problem 1006

Internal problem ID [4236]
Internal ﬁle name [OUTPUT/3729_Sunday_June_05_2022_10_33_26_AM_66498046/index.tex]

Book: Ordinary diﬀerential equations and their solutions. By George Moseley Murphy. 1960
Section: Various 34
Problem number: 1006.
ODE order: 1.
ODE degree: 2.

The type(s) of ODE detected by this program : "ﬁrst_order_ode_lie_symmetry_calculated"

\[ \boxed {4 y^{3} {y^{\prime }}^{2}-4 x y^{\prime }+y=0} \] Solving the given ode for \(y^{\prime }\) results in \(2\) diﬀerential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} y^{\prime }&=\frac {x +\sqrt {x^{2}-y^{4}}}{2 y^{3}} \tag {1} \\ y^{\prime }&=-\frac {-x +\sqrt {x^{2}-y^{4}}}{2 y^{3}} \tag {2} \end {align*}

Writing the ode as \begin {align*} y^{\prime }&=\frac {\sqrt {-\left (y^{2}-x \right ) \left (y^{2}+x \right )}+x}{2 y^{3}}\\ 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*}

The type of this ode is not in the lookup table. To determine \(\xi ,\eta \) then (A) is solved using ansatz. Making bivariate polynomials of degree 1 to use as anstaz gives \begin{align*} \tag{1E} \xi &= x a_{2}+y a_{3}+a_{1} \\ \tag{2E} \eta &= x b_{2}+y b_{3}+b_{1} \\ \end{align*} Where the unknown coeﬃcients are \[ \{a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}\} \] Substituting equations (1E,2E) and \(\omega \) into (A) gives \begin{equation} \tag{5E} b_{2}+\frac {\left (\sqrt {-\left (y^{2}-x \right ) \left (y^{2}+x \right )}+x \right ) \left (b_{3}-a_{2}\right )}{2 y^{3}}-\frac {{\left (\sqrt {-\left (y^{2}-x \right ) \left (y^{2}+x \right )}+x \right )}^{2} a_{3}}{4 y^{6}}-\frac {\left (\frac {x}{\sqrt {-\left (y^{2}-x \right ) \left (y^{2}+x \right )}}+1\right ) \left (x a_{2}+y a_{3}+a_{1}\right )}{2 y^{3}}-\left (\frac {-2 y \left (y^{2}+x \right )-2 \left (y^{2}-x \right ) y}{4 y^{3} \sqrt {-\left (y^{2}-x \right ) \left (y^{2}+x \right )}}-\frac {3 \left (\sqrt {-\left (y^{2}-x \right ) \left (y^{2}+x \right )}+x \right )}{2 y^{4}}\right ) \left (x b_{2}+y b_{3}+b_{1}\right ) = 0 \end{equation} Putting the above in normal form gives \[ -\frac {-4 b_{2} y^{6} \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}+2 x \,y^{6} b_{2}-2 y^{7} a_{2}+4 y^{7} b_{3}+2 y^{6} b_{1}-6 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, x^{2} y^{2} b_{2}+4 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, x \,y^{3} a_{2}-8 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, x \,y^{3} b_{3}+2 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, y^{4} a_{3}-6 x^{3} y^{2} b_{2}+4 x^{2} y^{3} a_{2}-8 x^{2} y^{3} b_{3}-6 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, x \,y^{2} b_{1}+2 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, y^{3} a_{1}-6 x^{2} y^{2} b_{1}+2 x \,y^{3} a_{1}+{\left (\left (-y^{2}+x \right ) \left (y^{2}+x \right )\right )}^{\frac {3}{2}} a_{3}+\sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, x^{2} a_{3}+2 x^{3} a_{3}}{4 y^{6} \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}} = 0 \] Setting the numerator to zero gives \begin{equation} \tag{6E} 4 b_{2} y^{6} \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}-2 x \,y^{6} b_{2}+2 y^{7} a_{2}-4 y^{7} b_{3}-2 y^{6} b_{1}+6 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, x^{2} y^{2} b_{2}-4 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, x \,y^{3} a_{2}+8 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, x \,y^{3} b_{3}-2 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, y^{4} a_{3}+6 x^{3} y^{2} b_{2}-4 x^{2} y^{3} a_{2}+8 x^{2} y^{3} b_{3}+6 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, x \,y^{2} b_{1}-2 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, y^{3} a_{1}+6 x^{2} y^{2} b_{1}-2 x \,y^{3} a_{1}-{\left (\left (-y^{2}+x \right ) \left (y^{2}+x \right )\right )}^{\frac {3}{2}} a_{3}-\sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, x^{2} a_{3}-2 x^{3} a_{3} = 0 \end{equation} Simplifying the above gives \begin{equation} \tag{6E} 4 b_{2} y^{6} \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}+4 x \,y^{6} b_{2}+4 y^{7} b_{3}+4 y^{6} b_{1}-6 \left (y^{2}-x \right ) \left (y^{2}+x \right ) x \,y^{2} b_{2}+2 \left (y^{2}-x \right ) \left (y^{2}+x \right ) y^{3} a_{2}-8 \left (y^{2}-x \right ) \left (y^{2}+x \right ) y^{3} b_{3}+6 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, x^{2} y^{2} b_{2}-4 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, x \,y^{3} a_{2}+8 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, x \,y^{3} b_{3}-2 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, y^{4} a_{3}-2 x^{2} y^{3} a_{2}-2 x \,y^{4} a_{3}-6 \left (y^{2}-x \right ) \left (y^{2}+x \right ) y^{2} b_{1}+6 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, x \,y^{2} b_{1}-2 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, y^{3} a_{1}-2 x \,y^{3} a_{1}-{\left (\left (-y^{2}+x \right ) \left (y^{2}+x \right )\right )}^{\frac {3}{2}} a_{3}+2 \left (y^{2}-x \right ) \left (y^{2}+x \right ) x a_{3}-\sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, x^{2} a_{3} = 0 \end{equation} Since the PDE has radicals, simplifying gives \[ -2 x \,y^{6} b_{2}+4 b_{2} y^{6} \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}+2 y^{7} a_{2}-4 y^{7} b_{3}-2 y^{6} b_{1}+6 x^{3} y^{2} b_{2}+6 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, x^{2} y^{2} b_{2}-4 x^{2} y^{3} a_{2}+8 x^{2} y^{3} b_{3}-4 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, x \,y^{3} a_{2}+8 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, x \,y^{3} b_{3}-\sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, y^{4} a_{3}+6 x^{2} y^{2} b_{1}+6 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, x \,y^{2} b_{1}-2 x \,y^{3} a_{1}-2 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, y^{3} a_{1}-2 x^{3} a_{3}-2 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, x^{2} a_{3} = 0 \] Looking at the above PDE shows the following are all the terms with \(\{x, y\}\) in them. \[ \left \{x, y, \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\right \} \] The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them \[ \left \{x = v_{1}, y = v_{2}, \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )} = v_{3}\right \} \] The above PDE (6E) now becomes \begin{equation} \tag{7E} 2 v_{2}^{7} a_{2}-2 v_{1} v_{2}^{6} b_{2}+4 b_{2} v_{2}^{6} v_{3}-4 v_{2}^{7} b_{3}-2 v_{2}^{6} b_{1}-4 v_{1}^{2} v_{2}^{3} a_{2}-4 v_{3} v_{1} v_{2}^{3} a_{2}-v_{3} v_{2}^{4} a_{3}+6 v_{1}^{3} v_{2}^{2} b_{2}+6 v_{3} v_{1}^{2} v_{2}^{2} b_{2}+8 v_{1}^{2} v_{2}^{3} b_{3}+8 v_{3} v_{1} v_{2}^{3} b_{3}-2 v_{1} v_{2}^{3} a_{1}-2 v_{3} v_{2}^{3} a_{1}+6 v_{1}^{2} v_{2}^{2} b_{1}+6 v_{3} v_{1} v_{2}^{2} b_{1}-2 v_{1}^{3} a_{3}-2 v_{3} v_{1}^{2} a_{3} = 0 \end{equation} Collecting the above on the terms \(v_i\) introduced, and these are \[ \{v_{1}, v_{2}, v_{3}\} \] Equation (7E) now becomes \begin{equation} \tag{8E} 6 v_{1}^{3} v_{2}^{2} b_{2}-2 v_{1}^{3} a_{3}+\left (-4 a_{2}+8 b_{3}\right ) v_{1}^{2} v_{2}^{3}+6 v_{3} v_{1}^{2} v_{2}^{2} b_{2}+6 v_{1}^{2} v_{2}^{2} b_{1}-2 v_{3} v_{1}^{2} a_{3}-2 v_{1} v_{2}^{6} b_{2}+\left (-4 a_{2}+8 b_{3}\right ) v_{1} v_{2}^{3} v_{3}-2 v_{1} v_{2}^{3} a_{1}+6 v_{3} v_{1} v_{2}^{2} b_{1}+\left (2 a_{2}-4 b_{3}\right ) v_{2}^{7}+4 b_{2} v_{2}^{6} v_{3}-2 v_{2}^{6} b_{1}-v_{3} v_{2}^{4} a_{3}-2 v_{3} v_{2}^{3} a_{1} = 0 \end{equation} Setting each coeﬃcients in (8E) to zero gives the following equations to solve \begin {align*} -2 a_{1}&=0\\ -2 a_{3}&=0\\ -a_{3}&=0\\ -2 b_{1}&=0\\ 6 b_{1}&=0\\ -2 b_{2}&=0\\ 4 b_{2}&=0\\ 6 b_{2}&=0\\ -4 a_{2}+8 b_{3}&=0\\ 2 a_{2}-4 b_{3}&=0 \end {align*}

