}{dy}=0 \tag {3} \end {equation} Collecting on all terms that depend on $$x$$ gives\begin {equation} x\frac {df}{dx}\left ( 2g\sqrt {1+y}\right ) +f\left ( 2y\sqrt {1+y}\frac {dg}{dy}-2g\sqrt {1+y}+2\frac {dg}{dy}\sqrt {1+y}+2y\frac {dg}{dy}+2\frac {dg}{dy}-g\right ) +xf\left ( 2\frac {dg}{dy}y-g+2\frac {dg}{dy}y\right ) =0 \tag {3B} \end {equation} Each term must be zero, hence this gives these trials\begin {align*} 2g\sqrt {1+y} & =0\\ 2\frac {dg}{dy}y-g+2\frac {dg}{dy}y & =0\\ 2y\sqrt {1+y}\frac {dg}{dy}-2g\sqrt {1+y}+2\frac {dg}{dy}\sqrt {1+y}+2y\frac {dg}{dy}+2\frac {dg}{dy}-g & =0 \end {align*}

Starting with the ﬁrst one above $$2g\sqrt {1+y}=0$$ which gives $$g=0$$ which does not match the ansatz. Now we try the second one above, which gives $\frac {dg}{dy}=\frac {g}{2+2y}$ Solving gives\begin {equation} g=c_{1}\sqrt {1+y} \tag {6} \end {equation} Which meets the  requirements of the ansatz. Now we need to use the above to generate $$f\left ( x\right )$$. We do not need to try the third one above unless this fails. Substituting (6) into (3) gives\begin {align} c_{2}\left ( 2\frac {df}{dx}xy+2\frac {df}{dx}x+2\frac {df}{dx}y-fy+2\frac {df}{dx}-f\right ) & =0\nonumber \\ 2\frac {df}{dx}xy+2\frac {df}{dx}x+2\frac {df}{dx}y-fy+2\frac {df}{dx}-f & =0 \tag {7} \end {align}

Collecting on $$y$$ gives$c_{1}\left ( 1+y\right ) \left ( 2\frac {df}{dx}x+2\frac {df}{dx}-f\right ) =0$ Hence $$2\frac {df}{dx}x+2\frac {df}{dx}-f$$ must be zero. This gives as solution \begin {align*} f\left ( x\right ) & =c_{2}\sqrt {1+x}\\ \frac {df}{dx} & =c_{2}\frac {1}{2\sqrt {1+x}} \end {align*}

Substituting the above into (7) to verify gives\begin {align*} 2\left ( c_{2}\frac {1}{2\sqrt {1+x}}\right ) xy+2\left ( c_{2}\frac {1}{2\sqrt {1+x}}\right ) x+2\left ( c_{2}\frac {1}{2\sqrt {1+x}}\right ) y-\left ( c_{2}\sqrt {1+x}\right ) y+2\left ( c_{2}\frac {1}{2\sqrt {1+x}}\right ) -c_{2}\sqrt {1+x} & =0\\ c_{2}\frac {1}{\sqrt {1+x}}xy+c_{2}\frac {1}{\sqrt {1+x}}x+c_{2}\frac {1}{\sqrt {1+x}}y-c_{2}\sqrt {1+x}y+c_{2}\frac {1}{\sqrt {1+x}}-c_{2}\sqrt {1+x} & =0\\ c_{2}\left ( \frac {1}{\sqrt {1+x}}xy+\frac {1}{\sqrt {1+x}}x+\frac {1}{\sqrt {1+x}}y-\sqrt {1+x}y+\frac {1}{\sqrt {1+x}}-\sqrt {1+x}\right ) & =0\\ 0 & =0 \end {align*}

Veriﬁed, Hence we have found $$f\left ( x\right ) ,g\left ( y\right )$$. Therefore\begin {align*} \xi & =0\\ \eta & =f\left ( x\right ) g\left ( y\right ) \\ & =\sqrt {1+x}\sqrt {1+y} \end {align*}

Where we set $$c_{1}=c_{2}=1$$. The integrating factor is therefore \begin {align*} \mu \left ( x,y\right ) & =\frac {1}{\eta -\xi \omega }\\ & =\frac {1}{\sqrt {1+x}\sqrt {1+y}} \end {align*}

The next step is to determine the canonical coordinates $$R,S$$. Where $$R$$ is the independent variable and $$S$$ is the dependent variable. This is done by using the standard characteristic equation by writing\begin {equation} \frac {dx}{\xi }=\frac {dy}{\eta }=dS\nonumber \end {equation} For the special case $$\xi =0\,$$ we have $$R=x$$. $$S\left ( x,y\right )$$ is now found from the last two pair of equations which gives\begin {align*} dS & =\frac {dy}{\eta }\\ dS & =\frac {dy}{\sqrt {1+x}\sqrt {1+y}}\\ S & =2\frac {\sqrt {1+y}}{\sqrt {1+x}} \end {align*}

Hence (constant of integration is set to zero)\begin {align} R & =x\tag {2}\\ S & =2\frac {\sqrt {1+y}}{\sqrt {1+x}}\nonumber \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. $$S_{x}=-\frac {\sqrt {1+y}}{\left ( 1+x\right ) ^{\frac {3}{2}}},R_{x}=1,S_{y}=\frac {1}{\sqrt {1+x}\sqrt {1+y}},R_{y}=0$$. Substituting into the above gives\begin {align*} \frac {dS}{dR} & =-\frac {\sqrt {1+y}}{\left ( 1+x\right ) ^{\frac {3}{2}}}+\omega \left ( x,y\right ) \frac {1}{\sqrt {1+x}\sqrt {1+y}}\\ & =-\frac {\sqrt {1+y}}{\left ( 1+x\right ) ^{\frac {3}{2}}}+\left ( \frac {x\sqrt {1+y}+\sqrt {1+y}+1+y}{1+x}\right ) \frac {1}{\sqrt {1+x}\sqrt {1+y}}\\ & =\frac {1}{\sqrt {x+1}}\\ & =\frac {1}{\sqrt {R+1}} \end {align*}

Hence $\frac {dS}{dR}=\frac {1}{\sqrt {R+1}}$ This is quadrature. Solving gives $S=2\sqrt {R+1}+c_{1}$ Convecting back to $$x,y\,$$ gives$2\frac {\sqrt {1+y}}{\sqrt {1+x}}=2\sqrt {x+1}+c_{1}$

##### 1.10.9 Example $$y^{\prime }=\frac {-y}{2x-ye^{y}}$$

Solve \begin {align*} y^{\prime } & =\frac {-y}{2x-ye^{y}}\\ 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 & =g\left ( y\right ) \\ \eta & =0 \end {align*}

Substituting this into (1) gives$-\omega ^{2}\frac {dg}{dy}-\omega _{x}g=0$ But $$\omega ^{2}=\frac {y^{2}}{\left ( 2x-ye^{y}\right ) ^{2}},\omega _{x}=\frac {d}{dx}\left ( \frac {-y}{2x-ye^{y}}\right ) =\frac {2y}{\left ( 2x-ye^{y}\right ) ^{2}}$$. The above becomes\begin {align*} -\frac {y^{2}}{\left ( 2x-ye^{y}\right ) ^{2}}\frac {dg}{dy}-\frac {2y}{\left ( 2x-ye^{y}\right ) ^{2}}g & =0\\ -y^{2}\frac {dg}{dy}-2yg & =0\\ \frac {dg}{dy}+\frac {2}{y}g & =0 \end {align*}

