problem 965

33.3 problem 965

Internal problem ID [4198]
Internal ﬁle name [OUTPUT/3691_Sunday_June_05_2022_10_13_22_AM_25934494/index.tex]

Book: Ordinary diﬀerential equations and their solutions. By George Moseley Murphy. 1960
Section: Various 33
Problem number: 965.
ODE order: 1.
ODE degree: 2.

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

Maple gives the following as the ode type

[_rational]

Unable to solve or complete the solution.

\[ \boxed {x y {y^{\prime }}^{2}+\left (a +x^{2}-y^{2}\right ) y^{\prime }-y x=0} \] 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 {-a -x^{2}+y^{2}+\sqrt {y^{4}+2 y^{2} x^{2}+x^{4}-2 y^{2} a +2 a \,x^{2}+a^{2}}}{2 y x} \tag {1} \\ y^{\prime }&=-\frac {-y^{2}+x^{2}+\sqrt {y^{4}+2 y^{2} x^{2}+x^{4}-2 y^{2} a +2 a \,x^{2}+a^{2}}+a}{2 y x} \tag {2} \end {align*}

Now each one of the above ODE is solved.

Solving equation (1)

Writing the ode as \begin {align*} y^{\prime }&=\frac {-a -x^{2}+y^{2}+\sqrt {x^{4}+2 x^{2} y^{2}+y^{4}+2 a \,x^{2}-2 y^{2} a +a^{2}}}{2 y x}\\ 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 3 to use as anstaz gives \begin{align*} \tag{1E} \xi &= x^{3} a_{7}+x^{2} y a_{8}+x \,y^{2} a_{9}+y^{3} a_{10}+x^{2} a_{4}+x y a_{5}+y^{2} a_{6}+x a_{2}+y a_{3}+a_{1} \\ \tag{2E} \eta &= x^{3} b_{7}+x^{2} y b_{8}+x \,y^{2} b_{9}+y^{3} b_{10}+x^{2} b_{4}+x y b_{5}+y^{2} b_{6}+x b_{2}+y b_{3}+b_{1} \\ \end{align*} Where the unknown coeﬃcients are \[ \{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}, a_{8}, a_{9}, a_{10}, b_{1}, b_{2}, b_{3}, b_{4}, b_{5}, b_{6}, b_{7}, b_{8}, b_{9}, b_{10}\} \] Substituting equations (1E,2E) and \(\omega \) into (A) gives \begin{equation} \tag{5E} 3 x^{2} b_{7}+2 x y b_{8}+y^{2} b_{9}+2 x b_{4}+y b_{5}+b_{2}+\frac {\left (-a -x^{2}+y^{2}+\sqrt {x^{4}+2 x^{2} y^{2}+y^{4}+2 a \,x^{2}-2 y^{2} a +a^{2}}\right ) \left (-3 x^{2} a_{7}+x^{2} b_{8}-2 x y a_{8}+2 x y b_{9}-y^{2} a_{9}+3 y^{2} b_{10}-2 x a_{4}+x b_{5}-y a_{5}+2 y b_{6}-a_{2}+b_{3}\right )}{2 y x}-\frac {\left (-a -x^{2}+y^{2}+\sqrt {x^{4}+2 x^{2} y^{2}+y^{4}+2 a \,x^{2}-2 y^{2} a +a^{2}}\right )^{2} \left (x^{2} a_{8}+2 x y a_{9}+3 y^{2} a_{10}+x a_{5}+2 y a_{6}+a_{3}\right )}{4 y^{2} x^{2}}-\left (-\frac {-a -x^{2}+y^{2}+\sqrt {x^{4}+2 x^{2} y^{2}+y^{4}+2 a \,x^{2}-2 y^{2} a +a^{2}}}{2 y \,x^{2}}+\frac {-2 x +\frac {4 x^{3}+4 x \,y^{2}+4 a x}{2 \sqrt {x^{4}+2 x^{2} y^{2}+y^{4}+2 a \,x^{2}-2 y^{2} a +a^{2}}}}{2 y x}\right ) \left (x^{3} a_{7}+x^{2} y a_{8}+x \,y^{2} a_{9}+y^{3} a_{10}+x^{2} a_{4}+x y a_{5}+y^{2} a_{6}+x a_{2}+y a_{3}+a_{1}\right )-\left (-\frac {-a -x^{2}+y^{2}+\sqrt {x^{4}+2 x^{2} y^{2}+y^{4}+2 a \,x^{2}-2 y^{2} a +a^{2}}}{2 y^{2} x}+\frac {2 y +\frac {4 x^{2} y +4 y^{3}-4 a y}{2 \sqrt {x^{4}+2 x^{2} y^{2}+y^{4}+2 a \,x^{2}-2 y^{2} a +a^{2}}}}{2 y x}\right ) \left (x^{3} b_{7}+x^{2} y b_{8}+x \,y^{2} b_{9}+y^{3} b_{10}+x^{2} b_{4}+x y 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 \[ \text {Expression too large to display} \] Setting the numerator to zero gives \begin{equation} \tag{6E} \text {Expression too large to display} \end{equation} Simplifying the above gives \begin{equation} \tag{6E} \text {Expression too large to display} \end{equation} Since the PDE has radicals, simplifying gives \[ \text {Expression too large to display} \] Looking at the above PDE shows the following are all the terms with \(\{x, y\}\) in them. \[ \left \{x, y, \sqrt {x^{4}+2 x^{2} y^{2}+y^{4}+2 a \,x^{2}-2 y^{2} a +a^{2}}\right \} \] The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them \[ \left \{x = v_{1}, y = v_{2}, \sqrt {x^{4}+2 x^{2} y^{2}+y^{4}+2 a \,x^{2}-2 y^{2} a +a^{2}} = v_{3}\right \} \] The above PDE (6E) now becomes \begin{equation} \tag{7E} \text {Expression too large to display} \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} \text {Expression too large to display} \end{equation} Setting each coeﬃcients in (8E) to zero gives the following equations to solve \begin {align*} -2 a_{6}&=0\\ -4 a_{10}&=0\\ 10 a a_{10}&=0\\ 14 a a_{10}&=0\\ -2 a^{2} a_{3}&=0\\ 2 a^{3} a_{3}&=0\\ -2 a_{4}-4 a_{6}&=0\\ -2 a_{5}-2 b_{4}&=0\\ -2 a_{5}+2 b_{6}&=0\\ -2 a_{5}+6 b_{6}&=0\\ 2 a_{5}+2 b_{4}&=0\\ -2 a_{8}-2 b_{7}&=0\\ 2 a_{8}+2 b_{7}&=0\\ -4 a_{9}+4 b_{10}&=0\\ -8 a_{4}+2 a_{6}+4 b_{5}&=0\\ -6 a_{4}+4 a_{6}+4 b_{5}&=0\\ -2 a_{4}+2 a_{6}+4 b_{5}&=0\\ 6 a_{4}-4 a_{6}-4 b_{5}&=0\\ -6 a_{5}-2 b_{4}+8 b_{6}&=0\\ 4 a_{5}+6 b_{4}-6 b_{6}&=0\\ -4 a_{7}+12 b_{10}-8 a_{9}&=0\\ 8 a_{7}-4 b_{8}-4 a_{9}&=0\\ -4 a_{8}+2 a_{10}+6 b_{9}&=0\\ -4 a_{8}+2 b_{9}-6 a_{10}&=0\\ 4 b_{8}+4 a_{9}-8 a_{7}&=0\\ 6 b_{9}+6 a_{10}-4 a_{8}&=0\\ 8 b_{10}+4 b_{8}-12 a_{7}&=0\\ -4 a_{7}+4 a_{9}-8 b_{10}+8 b_{8}&=0\\ 6 a_{8}-6 b_{9}-6 a_{10}+10 b_{7}&=0\\ -2 b_{7}+8 b_{9}-10 a_{8}+4 a_{10}&=0\\ 6 a a_{6}+2 a_{1}&=0\\ 8 a a_{6}+2 a_{1}&=0\\ -10 a^{2} a_{6}-4 a a_{1}&=0\\ -4 a^{2} a_{6}-2 a a_{1}&=0\\ 4 a^{3} a_{6}+2 a^{2} a_{1}&=0\\ -16 a^{2} a_{10}+2 a a_{3}&=0\\ -6 a^{2} a_{10}+2 a a_{3}&=0\\ 6 a^{3} a_{10}-4 a^{2} a_{3}&=0\\ -2 a^{2} a_{5}-2 a b_{1}&=0\\ 2 a^{3} a_{5}+2 a^{2} b_{1}&=0\\ -8 a a_{5}+12 a b_{6}&=0\\ -6 a^{2} a_{5}+6 a^{2} b_{6}&=0\\ -4 a^{2} a_{9}-4 a b_{3}&=0\\ 4 a^{3} a_{9}+4 a^{2} b_{3}&=0\\ 8 a a_{9}-8 a b_{10}&=0\\ 12 a a_{9}-12 a b_{10}&=0\\ -2 a^{2} a_{8}-4 a a_{3}-2 a b_{2}&=0\\ 2 a^{3} a_{8}+6 a^{2} a_{3}+2 a^{2} b_{2}&=0\\ 4 a a_{4}-8 a a_{6}-4 a b_{5}&=0\\ -2 a^{2} a_{4}+12 a^{2} a_{6}+4 a^{2} b_{5}&=0\\ -4 a a_{5}-2 a b_{4}-2 b_{1}&=0\\ 6 a a_{5}+4 a b_{4}+2 b_{1}&=0\\ 6 a^{2} a_{5}+2 a^{2} b_{4}+4 a b_{1}&=0\\ 4 a a_{5}-6 a b_{6}-2 b_{1}&=0\\ 6 a a_{5}-8 a b_{6}-2 b_{1}&=0\\ -12 a a_{8}+18 a a_{10}+12 a b_{9}&=0\\ -12 a^{2} a_{9}+8 a^{2} b_{10}-4 a b_{3}&=0\\ -8 a a_{4}+12 a a_{6}+8 a b_{5}-2 a_{1}&=0\\ 2 a a_{4}-8 a a_{6}-4 a b_{5}+2 a_{1}&=0\\ -8 a^{2} a_{8}+18 a^{2} a_{10}+6 a^{2} b_{9}-4 a a_{3}&=0\\ -4 a a_{8}-2 a b_{7}-2 a_{3}-2 b_{2}&=0\\ 6 a a_{8}+4 a b_{7}+2 a_{3}+2 b_{2}&=0\\ 6 a^{2} a_{8}+2 a^{2} b_{7}+6 a a_{3}+4 a b_{2}&=0\\ -12 a a_{7}+12 a a_{9}+8 a b_{8}-4 a_{2}+4 b_{3}&=0\\ 4 a a_{7}-8 a a_{9}-4 a b_{8}+4 a_{2}-4 b_{3}&=0\\ -4 a^{2} a_{7}+12 a^{2} a_{9}+4 a^{2} b_{8}-4 a a_{2}+8 a b_{3}&=0\\ 6 a a_{8}-12 a a_{10}-6 a b_{9}+2 a_{3}+2 b_{2}&=0\\ 10 a a_{8}-12 a a_{10}-8 a b_{9}-2 a_{3}-2 b_{2}&=0\\ 8 a a_{7}-12 a a_{9}-4 a b_{8}+16 a b_{10}-4 a_{2}+4 b_{3}&=0 \end {align*}

