1.391 problem 392
Internal
problem
ID
[9373]
Book
:
Differential
Gleichungen,
E.
Kamke,
3rd
ed.
Chelsea
Pub.
NY,
1948
Section
:
Chapter
1,
linear
first
order
Problem
number
:
392
Date
solved
:
Thursday, October 17, 2024 at 02:20:36 PM
CAS
classification
:
[[_1st_order, `_with_symmetry_[F(x),G(y)]`]]
Solve
\begin{align*} {y^{\prime }}^{2}-x y y^{\prime }+y^{2} \ln \left (a y\right )&=0 \end{align*}
Solving for the derivative gives these ODE’s to solve
\begin{align*}
\tag{1} y^{\prime }&=\left (\frac {x}{2}+\frac {\sqrt {x^{2}-4 \ln \left (a y\right )}}{2}\right ) y \\
\tag{2} y^{\prime }&=\left (\frac {x}{2}-\frac {\sqrt {x^{2}-4 \ln \left (a y\right )}}{2}\right ) y \\
\end{align*}
Now each of the above is solved
separately.
Solving Eq. (1)
Writing the ode as
\begin{align*} y^{\prime }&=\frac {\left (x +\sqrt {x^{2}-4 \ln \left (a y \right )}\right ) y}{2}\\ y^{\prime }&= \omega \left ( x,y\right ) \end{align*}
The condition of Lie symmetry is the linearized PDE given by
\begin{align*} \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0\tag {A} \end{align*}
To determine \(\xi ,\eta \) then (A) is solved using ansatz. Making bivariate polynomials of
degree 2 to use as anstaz gives
\begin{align*}
\tag{1E} \xi &= x^{2} a_{4}+y x a_{5}+y^{2} a_{6}+x a_{2}+y a_{3}+a_{1} \\
\tag{2E} \eta &= x^{2} b_{4}+y x b_{5}+y^{2} b_{6}+x b_{2}+y b_{3}+b_{1} \\
\end{align*}
Where the unknown coefficients are
\[
\{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, b_{1}, b_{2}, b_{3}, b_{4}, b_{5}, b_{6}\}
\]
Substituting
equations (1E,2E) and \(\omega \) into (A) gives
\begin{equation}
\tag{5E} 2 x b_{4}+y b_{5}+b_{2}+\frac {\left (x +\sqrt {x^{2}-4 \ln \left (a y \right )}\right ) y \left (-2 x a_{4}+x b_{5}-y a_{5}+2 y b_{6}-a_{2}+b_{3}\right )}{2}-\frac {\left (x +\sqrt {x^{2}-4 \ln \left (a y \right )}\right )^{2} y^{2} \left (x a_{5}+2 y a_{6}+a_{3}\right )}{4}-\frac {\left (1+\frac {x}{\sqrt {x^{2}-4 \ln \left (a y \right )}}\right ) y \left (x^{2} a_{4}+y x a_{5}+y^{2} a_{6}+x a_{2}+y a_{3}+a_{1}\right )}{2}-\left (-\frac {1}{\sqrt {x^{2}-4 \ln \left (a y \right )}}+\frac {x}{2}+\frac {\sqrt {x^{2}-4 \ln \left (a y \right )}}{2}\right ) \left (x^{2} b_{4}+y x b_{5}+y^{2} b_{6}+x b_{2}+y b_{3}+b_{1}\right ) = 0
\end{equation}
Putting the above in normal form gives
\[
-\frac {-4 y x b_{5}-4 x b_{2}-4 y b_{3}-4 b_{1}-4 b_{2} \sqrt {x^{2}-4 \ln \left (a y \right )}+2 x^{3} b_{2}+2 x^{2} b_{1}-8 \ln \left (a y \right ) b_{1}-2 x^{2} y^{2} b_{6}+2 x^{4} y^{2} a_{5}+4 x^{3} y^{3} a_{6}-8 \ln \left (a y \right ) x^{2} b_{4}-8 \ln \left (a y \right ) y^{2} a_{5}+8 \ln \left (a y \right ) y^{2} b_{6}+2 x \,y^{3} a_{6}+2 \left (x^{2}-4 \ln \left (a y \right )\right )^{{3}/{2}} y^{3} a_{6}+2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{3} b_{4}+2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, y^{3} a_{6}-8 x b_{4} \sqrt {x^{2}-4 \ln \left (a y \right )}-4 y b_{5} \sqrt {x^{2}-4 \ln \left (a y \right )}+6 x^{3} y a_{4}+4 x^{2} y^{2} a_{5}+2 x^{4} b_{4}-4 x^{2} b_{4}-4 y^{2} b_{6}-8 \ln \left (a y \right ) x^{2} y^{2} a_{5}-16 \ln \left (a y \right ) x \,y^{3} a_{6}-16 \ln \left (a y \right ) x y a_{4}+\left (x^{2}-4 \ln \left (a y \right )\right )^{{3}/{2}} x \,y^{2} a_{5}+\sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{3} y^{2} a_{5}+2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{2} y^{3} a_{6}+6 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{2} y a_{4}+4 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x \,y^{2} a_{5}-2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x \,y^{2} b_{6}+\sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{2} y^{2} a_{3}+4 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x y a_{2}-8 \ln \left (a y \right ) x \,y^{2} a_{3}+\left (x^{2}-4 \ln \left (a y \right )\right )^{{3}/{2}} y^{2} a_{3}+2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{2} b_{2}+2 y^{2} a_{3} \sqrt {x^{2}-4 \ln \left (a y \right )}+2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x b_{1}+2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, y a_{1}+2 x^{3} y^{2} a_{3}-8 \ln \left (a y \right ) x b_{2}-8 \ln \left (a y \right ) y a_{2}+4 x^{2} y a_{2}+2 x \,y^{2} a_{3}+2 x y a_{1}}{4 \sqrt {x^{2}-4 \ln \left (a y \right )}} = 0
\]
Setting the numerator to zero gives
\begin{equation}
\tag{6E} 4 y x b_{5}+4 x b_{2}+4 y b_{3}+4 b_{1}+4 b_{2} \sqrt {x^{2}-4 \ln \left (a y \right )}-2 x^{3} b_{2}-2 x^{2} b_{1}+8 \ln \left (a y \right ) b_{1}+2 x^{2} y^{2} b_{6}-2 x^{4} y^{2} a_{5}-4 x^{3} y^{3} a_{6}+8 \ln \left (a y \right ) x^{2} b_{4}+8 \ln \left (a y \right ) y^{2} a_{5}-8 \ln \left (a y \right ) y^{2} b_{6}-2 x \,y^{3} a_{6}-2 \left (x^{2}-4 \ln \left (a y \right )\right )^{{3}/{2}} y^{3} a_{6}-2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{3} b_{4}-2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, y^{3} a_{6}+8 x b_{4} \sqrt {x^{2}-4 \ln \left (a y \right )}+4 y b_{5} \sqrt {x^{2}-4 \ln \left (a y \right )}-6 x^{3} y a_{4}-4 x^{2} y^{2} a_{5}-2 x^{4} b_{4}+4 x^{2} b_{4}+4 y^{2} b_{6}+8 \ln \left (a y \right ) x^{2} y^{2} a_{5}+16 \ln \left (a y \right ) x \,y^{3} a_{6}+16 \ln \left (a y \right ) x y a_{4}-\left (x^{2}-4 \ln \left (a y \right )\right )^{{3}/{2}} x \,y^{2} a_{5}-\sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{3} y^{2} a_{5}-2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{2} y^{3} a_{6}-6 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{2} y a_{4}-4 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x \,y^{2} a_{5}+2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x \,y^{2} b_{6}-\sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{2} y^{2} a_{3}-4 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x y a_{2}+8 \ln \left (a y \right ) x \,y^{2} a_{3}-\left (x^{2}-4 \ln \left (a y \right )\right )^{{3}/{2}} y^{2} a_{3}-2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{2} b_{2}-2 y^{2} a_{3} \sqrt {x^{2}-4 \ln \left (a y \right )}-2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x b_{1}-2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, y a_{1}-2 x^{3} y^{2} a_{3}+8 \ln \left (a y \right ) x b_{2}+8 \ln \left (a y \right ) y a_{2}-4 x^{2} y a_{2}-2 x \,y^{2} a_{3}-2 x y a_{1} = 0
\end{equation}
Simplifying the above gives
\begin{equation}
\tag{6E} 4 y x b_{5}+4 x b_{2}+4 y b_{3}+4 b_{1}-2 \left (x^{2}-4 \ln \left (a y \right )\right ) b_{1}+4 b_{2} \sqrt {x^{2}-4 \ln \left (a y \right )}-2 x \,y^{3} a_{6}-2 \left (x^{2}-4 \ln \left (a y \right )\right )^{{3}/{2}} y^{3} a_{6}-2 \left (x^{2}-4 \ln \left (a y \right )\right ) x^{2} b_{4}-2 \left (x^{2}-4 \ln \left (a y \right )\right ) y^{2} a_{5}+2 \left (x^{2}-4 \ln \left (a y \right )\right ) y^{2} b_{6}-2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{3} b_{4}-2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, y^{3} a_{6}+8 x b_{4} \sqrt {x^{2}-4 \ln \left (a y \right )}+4 y b_{5} \sqrt {x^{2}-4 \ln \left (a y \right )}-2 x^{3} y a_{4}-2 x^{2} y^{2} a_{5}+4 x^{2} b_{4}+4 y^{2} b_{6}-\left (x^{2}-4 \ln \left (a y \right )\right )^{{3}/{2}} x \,y^{2} a_{5}-2 \left (x^{2}-4 \ln \left (a y \right )\right ) x^{2} y^{2} a_{5}-4 \left (x^{2}-4 \ln \left (a y \right )\right ) x \,y^{3} a_{6}-\sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{3} y^{2} a_{5}-2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{2} y^{3} a_{6}-4 \left (x^{2}-4 \ln \left (a y \right )\right ) x y a_{4}-6 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{2} y a_{4}-4 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x \,y^{2} a_{5}+2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x \,y^{2} b_{6}-2 \left (x^{2}-4 \ln \left (a y \right )\right ) x \,y^{2} a_{3}-\sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{2} y^{2} a_{3}-4 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x y a_{2}-\left (x^{2}-4 \ln \left (a y \right )\right )^{{3}/{2}} y^{2} a_{3}-2 \left (x^{2}-4 \ln \left (a y \right )\right ) x b_{2}-2 \left (x^{2}-4 \ln \left (a y \right )\right ) y a_{2}-2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{2} b_{2}-2 y^{2} a_{3} \sqrt {x^{2}-4 \ln \left (a y \right )}-2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x b_{1}-2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, y a_{1}-2 x^{2} y a_{2}-2 x \,y^{2} a_{3}-2 x y a_{1} = 0
\end{equation}
Since the PDE has
radicals, simplifying gives
\[
4 y x b_{5}+4 x b_{2}+4 y b_{3}+4 b_{1}+4 b_{2} \sqrt {x^{2}-4 \ln \left (a y \right )}-2 x^{3} b_{2}-2 x^{2} b_{1}+8 \ln \left (a y \right ) b_{1}+2 x^{2} y^{2} b_{6}-2 x^{4} y^{2} a_{5}-4 x^{3} y^{3} a_{6}+8 \ln \left (a y \right ) x^{2} b_{4}+8 \ln \left (a y \right ) y^{2} a_{5}-8 \ln \left (a y \right ) y^{2} b_{6}-2 x \,y^{3} a_{6}-2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{3} b_{4}-2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, y^{3} a_{6}+8 x b_{4} \sqrt {x^{2}-4 \ln \left (a y \right )}+4 y b_{5} \sqrt {x^{2}-4 \ln \left (a y \right )}-6 x^{3} y a_{4}-4 x^{2} y^{2} a_{5}-2 x^{4} b_{4}+8 \ln \left (a y \right ) \sqrt {x^{2}-4 \ln \left (a y \right )}\, y^{3} a_{6}+4 x^{2} b_{4}+4 y^{2} b_{6}+8 \ln \left (a y \right ) x^{2} y^{2} a_{5}+16 \ln \left (a y \right ) x \,y^{3} a_{6}+16 \ln \left (a y \right ) x y a_{4}-2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{3} y^{2} a_{5}-4 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{2} y^{3} a_{6}-6 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{2} y a_{4}-4 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x \,y^{2} a_{5}+2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x \,y^{2} b_{6}+4 \ln \left (a y \right ) \sqrt {x^{2}-4 \ln \left (a y \right )}\, y^{2} a_{3}-2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{2} y^{2} a_{3}-4 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x y a_{2}+8 \ln \left (a y \right ) x \,y^{2} a_{3}+4 \ln \left (a y \right ) x \sqrt {x^{2}-4 \ln \left (a y \right )}\, y^{2} a_{5}-2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{2} b_{2}-2 y^{2} a_{3} \sqrt {x^{2}-4 \ln \left (a y \right )}-2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x b_{1}-2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, y a_{1}-2 x^{3} y^{2} a_{3}+8 \ln \left (a y \right ) x b_{2}+8 \ln \left (a y \right ) y a_{2}-4 x^{2} y a_{2}-2 x \,y^{2} a_{3}-2 x y a_{1} = 0
\]
Looking at the above PDE shows the following are all
the terms with \(\{x, y\}\) in them.