This is linear ode. The solution is $g=\frac {c_{1}}{y^{2}}$ Hence\begin {align*} \xi & =\frac {1}{y^{2}}\\ \eta & =0 \end {align*}

But taking $$c_{1}=1$$.  The integrating factor is therefore \begin {align*} \mu \left ( x,y\right ) & =\frac {1}{\eta -\xi \omega }\\ & =\frac {1}{-\frac {1}{y^{2}}\left ( \frac {-y}{2x-ye^{y}}\right ) }\\ & =y\left ( 2x-ye^{y}\right ) \end {align*}

The next step is to determine the canonical coordinates $$R,S$$. Where $$R$$ is the independent variable and $$S$$ is the dependent variable. This is done by using the standard characteristic equation by writing\begin {equation} \frac {dx}{\xi }=\frac {dy}{\eta }=dS\nonumber \end {equation} Since$$\ \eta =0$$, then in this special case $$R=c_{1}=y$$. To ﬁnd $$S$$ we use $$dS=\frac {dx}{\xi }$$ or $$dS=y^{2}dx$$. Hence $$S=c_{1}^{2}x+c_{2}=c_{1}^{2}x$$ by taking $$c_{2}=0$$. Therefore  $$S=y^{2}x$$ since $$c_{1}=y$$.\begin {align} R & =y\tag {2}\\ S & =y^{2}x\nonumber \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. $$S_{x}=y^{2},R_{x}=0,S_{y}=2yx,R_{y}=1$$. Substituting into the above gives \begin {align*} \frac {dS}{dR} & =\frac {y^{2}+\omega \left ( x,y\right ) 2yx}{\omega \left ( x,y\right ) }\\ & =\frac {y^{2}+\left ( \frac {-y}{2x-ye^{y}}\right ) 2yx}{\left ( \frac {-y}{2x-ye^{y}}\right ) }\\ & =y^{2}e^{y} \end {align*}

Now we need to express the RHS in terms of $$R,S.$$ From (2) we see that $$y=R$$, hence the above becomes$\frac {dS}{dR}=R^{2}e^{R}$ This is quadrature. Solving gives $S=\left ( R^{2}-2R+2\right ) e^{R}+c_{1}$ Convecting back to $$x,y\,$$ gives$y^{2}x=\left ( y^{2}-2y+2\right ) e^{y}+c_{1}$

##### 1.10.10 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.

##### 1.10.11 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*}

##### 1.10.12 Example $$y^{\prime }=4\left ( yx\right ) ^{\frac {1}{3}}$$

Solve \begin {align*} y^{\prime } & =4\left ( yx\right ) ^{\frac {1}{3}}\\ 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$\left ( -16a_{1}+8b_{1}\right ) yx-4xb_{0}-4ya_{0}=0$ Setting all coeﬃcients to zero gives\begin {align*} -16a_{1}+8b_{1} & =0\\ b_{0} & =0\\ a_{0} & =0 \end {align*}

Hence $$a_{1}=\frac {1}{2}b_{1}$$. Letting $$b_{1}=1$$ then $$a_{1}=\frac {1}{2}$$ and the inﬁnitesimals are \begin {align*} \xi & =\frac {1}{2}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}{2}x\left ( 4\left ( yx\right ) ^{\frac {1}{3}}\right ) }\\ & =\frac {1}{y-2x\left ( xy\right ) ^{\frac {1}{3}}} \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$\frac {dy}{dx}=\frac {\eta }{\xi }=\frac {2y}{x}$ Solving gives$y=c_{1}x^{2}$ Hence \begin {equation} R=c_{1}=\frac {y}{x^{2}} \tag {2} \end {equation} And $$S$$ is found from $dS=\frac {dx}{\xi }=2\frac {dx}{x}$ Integrating gives\begin {align*} S & =2\ln x+c_{1}\\ & =2\ln x \end {align*}

By choosing $$c_{1}=0$$. Now the ODE $$\frac {dS}{dR}=F\left ( R\right )$$ is found 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 $$S_{x}=\frac {2}{x},R_{x}=-2\frac {y}{x^{3}},S_{y}=0,R_{y}=\frac {2}{x^{2}}$$. Substituting these into the above and simplifying gives\begin {align*} \frac {dS}{dR} & =\frac {x^{2}}{2x\left ( yx\right ) ^{\frac {1}{3}}-y}\\ & =\frac {1}{2\frac {1}{x}\left ( yx\right ) ^{\frac {1}{3}}-\frac {y}{x^{2}}}\\ & =\frac {1}{2y^{\frac {1}{3}}x^{-\frac {2}{3}}-\frac {y}{x^{2}}}\\ & =\frac {1}{2\left ( \frac {y}{x^{2}}\right ) ^{\frac {1}{3}}-\frac {y}{x^{2}}}\\ & =\frac {1}{2\left ( R\right ) ^{\frac {1}{3}}-R} \end {align*}

Hence$\frac {dS}{dR}=\frac {1}{2R^{\frac {1}{3}}-R}$ Which is a quadrature. Solving gives\begin {align*} \int dS & =\int \frac {1}{2R^{\frac {1}{3}}-R}dR\\ S & =-\frac {3}{2}\ln \left ( -2+R^{\frac {2}{3}}\right ) +c_{1} \end {align*}

Converting back to $$x,y$$ gives$2\ln x=-\frac {3}{2}\ln \left ( -2+\left ( \frac {y}{x^{2}}\right ) ^{\frac {2}{3}}\right ) +c_{1}$ The above can be simpliﬁed more if needed to solve for $$y\left ( x\right )$$ explicitly.

##### 1.10.13 Example $$y^{\prime }=2y+3e^{2x}$$

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

From the lookup table, since this is linear ode $$y^{\prime }=f\left ( x\right ) y+g\left ( x\right )$$ then \begin {align*} \xi & =0\\ \eta & =e^{\int fdx}\\ & =e^{\int 2dx}\\ & =e^{2x}. \end {align*}

If we were to use the integrating factor method, then \begin {align*} \mu \left ( x,y\right ) & =\frac {1}{\eta -\xi \omega }\\ & =\frac {1}{e^{2x}}\\ & =e^{-2x} \end {align*}

Then the general solution is \begin {align*} \int \mu \left ( x,y\right ) \left ( dy-\omega dx\right ) & =c_{1}\\ \int e^{-2x}\left ( dy-\left ( 2y+3e^{2x}\right ) dx\right ) & =c_{1}\\ \int e^{-2x}dy-\left ( 2ye^{-2x}+3\right ) dx & =c_{1}\\ \int e^{-2x}dy-2ye^{-2x}dx & =\int 3dx+c_{1}\\ \int d\left ( e^{-2x}y\right ) & =\int 3dx+c_{1} \end {align*}

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

But if we were to use the basic Lie symmetry method, then 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} Since $$\xi =0$$ then this is the special case where $$R=x$$. And $$S$$ is found from $dS=\frac {dy}{\eta }=e^{-2x}dy$ Integrating gives\begin {align*} S & =e^{-2x}y+c_{1}\\ & =e^{-2x}y \end {align*}

By choosing $$c_{1}=0$$. Now the ODE $$\frac {dS}{dR}=F\left ( R\right )$$ is found 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 $$S_{x}=-2e^{-2x}y,R_{x}=1,S_{y}=e^{-2x},R_{y}=0$$. Substituting these into the above and simplifying gives\begin {align*} \frac {dS}{dR} & =-2e^{-2x}y+\left ( 2y+3e^{2x}\right ) e^{-2x}\\ & =-2e^{-2x}y+2ye^{-2x}+3\\ & =3 \end {align*}

Which is a quadrature. Solving gives\begin {align*} \int dS & =\int 3dR\\ S & =3R+c_{1} \end {align*}

Converting back to $$x,y$$ gives\begin {align*} e^{-2x}y & =3x+c_{1}\\ y & =\left ( 3x+c_{1}\right ) e^{2x} \end {align*}

Of course, this ode is ﬁrst order linear and can be solved much easier using integrating factor method. But this is just to illustrate the Lie symmetry method.

