Internal problem ID [4229]
Internal file name [OUTPUT/3722_Sunday_June_05_2022_10_25_58_AM_91433928/index.tex]
Book: Ordinary differential equations and their solutions. By George Moseley Murphy.
1960
Section: Various 34
Problem number: 997.
ODE order: 1.
ODE degree: 2.
The type(s) of ODE detected by this program : "dAlembert"
Maple gives the following as the ode type
[[_homogeneous, `class A`], _rational, _dAlembert]
\[ \boxed {\left (-a^{2}+1\right ) y^{2} {y^{\prime }}^{2}-3 a^{2} x y y^{\prime }+y^{2}=a^{2} x^{2}} \]
Let \(p=y^{\prime }\) the ode becomes \begin {align*} \left (-a^{2}+1\right ) y^{2} p^{2}-3 a^{2} x y p +y^{2} = a^{2} x^{2} \end {align*}
Solving for \(y\) from the above results in \begin {align*} y &= \frac {\left (-3 a p +\sqrt {5 a^{2} p^{2}+4 p^{2}+4}\right ) x a}{2 a^{2} p^{2}-2 p^{2}-2}\tag {1A}\\ y &= -\frac {\left (3 a p +\sqrt {5 a^{2} p^{2}+4 p^{2}+4}\right ) x a}{2 \left (a^{2} p^{2}-p^{2}-1\right )}\tag {2A} \end {align*}
This has the form \begin {align*} y=xf(p)+g(p)\tag {*} \end {align*}
Where \(f,g\) are functions of \(p=y'(x)\). Each of the above ode’s is dAlembert ode which is now solved. Solving ode 1A Taking derivative of (*) w.r.t. \(x\) gives \begin {align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end {align*}
Comparing the form \(y=x f + g\) to (1A) shows that \begin {align*} f &= \frac {\left (-3 a p +\sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\right ) a}{2 a^{2} p^{2}-2 p^{2}-2}\\ g &= 0 \end {align*}
Hence (2) becomes \begin {align*} p -\frac {\left (-3 a p +\sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\right ) a}{2 a^{2} p^{2}-2 p^{2}-2} = x \left (\frac {\left (-3 a +\frac {\left (5 a^{2}+4\right ) p}{\sqrt {4+\left (5 a^{2}+4\right ) p^{2}}}\right ) a}{2 a^{2} p^{2}-2 p^{2}-2}-\frac {\left (-3 a p +\sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\right ) a \left (4 a^{2} p -4 p \right )}{\left (2 a^{2} p^{2}-2 p^{2}-2\right )^{2}}\right ) p^{\prime }\left (x \right )\tag {2A} \end {align*}
The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives \begin {align*} p -\frac {\left (-3 a p +\sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\right ) a}{2 a^{2} p^{2}-2 p^{2}-2} = 0 \end {align*}
Solving for \(p\) from the above gives \begin {align*} p&=\frac {\sqrt {2}\, \sqrt {\frac {-3 a^{2}+1+\sqrt {5 a^{4}-2 a^{2}+1}}{a^{2}-1}}}{2}\\ p&=-\frac {\sqrt {2}\, \sqrt {\frac {-3 a^{2}+1+\sqrt {5 a^{4}-2 a^{2}+1}}{a^{2}-1}}}{2}\\ p&=\frac {\sqrt {-\frac {2 \left (3 a^{2}+\sqrt {5 a^{4}-2 a^{2}+1}-1\right )}{a^{2}-1}}}{2}\\ p&=-\frac {\sqrt {-\frac {2 \left (3 a^{2}+\sqrt {5 a^{4}-2 a^{2}+1}-1\right )}{a^{2}-1}}}{2} \end {align*}