Solving the above equations for the unknowns gives \begin {align*} a_{1}&=0\\ a_{2}&=a b_{10}\\ a_{3}&=0\\ a_{4}&=0\\ a_{5}&=0\\ a_{6}&=0\\ a_{7}&=b_{10}\\ a_{8}&=0\\ a_{9}&=b_{10}\\ a_{10}&=0\\ b_{1}&=0\\ b_{2}&=0\\ b_{3}&=-a b_{10}\\ b_{4}&=0\\ b_{5}&=0\\ b_{6}&=0\\ b_{7}&=0\\ b_{8}&=b_{10}\\ b_{9}&=0\\ b_{10}&=b_{10} \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 &= x^{3}+x \,y^{2}+a x \\ \eta &= x^{2} y +y^{3}-a y \\ \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 \(S(R)\). Therefore \begin {align*} \frac {dy}{dx} &= \frac {\eta }{\xi }\\ &= \frac {x^{2} y +y^{3}-a y}{x^{3}+x \,y^{2}+a x}\\ &= -\frac {y \left (-x^{2}-y^{2}+a \right )}{x \left (x^{2}+y^{2}+a \right )} \end {align*}

This is easily solved to give \begin {align*} \frac {1}{\frac {1}{y^{2}}+\frac {1}{x^{2}+a}} = -\frac {x \sqrt {x^{2}+a}}{\sqrt {c_{1} -\frac {4 a}{x^{2}+a}}}+\frac {x^{2}}{2}+\frac {a}{2} \end {align*}