##### 1.10.14 Example $$y^{\prime }=\frac {1}{3}\frac {2y+y^{3}-x^{2}}{x}$$

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

Using Maple the inﬁnitesimals are \begin {align*} \xi & =\frac {3}{2x^{\frac {1}{3}}}\\ \eta & =\frac {y}{x^{\frac {4}{3}}} \end {align*}

(Will need to show how to obtain these). Lets solve this using the integration factor method ﬁrst. The integrating factor is given by\begin {align*} \mu \left ( x,y\right ) & =\frac {1}{\eta -\xi \omega }\\ & =\frac {1}{\frac {y}{x^{\frac {4}{3}}}-\frac {3}{2x^{\frac {1}{3}}}\left ( \frac {1}{3}\frac {2y+y^{3}-x^{2}}{x}\right ) }\\ & =2\frac {x^{\frac {4}{3}}}{x^{2}-y^{3}} \end {align*}

Then the general solution is \begin {align*} \int \mu \left ( x,y\right ) \left ( dy-\omega dx\right ) & =c_{1}\\ \int 2\frac {x^{\frac {4}{3}}}{x^{2}-y^{3}}\left ( dy-\left ( \frac {1}{3}\frac {2y+y^{3}-x^{2}}{x}\right ) dx\right ) & =c_{1}\\ \int \left ( 2\frac {x^{\frac {4}{3}}}{x^{2}-y^{3}}dy-\left ( 2\frac {x^{\frac {4}{3}}}{x^{2}-y^{3}}\right ) \left ( \frac {1}{3}\frac {2y+y^{3}-x^{2}}{x}\right ) dx\right ) & =c_{1}\\ \int \left ( 2\frac {x^{\frac {4}{3}}}{x^{2}-y^{3}}dy-\left ( \frac {2}{3}\frac {x^{\frac {1}{3}}}{x^{2}-y^{3}}\right ) \left ( 2y+y^{3}-x^{2}\right ) dx\right ) & =c_{1} \end {align*}

Hence we need to ﬁnd $$F\left ( x,y\right )$$ s.t. $$dF=\left ( 2\frac {x^{\frac {4}{3}}}{x^{2}-y^{3}}dy-\left ( \frac {2}{3}\frac {x^{\frac {1}{3}}}{x^{2}-y^{3}}\right ) \left ( 2y+y^{3}-x^{2}\right ) dx\right )$$ which will make the solution $$F=c$$. Therefore\begin {align*} dF & =\frac {\partial F}{\partial x}dx+\frac {\partial F}{\partial y}dy\\ & =2\frac {x^{\frac {4}{3}}}{x^{2}-y^{3}}dy-\left ( \frac {2}{3}\frac {x^{\frac {1}{3}}}{x^{2}-y^{3}}\right ) \left ( 2y+y^{3}-x^{2}\right ) dx \end {align*}

Hence\begin {align} \frac {\partial F}{\partial x} & =-\frac {2}{3}\frac {x^{\frac {1}{3}}\left ( 2y+y^{3}-x^{2}\right ) }{x^{2}-y^{3}}\tag {1}\\ \frac {\partial F}{\partial y} & =2\frac {x^{\frac {4}{3}}}{x^{2}-y^{3}} \tag {2} \end {align}

Integrating (1) gives\begin {align} F & =\left ( \int -\frac {2}{3}\frac {x^{\frac {1}{3}}\left ( 2y+y^{3}-x^{2}\right ) }{x^{2}-y^{3}}dx\right ) +g\left ( y\right ) \nonumber \\ & =\frac {1}{2}x^{\frac {4}{3}}+\frac {1}{3}\ln \left ( x^{\frac {4}{3}}+x^{\frac {2}{3}}y+y^{2}\right ) -\frac {2}{3}\sqrt {3}\arctan \left ( \frac {1}{3}\frac {\left ( 2x^{\frac {2}{3}}+y\right ) \sqrt {3}}{y}\right ) -\frac {2}{3}\ln \left ( x^{\frac {2}{3}}-y\right ) +g\left ( y\right ) \tag {3} \end {align}

Where $$g\left ( y\right )$$ acts as the integration constant but $$F$$ depends on $$x,y$$ it becomes an arbitrary function. Taking derivative of the above w.r.t. $$y$$ gives\begin {equation} \frac {\partial F}{\partial y}=2\frac {x^{\frac {4}{3}}}{x^{2}-y^{3}}+g^{\prime }\left ( y\right ) \tag {4} \end {equation} Equating (4,2) gives\begin {align*} 2\frac {x^{\frac {4}{3}}}{x^{2}-y^{3}} & =2\frac {x^{\frac {4}{3}}}{x^{2}-y^{3}}+g^{\prime }\left ( y\right ) \\ 0 & =g^{\prime }\left ( y\right ) \\ g\left ( y\right ) & =c_{1} \end {align*}

Hence (3) becomes$F=\frac {1}{2}x^{\frac {4}{3}}+\frac {1}{3}\ln \left ( x^{\frac {4}{3}}+x^{\frac {2}{3}}y+y^{2}\right ) -\frac {2}{3}\sqrt {3}\arctan \left ( \frac {1}{3}\frac {\left ( 2x^{\frac {2}{3}}+y\right ) \sqrt {3}}{y}\right ) -\frac {2}{3}\ln \left ( x^{\frac {2}{3}}-y\right ) +c_{1}$ Therefore the solution is \begin {align*} F & =c\\ \frac {1}{2}x^{\frac {4}{3}}+\frac {1}{3}\ln \left ( x^{\frac {4}{3}}+x^{\frac {2}{3}}y+y^{2}\right ) -\frac {2}{3}\sqrt {3}\arctan \left ( \frac {1}{3}\frac {\left ( 2x^{\frac {2}{3}}+y\right ) \sqrt {3}}{y}\right ) -\frac {2}{3}\ln \left ( x^{\frac {2}{3}}-y\right ) & =c_{2} \end {align*}

Where constants $$c_{1},c\,$$ were combined into $$c_{2}$$. Now this ode will be solved using direct symmetry by converting to canonical coordinates.  This is done by using the standard characteristic equation by writing\begin {align*} \frac {dx}{\xi } & =\frac {dy}{\eta }=dS\\ \frac {dx}{\frac {3}{2x^{\frac {1}{3}}}} & =\frac {dy}{\frac {y}{x^{\frac {4}{3}}}}=dS \end {align*}