Substituting these in (1A) gives \begin {align*} y&=\frac {-3 a^{2} x \sqrt {-\frac {2 \left (3 a^{2}+\sqrt {5 a^{4}-2 a^{2}+1}-1\right )}{a^{2}-1}}-a x \sqrt {-\frac {2 \left (15 a^{4}+5 \sqrt {5 a^{4}-2 a^{2}+1}\, a^{2}-a^{2}+4 \sqrt {5 a^{4}-2 a^{2}+1}+4\right )}{a^{2}-1}}}{6 a^{2}+2 \sqrt {5 a^{4}-2 a^{2}+1}+2}\\ y&=\frac {3 a^{2} x \sqrt {-\frac {2 \left (3 a^{2}+\sqrt {5 a^{4}-2 a^{2}+1}-1\right )}{a^{2}-1}}-a x \sqrt {-\frac {2 \left (15 a^{4}+5 \sqrt {5 a^{4}-2 a^{2}+1}\, a^{2}-a^{2}+4 \sqrt {5 a^{4}-2 a^{2}+1}+4\right )}{a^{2}-1}}}{6 a^{2}+2 \sqrt {5 a^{4}-2 a^{2}+1}+2}\\ y&=\frac {3 \sqrt {2}\, a^{2} x \sqrt {\frac {-3 a^{2}+1+\sqrt {5 a^{4}-2 a^{2}+1}}{a^{2}-1}}+\sqrt {2}\, a x \sqrt {\frac {-15 a^{4}+5 \sqrt {5 a^{4}-2 a^{2}+1}\, a^{2}+a^{2}+4 \sqrt {5 a^{4}-2 a^{2}+1}-4}{a^{2}-1}}}{-6 a^{2}+2 \sqrt {5 a^{4}-2 a^{2}+1}-2}\\ y&=\frac {-3 \sqrt {2}\, a^{2} x \sqrt {\frac {-3 a^{2}+1+\sqrt {5 a^{4}-2 a^{2}+1}}{a^{2}-1}}+\sqrt {2}\, a x \sqrt {\frac {-15 a^{4}+5 \sqrt {5 a^{4}-2 a^{2}+1}\, a^{2}+a^{2}+4 \sqrt {5 a^{4}-2 a^{2}+1}-4}{a^{2}-1}}}{-6 a^{2}+2 \sqrt {5 a^{4}-2 a^{2}+1}-2} \end {align*}
The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in \begin {align*} p^{\prime }\left (x \right ) = \frac {p \left (x \right )-\frac {\left (-3 a p \left (x \right )+\sqrt {4+\left (5 a^{2}+4\right ) p \left (x \right )^{2}}\right ) a}{2 a^{2} p \left (x \right )^{2}-2 p \left (x \right )^{2}-2}}{x \left (\frac {\left (-3 a +\frac {\left (5 a^{2}+4\right ) p \left (x \right )}{\sqrt {4+\left (5 a^{2}+4\right ) p \left (x \right )^{2}}}\right ) a}{2 a^{2} p \left (x \right )^{2}-2 p \left (x \right )^{2}-2}-\frac {\left (-3 a p \left (x \right )+\sqrt {4+\left (5 a^{2}+4\right ) p \left (x \right )^{2}}\right ) a \left (4 a^{2} p \left (x \right )-4 p \left (x \right )\right )}{\left (2 a^{2} p \left (x \right )^{2}-2 p \left (x \right )^{2}-2\right )^{2}}\right )}\tag {3} \end {align*}
This ODE is now solved for \(p \left (x \right )\).
Inverting the above ode gives \begin {align*} \frac {d}{d p}x \left (p \right ) = \frac {x \left (p \right ) \left (\frac {\left (-3 a +\frac {\left (5 a^{2}+4\right ) p}{\sqrt {4+\left (5 a^{2}+4\right ) p^{2}}}\right ) a}{2 a^{2} p^{2}-2 p^{2}-2}-\frac {\left (-3 a p +\sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\right ) a \left (4 a^{2} p -4 p \right )}{\left (2 a^{2} p^{2}-2 p^{2}-2\right )^{2}}\right )}{p -\frac {\left (-3 a p +\sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\right ) a}{2 a^{2} p^{2}-2 p^{2}-2}}\tag {4} \end {align*}
This ODE is now solved for \(x \left (p \right )\).