Solving the above equations for the unknowns gives \begin {align*} a_{1}&=0\\ a_{2}&=2 b_{3}\\ a_{3}&=0\\ b_{1}&=0\\ b_{2}&=0\\ b_{3}&=b_{3} \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 x \\ \eta &= y \\ \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 - \left (\frac {\sqrt {-\left (y^{2}-x \right ) \left (y^{2}+x \right )}+x}{2 y^{3}}\right ) \left (2 x\right ) \\ &= \frac {y^{4}-\sqrt {-y^{4}+x^{2}}\, x -x^{2}}{y^{3}}\\ \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}{\frac {y^{4}-\sqrt {-y^{4}+x^{2}}\, x -x^{2}}{y^{3}}}} dy \end {align*}

Which results in \begin {align*} S&= \ln \left (y \right )-\frac {\ln \left (y^{2}-x \right )}{4}-\frac {\ln \left (y^{2}+x \right )}{4}+\frac {\ln \left (y^{4}-x^{2}\right )}{4}+\frac {x \ln \left (\frac {2 x^{2}+2 \sqrt {x^{2}}\, \sqrt {-y^{4}+x^{2}}}{y^{2}}\right )}{2 \sqrt {x^{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 {\sqrt {-\left (y^{2}-x \right ) \left (y^{2}+x \right )}+x}{2 y^{3}} \end {align*}

Evaluating all the partial derivatives gives \begin {align*} R_{x} &= 1\\ R_{y} &= 0\\ S_{x} &= \frac {x +\sqrt {-y^{4}+x^{2}}}{2 \sqrt {-y^{4}+x^{2}}\, x}\\ S_{y} &= -\frac {y^{3}}{\sqrt {-y^{4}+x^{2}}\, \left (x +\sqrt {-y^{4}+x^{2}}\right )} \end {align*}

Substituting all the above in (2) and simplifying gives the ode in canonical coordinates. \begin {align*} \frac {dS}{dR} &= \frac {1}{2 x}\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 R} \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 \(R,S\). Integrating the above gives \begin {align*} S \left (R \right ) = \frac {\ln \left (R \right )}{2}+c_{1}\tag {4} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} \frac {\ln \left (2\right )}{2}+\frac {\ln \left (x \right )}{2}+\frac {\ln \left (x +\sqrt {x^{2}-y^{4}}\right )}{2} &= \frac {\ln \left (x \right )}{2}+c_{1} \\ \end{align*}

Veriﬁcation of solutions

\[ \frac {\ln \left (2\right )}{2}+\frac {\ln \left (x \right )}{2}+\frac {\ln \left (x +\sqrt {x^{2}-y^{4}}\right )}{2} = \frac {\ln \left (x \right )}{2}+c_{1} \] Veriﬁed OK.