First pair of ode’s give$\frac {dy}{dx}=\frac {\frac {y}{x^{\frac {4}{3}}}}{\frac {3}{2x^{\frac {1}{3}}}}=\frac {2}{3x}y$ Hence $y=c_{1}x^{\frac {2}{3}}$ Therefore$R=yx^{-\frac {2}{3}}$ And $dS=\frac {dx}{\xi }=\frac {2}{3}x^{\frac {1}{3}}dx$ Integrating gives\begin {align*} S & =\int \frac {2}{3}x^{\frac {1}{3}}dx\\ & =\frac {1}{2}x^{\frac {4}{3}}+c_{1}\\ & =\frac {1}{2}x^{\frac {4}{3}} \end {align*}

By choosing $$c_{1}=0$$. Now the ODE $$\frac {dS}{dR}=F\left ( R\right )$$ is found 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 $$S_{x}=\frac {2}{3}x^{\frac {1}{3}},R_{x}=-\frac {2}{3}yx^{-\frac {5}{3}},S_{y}=0,R_{y}=x^{-\frac {2}{3}}$$. Substituting these into the above and simplifying gives\begin {align*} \frac {dS}{dR} & =\frac {\frac {2}{3}x^{\frac {1}{3}}}{-\frac {2}{3}yx^{-\frac {5}{3}}+\omega \left ( x,y\right ) x^{-\frac {2}{3}}}\\ & =\frac {\frac {2}{3}x^{\frac {1}{3}}}{-\frac {2}{3}yx^{-\frac {5}{3}}+\left ( \frac {1}{3}\frac {2y+y^{3}-x^{2}}{x}\right ) x^{-\frac {2}{3}}}\\ & =-2\frac {x^{2}}{x^{2}-y^{3}} \end {align*}

But $$R=yx^{-\frac {2}{3}}$$ or $$y=Rx^{\frac {2}{3}}$$. The above becomes\begin {align*} \frac {dS}{dR} & =-2\frac {x^{2}}{x^{2}-R^{3}x^{2}}\\ & =\frac {-2}{1-R^{3}} \end {align*}

Which is a quadrature. Solving gives\begin {align*} \int dS & =\int \frac {-2}{1-R^{3}}dR\\ S & =-\frac {1}{3}\ln \left ( R^{2}+x+1\right ) -\frac {2}{3}\sqrt {3}\arctan \left ( \frac {1}{3}\left ( 1+2R\right ) \sqrt {3}\right ) +\frac {2}{3}\ln \left ( R-1\right ) +c_{1} \end {align*}

Converting back to $$x,y$$ gives\begin {align*} \frac {1}{2}x^{\frac {4}{3}} & =-\frac {1}{3}\ln \left ( \left ( yx^{-\frac {2}{3}}\right ) ^{2}+x+1\right ) -\frac {2}{3}\sqrt {3}\arctan \left ( \frac {1}{3}\left ( 1+2\left ( yx^{-\frac {2}{3}}\right ) \right ) \sqrt {3}\right ) +\frac {2}{3}\ln \left ( \left ( yx^{-\frac {2}{3}}\right ) -1\right ) +c_{1}\\ \frac {1}{2}x^{\frac {4}{3}} & =-\frac {1}{3}\ln \left ( y^{2}x^{-\frac {4}{3}}+x+1\right ) -\frac {2}{3}\sqrt {3}\arctan \left ( \frac {1}{3}\left ( 1+2yx^{-\frac {2}{3}}\right ) \sqrt {3}\right ) +\frac {2}{3}\ln \left ( yx^{-\frac {2}{3}}-1\right ) +c_{1} \end {align*}

##### 1.10.15 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 $$\left ( R,S\right )$$ are now found. Using\begin {align} \frac {dx}{\xi } & =\frac {dy}{\eta }=dS\nonumber \\ \frac {dx}{x} & =\frac {dy}{y}=dS \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*} R & =c\\ & =\frac {y}{x} \end {align*}

Now we ﬁnd $$S$$ from the last pair of equations\begin {align*} \frac {dy}{y} & =dS\\ S & =\ln y \end {align*}

What is left is to ﬁnd $$\frac {dS}{dR}$$. This is given by$\frac {dS}{dR}=G\left ( R\right )$ To ﬁnd $$G\left ( R\right )$$, we use  $$dS=S_{x}dx+S_{y}dy=\frac {1}{y}dy$$ and $$dR=R_{x}dx+R_{y}dy=-\frac {y}{x^{2}}dx+\frac {1}{x}dy$$. Hence\begin {align*} \frac {dS}{dR} & =\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}}{-R^{2}+R\frac {dy}{dx}} \end {align*}

But $$\frac {dy}{dx}=3-2\frac {y}{x}=3-2R$$, hence\begin {align*} \frac {dS}{dR} & =\frac {3-2R}{-R^{2}+R\left ( 3-2R\right ) }\\ & =\frac {3-2R}{3\left ( R-R^{2}\right ) } \end {align*}

Which is a quadrature. In Lie method, for ﬁrst order ode, we always obtain $$\frac {dS}{dR}=G\left ( R\right )$$. Integrating the above gives\begin {align*} \int dS & =\int \frac {3-2R}{3\left ( R-R^{2}\right ) }dR\\ S & =\ln R-\frac {1}{3}\ln \left ( R-1\right ) +c_{1} \end {align*}

Final step is to replace $$R,S$$ 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*}

##### 1.10.16 Example $$y^{\prime }=\frac {-3+\frac {y}{x}}{-1-\frac {y}{x}}$$

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*}

Canonical coordinates $$\left ( R,S\right )$$ are found similar to the above which gives\begin {align*} R & =\frac {y}{x}\\ S & =\ln y \end {align*}

What is left is to ﬁnd $$\frac {dS}{dR}$$. This is given by$\frac {dS}{dR}=G\left ( R\right )$ Which is the same as above$\frac {dS}{dR}=\frac {\frac {dy}{dx}}{-R^{2}+R\frac {dy}{dx}}$ But in this problem, the only diﬀerence is that $$\frac {dy}{dx}=\frac {-3+\frac {y}{x}}{-1-\frac {y}{x}}=\frac {-3+R}{-1-R}$$, hence\begin {align*} \frac {dS}{dR} & =\frac {\frac {-3+R}{-1-R}}{-R^{2}+R\left ( \frac {-3+R}{-1-R}\right ) }\\ & =\frac {1}{R}\frac {R-3}{R^{2}+2R-3} \end {align*}

Which is a quadrature. In Lie method, for ﬁrst order ode, we always obtain $$\frac {dS}{dR}=G\left ( R\right )$$. Integrating the above gives\begin {align*} \int dS & =\int \frac {1}{R}\left ( \frac {R-3}{R^{2}+2R-3}\right ) dR\\ S & =\ln \left ( R\right ) -\frac {1}{2}\ln \left ( R+3\right ) -\frac {1}{2}\ln \left ( R-1\right ) +c_{1} \end {align*}

Final step is to replace $$R,S$$ back with $$x,y$$ which gives$\ln y=\ln \left ( \frac {y}{x}\right ) -\frac {1}{2}\ln \left ( \frac {y}{x}+3\right ) -\frac {1}{2}\ln \left ( \frac {y}{x}-1\right ) +c_{1}$ This can be solved for $$y$$ if an explicit solution is needed.