\[
\left \{x, y, \sqrt {x^{2}-4 \ln \left (a y \right )}, \ln \left (a y \right )\right \}
\]
The following substitution is now made to be able to
collect on all terms with \(\{x, y\}\) in them
\[
\left \{x = v_{1}, y = v_{2}, \sqrt {x^{2}-4 \ln \left (a y \right )} = v_{3}, \ln \left (a y \right ) = v_{4}\right \}
\]
The above PDE (6E) now becomes
\begin{equation}
\tag{7E} -2 v_{1}^{4} v_{2}^{2} a_{5}-2 v_{3} v_{1}^{3} v_{2}^{2} a_{5}-4 v_{1}^{3} v_{2}^{3} a_{6}-4 v_{3} v_{1}^{2} v_{2}^{3} a_{6}-2 v_{1}^{3} v_{2}^{2} a_{3}-2 v_{3} v_{1}^{2} v_{2}^{2} a_{3}+8 v_{4} v_{1}^{2} v_{2}^{2} a_{5}+4 v_{4} v_{1} v_{3} v_{2}^{2} a_{5}+16 v_{4} v_{1} v_{2}^{3} a_{6}+8 v_{4} v_{3} v_{2}^{3} a_{6}+8 v_{4} v_{1} v_{2}^{2} a_{3}+4 v_{4} v_{3} v_{2}^{2} a_{3}-6 v_{1}^{3} v_{2} a_{4}-6 v_{3} v_{1}^{2} v_{2} a_{4}-4 v_{1}^{2} v_{2}^{2} a_{5}-4 v_{3} v_{1} v_{2}^{2} a_{5}-2 v_{1} v_{2}^{3} a_{6}-2 v_{3} v_{2}^{3} a_{6}-2 v_{1}^{4} b_{4}-2 v_{3} v_{1}^{3} b_{4}+2 v_{1}^{2} v_{2}^{2} b_{6}+2 v_{3} v_{1} v_{2}^{2} b_{6}-4 v_{1}^{2} v_{2} a_{2}-4 v_{3} v_{1} v_{2} a_{2}-2 v_{1} v_{2}^{2} a_{3}-2 v_{2}^{2} a_{3} v_{3}+16 v_{4} v_{1} v_{2} a_{4}+8 v_{4} v_{2}^{2} a_{5}-2 v_{1}^{3} b_{2}-2 v_{3} v_{1}^{2} b_{2}+8 v_{4} v_{1}^{2} b_{4}-8 v_{4} v_{2}^{2} b_{6}-2 v_{1} v_{2} a_{1}-2 v_{3} v_{2} a_{1}+8 v_{4} v_{2} a_{2}-2 v_{1}^{2} b_{1}-2 v_{3} v_{1} b_{1}+8 v_{4} v_{1} b_{2}+4 v_{1}^{2} b_{4}+8 v_{1} b_{4} v_{3}+4 v_{2} v_{1} b_{5}+4 v_{2} b_{5} v_{3}+4 v_{2}^{2} b_{6}+8 v_{4} b_{1}+4 v_{1} b_{2}+4 b_{2} v_{3}+4 v_{2} b_{3}+4 b_{1} = 0
\end{equation}
Collecting
the above on the terms \(v_i\) introduced, and these are
\[
\{v_{1}, v_{2}, v_{3}, v_{4}\}
\]
Equation (7E) now becomes
\begin{equation}
\tag{8E} 8 v_{4} v_{3} v_{2}^{3} a_{6}+8 v_{4} v_{1}^{2} v_{2}^{2} a_{5}+16 v_{4} v_{1} v_{2}^{3} a_{6}+16 v_{4} v_{1} v_{2} a_{4}-2 v_{3} v_{1}^{3} v_{2}^{2} a_{5}-4 v_{3} v_{1}^{2} v_{2}^{3} a_{6}-6 v_{3} v_{1}^{2} v_{2} a_{4}+4 v_{4} v_{3} v_{2}^{2} a_{3}-2 v_{3} v_{1}^{2} v_{2}^{2} a_{3}-4 v_{3} v_{1} v_{2} a_{2}+8 v_{4} v_{1} v_{2}^{2} a_{3}+\left (-4 a_{5}+2 b_{6}\right ) v_{1}^{2} v_{2}^{2}+\left (-2 a_{1}+4 b_{5}\right ) v_{1} v_{2}+\left (-2 b_{1}+8 b_{4}\right ) v_{1} v_{3}+\left (8 a_{5}-8 b_{6}\right ) v_{2}^{2} v_{4}+\left (-2 a_{1}+4 b_{5}\right ) v_{2} v_{3}+4 b_{1}+4 v_{1} b_{2}+4 v_{2} b_{3}+4 b_{2} v_{3}-2 v_{1}^{3} b_{2}+8 v_{4} b_{1}-2 v_{1}^{4} b_{4}+4 v_{2}^{2} b_{6}+4 v_{4} v_{1} v_{3} v_{2}^{2} a_{5}+\left (-2 b_{1}+4 b_{4}\right ) v_{1}^{2}-2 v_{3} v_{1}^{3} b_{4}-2 v_{3} v_{2}^{3} a_{6}-6 v_{1}^{3} v_{2} a_{4}-2 v_{3} v_{1}^{2} b_{2}-2 v_{2}^{2} a_{3} v_{3}-2 v_{1}^{3} v_{2}^{2} a_{3}+8 v_{4} v_{1} b_{2}+8 v_{4} v_{2} a_{2}-4 v_{1}^{2} v_{2} a_{2}-2 v_{1} v_{2}^{2} a_{3}-2 v_{1}^{4} v_{2}^{2} a_{5}-4 v_{1}^{3} v_{2}^{3} a_{6}+8 v_{4} v_{1}^{2} b_{4}-2 v_{1} v_{2}^{3} a_{6}+\left (-4 a_{5}+2 b_{6}\right ) v_{1} v_{2}^{2} v_{3} = 0
\end{equation}
Setting each coefficients in (8E) to zero gives the following equations to solve
\begin{align*} -4 a_{2}&=0\\ 8 a_{2}&=0\\ -2 a_{3}&=0\\ 4 a_{3}&=0\\ 8 a_{3}&=0\\ -6 a_{4}&=0\\ 16 a_{4}&=0\\ -2 a_{5}&=0\\ 4 a_{5}&=0\\ 8 a_{5}&=0\\ -4 a_{6}&=0\\ -2 a_{6}&=0\\ 8 a_{6}&=0\\ 16 a_{6}&=0\\ 4 b_{1}&=0\\ 8 b_{1}&=0\\ -2 b_{2}&=0\\ 4 b_{2}&=0\\ 8 b_{2}&=0\\ 4 b_{3}&=0\\ -2 b_{4}&=0\\ 8 b_{4}&=0\\ 4 b_{6}&=0\\ -2 a_{1}+4 b_{5}&=0\\ -4 a_{5}+2 b_{6}&=0\\ 8 a_{5}-8 b_{6}&=0\\ -2 b_{1}+4 b_{4}&=0\\ -2 b_{1}+8 b_{4}&=0 \end{align*}
Solving the above equations for the unknowns gives
\begin{align*} a_{1}&=2 b_{5}\\ a_{2}&=0\\ a_{3}&=0\\ a_{4}&=0\\ a_{5}&=0\\ a_{6}&=0\\ b_{1}&=0\\ b_{2}&=0\\ b_{3}&=0\\ b_{4}&=0\\ b_{5}&=b_{5}\\ b_{6}&=0 \end{align*}
Substituting the above solution in the anstaz (1E,2E) (using \(1\) as arbitrary value for any
unknown in the RHS) gives
\begin{align*}
\xi &= 2 \\
\eta &= y x \\
\end{align*}
Shifting is now applied to make \(\xi =0\) in order to simplify the rest of
the computation
\begin{align*} \eta &= \eta - \omega \left (x,y\right ) \xi \\ &= y x - \left (\frac {\left (x +\sqrt {x^{2}-4 \ln \left (a y \right )}\right ) y}{2}\right ) \left (2\right ) \\ &= -y \sqrt {x^{2}-4 \ln \left (a y \right )}\\ \xi &= 0 \end{align*}
The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( x,y\right ) \to \left ( R,S \right )\)
where \(\left ( R,S \right )\) are the canonical coordinates which make the original ode become a quadrature and
hence solved by integration.