Writing the ode as \begin {align*} y^{\prime }&=-\frac {\sqrt {-\left (y^{2}-x \right ) \left (y^{2}+x \right )}-x}{2 y^{3}}\\ y^{\prime }&= \omega \left ( x,y\right ) \end {align*}

The type of this ode is not in the lookup table. To determine \(\xi ,\eta \) then (A) is solved using ansatz. Making bivariate polynomials of degree 1 to use as anstaz gives \begin{align*} \tag{1E} \xi &= x a_{2}+y a_{3}+a_{1} \\ \tag{2E} \eta &= x b_{2}+y b_{3}+b_{1} \\ \end{align*} Where the unknown coeﬃcients are \[ \{a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}\} \] Substituting equations (1E,2E) and \(\omega \) into (A) gives \begin{equation} \tag{5E} b_{2}-\frac {\left (\sqrt {-\left (y^{2}-x \right ) \left (y^{2}+x \right )}-x \right ) \left (b_{3}-a_{2}\right )}{2 y^{3}}-\frac {{\left (\sqrt {-\left (y^{2}-x \right ) \left (y^{2}+x \right )}-x \right )}^{2} a_{3}}{4 y^{6}}+\frac {\left (\frac {x}{\sqrt {-\left (y^{2}-x \right ) \left (y^{2}+x \right )}}-1\right ) \left (x a_{2}+y a_{3}+a_{1}\right )}{2 y^{3}}-\left (-\frac {-2 y \left (y^{2}+x \right )-2 \left (y^{2}-x \right ) y}{4 y^{3} \sqrt {-\left (y^{2}-x \right ) \left (y^{2}+x \right )}}+\frac {\frac {3 \sqrt {-\left (y^{2}-x \right ) \left (y^{2}+x \right )}}{2}-\frac {3 x}{2}}{y^{4}}\right ) \left (x b_{2}+y b_{3}+b_{1}\right ) = 0 \end{equation} Putting the above in normal form gives \[ -\frac {-4 b_{2} y^{6} \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}-2 x \,y^{6} b_{2}+2 y^{7} a_{2}-4 y^{7} b_{3}-2 y^{6} b_{1}-6 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, x^{2} y^{2} b_{2}+4 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, x \,y^{3} a_{2}-8 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, x \,y^{3} b_{3}+2 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, y^{4} a_{3}+6 x^{3} y^{2} b_{2}-4 x^{2} y^{3} a_{2}+8 x^{2} y^{3} b_{3}-6 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, x \,y^{2} b_{1}+2 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, y^{3} a_{1}+6 x^{2} y^{2} b_{1}-2 x \,y^{3} a_{1}+{\left (\left (-y^{2}+x \right ) \left (y^{2}+x \right )\right )}^{\frac {3}{2}} a_{3}+\sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, x^{2} a_{3}-2 x^{3} a_{3}}{4 y^{6} \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}} = 0 \] Setting the numerator to zero gives \begin{equation} \tag{6E} 4 b_{2} y^{6} \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}+2 x \,y^{6} b_{2}-2 y^{7} a_{2}+4 y^{7} b_{3}+2 y^{6} b_{1}+6 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, x^{2} y^{2} b_{2}-4 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, x \,y^{3} a_{2}+8 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, x \,y^{3} b_{3}-2 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, y^{4} a_{3}-6 x^{3} y^{2} b_{2}+4 x^{2} y^{3} a_{2}-8 x^{2} y^{3} b_{3}+6 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, x \,y^{2} b_{1}-2 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, y^{3} a_{1}-6 x^{2} y^{2} b_{1}+2 x \,y^{3} a_{1}-{\left (\left (-y^{2}+x \right ) \left (y^{2}+x \right )\right )}^{\frac {3}{2}} a_{3}-\sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, x^{2} a_{3}+2 x^{3} a_{3} = 0 \end{equation} Simplifying the above gives \begin{equation} \tag{6E} 4 b_{2} y^{6} \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}-4 x \,y^{6} b_{2}-4 y^{7} b_{3}-4 y^{6} b_{1}+6 \left (y^{2}-x \right ) \left (y^{2}+x \right ) x \,y^{2} b_{2}-2 \left (y^{2}-x \right ) \left (y^{2}+x \right ) y^{3} a_{2}+8 \left (y^{2}-x \right ) \left (y^{2}+x \right ) y^{3} b_{3}+6 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, x^{2} y^{2} b_{2}-4 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, x \,y^{3} a_{2}+8 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, x \,y^{3} b_{3}-2 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, y^{4} a_{3}+2 x^{2} y^{3} a_{2}+2 x \,y^{4} a_{3}+6 \left (y^{2}-x \right ) \left (y^{2}+x \right ) y^{2} b_{1}+6 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, x \,y^{2} b_{1}-2 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, y^{3} a_{1}+2 x \,y^{3} a_{1}-{\left (\left (-y^{2}+x \right ) \left (y^{2}+x \right )\right )}^{\frac {3}{2}} a_{3}-2 \left (y^{2}-x \right ) \left (y^{2}+x \right ) x a_{3}-\sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, x^{2} a_{3} = 0 \end{equation} Since the PDE has radicals, simplifying gives \[ 2 x \,y^{6} b_{2}+4 b_{2} y^{6} \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}-2 y^{7} a_{2}+4 y^{7} b_{3}+2 y^{6} b_{1}-6 x^{3} y^{2} b_{2}+6 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, x^{2} y^{2} b_{2}+4 x^{2} y^{3} a_{2}-8 x^{2} y^{3} b_{3}-4 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, x \,y^{3} a_{2}+8 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, x \,y^{3} b_{3}-\sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, y^{4} a_{3}-6 x^{2} y^{2} b_{1}+6 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, x \,y^{2} b_{1}+2 x \,y^{3} a_{1}-2 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, y^{3} a_{1}+2 x^{3} a_{3}-2 \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\, x^{2} a_{3} = 0 \] Looking at the above PDE shows the following are all the terms with \(\{x, y\}\) in them. \[ \left \{x, y, \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )}\right \} \] The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them \[ \left \{x = v_{1}, y = v_{2}, \sqrt {\left (-y^{2}+x \right ) \left (y^{2}+x \right )} = v_{3}\right \} \] The above PDE (6E) now becomes \begin{equation} \tag{7E} -2 v_{2}^{7} a_{2}+2 v_{1} v_{2}^{6} b_{2}+4 b_{2} v_{2}^{6} v_{3}+4 v_{2}^{7} b_{3}+2 v_{2}^{6} b_{1}+4 v_{1}^{2} v_{2}^{3} a_{2}-4 v_{3} v_{1} v_{2}^{3} a_{2}-v_{3} v_{2}^{4} a_{3}-6 v_{1}^{3} v_{2}^{2} b_{2}+6 v_{3} v_{1}^{2} v_{2}^{2} b_{2}-8 v_{1}^{2} v_{2}^{3} b_{3}+8 v_{3} v_{1} v_{2}^{3} b_{3}+2 v_{1} v_{2}^{3} a_{1}-2 v_{3} v_{2}^{3} a_{1}-6 v_{1}^{2} v_{2}^{2} b_{1}+6 v_{3} v_{1} v_{2}^{2} b_{1}+2 v_{1}^{3} a_{3}-2 v_{3} v_{1}^{2} a_{3} = 0 \end{equation} Collecting the above on the terms \(v_i\) introduced, and these are \[ \{v_{1}, v_{2}, v_{3}\} \] Equation (7E) now becomes \begin{equation} \tag{8E} -6 v_{1}^{3} v_{2}^{2} b_{2}+2 v_{1}^{3} a_{3}+\left (4 a_{2}-8 b_{3}\right ) v_{1}^{2} v_{2}^{3}+6 v_{3} v_{1}^{2} v_{2}^{2} b_{2}-6 v_{1}^{2} v_{2}^{2} b_{1}-2 v_{3} v_{1}^{2} a_{3}+2 v_{1} v_{2}^{6} b_{2}+\left (-4 a_{2}+8 b_{3}\right ) v_{1} v_{2}^{3} v_{3}+2 v_{1} v_{2}^{3} a_{1}+6 v_{3} v_{1} v_{2}^{2} b_{1}+\left (-2 a_{2}+4 b_{3}\right ) v_{2}^{7}+4 b_{2} v_{2}^{6} v_{3}+2 v_{2}^{6} b_{1}-v_{3} v_{2}^{4} a_{3}-2 v_{3} v_{2}^{3} a_{1} = 0 \end{equation} Setting each coeﬃcients in (8E) to zero gives the following equations to solve \begin {align*} -2 a_{1}&=0\\ 2 a_{1}&=0\\ -2 a_{3}&=0\\ -a_{3}&=0\\ 2 a_{3}&=0\\ -6 b_{1}&=0\\ 2 b_{1}&=0\\ 6 b_{1}&=0\\ -6 b_{2}&=0\\ 2 b_{2}&=0\\ 4 b_{2}&=0\\ 6 b_{2}&=0\\ -4 a_{2}+8 b_{3}&=0\\ -2 a_{2}+4 b_{3}&=0\\ 4 a_{2}-8 b_{3}&=0 \end {align*}