##### 1.10.17 Example $$y^{\prime }=\frac {1+3\left ( \frac {y}{x}\right ) ^{2}}{2\frac {y}{x}}$$

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 canonical ode is$\frac {dS}{dR}=\frac {\frac {dy}{dx}}{-R^{2}+R\frac {dy}{dx}}$ The above is the same ode in canonical coordinates for any ode of the form $$y^{\prime }=F\left ( \frac {y}{x}\right )$$. We just need to express $$y^{\prime }$$ as function of $$R$$. In this case the above becomes\begin {align*} \frac {dS}{dR} & =\frac {\frac {1+3R^{2}}{2R}}{-R^{2}+R\left ( \frac {1+3R^{2}}{2R}\right ) }\\ & =\frac {3R^{2}+1}{R^{3}+R} \end {align*}

Integrating gives$S=\ln \left ( R\left ( R^{2}+1\right ) \right ) +c_{1}$ Final step is to replace $$R,S$$ back with $$x,y$$ which gives\begin {align*} \ln y & =\ln \left ( \frac {y}{x}\left ( \left ( \frac {y}{x}\right ) ^{2}+1\right ) \right ) +c_{1}\\ y & =c_{2}\frac {y}{x}\left ( \left ( \frac {y}{x}\right ) ^{2}+1\right ) \\ 1 & =\frac {c_{2}}{x}\left ( \left ( \frac {y}{x}\right ) ^{2}+1\right ) \\ \frac {y^{2}}{x^{2}} & =c_{3}x-1\\ y^{2} & =c_{3}x^{3}-x^{2} \end {align*}

Hence\begin {align*} y & =\pm \sqrt {c_{3}x^{3}-x^{2}}\\ & =\pm x\sqrt {c_{3}x-1} \end {align*}

Finding $$\xi ,\eta$$ from symmetry condition for the above ode This shows how to ﬁnd $$\xi ,\eta$$ directly also. 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} Try Ansatz \begin {align*} \xi & =c_{0}+c_{1}x\\ \eta & =c_{2}+c_{3}y \end {align*}

And given\begin {align*} \omega & =\frac {1}{2}\frac {x^{2}+3y^{2}}{xy}\\ \omega ^{2} & =\frac {1}{4}\frac {\left ( x^{2}+3y^{2}\right ) ^{2}}{x^{2}y^{2}}\\ \omega _{x} & =\frac {1}{2}\frac {x^{2}-3y^{2}}{yx^{2}}\\ \omega _{y} & =\frac {1}{2}\frac {3y^{2}-x^{2}}{xy^{2}} \end {align*}

Hence (14) becomes$\eta _{x}+\frac {1}{2}\frac {x^{2}+3y^{2}}{xy}\eta _{y}-\frac {1}{2}\frac {x^{2}-3y^{2}}{yx^{2}}\xi -\frac {1}{2}\frac {3y^{2}-x^{2}}{xy^{2}}\eta =0$ Therefore the above becomes$\frac {1}{2}\frac {x^{2}+3y^{2}}{xy}c_{3}-\frac {1}{2}\frac {x^{2}-3y^{2}}{yx^{2}}\left ( c_{0}+c_{1}x\right ) -\frac {1}{2}\frac {3y^{2}-x^{2}}{xy^{2}}\left ( c_{2}+c_{3}y\right ) =0$ Using the computer the above simpliﬁes to $\frac {x}{y}\left ( c_{3}-c_{1}\right ) +\frac {1}{2}c_{2}\frac {x}{y^{2}}-\frac {1}{y}\left ( \frac {1}{2}c_{0}\right ) -\frac {1}{x}\frac {3}{2}c_{2}+\frac {3}{2}c_{0}\frac {y}{x^{2}}=0$ Hence \begin {align*} c_{3}-c_{1} & =0\\ \frac {1}{2}c_{2} & =0\\ -\frac {1}{2}c_{0} & =0\\ -\frac {3}{2}c_{2} & =0\\ \frac {3}{2}c_{0} & =0 \end {align*}

Solving gives $$c_{0}=0,c_{2}=0$$ and $$c_{3}=c_{1}$$. Hence the solution is \begin {align*} \xi & =c_{1}x\\ \eta & =c_{3}y \end {align*}

Let $$c_{1}=1$$, therefore $$c_{3}=1$$ and we obtain\begin {align*} \xi & =x\\ \eta & =y \end {align*}

Which is the result we used in solving the above problem. Notice that any scaler will also work. Hence\begin {align*} \xi & =5x\\ \eta & =5y \end {align*}

And\begin {align*} \xi & =10x\\ \eta & =10y \end {align*}

This will also give same solution.

##### 1.10.18 Example $$y^{\prime }=\frac {y}{x}+\frac {1}{x}F\left ( \frac {y}{x}\right )$$

This is homogeneous class D  $$y^{\prime }=\frac {y}{x}+g\left ( x\right ) F\left ( \frac {y}{x}\right )$$. Hence from lookup table\begin {align*} \xi & =x^{2}\\ \eta & =xy \end {align*}

Now we just need to ﬁnd canonical coordinates $$\left ( R,S\right )$$ since $$\xi ,\eta$$ are known. Using\begin {align} \frac {dx}{\xi } & =\frac {dy}{\eta }=dS\nonumber \\ \frac {dx}{x^{2}} & =\frac {dy}{xy}=dS \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*} R & =c\\ & =\frac {y}{x} \end {align*}

Now we ﬁnd $$S$$ from the last pair of equations (we could also use the ﬁrst and last equations in (1)).\begin {align*} \frac {dy}{xy} & =dS\\ S & =\frac {1}{x}\ln y \end {align*}

What is left is to ﬁnd $$\frac {dS}{dR}$$. This is given by\begin {align*} \frac {dS}{dR} & =G\left ( R\right ) \\ & =\frac {S_{x}+S_{y}y^{\prime }}{R_{x}+R_{y}y^{\prime }} \end {align*}

To ﬁnd $$G\left ( R\right )$$, we use  $$S_{x}=\frac {-1}{x^{2}}\ln y,S_{y}=\frac {1}{xy}$$ and $$R_{x}=-\frac {y}{x^{2}},R_{y}=\frac {1}{x}$$. Hence\begin {align*} \frac {dS}{dR} & =\frac {\frac {-1}{x^{2}}\ln y+\frac {1}{xy}y^{\prime }}{-\frac {y}{x^{2}}+\frac {1}{x}y^{\prime }}\\ & =\frac {-\ln y-\frac {x}{y}y^{\prime }}{y+xy^{\prime }}\\ & =\frac {-\ln y-\frac {1}{R}y^{\prime }}{y+xy^{\prime }} \end {align*}

But $$y^{\prime }=\frac {y}{x}+\frac {1}{x}F\left ( \frac {y}{x}\right ) =R+\frac {1}{x}F\left ( R\right )$$. The above becomes\begin {align*} \frac {dS}{dR} & =\frac {-\ln y-\frac {1}{R}\left ( R+\frac {1}{x}F\left ( R\right ) \right ) }{y+x\left ( R+\frac {1}{x}F\left ( R\right ) \right ) }\\ & =\frac {-\ln y-1-\frac {1}{xR}F\left ( R\right ) }{y+xR+F\left ( R\right ) }\\ & =\frac {-\ln y-1-\frac {1}{x\frac {y}{x}}F\left ( R\right ) }{y+x\frac {y}{x}+F\left ( R\right ) }\\ & =\frac {-\ln y-1-\frac {1}{y}F\left ( R\right ) }{2y+F\left ( R\right ) } \end {align*}