The characteristic pde which is used to find the canonical coordinates is
\begin{align*} \frac {d x}{\xi } &= \frac {d y}{\eta } = dS \tag {1} \end{align*}
The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial y}\right ) S(x,y) = 1\). Starting with the first pair of ode’s in (1)
gives an ode to solve for the independent variable \(R\) in the canonical coordinates, where \(S(R)\). Since
\(\xi =0\) then in this special case
\begin{align*} R = x \end{align*}
\(S\) is found from
\begin{align*} S &= \int { \frac {1}{\eta }} dy\\ &= \int { \frac {1}{-y \sqrt {x^{2}-4 \ln \left (a y \right )}}} dy \end{align*}
Which results in
\begin{align*} S&= \frac {\sqrt {x^{2}-4 \ln \left (a y \right )}}{2} \end{align*}
Now that \(R,S\) are found, we need to setup the ode in these coordinates. This is done by
evaluating
\begin{align*} \frac {dS}{dR} &= \frac { S_{x} + \omega (x,y) S_{y} }{ R_{x} + \omega (x,y) R_{y} }\tag {2} \end{align*}
Where in the above \(R_{x},R_{y},S_{x},S_{y}\) are all partial derivatives and \(\omega (x,y)\) is the right hand side of the original ode
given by
\begin{align*} \omega (x,y) &= \frac {\left (x +\sqrt {x^{2}-4 \ln \left (a y \right )}\right ) y}{2} \end{align*}
Evaluating all the partial derivatives gives
\begin{align*} R_{x} &= 1\\ R_{y} &= 0\\ S_{x} &= \frac {x}{2 \sqrt {x^{2}-4 \ln \left (a \right )-4 \ln \left (y \right )}}\\ S_{y} &= -\frac {1}{\sqrt {x^{2}-4 \ln \left (a \right )-4 \ln \left (y \right )}\, y} \end{align*}
Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.
\begin{align*} \frac {dS}{dR} &= -\frac {\sqrt {x^{2}-4 \ln \left (a y \right )}}{2 \sqrt {x^{2}-4 \ln \left (a \right )-4 \ln \left (y \right )}}\tag {2A} \end{align*}
We now need to express the RHS as function of \(R\) only. This is done by solving for \(x,y\) in terms of
\(R,S\) from the result obtained earlier and simplifying. This gives
\begin{align*} \frac {dS}{dR} &= -{\frac {1}{2}} \end{align*}
The above is a quadrature ode. This is the whole point of Lie symmetry method. It converts
an ode, no matter how complicated it is, to one that can be solved by integration when the
ode is in the canonical coordiates \(R,S\).
Since the ode has the form \(\frac {d}{d R}S \left (R \right )=f(R)\), then we only need to integrate \(f(R)\).
\begin{align*} \int {dS} &= \int {-{\frac {1}{2}}\, dR}\\ S \left (R \right ) &= -\frac {R}{2} + c_2 \end{align*}
To complete the solution, we just need to transform the above back to \(x,y\) coordinates. This
results in
\begin{align*} \frac {\sqrt {x^{2}-4 \ln \left (a \right )-4 \ln \left (y\right )}}{2} = -\frac {x}{2}+c_2 \end{align*}
Which gives
\begin{align*} y = \frac {{\mathrm e}^{-c_2^{2}+c_2 x}}{a} \end{align*}
We now need to find the singular solutions, these are found by finding for what values \((\left (\frac {x}{2}+\frac {\sqrt {x^{2}-4 \ln \left (a y \right )}}{2}\right ) y)\) is
zero. These give
\begin{align*}
y&=\frac {1}{a} \\
\end{align*}
Now we go over each such singular solution and check if it verifies the ode
itself and any initial conditions given. If it does not then the singular solution will not be
used.
The solution \(y = \frac {1}{a}\) satisfies the ode and initial conditions.