Entering Linear first order ODE solver. In canonical form a linear first order is \begin {align*} \frac {d}{d p}x \left (p \right ) + p(p)x \left (p \right ) &= q(p) \end {align*}
Where here \begin {align*} p(p) &=\frac {a \left (-5 a^{4} p^{3}+3 p^{2} \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a^{3}+a^{2} p^{3}-3 p^{2} \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a -13 a^{2} p +4 p^{3}+3 \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a +4 p \right )}{\left (-2 a^{2} p^{3}-3 a^{2} p +2 p^{3}+\sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a +2 p \right ) \left (a^{2} p^{2}-p^{2}-1\right ) \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}}\\ q(p) &=0 \end {align*}
Hence the ode is \begin {align*} \frac {d}{d p}x \left (p \right )+\frac {x \left (p \right ) a \left (-5 a^{4} p^{3}+3 p^{2} \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a^{3}+a^{2} p^{3}-3 p^{2} \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a -13 a^{2} p +4 p^{3}+3 \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a +4 p \right )}{\left (-2 a^{2} p^{3}-3 a^{2} p +2 p^{3}+\sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a +2 p \right ) \left (a^{2} p^{2}-p^{2}-1\right ) \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}} = 0 \end {align*}
The integrating factor \(\mu \) is \[ \mu = {\mathrm e}^{\int \frac {a \left (-5 a^{4} p^{3}+3 p^{2} \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a^{3}+a^{2} p^{3}-3 p^{2} \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a -13 a^{2} p +4 p^{3}+3 \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a +4 p \right )}{\left (-2 a^{2} p^{3}-3 a^{2} p +2 p^{3}+\sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a +2 p \right ) \left (a^{2} p^{2}-p^{2}-1\right ) \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}}d p} \] The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}p}} \mu x &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}p}} \left ({\mathrm e}^{\int \frac {a \left (-5 a^{4} p^{3}+3 p^{2} \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a^{3}+a^{2} p^{3}-3 p^{2} \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a -13 a^{2} p +4 p^{3}+3 \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a +4 p \right )}{\left (-2 a^{2} p^{3}-3 a^{2} p +2 p^{3}+\sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a +2 p \right ) \left (a^{2} p^{2}-p^{2}-1\right ) \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}}d p} x\right ) &= 0 \end {align*}
Integrating gives \begin {align*} {\mathrm e}^{\int \frac {a \left (-5 a^{4} p^{3}+3 p^{2} \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a^{3}+a^{2} p^{3}-3 p^{2} \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a -13 a^{2} p +4 p^{3}+3 \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a +4 p \right )}{\left (-2 a^{2} p^{3}-3 a^{2} p +2 p^{3}+\sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a +2 p \right ) \left (a^{2} p^{2}-p^{2}-1\right ) \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}}d p} x &= c_{2} \end {align*}
Dividing both sides by the integrating factor \(\mu ={\mathrm e}^{\int \frac {a \left (-5 a^{4} p^{3}+3 p^{2} \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a^{3}+a^{2} p^{3}-3 p^{2} \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a -13 a^{2} p +4 p^{3}+3 \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a +4 p \right )}{\left (-2 a^{2} p^{3}-3 a^{2} p +2 p^{3}+\sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a +2 p \right ) \left (a^{2} p^{2}-p^{2}-1\right ) \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}}d p}\) results in \begin {align*} x \left (p \right ) &= c_{2} {\mathrm e}^{-a \left (\int \frac {\left (\left (-3 a^{3}+3 a \right ) p^{2}-3 a \right ) \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}+\left (5 a^{4}-a^{2}-4\right ) p^{3}+\left (13 a^{2}-4\right ) p}{\left (2 a^{2} p^{3}+3 a^{2} p -2 p^{3}-\sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a -2 p \right ) \left (a^{2} p^{2}-p^{2}-1\right ) \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}}d p \right )} \end {align*}
Since the solution \(x \left (p \right )\) has unresolved integral, unable to continue.
Solving ode 2A Taking derivative of (*) w.r.t. \(x\) gives \begin {align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end {align*}