Where now the coordinate \(R\) is taken as the constant of integration. Hence \begin {align*} R &= \frac {4 x^{4}+8 x^{2} y^{2}+4 y^{4}+8 a \,x^{2}-8 y^{2} a +4 a^{2}}{x^{4}-2 x^{2} y^{2}+y^{4}+2 a \,x^{2}-2 y^{2} a +a^{2}} \end {align*}

Since \(\xi \) depends on \(y\) and \(\eta \) depends on \(x\) then we can use either one to ﬁnd \(S\). Let us use \begin {align*} dS &= \frac {dx}{\xi } \\ &= \frac {dx}{x^{3}+x \,y^{2}+a x} \end {align*}

But we have now to replace \(y\) in \(\xi \) from its value from the solution of \(\frac {dy}{dx}=\frac {\eta }{\xi }\) found above. This results in \begin {align*} \xi &= x^{3}+x {\left (-\frac {x \sqrt {x^{2}+a}}{\sqrt {c_{1} -\frac {4 a}{x^{2}+a}}}+\frac {x^{2}}{2}+\frac {a}{2}\right )}^{2}+a x \end {align*}

Integrating gives \begin {align*} S &= \frac {dx}{x^{3}+x {\left (-\frac {x \sqrt {x^{2}+a}}{\sqrt {c_{1} -\frac {4 a}{x^{2}+a}}}+\frac {x^{2}}{2}+\frac {a}{2}\right )}^{2}+a x}\\ &= \text {Expression too large to display} \end {align*}

Where the constant of integration is set to zero as we just need one solution. Replacing back \(c_{1} = \frac {4 x^{4}+8 x^{2} y^{2}+4 y^{4}+8 a \,x^{2}-8 y^{2} a +4 a^{2}}{x^{4}-2 x^{2} y^{2}+y^{4}+2 a \,x^{2}-2 y^{2} a +a^{2}}\) then the above becomes \begin {align*} S &= \text {Expression too large to display} \end {align*}

Unable to determine ODE type.

Solving equation (2)