Solving Eq. (2)
Writing the ode as
\begin{align*} y^{\prime }&=-\frac {\left (-x +\sqrt {x^{2}-4 \ln \left (a y \right )}\right ) y}{2}\\ y^{\prime }&= \omega \left ( x,y\right ) \end{align*}
The condition of Lie symmetry is the linearized PDE given by
\begin{align*} \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0\tag {A} \end{align*}
To determine \(\xi ,\eta \) then (A) is solved using ansatz. Making bivariate polynomials of
degree 2 to use as anstaz gives
\begin{align*}
\tag{1E} \xi &= x^{2} a_{4}+y x a_{5}+y^{2} a_{6}+x a_{2}+y a_{3}+a_{1} \\
\tag{2E} \eta &= x^{2} b_{4}+y x b_{5}+y^{2} b_{6}+x b_{2}+y b_{3}+b_{1} \\
\end{align*}
Where the unknown coefficients are
\[
\{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, b_{1}, b_{2}, b_{3}, b_{4}, b_{5}, b_{6}\}
\]
Substituting
equations (1E,2E) and \(\omega \) into (A) gives
\begin{equation}
\tag{5E} 2 x b_{4}+y b_{5}+b_{2}-\frac {\left (-x +\sqrt {x^{2}-4 \ln \left (a y \right )}\right ) y \left (-2 x a_{4}+x b_{5}-y a_{5}+2 y b_{6}-a_{2}+b_{3}\right )}{2}-\frac {\left (-x +\sqrt {x^{2}-4 \ln \left (a y \right )}\right )^{2} y^{2} \left (x a_{5}+2 y a_{6}+a_{3}\right )}{4}+\frac {\left (-1+\frac {x}{\sqrt {x^{2}-4 \ln \left (a y \right )}}\right ) y \left (x^{2} a_{4}+y x a_{5}+y^{2} a_{6}+x a_{2}+y a_{3}+a_{1}\right )}{2}-\left (\frac {1}{\sqrt {x^{2}-4 \ln \left (a y \right )}}+\frac {x}{2}-\frac {\sqrt {x^{2}-4 \ln \left (a y \right )}}{2}\right ) \left (x^{2} b_{4}+y x b_{5}+y^{2} b_{6}+x b_{2}+y b_{3}+b_{1}\right ) = 0
\end{equation}
Putting the above in normal form gives
\[
-\frac {-2 x^{3} b_{2}-2 x^{2} b_{1}+8 \ln \left (a y \right ) b_{1}-4 b_{2} \sqrt {x^{2}-4 \ln \left (a y \right )}+4 x b_{2}+4 y b_{3}+4 x^{2} b_{4}+4 y^{2} b_{6}-4 x^{2} y a_{2}-2 x \,y^{2} a_{3}-2 x y a_{1}+\left (x^{2}-4 \ln \left (a y \right )\right )^{{3}/{2}} y^{2} a_{3}+2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{2} b_{2}+2 y^{2} a_{3} \sqrt {x^{2}-4 \ln \left (a y \right )}+2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x b_{1}+2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, y a_{1}-2 x^{3} y^{2} a_{3}+8 \ln \left (a y \right ) x b_{2}+8 \ln \left (a y \right ) y a_{2}+4 b_{1}-6 x^{3} y a_{4}-4 x^{2} y^{2} a_{5}-2 x \,y^{3} a_{6}+2 \left (x^{2}-4 \ln \left (a y \right )\right )^{{3}/{2}} y^{3} a_{6}+2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{3} b_{4}+2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, y^{3} a_{6}-8 x b_{4} \sqrt {x^{2}-4 \ln \left (a y \right )}-4 y b_{5} \sqrt {x^{2}-4 \ln \left (a y \right )}+\sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{2} y^{2} a_{3}+4 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x y a_{2}+8 \ln \left (a y \right ) x \,y^{2} a_{3}+8 \ln \left (a y \right ) x^{2} y^{2} a_{5}+16 \ln \left (a y \right ) x \,y^{3} a_{6}+16 \ln \left (a y \right ) x y a_{4}+\left (x^{2}-4 \ln \left (a y \right )\right )^{{3}/{2}} x \,y^{2} a_{5}+\sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{3} y^{2} a_{5}+2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{2} y^{3} a_{6}+6 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{2} y a_{4}+4 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x \,y^{2} a_{5}-2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x \,y^{2} b_{6}+4 y x b_{5}-2 x^{4} b_{4}-2 x^{4} y^{2} a_{5}-4 x^{3} y^{3} a_{6}+2 x^{2} y^{2} b_{6}+8 \ln \left (a y \right ) x^{2} b_{4}+8 \ln \left (a y \right ) y^{2} a_{5}-8 \ln \left (a y \right ) y^{2} b_{6}}{4 \sqrt {x^{2}-4 \ln \left (a y \right )}} = 0
\]
Setting the numerator to zero gives
\begin{equation}
\tag{6E} 2 x^{3} b_{2}+2 x^{2} b_{1}-8 \ln \left (a y \right ) b_{1}+4 b_{2} \sqrt {x^{2}-4 \ln \left (a y \right )}-4 x b_{2}-4 y b_{3}-4 x^{2} b_{4}-4 y^{2} b_{6}+4 x^{2} y a_{2}+2 x \,y^{2} a_{3}+2 x y a_{1}-\left (x^{2}-4 \ln \left (a y \right )\right )^{{3}/{2}} y^{2} a_{3}-2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{2} b_{2}-2 y^{2} a_{3} \sqrt {x^{2}-4 \ln \left (a y \right )}-2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x b_{1}-2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, y a_{1}+2 x^{3} y^{2} a_{3}-8 \ln \left (a y \right ) x b_{2}-8 \ln \left (a y \right ) y a_{2}-4 b_{1}+6 x^{3} y a_{4}+4 x^{2} y^{2} a_{5}+2 x \,y^{3} a_{6}-2 \left (x^{2}-4 \ln \left (a y \right )\right )^{{3}/{2}} y^{3} a_{6}-2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{3} b_{4}-2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, y^{3} a_{6}+8 x b_{4} \sqrt {x^{2}-4 \ln \left (a y \right )}+4 y b_{5} \sqrt {x^{2}-4 \ln \left (a y \right )}-\sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{2} y^{2} a_{3}-4 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x y a_{2}-8 \ln \left (a y \right ) x \,y^{2} a_{3}-8 \ln \left (a y \right ) x^{2} y^{2} a_{5}-16 \ln \left (a y \right ) x \,y^{3} a_{6}-16 \ln \left (a y \right ) x y a_{4}-\left (x^{2}-4 \ln \left (a y \right )\right )^{{3}/{2}} x \,y^{2} a_{5}-\sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{3} y^{2} a_{5}-2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{2} y^{3} a_{6}-6 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{2} y a_{4}-4 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x \,y^{2} a_{5}+2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x \,y^{2} b_{6}-4 y x b_{5}+2 x^{4} b_{4}+2 x^{4} y^{2} a_{5}+4 x^{3} y^{3} a_{6}-2 x^{2} y^{2} b_{6}-8 \ln \left (a y \right ) x^{2} b_{4}-8 \ln \left (a y \right ) y^{2} a_{5}+8 \ln \left (a y \right ) y^{2} b_{6} = 0
\end{equation}
Simplifying the above gives
\begin{equation}
\tag{6E} 2 \left (x^{2}-4 \ln \left (a y \right )\right ) b_{1}+4 b_{2} \sqrt {x^{2}-4 \ln \left (a y \right )}-4 x b_{2}-4 y b_{3}-4 x^{2} b_{4}-4 y^{2} b_{6}+2 x^{2} y a_{2}+2 x \,y^{2} a_{3}+2 x y a_{1}-\left (x^{2}-4 \ln \left (a y \right )\right )^{{3}/{2}} y^{2} a_{3}+2 \left (x^{2}-4 \ln \left (a y \right )\right ) x b_{2}+2 \left (x^{2}-4 \ln \left (a y \right )\right ) y a_{2}-2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{2} b_{2}-2 y^{2} a_{3} \sqrt {x^{2}-4 \ln \left (a y \right )}-2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x b_{1}-2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, y a_{1}-4 b_{1}+2 x^{3} y a_{4}+2 x^{2} y^{2} a_{5}+2 x \,y^{3} a_{6}-2 \left (x^{2}-4 \ln \left (a y \right )\right )^{{3}/{2}} y^{3} a_{6}+2 \left (x^{2}-4 \ln \left (a y \right )\right ) x^{2} b_{4}+2 \left (x^{2}-4 \ln \left (a y \right )\right ) y^{2} a_{5}-2 \left (x^{2}-4 \ln \left (a y \right )\right ) y^{2} b_{6}-2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{3} b_{4}-2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, y^{3} a_{6}+8 x b_{4} \sqrt {x^{2}-4 \ln \left (a y \right )}+4 y b_{5} \sqrt {x^{2}-4 \ln \left (a y \right )}+2 \left (x^{2}-4 \ln \left (a y \right )\right ) x \,y^{2} a_{3}-\sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{2} y^{2} a_{3}-4 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x y a_{2}-\left (x^{2}-4 \ln \left (a y \right )\right )^{{3}/{2}} x \,y^{2} a_{5}+2 \left (x^{2}-4 \ln \left (a y \right )\right ) x^{2} y^{2} a_{5}+4 \left (x^{2}-4 \ln \left (a y \right )\right ) x \,y^{3} a_{6}-\sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{3} y^{2} a_{5}-2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{2} y^{3} a_{6}+4 \left (x^{2}-4 \ln \left (a y \right )\right ) x y a_{4}-6 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{2} y a_{4}-4 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x \,y^{2} a_{5}+2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x \,y^{2} b_{6}-4 y x b_{5} = 0
\end{equation}
Since the PDE has
radicals, simplifying gives
\[
2 x^{3} b_{2}+2 x^{2} b_{1}-8 \ln \left (a y \right ) b_{1}+4 b_{2} \sqrt {x^{2}-4 \ln \left (a y \right )}-4 x b_{2}-4 y b_{3}-4 x^{2} b_{4}-4 y^{2} b_{6}+4 x^{2} y a_{2}+2 x \,y^{2} a_{3}+2 x y a_{1}-2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{2} b_{2}-2 y^{2} a_{3} \sqrt {x^{2}-4 \ln \left (a y \right )}-2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x b_{1}-2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, y a_{1}+2 x^{3} y^{2} a_{3}-8 \ln \left (a y \right ) x b_{2}-8 \ln \left (a y \right ) y a_{2}-4 b_{1}+6 x^{3} y a_{4}+4 x^{2} y^{2} a_{5}+2 x \,y^{3} a_{6}-2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{3} b_{4}-2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, y^{3} a_{6}+8 x b_{4} \sqrt {x^{2}-4 \ln \left (a y \right )}+4 y b_{5} \sqrt {x^{2}-4 \ln \left (a y \right )}+4 \ln \left (a y \right ) x \sqrt {x^{2}-4 \ln \left (a y \right )}\, y^{2} a_{5}+4 \ln \left (a y \right ) \sqrt {x^{2}-4 \ln \left (a y \right )}\, y^{2} a_{3}-2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{2} y^{2} a_{3}-4 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x y a_{2}-8 \ln \left (a y \right ) x \,y^{2} a_{3}+8 \ln \left (a y \right ) \sqrt {x^{2}-4 \ln \left (a y \right )}\, y^{3} a_{6}-8 \ln \left (a y \right ) x^{2} y^{2} a_{5}-16 \ln \left (a y \right ) x \,y^{3} a_{6}-16 \ln \left (a y \right ) x y a_{4}-2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{3} y^{2} a_{5}-4 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{2} y^{3} a_{6}-6 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x^{2} y a_{4}-4 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x \,y^{2} a_{5}+2 \sqrt {x^{2}-4 \ln \left (a y \right )}\, x \,y^{2} b_{6}-4 y x b_{5}+2 x^{4} b_{4}+2 x^{4} y^{2} a_{5}+4 x^{3} y^{3} a_{6}-2 x^{2} y^{2} b_{6}-8 \ln \left (a y \right ) x^{2} b_{4}-8 \ln \left (a y \right ) y^{2} a_{5}+8 \ln \left (a y \right ) y^{2} b_{6} = 0
