3.4.10.5 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.