Writing the ode as \begin {align*} y^{\prime }&=-\frac {-y^{2}+x^{2}+\sqrt {x^{4}+2 x^{2} y^{2}+y^{4}+2 a \,x^{2}-2 y^{2} a +a^{2}}+a}{2 y x}\\ 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 3 to use as anstaz gives \begin{align*} \tag{1E} \xi &= x^{3} a_{7}+x^{2} y a_{8}+x \,y^{2} a_{9}+y^{3} a_{10}+x^{2} a_{4}+x y a_{5}+y^{2} a_{6}+x a_{2}+y a_{3}+a_{1} \\ \tag{2E} \eta &= x^{3} b_{7}+x^{2} y b_{8}+x \,y^{2} b_{9}+y^{3} b_{10}+x^{2} b_{4}+x y b_{5}+y^{2} b_{6}+x b_{2}+y b_{3}+b_{1} \\ \end{align*} Where the unknown coeﬃcients are \[ \{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}, a_{8}, a_{9}, a_{10}, b_{1}, b_{2}, b_{3}, b_{4}, b_{5}, b_{6}, b_{7}, b_{8}, b_{9}, b_{10}\} \] Substituting equations (1E,2E) and \(\omega \) into (A) gives \begin{equation} \tag{5E} 3 x^{2} b_{7}+2 x y b_{8}+y^{2} b_{9}+2 x b_{4}+y b_{5}+b_{2}-\frac {\left (-y^{2}+x^{2}+\sqrt {x^{4}+2 x^{2} y^{2}+y^{4}+2 a \,x^{2}-2 y^{2} a +a^{2}}+a \right ) \left (-3 x^{2} a_{7}+x^{2} b_{8}-2 x y a_{8}+2 x y b_{9}-y^{2} a_{9}+3 y^{2} b_{10}-2 x a_{4}+x b_{5}-y a_{5}+2 y b_{6}-a_{2}+b_{3}\right )}{2 y x}-\frac {\left (-y^{2}+x^{2}+\sqrt {x^{4}+2 x^{2} y^{2}+y^{4}+2 a \,x^{2}-2 y^{2} a +a^{2}}+a \right )^{2} \left (x^{2} a_{8}+2 x y a_{9}+3 y^{2} a_{10}+x a_{5}+2 y a_{6}+a_{3}\right )}{4 y^{2} x^{2}}-\left (-\frac {2 x +\frac {4 x^{3}+4 x \,y^{2}+4 a x}{2 \sqrt {x^{4}+2 x^{2} y^{2}+y^{4}+2 a \,x^{2}-2 y^{2} a +a^{2}}}}{2 y x}+\frac {-y^{2}+x^{2}+\sqrt {x^{4}+2 x^{2} y^{2}+y^{4}+2 a \,x^{2}-2 y^{2} a +a^{2}}+a}{2 x^{2} y}\right ) \left (x^{3} a_{7}+x^{2} y a_{8}+x \,y^{2} a_{9}+y^{3} a_{10}+x^{2} a_{4}+x y a_{5}+y^{2} a_{6}+x a_{2}+y a_{3}+a_{1}\right )-\left (-\frac {-2 y +\frac {4 x^{2} y +4 y^{3}-4 a y}{2 \sqrt {x^{4}+2 x^{2} y^{2}+y^{4}+2 a \,x^{2}-2 y^{2} a +a^{2}}}}{2 y x}+\frac {-y^{2}+x^{2}+\sqrt {x^{4}+2 x^{2} y^{2}+y^{4}+2 a \,x^{2}-2 y^{2} a +a^{2}}+a}{2 y^{2} x}\right ) \left (x^{3} b_{7}+x^{2} y b_{8}+x \,y^{2} b_{9}+y^{3} b_{10}+x^{2} b_{4}+x y 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 \[ \text {Expression too large to display} \] Setting the numerator to zero gives \begin{equation} \tag{6E} \text {Expression too large to display} \end{equation} Simplifying the above gives \begin{equation} \tag{6E} \text {Expression too large to display} \end{equation} Since the PDE has radicals, simplifying gives \[ \text {Expression too large to display} \] Looking at the above PDE shows the following are all the terms with \(\{x, y\}\) in them. \[ \left \{x, y, \sqrt {x^{4}+2 x^{2} y^{2}+y^{4}+2 a \,x^{2}-2 y^{2} a +a^{2}}\right \} \] The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them \[ \left \{x = v_{1}, y = v_{2}, \sqrt {x^{4}+2 x^{2} y^{2}+y^{4}+2 a \,x^{2}-2 y^{2} a +a^{2}} = v_{3}\right \} \] The above PDE (6E) now becomes \begin{equation} \tag{7E} \text {Expression too large to display} \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} \text {Expression too large to display} \end{equation} Setting each coeﬃcients in (8E) to zero gives the following equations to solve \begin {align*} -2 a_{6}&=0\\ 2 a_{6}&=0\\ -4 