Something is wrong. $$\frac {dS}{dR}$$ should only be a function of $$R$$. Need to ﬁnd out why. Let me try the other pair of equations from (1) to solve for $$S$$ and see what happens.\begin {align*} \frac {dx}{x^{2}} & =dS\\ S & =-\frac {1}{x} \end {align*}

What is left is to ﬁnd $$\frac {dS}{dR}$$. This is given by\begin {align*} \frac {dS}{dR} & =G\left ( R\right ) \\ & =\frac {S_{x}+S_{y}y^{\prime }}{R_{x}+R_{y}y^{\prime }} \end {align*}

To ﬁnd $$G\left ( R\right )$$, we use  $$S_{x}=\frac {1}{x^{2}},S_{y}=0$$ and $$R_{x}=-\frac {y}{x^{2}},R_{y}=\frac {1}{x}$$. Hence\begin {align*} \frac {dS}{dR} & =\frac {\frac {1}{x^{2}}}{-\frac {y}{x^{2}}+\frac {1}{x}y^{\prime }}\\ & =\frac {1}{-y+xy^{\prime }} \end {align*}

But $$y^{\prime }=\frac {y}{x}+\frac {1}{x}F\left ( \frac {y}{x}\right ) =R+\frac {1}{x}F\left ( R\right )$$. The above becomes\begin {align*} \frac {dS}{dR} & =\frac {1}{-y+x\left ( R+\frac {1}{x}F\left ( R\right ) \right ) }\\ & =\frac {1}{-y+xR+F\left ( R\right ) }\\ & =\frac {1}{-y+x\frac {y}{x}+F\left ( R\right ) }\\ & =\frac {1}{F\left ( R\right ) } \end {align*}

This worked. But why the ﬁrst choice did not work? OK, let me continue now. Integrating the above gives$S=\int \frac {1}{F\left ( R\right ) }dR+c$ But $$S=-\frac {1}{x}$$, hence \begin {align*} -\frac {1}{x} & =\int ^{\frac {y}{x}}\frac {1}{F\left ( r\right ) }dr+c\\ 0 & =\int ^{\frac {y}{x}}\frac {1}{F\left ( r\right ) }dr+c+\frac {1}{x} \end {align*}

This example shows that when solving for $$S$$ from$\frac {dx}{x^{2}}=\frac {dy}{xy}=dS$ There are two choice. One is $$dS=\frac {dy}{xy}$$ and the other $$dS=\frac {dx}{x^{2}}$$. Using the ﬁrst choice did not work here (unless I made a mistake, but do not see it)., Only the second choice worked because we must end up with $$\frac {dS}{dR}=G\left ( R\right )$$ where RHS is function of $$R$$ only. I need to look more into this. In theory, any choice should have worked.

##### 1.10.19 Example $$y^{\prime }=\frac {y}{x}+\frac {1}{x}e^{-\frac {y}{x}}$$

This is homogeneous class D  $$y^{\prime }=\frac {y}{x}+g\left ( x\right ) F\left ( \frac {y}{x}\right )$$. Hence from lookup table\begin {align*} \xi & =x^{2}\\ \eta & =xy \end {align*}

From above we found the solution to be$S=\int \frac {1}{F\left ( R\right ) }dR+c$ In this case $$F\left ( R\right ) =e^{-R}$$. Hence\begin {align*} S & =\int e^{R}dR+c\\ S & =e^{R}+c \end {align*}

Now we just need to ﬁnd canonical coordinates $$\left ( R,S\right )$$ since $$\xi ,\eta$$ are known. From above\begin {align*} R & =\frac {y}{x}\\ S & =-\frac {1}{x} \end {align*}

Hence the solution becomes\begin {align*} -\frac {1}{x} & =e^{\frac {y}{x}}+c\\ e^{\frac {y}{x}} & =c_{2}-\frac {1}{x}\\ \frac {y}{x} & =\ln \left ( c_{2}-\frac {1}{x}\right ) \\ y & =x\ln \left ( c_{2}-\frac {1}{x}\right ) \end {align*}

The nice thing about this method is that once we solve for one pattern of an ode, then the same solution in canonical coordinates is used, the only change need is to plug-in in the RHS of the original ode in the solution and integrate.

##### 1.10.20 Example $$y^{\prime }=\frac {1-y^{2}+x^{2}}{1+y^{2}-x^{2}}$$

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

Using anstaz’s it is found that \begin {align*} \xi & =x-y\\ \eta & =y-x \end {align*}

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

The ﬁrst two give$\frac {dy}{dx}=\frac {\eta }{\xi }=\frac {y-x}{x-y}=-1$ Hence\begin {equation} y=-x+c_{1} \tag {2} \end {equation} Therefore\begin {align*} R & =c_{1}\\ & =y+x \end {align*}

To ﬁnd $$S$$, since both $$\xi ,\eta$$ depend on both $$x,y$$, then $$\frac {dy}{\eta }=dS\,$$ or $$\frac {dx}{\xi }=dS$$ can be used. Lets try both to show same answer results.\begin {align*} \frac {dy}{\eta } & =dS\\ dS & =\frac {dy}{y-x} \end {align*}

But from (2), $$x=c_{1}-y$$. The above becomes\begin {align*} dS & =\frac {dy}{y-\left ( c_{1}-y\right ) }\\ & =\frac {dy}{2y-c_{1}} \end {align*}

Hence$S=\frac {1}{2}\ln \left ( 2y-c_{1}\right )$ But $$c_{1}=y+x$$. So the above becomes\begin {align} S & =\frac {1}{2}\ln \left ( 2y-\left ( y+x\right ) \right ) \nonumber \\ & =\frac {1}{2}\ln \left ( y-x\right ) \tag {3} \end {align}

Let us now try the other ode\begin {align*} \frac {dx}{\xi } & =dS\\ dS & =\frac {dx}{x-y} \end {align*}

But from (2) $$y=-x+c_{1}$$. The above becomes\begin {align*} dS & =\frac {dx}{x-\left ( -x+c_{1}\right ) }\\ & =\frac {dx}{2x-c_{1}} \end {align*}

Therefore$S=\frac {1}{2}\ln \left ( 2x-c_{1}\right )$ But $$c_{1}=y+x$$. Therefore\begin {align} S & =\frac {1}{2}\ln \left ( 2x-\left ( y+x\right ) \right ) \nonumber \\ & =\frac {1}{2}\ln \left ( x-y\right ) \tag {4} \end {align}

The constant of integration is set to zero when ﬁnding $$S$$. What is left is to ﬁnd $$\frac {dS}{dR}$$. This is given by\begin {equation} \frac {dS}{dR}=\frac {S_{x}+S_{y}\omega }{R_{x}+R_{y}\omega } \tag {5} \end {equation} But, and using (4) for $$S$$ we have\begin {align*} R_{x} & =1\\ R_{y} & =1\\ S_{x} & =\frac {-1}{y-x}\\ S_{y} & =\frac {1}{y-x} \end {align*}