\]
Looking at the above PDE shows the following are all
the terms with \(\{x, y\}\) in them.
\[
\left \{x, y, \sqrt {x^{2}-4 \ln \left (a y \right )}, \ln \left (a y \right )\right \}
\]
The following substitution is now made to be able to
collect on all terms with \(\{x, y\}\) in them
\[
\left \{x = v_{1}, y = v_{2}, \sqrt {x^{2}-4 \ln \left (a y \right )} = v_{3}, \ln \left (a y \right ) = v_{4}\right \}
\]
The above PDE (6E) now becomes
\begin{equation}
\tag{7E} 2 v_{1}^{4} v_{2}^{2} a_{5}-2 v_{3} v_{1}^{3} v_{2}^{2} a_{5}+4 v_{1}^{3} v_{2}^{3} a_{6}-4 v_{3} v_{1}^{2} v_{2}^{3} a_{6}+2 v_{1}^{3} v_{2}^{2} a_{3}-2 v_{3} v_{1}^{2} v_{2}^{2} a_{3}-8 v_{4} v_{1}^{2} v_{2}^{2} a_{5}+4 v_{4} v_{1} v_{3} v_{2}^{2} a_{5}-16 v_{4} v_{1} v_{2}^{3} a_{6}+8 v_{4} v_{3} v_{2}^{3} a_{6}-8 v_{4} v_{1} v_{2}^{2} a_{3}+4 v_{4} v_{3} v_{2}^{2} a_{3}+6 v_{1}^{3} v_{2} a_{4}-6 v_{3} v_{1}^{2} v_{2} a_{4}+4 v_{1}^{2} v_{2}^{2} a_{5}-4 v_{3} v_{1} v_{2}^{2} a_{5}+2 v_{1} v_{2}^{3} a_{6}-2 v_{3} v_{2}^{3} a_{6}+2 v_{1}^{4} b_{4}-2 v_{3} v_{1}^{3} b_{4}-2 v_{1}^{2} v_{2}^{2} b_{6}+2 v_{3} v_{1} v_{2}^{2} b_{6}+4 v_{1}^{2} v_{2} a_{2}-4 v_{3} v_{1} v_{2} a_{2}+2 v_{1} v_{2}^{2} a_{3}-2 v_{2}^{2} a_{3} v_{3}-16 v_{4} v_{1} v_{2} a_{4}-8 v_{4} v_{2}^{2} a_{5}+2 v_{1}^{3} b_{2}-2 v_{3} v_{1}^{2} b_{2}-8 v_{4} v_{1}^{2} b_{4}+8 v_{4} v_{2}^{2} b_{6}+2 v_{1} v_{2} a_{1}-2 v_{3} v_{2} a_{1}-8 v_{4} v_{2} a_{2}+2 v_{1}^{2} b_{1}-2 v_{3} v_{1} b_{1}-8 v_{4} v_{1} b_{2}-4 v_{1}^{2} b_{4}+8 v_{1} b_{4} v_{3}-4 v_{2} v_{1} b_{5}+4 v_{2} b_{5} v_{3}-4 v_{2}^{2} b_{6}-8 v_{4} b_{1}-4 v_{1} b_{2}+4 b_{2} v_{3}-4 v_{2} b_{3}-4 b_{1} = 0
\end{equation}
Collecting
the above on the terms \(v_i\) introduced, and these are
\[
\{v_{1}, v_{2}, v_{3}, v_{4}\}
\]
Equation (7E) now becomes
\begin{equation}
\tag{8E} 4 v_{1}^{2} v_{2} a_{2}+2 v_{1} v_{2}^{2} a_{3}-2 v_{3} v_{1}^{2} b_{2}-2 v_{2}^{2} a_{3} v_{3}+2 v_{1}^{3} v_{2}^{2} a_{3}-8 v_{4} v_{1} b_{2}-8 v_{4} v_{2} a_{2}+6 v_{1}^{3} v_{2} a_{4}+2 v_{1} v_{2}^{3} a_{6}-2 v_{3} v_{1}^{3} b_{4}-2 v_{3} v_{2}^{3} a_{6}+2 v_{1}^{4} v_{2}^{2} a_{5}+4 v_{1}^{3} v_{2}^{3} a_{6}-8 v_{4} v_{1}^{2} b_{4}+2 v_{1}^{3} b_{2}-8 v_{4} b_{1}+4 b_{2} v_{3}-4 v_{1} b_{2}-4 v_{2} b_{3}-4 v_{2}^{2} b_{6}+2 v_{1}^{4} b_{4}-4 b_{1}+\left (4 a_{5}-2 b_{6}\right ) v_{1}^{2} v_{2}^{2}+\left (2 a_{1}-4 b_{5}\right ) v_{1} v_{2}+\left (-2 b_{1}+8 b_{4}\right ) v_{1} v_{3}+\left (-8 a_{5}+8 b_{6}\right ) v_{2}^{2} v_{4}+\left (-2 a_{1}+4 b_{5}\right ) v_{2} v_{3}+4 v_{4} v_{1} v_{3} v_{2}^{2} a_{5}+4 v_{4} v_{3} v_{2}^{2} a_{3}+\left (2 b_{1}-4 b_{4}\right ) v_{1}^{2}-2 v_{3} v_{1}^{2} v_{2}^{2} a_{3}-4 v_{3} v_{1} v_{2} a_{2}-8 v_{4} v_{1} v_{2}^{2} a_{3}+8 v_{4} v_{3} v_{2}^{3} a_{6}-8 v_{4} v_{1}^{2} v_{2}^{2} a_{5}-16 v_{4} v_{1} v_{2}^{3} a_{6}-16 v_{4} v_{1} v_{2} a_{4}-2 v_{3} v_{1}^{3} v_{2}^{2} a_{5}-4 v_{3} v_{1}^{2} v_{2}^{3} a_{6}-6 v_{3} v_{1}^{2} v_{2} a_{4}+\left (-4 a_{5}+2 b_{6}\right ) v_{1} v_{2}^{2} v_{3} = 0
\end{equation}
Setting each coefficients in (8E) to zero gives the following equations to solve
\begin{align*} -8 a_{2}&=0\\ -4 a_{2}&=0\\ 4 a_{2}&=0\\ -8 a_{3}&=0\\ -2 a_{3}&=0\\ 2 a_{3}&=0\\ 4 a_{3}&=0\\ -16 a_{4}&=0\\ -6 a_{4}&=0\\ 6 a_{4}&=0\\ -8 a_{5}&=0\\ -2 a_{5}&=0\\ 2 a_{5}&=0\\ 4 a_{5}&=0\\ -16 a_{6}&=0\\ -4 a_{6}&=0\\ -2 a_{6}&=0\\ 2 a_{6}&=0\\ 4 a_{6}&=0\\ 8 a_{6}&=0\\ -8 b_{1}&=0\\ -4 b_{1}&=0\\ -8 b_{2}&=0\\ -4 b_{2}&=0\\ -2 b_{2}&=0\\ 2 b_{2}&=0\\ 4 b_{2}&=0\\ -4 b_{3}&=0\\ -8 b_{4}&=0\\ -2 b_{4}&=0\\ 2 b_{4}&=0\\ -4 b_{6}&=0\\ -2 a_{1}+4 b_{5}&=0\\ 2 a_{1}-4 b_{5}&=0\\ -8 a_{5}+8 b_{6}&=0\\ -4 a_{5}+2 b_{6}&=0\\ 4 a_{5}-2 b_{6}&=0\\ -2 b_{1}+8 b_{4}&=0\\ 2 b_{1}-4 b_{4}&=0 \end{align*}
Solving the above equations for the unknowns gives
\begin{align*} a_{1}&=2 b_{5}\\ a_{2}&=0\\ a_{3}&=0\\ a_{4}&=0\\ a_{5}&=0\\ a_{6}&=0\\ b_{1}&=0\\ b_{2}&=0\\ b_{3}&=0\\ b_{4}&=0\\ b_{5}&=b_{5}\\ b_{6}&=0 \end{align*}
Substituting the above solution in the anstaz (1E,2E) (using \(1\) as arbitrary value for any
unknown in the RHS) gives
\begin{align*}
\xi &= 2 \\
\eta &= y x \\
\end{align*}
Shifting is now applied to make \(\xi =0\) in order to simplify the rest of
the computation
\begin{align*} \eta &= \eta - \omega \left (x,y\right ) \xi \\ &= y x - \left (-\frac {\left (-x +\sqrt {x^{2}-4 \ln \left (a y \right )}\right ) y}{2}\right ) \left (2\right ) \\ &= y \sqrt {x^{2}-4 \ln \left (a y \right )}\\ \xi &= 0 \end{align*}
The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( x,y\right ) \to \left ( R,S \right )\)
where \(\left ( R,S \right )\) are the canonical coordinates which make the original ode become a quadrature and
hence solved by integration.