Solving the above equations for the unknowns gives \begin {align*} a_{1}&=0\\ a_{2}&=2 b_{3}\\ a_{3}&=0\\ b_{1}&=0\\ b_{2}&=0\\ b_{3}&=b_{3} \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 x \\ \eta &= y \\ \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 - \left (-\frac {\sqrt {-\left (y^{2}-x \right ) \left (y^{2}+x \right )}-x}{2 y^{3}}\right ) \left (2 x\right ) \\ &= \frac {y^{4}+\sqrt {-y^{4}+x^{2}}\, x -x^{2}}{y^{3}}\\ \xi &= 0 \end {align*}

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*}

\(S\) is found from \begin {align*} S &= \int { \frac {1}{\eta }} dy\\ &= \int { \frac {1}{\frac {y^{4}+\sqrt {-y^{4}+x^{2}}\, x -x^{2}}{y^{3}}}} dy \end {align*}

Which results in \begin {align*} S&= \frac {\ln \left (y^{4}-x^{2}\right )}{4}+\ln \left (y \right )-\frac {\ln \left (y^{2}-x \right )}{4}-\frac {\ln \left (y^{2}+x \right )}{4}-\frac {x \ln \left (\frac {2 x^{2}+2 \sqrt {x^{2}}\, \sqrt {-y^{4}+x^{2}}}{y^{2}}\right )}{2 \sqrt {x^{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 {\sqrt {-\left (y^{2}-x \right ) \left (y^{2}+x \right )}-x}{2 y^{3}} \end {align*}

Evaluating all the partial derivatives gives \begin {align*} R_{x} &= 1\\ R_{y} &= 0\\ S_{x} &= -\frac {x +\sqrt {-y^{4}+x^{2}}}{2 \sqrt {-y^{4}+x^{2}}\, x}\\ S_{y} &= \frac {-y^{4}+2 x^{2}+2 \sqrt {-y^{4}+x^{2}}\, x}{y \sqrt {-y^{4}+x^{2}}\, \left (x +\sqrt {-y^{4}+x^{2}}\right )} \end {align*}

Substituting all the above in (2) and simplifying gives the ode in canonical coordinates. \begin {align*} \frac {dS}{dR} &= -\frac {-y^{4}+\sqrt {-y^{4}+x^{2}}\, x +x^{2}}{2 x \sqrt {-y^{4}+x^{2}}\, \left (x +\sqrt {-y^{4}+x^{2}}\right )}\tag {2A} \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 \(R,S\). Integrating the above gives \begin {align*} S \left (R \right ) = -\frac {\ln \left (R \right )}{2}+c_{1}\tag {4} \end {align*}