a_{10}&=0\\ 4 a_{10}&=0\\ -14 a a_{10}&=0\\ 10 a a_{10}&=0\\ -2 a^{2} a_{3}&=0\\ -2 a^{3} a_{3}&=0\\ 2 a_{4}+4 a_{6}&=0\\ -2 a_{5}-2 b_{4}&=0\\ -2 a_{5}+2 b_{6}&=0\\ 2 a_{5}-6 b_{6}&=0\\ 2 a_{5}-2 b_{6}&=0\\ -4 a_{9}+4 b_{10}&=0\\ -2 b_{7}-2 a_{8}&=0\\ -4 b_{10}+4 a_{9}&=0\\ -2 a_{4}+2 a_{6}+4 b_{5}&=0\\ 6 a_{4}-4 a_{6}-4 b_{5}&=0\\ 8 a_{4}-2 a_{6}-4 b_{5}&=0\\ 4 a_{5}+6 b_{4}-6 b_{6}&=0\\ 6 a_{5}+2 b_{4}-8 b_{6}&=0\\ 8 a_{7}-4 b_{8}-4 a_{9}&=0\\ 12 a_{7}-4 b_{8}-8 b_{10}&=0\\ 4 a_{8}-6 b_{9}-6 a_{10}&=0\\ 8 a_{9}-12 b_{10}+4 a_{7}&=0\\ 6 a_{10}-2 b_{9}+4 a_{8}&=0\\ 6 b_{9}-4 a_{8}+2 a_{10}&=0\\ 10 a_{8}-4 a_{10}+2 b_{7}-8 b_{9}&=0\\ 10 b_{7}+6 a_{8}-6 b_{9}-6 a_{10}&=0\\ 8 b_{8}-4 a_{7}+4 a_{9}-8 b_{10}&=0\\ -8 a a_{6}-2 a_{1}&=0\\ 6 a a_{6}+2 a_{1}&=0\\ -4 a^{2} a_{6}-2 a a_{1}&=0\\ 10 a^{2} a_{6}+4 a a_{1}&=0\\ -4 a^{3} a_{6}-2 a^{2} a_{1}&=0\\ -6 a^{2} a_{10}+2 a a_{3}&=0\\ 16 a^{2} a_{10}-2 a a_{3}&=0\\ -6 a^{3} a_{10}+4 a^{2} a_{3}&=0\\ -2 a^{2} a_{5}-2 a b_{1}&=0\\ -2 a^{3} a_{5}-2 a^{2} b_{1}&=0\\ 8 a a_{5}-12 a b_{6}&=0\\ 6 a^{2} a_{5}-6 a^{2} b_{6}&=0\\ -4 a^{2} a_{9}-4 a b_{3}&=0\\ -4 a^{3} a_{9}-4 a^{2} b_{3}&=0\\ -12 a a_{9}+12 a b_{10}&=0\\ 8 a a_{9}-8 a b_{10}&=0\\ -2 a^{2} a_{8}-4 a a_{3}-2 a b_{2}&=0\\ -2 a^{3} a_{8}-6 a^{2} a_{3}-2 a^{2} b_{2}&=0\\ -4 a a_{4}+8 a a_{6}+4 a b_{5}&=0\\ 2 a^{2} a_{4}-12 a^{2} a_{6}-4 a^{2} b_{5}&=0\\ -6 a a_{5}-4 a b_{4}-2 b_{1}&=0\\ -4 a a_{5}-2 a b_{4}-2 b_{1}&=0\\ -6 a^{2} a_{5}-2 a^{2} b_{4}-4 a b_{1}&=0\\ -6 a a_{5}+8 a b_{6}+2 b_{1}&=0\\ 4 a a_{5}-6 a b_{6}-2 b_{1}&=0\\ 12 a a_{8}-18 a a_{10}-12 a b_{9}&=0\\ 12 a^{2} a_{9}-8 a^{2} b_{10}+4 a b_{3}&=0\\ 2 a a_{4}-8 a a_{6}-4 a b_{5}+2 a_{1}&=0\\ 8 a a_{4}-12 a a_{6}-8 a b_{5}+2 a_{1}&=0\\ 8 a^{2} a_{8}-18 a^{2} a_{10}-6 a^{2} b_{9}+4 a a_{3}&=0\\ -6 a a_{8}-4 a b_{7}-2 a_{3}-2 b_{2}&=0\\ -4 a a_{8}-2 a b_{7}-2 a_{3}-2 b_{2}&=0\\ -6 a^{2} a_{8}-2 a^{2} b_{7}-6 a a_{3}-4 a b_{2}&=0\\ 4 a a_{7}-8 a a_{9}-4 a b_{8}+4 a_{2}-4 b_{3}&=0\\ 12 a a_{7}-12 a a_{9}-8 a b_{8}+4 a_{2}-4 b_{3}&=0\\ 4 a^{2} a_{7}-12 a^{2} a_{9}-4 a^{2} b_{8}+4 a a_{2}-8 a b_{3}&=0\\ -10 a a_{8}+12 a a_{10}+8 a b_{9}+2 a_{3}+2 b_{2}&=0\\ 6 a a_{8}-12 a a_{10}-6 a b_{9}+2 a_{3}+2 b_{2}&=0\\ -8 a a_{7}+12 a a_{9}+4 a b_{8}-16 a b_{10}+4 a_{2}-4 b_{3}&=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*}

Unable to determine ODE type.

33.3.1 Maple step by step solution

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

✗ Solution by Maple

dsolve(x*y(x)*diff(y(x),x)^2+(a+x^2-y(x)^2)*diff(y(x),x)-x*y(x) = 0,y(x), singsol=all)

\[ \text {No solution found} \]

✓ Solution by Mathematica

Time used: 0.419 (sec). Leaf size: 112

DSolve[x y[x] (y'[x])^2+(a+x^2-y[x]^2)y'[x]-x y[x]==0,y[x],x,IncludeSingularSolutions -> True]

\begin{align*} y(x)\to \sqrt {c_1 \left (x^2+\frac {a}{1+c_1}\right )} \\ y(x)\to -\sqrt {\left (\sqrt {a}-i x\right )^2} \\ y(x)\to \sqrt {\left (\sqrt {a}-i x\right )^2} \\ y(x)\to -\sqrt {\left (\sqrt {a}+i x\right )^2} \\ y(x)\to \sqrt {\left (\sqrt {a}+i x\right )^2} \\ \end{align*}

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