The characteristic pde which is used to find the canonical coordinates is
\begin{align*} \frac {d x}{\xi } &= \frac {d y}{\eta } = dS \tag {1} \end{align*}
The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial y}\right ) S(x,y) = 1\). Starting with the first pair of ode’s in (1)
gives an ode to solve for the independent variable \(R\) in the canonical coordinates, where \(S(R)\). Since
\(\xi =0\) then in this special case
\begin{align*} R = x \end{align*}
\(S\) is found from
\begin{align*} S &= \int { \frac {1}{\eta }} dy\\ &= \int { \frac {1}{y \sqrt {x^{2}-4 \ln \left (a y \right )}}} dy \end{align*}
Which results in
\begin{align*} S&= -\frac {\sqrt {x^{2}-4 \ln \left (a y \right )}}{2} \end{align*}
Now that \(R,S\) are found, we need to setup the ode in these coordinates. This is done by
evaluating
\begin{align*} \frac {dS}{dR} &= \frac { S_{x} + \omega (x,y) S_{y} }{ R_{x} + \omega (x,y) R_{y} }\tag {2} \end{align*}
Where in the above \(R_{x},R_{y},S_{x},S_{y}\) are all partial derivatives and \(\omega (x,y)\) is the right hand side of the original ode
given by
\begin{align*} \omega (x,y) &= -\frac {\left (-x +\sqrt {x^{2}-4 \ln \left (a y \right )}\right ) y}{2} \end{align*}
Evaluating all the partial derivatives gives
\begin{align*} R_{x} &= 1\\ R_{y} &= 0\\ S_{x} &= -\frac {x}{2 \sqrt {x^{2}-4 \ln \left (a \right )-4 \ln \left (y \right )}}\\ S_{y} &= \frac {1}{\sqrt {x^{2}-4 \ln \left (a \right )-4 \ln \left (y \right )}\, y} \end{align*}
Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.
\begin{align*} \frac {dS}{dR} &= -\frac {\sqrt {x^{2}-4 \ln \left (a y \right )}}{2 \sqrt {x^{2}-4 \ln \left (a \right )-4 \ln \left (y \right )}}\tag {2A} \end{align*}
We now need to express the RHS as function of \(R\) only. This is done by solving for \(x,y\) in terms of
\(R,S\) from the result obtained earlier and simplifying. This gives
\begin{align*} \frac {dS}{dR} &= -{\frac {1}{2}} \end{align*}
The above is a quadrature ode. This is the whole point of Lie symmetry method. It converts
an ode, no matter how complicated it is, to one that can be solved by integration when the
ode is in the canonical coordiates \(R,S\).
Since the ode has the form \(\frac {d}{d R}S \left (R \right )=f(R)\), then we only need to integrate \(f(R)\).
\begin{align*} \int {dS} &= \int {-{\frac {1}{2}}\, dR}\\ S \left (R \right ) &= -\frac {R}{2} + c_4 \end{align*}
To complete the solution, we just need to transform the above back to \(x,y\) coordinates. This
results in
\begin{align*} -\frac {\sqrt {x^{2}-4 \ln \left (a \right )-4 \ln \left (y\right )}}{2} = -\frac {x}{2}+c_4 \end{align*}
Which gives
\begin{align*} y = \frac {{\mathrm e}^{-c_4^{2}+c_4 x}}{a} \end{align*}
We now need to find the singular solutions, these are found by finding for what values \((\left (\frac {x}{2}-\frac {\sqrt {x^{2}-4 \ln \left (a y \right )}}{2}\right ) y)\) is
zero. These give
\begin{align*}
y&=\frac {1}{a} \\
\end{align*}
Now we go over each such singular solution and check if it verifies the ode
itself and any initial conditions given. If it does not then the singular solution will not be
used.
The solution \(y = \frac {1}{a}\) satisfies the ode and initial conditions.
1.391.1 Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y \left (x \right )\right )^{2}-x y \left (x \right ) \left (\frac {d}{d x}y \left (x \right )\right )+y \left (x \right )^{2} \ln \left (a y \left (x \right )\right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d x}y \left (x \right )=\left (\frac {x}{2}-\frac {\sqrt {x^{2}-4 \ln \left (a y \left (x \right )\right )}}{2}\right ) y \left (x \right ), \frac {d}{d x}y \left (x \right )=\left (\frac {x}{2}+\frac {\sqrt {x^{2}-4 \ln \left (a y \left (x \right )\right )}}{2}\right ) y \left (x \right )\right ] \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}y \left (x \right )=\left (\frac {x}{2}-\frac {\sqrt {x^{2}-4 \ln \left (a y \left (x \right )\right )}}{2}\right ) y \left (x \right ) \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}y \left (x \right )=\left (\frac {x}{2}+\frac {\sqrt {x^{2}-4 \ln \left (a y \left (x \right )\right )}}{2}\right ) y \left (x \right ) \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\mathit {workingODE} , \mathit {workingODE}\right \} \end {array} \]
1.391.2 Maple trace
`Methods for first order ODEs:
-> Solving 1st order ODE of high degree, 1st attempt
trying 1st order WeierstrassP solution for high degree ODE
trying 1st order WeierstrassPPrime solution for high degree ODE
trying 1st order JacobiSN solution for high degree ODE
trying 1st order ODE linearizable_by_differentiation
trying differential order: 1; missing variables
trying simple symmetries for implicit equations
Successful isolation of dy/dx: 2 solutions were found. Trying to solve each resulting ODE.
*** Sublevel 2 ***
Methods for first order ODEs:
--- Trying classification methods ---
trying homogeneous types:
trying exact
Looking for potential symmetries
trying an equivalence to an Abel ODE
trying 1st order ODE linearizable_by_differentiation
-> Solving 1st order ODE of high degree, Lie methods, 1st trial
`, `-> Computing symmetries using: way = 3`[1, 1/2*x*y]
1.391.3 Maple dsolve solution
Solving time : 0.147 (sec)
Leaf size : 43
dsolve(diff(y(x),x)^2-y(x)*diff(y(x),x)*x+y(x)^2*ln(a*y(x)) = 0,
y(x),singsol=all)
\begin{align*}
y &= \frac {{\mathrm e}^{\frac {x^{2}}{4}}}{a} \\
y &= \frac {{\mathrm e}^{c_{1} \left (-c_{1} +x \right )}}{a} \\
y &= \frac {{\mathrm e}^{-c_{1} \left (c_{1} +x \right )}}{a} \\
\end{align*}
1.391.4 Mathematica DSolve solution
Solving time : 0.29 (sec)
Leaf size : 30
DSolve[{Log[a*y[x]]*y[x]^2 - x*y[x]*D[y[x],x] + D[y[x],x]^2==0,{}},
y[x],x,IncludeSingularSolutions->True]
\begin{align*}
y(x)\to \frac {e^{\frac {1}{4} c_1 (2 x-c_1)}}{a} \\
y(x)\to 0 \\
\end{align*}