Which gives \begin {align*} y = {\mathrm e}^{\frac {\ln \left (2\right )}{4}+\frac {\ln \left (-2 \,{\mathrm e}^{4 c_{1}} {\mathrm e}^{-2 c_{1}}+2 \,{\mathrm e}^{2 c_{1}} {\mathrm e}^{-2 c_{1}} x \right )}{4}+\frac {c_{1}}{2}} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{\frac {\ln \left (2\right )}{4}+\frac {\ln \left (-2 \,{\mathrm e}^{4 c_{1}} {\mathrm e}^{-2 c_{1}}+2 \,{\mathrm e}^{2 c_{1}} {\mathrm e}^{-2 c_{1}} x \right )}{4}+\frac {c_{1}}{2}} \\ \end{align*}

Veriﬁcation of solutions

\[ y = {\mathrm e}^{\frac {\ln \left (2\right )}{4}+\frac {\ln \left (-2 \,{\mathrm e}^{4 c_{1}} {\mathrm e}^{-2 c_{1}}+2 \,{\mathrm e}^{2 c_{1}} {\mathrm e}^{-2 c_{1}} x \right )}{4}+\frac {c_{1}}{2}} \] Veriﬁed OK.

34.9.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 4 y^{3} {y^{\prime }}^{2}-4 x y^{\prime }+y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=-\frac {-x +\sqrt {x^{2}-y^{4}}}{2 y^{3}}, y^{\prime }=\frac {x +\sqrt {x^{2}-y^{4}}}{2 y^{3}}\right ] \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {-x +\sqrt {x^{2}-y^{4}}}{2 y^{3}} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {x +\sqrt {x^{2}-y^{4}}}{2 y^{3}} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\mathit {workingODE} , \mathit {workingODE}\right \} \end {array} \]

Maple trace

`Methods for first order ODEs: 
   *** Sublevel 2 *** 
   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 3 *** 
      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`[2*x, y]

\begin{align*} y \left (x \right ) &= \sqrt {-x} \\ y \left (x \right ) &= -\sqrt {-x} \\ y \left (x \right ) &= \sqrt {x} \\ y \left (x \right ) &= -\sqrt {x} \\ y \left (x \right ) &= 0 \\ y \left (x \right ) &= \operatorname {RootOf}\left (-\ln \left (x \right )-2 \left (\int _{}^{\textit {\_Z}}\frac {\textit {\_a}^{4}+\sqrt {-\textit {\_a}^{4}+1}-1}{\textit {\_a} \left (\textit {\_a}^{4}-1\right )}d \textit {\_a} \right )+c_{1} \right ) \sqrt {x} \\ \end{align*}

\begin{align*} y(x)\to -e^{\frac {c_1}{4}} \sqrt [4]{e^{c_1}-2 i x} \\ y(x)\to -i e^{\frac {c_1}{4}} \sqrt [4]{e^{c_1}-2 i x} \\ y(x)\to i e^{\frac {c_1}{4}} \sqrt [4]{e^{c_1}-2 i x} \\ y(x)\to e^{\frac {c_1}{4}} \sqrt [4]{e^{c_1}-2 i x} \\ y(x)\to -e^{\frac {c_1}{4}} \sqrt [4]{2 i x+e^{c_1}} \\ y(x)\to -i e^{\frac {c_1}{4}} \sqrt [4]{2 i x+e^{c_1}} \\ y(x)\to i e^{\frac {c_1}{4}} \sqrt [4]{2 i x+e^{c_1}} \\ y(x)\to e^{\frac {c_1}{4}} \sqrt [4]{2 i x+e^{c_1}} \\ y(x)\to 0 \\ y(x)\to -\sqrt {x} \\ y(x)\to -i \sqrt {x} \\ y(x)\to i \sqrt {x} \\ y(x)\to \sqrt {x} \\ \end{align*}

34.9 problem 1006

34.9.1 Maple step by step solution