Hence (2) becomes\begin {align*} \frac {dS}{dR} & =\frac {\frac {-1}{y-x}+\frac {1}{y-x}\omega }{1+\omega }\\ & =\frac {-\frac {\omega -1}{x-y}}{1+\omega }\\ & =\frac {1-\omega }{\left ( 1+\omega \right ) \left ( x-y\right ) }\\ & =\frac {1-\left ( \frac {1-y^{2}+x^{2}}{1+y^{2}-x^{2}}\right ) }{\left ( 1+\left ( \frac {1-y^{2}+x^{2}}{1+y^{2}-x^{2}}\right ) \right ) \left ( x-y\right ) }\\ & =-x-y\\ & =-\left ( x+y\right ) \\ & =-R \end {align*}

Hence \begin {align*} \frac {dS}{dR} & =-R\\ S & =-\frac {R^{2}}{2} \end {align*}

Converting back to $$x,y$$ gives$\ln \left ( y-x\right ) =-\frac {\left ( y+x\right ) ^{2}}{2}$

##### 1.10.21 Example $$y^{\prime }=-\frac {1}{4}xe^{-2y}+\frac {1}{4}\sqrt {\left ( e^{-2y}\right ) ^{2}x^{2}+4e^{-2y}}$$

\begin {align*} y^{\prime } & =-\frac {1}{4}xe^{-2y}+\frac {1}{4}\sqrt {\left ( e^{-2y}\right ) ^{2}x^{2}+4e^{-2y}}\\ & =\omega \left ( x,y\right ) \end {align*}

Using anstaz’s it is found that \begin {align*} \xi & =x\\ \eta & =1 \end {align*}

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

The ﬁrst two give$\frac {dy}{dx}=\frac {1}{x}$ Hence$y=\ln x+c_{1}$ Therefore\begin {align*} R & =c_{1}\\ & =y-\ln x \end {align*}

And $$S$$ is found from either $$\frac {dy}{\eta }=dS\,$$ or $$\frac {dx}{\xi }=dS$$. Since $$\eta =1$$, it is simpler to use $$\frac {dy}{\eta }=dS$$ instead.\begin {align*} \frac {dy}{\eta } & =dS\\ dy & =dS\\ S & =y \end {align*}

Where constant of integration is set to zero. What is left is to ﬁnd $$\frac {dS}{dR}$$. This is given by\begin {equation} \frac {dS}{dR}=\frac {S_{x}+S_{y}\omega }{R_{x}+R_{y}\omega } \tag {2} \end {equation} But\begin {align*} R_{x} & =-\frac {1}{x}\\ R_{y} & =1\\ S_{x} & =0\\ S_{y} & =1 \end {align*}

Hence (2) becomes\begin {align*} \frac {dS}{dR} & =\frac {\omega }{-\frac {1}{x}+\omega }=\frac {1}{-\frac {1}{x\omega }+1}\\ & =\frac {1}{1-\frac {1}{x\left ( -\frac {1}{4}xe^{-2y}+\frac {1}{4}\sqrt {\left ( e^{-2y}\right ) ^{2}x^{2}+4e^{-2y}}\right ) }} \end {align*}

But $$y=R+\ln x$$. The above becomes\begin {align*} \frac {dS}{dR} & =\frac {1}{1-\frac {1}{x\left ( -\frac {1}{4}xe^{-2\left ( R+\ln x\right ) }+\frac {1}{4}\sqrt {\left ( e^{-2\left ( R+\ln x\right ) }\right ) ^{2}x^{2}+4e^{-2R+\ln x}}\right ) }}\\ & =\frac {1}{1-\frac {1}{x\left ( -\frac {1}{4}\frac {xe^{-2R}}{x^{2}}+\frac {1}{4}\frac {1}{x}\sqrt {e^{-4R}+4e^{-2R}}\right ) }}\\ & =\frac {1}{1-\frac {1}{\left ( -\frac {1}{4}e^{-2R}+\frac {1}{4}\sqrt {e^{-4R}+4e^{-2R}}\right ) }} \end {align*}

Integrating gives$S=\frac {\sqrt {\frac {1+4e^{2R}}{e^{4R}}}e^{2R}\operatorname {arctanh}\left ( \frac {1}{\sqrt {1+4e^{2R}}}\right ) }{\sqrt {1+4e^{2R}}}$ Converting back to $$x,y$$ gives$y=\frac {\sqrt {\frac {1+4e^{2\left ( y-\ln x\right ) }}{e^{4\left ( y-\ln x\right ) }}}e^{2\left ( y-\ln x\right ) }\operatorname {arctanh}\left ( \frac {1}{\sqrt {1+4e^{2\left ( y-\ln x\right ) }}}\right ) }{\sqrt {1+4e^{2\left ( y-\ln x\right ) }}}$

##### 1.10.22 Example $$y^{\prime }=\frac {y-xf\left ( x^{2}+ay^{2}\right ) }{x+ayf\left ( x^{2}+ay^{2}\right ) }$$

\begin {align*} y^{\prime } & =\frac {y-xf\left ( x^{2}+ay^{2}\right ) }{x+ayf\left ( x^{2}+ay^{2}\right ) }\\ & =\omega \left ( x,y\right ) \end {align*}

Using anstaz’s it is found that \begin {align*} \xi & =-ay\\ \eta & =x \end {align*}

Hence\begin {align} \frac {dx}{\xi } & =\frac {dy}{\eta }=dS\nonumber \\ \frac {dx}{-ay} & =\frac {dy}{x}=dS \tag {1} \end {align}

The ﬁrst two give$\frac {dy}{dx}=\frac {x}{-ay}$ This is separable. Solving gives (taking one root)$y=\frac {\sqrt {a\left ( ac_{1}-x^{2}\right ) }}{a}$ Solving for $$c_{1}$$ gives$c_{1}=\frac {x^{2}+ay^{2}}{a}$ Hence$R=\frac {x^{2}+ay^{2}}{a}$ $$S$$ is found from either $$\frac {dy}{\eta }=dS\,$$ or $$\frac {dx}{\xi }=dS$$. Using $$\frac {dx}{-ay}=dS$$ then$\frac {dx}{-ay}=dS$ But $$y=\frac {\sqrt {a\left ( ac_{1}-x^{2}\right ) }}{a}$$. Hence\begin {align*} \frac {dx}{-a\frac {\sqrt {a\left ( ac_{1}-x^{2}\right ) }}{a}} & =dS\\ \frac {dx}{-\sqrt {a\left ( ac_{1}-x^{2}\right ) }} & =dS\\ -\frac {1}{\sqrt {a}}\arctan \left ( \frac {\sqrt {a}x}{\sqrt {c_{1}a^{2}-x^{2}a}}\right ) & =S\\ -\frac {1}{\sqrt {a}}\arctan \left ( \frac {\sqrt {a}x}{ay}\right ) & =S \end {align*}

Where constant of integration is set to zero. What is left is to ﬁnd $$\frac {dS}{dR}$$. This is given by\begin {equation} \frac {dS}{dR}=\frac {S_{x}+S_{y}\omega }{R_{x}+R_{y}\omega } \tag {2} \end {equation} But\begin {align*} R_{x} & =\frac {2x}{a}\\ R_{y} & =2y\\ S_{x} & =-\frac {y}{x^{2}y^{2}+a}\\ S_{y} & =-\frac {x}{a\left ( 1+\frac {x^{2}y^{2}}{a}\right ) } \end {align*}

Hence (2) becomes$\frac {dS}{dR}=\frac {-\frac {y}{x^{2}y^{2}+a}+\left ( -\frac {x}{a\left ( 1+\frac {x^{2}y^{2}}{a}\right ) }\right ) \omega }{\frac {2x}{a}+2y\omega }$ But $$R=\frac {x^{2}+ay^{2}}{a}$$. The above becomes$\frac {dS}{dR}=\frac {-\frac {y}{aR}+\left ( -\frac {x}{a\left ( 1+\frac {x^{2}y^{2}}{a}\right ) }\right ) \omega }{\frac {2x}{a}+2y\omega }$ To ﬁnish. Another hard part of this Lie method is to convert back $$\,\frac {dS}{dR}=\frac {S_{x}+S_{y}\omega }{R_{x}+R_{y}\omega }$$ so that the RHS is only a function of $$R$$. Need to ﬁnd a robust way to do this. This is now a weak point in my program as I have few ode’s that it can’t do it

#### 1.11 Alternative form for the similarity condition PDE

This section shows how to obtain eq. (8) in paper "Computer Algebra Solving of First Order ODEs Using Symmetry Methods" 1996 by Durate, Terrab, Mota. Which is an alternative equation to solve instead of the main Lie condition for symmetry we were looking at above.

