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.