Starting with the main linearized symmetry pde\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} Assuming anstaz \begin {equation} \eta =\xi \omega +\chi \tag {A} \end {equation} Hence \begin {align*} \eta _{x} & =\xi _{x}\omega +\xi \omega _{x}+\chi _{x}\\ \eta _{y} & =\xi _{y}\omega +\xi \omega _{y}+\chi _{y} \end {align*}

Then (14) becomes\begin {align} \left ( \xi _{x}\omega +\xi \omega _{x}+\chi _{x}\right ) +\omega \left ( \left ( \xi _{y}\omega +\xi \omega _{y}+\chi _{y}\right ) -\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\left ( \xi \omega +\chi \right ) & =0\nonumber \\ \xi _{x}\omega +\xi \omega _{x}+\chi _{x}+\xi _{y}\omega ^{2}+\xi \omega _{y}\omega +\chi _{y}\omega -\omega \xi _{x}-\omega ^{2}\xi _{y}-\omega _{x}\xi -\xi \omega \omega _{y}-\omega _{y}\chi & =0\nonumber \\ \xi _{x}\omega +\chi _{x}+\xi _{y}\omega ^{2}+\xi \omega _{y}\omega +\chi _{y}\omega -\omega \xi _{x}-\omega ^{2}\xi _{y}-\xi \omega \omega _{y}-\omega _{y}\chi & =0\nonumber \\ \chi _{x}+\xi _{y}\omega ^{2}+\xi \omega _{y}\omega +\chi _{y}\omega -\omega ^{2}\xi _{y}-\xi \omega \omega _{y}-\omega _{y}\chi & =0\nonumber \\ \chi _{x}+\xi \omega _{y}\omega +\chi _{y}\omega -\xi \omega \omega _{y}-\omega _{y}\chi & =0\nonumber \end {align}

Or\begin {equation} \chi _{x}+\chi _{y}\omega -\omega _{y}\chi =0 \tag {1} \end {equation} And hence (1) is now solved for $$\chi \left ( x,y\right )$$. If we are able to ﬁnd $$\chi$$ then we can use the anstaz $$\eta =\xi \omega +\chi$$. This leaves only one unknown $$\xi$$. The paper does not explain how to solve for this, $$\xi$$, which I assume is by using (14) again. The paper only said

The knowledge of $$\chi$$, in turn, allows one to set $$\xi$$ and $$\eta ~$$as desired using (A)

Which is not too clear how in practice this is done. I need to work an example showing this. The paper says that (1) is solved for $$\chi \left ( x,y\right )$$ by using bivariate polynomial anstaz. The degree can be set by a user, or Maple internally determines this.

### 2 Second order ODE

#### 2.1 Linearized PDE of the similarity condition

Obtaining the linearized PDE of the similarity condition for second order ode, which is used to solve for $$\xi ,\eta$$ follows similar method as given earlier for the ﬁrst order ode. The diﬀerence is that instead of $$y^{\prime }=\omega \left ( x,y\right )$$ the ode now $$y^{\prime \prime }=\omega \left ( x,y,y^{\prime }\right )$$.\begin {align} \frac {d^{2}y}{dx^{2}} & =\omega \left ( x,y,\frac {dy}{dx}\right ) \nonumber \\ y^{\prime \prime } & =\omega \left ( x,y,y^{\prime }\right ) \tag {A} \end {align}

The linearized similarity condition for second order ode when $$\omega =0$$ is$\eta _{xx}+\left ( 2\eta _{xy}-\xi _{xx}\right ) y^{\prime }+\left ( \eta _{yy}-2\xi _{xy}\right ) \left ( y^{\prime }\right ) ^{2}-\xi _{yy}\left ( y^{\prime }\right ) ^{3}=0$ Which is polynomial in $$y^{\prime }$$ hence all the coeﬃcients must be zero giving\begin {align*} 2\eta _{xy}-\xi _{xx} & =0\\ \eta _{yy}-2\xi _{xy} & =0\\ \xi _{yy} & =0\\ \eta _{xx} & =0 \end {align*}

And for general $$\omega \left ( x,y,y^{\prime }\right )$$, the linearized similarity condition is$-\eta \omega _{y}+\left ( -3y^{\prime }\xi _{y}-2\xi _{x}+\eta _{y}\right ) \omega -\xi \omega _{x}+\left ( -y^{\prime }\eta _{y}+\left ( y^{\prime }\right ) ^{2}\xi _{y}+y^{\prime }\xi _{x}-\eta _{x}\right ) \omega _{y^{\prime }}+\eta _{xx}-\xi _{yy}\left ( y^{\prime }\right ) ^{3}+\left ( \eta _{yy}-2\xi _{yx}\right ) \left ( y^{\prime }\right ) ^{2}+\left ( 2\eta _{yx}-\xi _{xx}\right ) y^{\prime }=0$ To continue

### 3 Analysis of Maple’s symgen methods for ﬁnding symmetries

This section gives an overview of Maple’s methods of solving for Lie symmetries. There are 16 total algorithms.

### 4 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?  For example $$y^{\prime }+y^{3}+xy^{2}=0$$ which is Abel ode type, Maple found no symmetry using all methods.

### 5 References

1. Symmetry Methods for Diﬀerential Equations. By PETER E. HYDON. CAMBRIDGE UNIVERSITY PRESS. 2000.
2. Symmetry and integration methods for diﬀerential equations. By Bluman and Anco. Springer publishing.
3. Introduction to symmetry analysis. By Brian J. Cantwell. See page 158 for table of known $$\xi ,\eta$$ per ﬁrst ode type.
4. Using Lie Symmetry to solve ﬁrst and second order Linear Diﬀerential Equation. Int. J. Adv. Appl. Math. andMech. 7(4) (2020) 91 – 99 (ISSN: 2347-2529).
5. Sym Mathematica software package. By Stilianos Dimas and Dimitris Tsoubelis.
6. SymmetryAnalysis Mathematica package. By Brian J. Cantwell.
7. Maple symgen and related commands in the ODEtools package.
8. Computer Algebra Solving of First Order ODEs Using Symmetry Methods. 1996 by Durate, Terrab, Mota. IF-UERJ-27/96
9. The Truth About Lie Symmetries: Solving Diﬀerential Equations With Symmetry Methods by Ruth A. Steinhour, 2013.
10. Many other references, listed in my links page under computer algebra section.