Comparing the form \(y=x f + g\) to (1A) shows that \begin {align*} f &= -\frac {\left (3 a p +\sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\right ) a}{2 a^{2} p^{2}-2 p^{2}-2}\\ g &= 0 \end {align*}
Hence (2) becomes \begin {align*} p +\frac {\left (3 a p +\sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\right ) a}{2 a^{2} p^{2}-2 p^{2}-2} = x \left (-\frac {\left (3 a +\frac {\left (5 a^{2}+4\right ) p}{\sqrt {4+\left (5 a^{2}+4\right ) p^{2}}}\right ) a}{2 a^{2} p^{2}-2 p^{2}-2}+\frac {\left (3 a p +\sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\right ) a \left (4 a^{2} p -4 p \right )}{\left (2 a^{2} p^{2}-2 p^{2}-2\right )^{2}}\right ) p^{\prime }\left (x \right )\tag {2A} \end {align*}
The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives \begin {align*} p +\frac {\left (3 a p +\sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\right ) a}{2 a^{2} p^{2}-2 p^{2}-2} = 0 \end {align*}
Solving for \(p\) from the above gives \begin {align*} p&=\frac {\sqrt {2}\, \sqrt {\frac {-3 a^{2}+1+\sqrt {5 a^{4}-2 a^{2}+1}}{a^{2}-1}}}{2}\\ p&=-\frac {\sqrt {2}\, \sqrt {\frac {-3 a^{2}+1+\sqrt {5 a^{4}-2 a^{2}+1}}{a^{2}-1}}}{2}\\ p&=\frac {\sqrt {-\frac {2 \left (3 a^{2}+\sqrt {5 a^{4}-2 a^{2}+1}-1\right )}{a^{2}-1}}}{2}\\ p&=-\frac {\sqrt {-\frac {2 \left (3 a^{2}+\sqrt {5 a^{4}-2 a^{2}+1}-1\right )}{a^{2}-1}}}{2} \end {align*}
Removing solutions for \(p\) which leads to undefined results and substituting these in (1A) gives \begin {align*} y&=\frac {3 a^{2} x \sqrt {-\frac {2 \left (3 a^{2}+\sqrt {5 a^{4}-2 a^{2}+1}-1\right )}{a^{2}-1}}+a x \sqrt {-\frac {2 \left (15 a^{4}+5 \sqrt {5 a^{4}-2 a^{2}+1}\, a^{2}-a^{2}+4 \sqrt {5 a^{4}-2 a^{2}+1}+4\right )}{a^{2}-1}}}{6 a^{2}+2 \sqrt {5 a^{4}-2 a^{2}+1}+2}\\ y&=\frac {3 \sqrt {2}\, a^{2} x \sqrt {\frac {-3 a^{2}+1+\sqrt {5 a^{4}-2 a^{2}+1}}{a^{2}-1}}-\sqrt {2}\, a x \sqrt {\frac {-15 a^{4}+5 \sqrt {5 a^{4}-2 a^{2}+1}\, a^{2}+a^{2}+4 \sqrt {5 a^{4}-2 a^{2}+1}-4}{a^{2}-1}}}{-6 a^{2}+2 \sqrt {5 a^{4}-2 a^{2}+1}-2}\\ y&=\frac {-3 \sqrt {2}\, a^{2} x \sqrt {\frac {-3 a^{2}+1+\sqrt {5 a^{4}-2 a^{2}+1}}{a^{2}-1}}-\sqrt {2}\, a x \sqrt {\frac {-15 a^{4}+5 \sqrt {5 a^{4}-2 a^{2}+1}\, a^{2}+a^{2}+4 \sqrt {5 a^{4}-2 a^{2}+1}-4}{a^{2}-1}}}{-6 a^{2}+2 \sqrt {5 a^{4}-2 a^{2}+1}-2} \end {align*}
The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in \begin {align*} p^{\prime }\left (x \right ) = \frac {p \left (x \right )+\frac {\left (3 a p \left (x \right )+\sqrt {4+\left (5 a^{2}+4\right ) p \left (x \right )^{2}}\right ) a}{2 a^{2} p \left (x \right )^{2}-2 p \left (x \right )^{2}-2}}{x \left (-\frac {\left (3 a +\frac {\left (5 a^{2}+4\right ) p \left (x \right )}{\sqrt {4+\left (5 a^{2}+4\right ) p \left (x \right )^{2}}}\right ) a}{2 a^{2} p \left (x \right )^{2}-2 p \left (x \right )^{2}-2}+\frac {\left (3 a p \left (x \right )+\sqrt {4+\left (5 a^{2}+4\right ) p \left (x \right )^{2}}\right ) a \left (4 a^{2} p \left (x \right )-4 p \left (x \right )\right )}{\left (2 a^{2} p \left (x \right )^{2}-2 p \left (x \right )^{2}-2\right )^{2}}\right )}\tag {3} \end {align*}
This ODE is now solved for \(p \left (x \right )\).
Inverting the above ode gives \begin {align*} \frac {d}{d p}x \left (p \right ) = \frac {x \left (p \right ) \left (-\frac {\left (3 a +\frac {\left (5 a^{2}+4\right ) p}{\sqrt {4+\left (5 a^{2}+4\right ) p^{2}}}\right ) a}{2 a^{2} p^{2}-2 p^{2}-2}+\frac {\left (3 a p +\sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\right ) a \left (4 a^{2} p -4 p \right )}{\left (2 a^{2} p^{2}-2 p^{2}-2\right )^{2}}\right )}{p +\frac {\left (3 a p +\sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\right ) a}{2 a^{2} p^{2}-2 p^{2}-2}}\tag {4} \end {align*}
This ODE is now solved for \(x \left (p \right )\).
Entering Linear first order ODE solver. In canonical form a linear first order is \begin {align*} \frac {d}{d p}x \left (p \right ) + p(p)x \left (p \right ) &= q(p) \end {align*}
Where here \begin {align*} p(p) &=-\frac {a \left (5 a^{4} p^{3}+3 p^{2} \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a^{3}-a^{2} p^{3}-3 p^{2} \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a -4 p^{3}+13 a^{2} p +3 \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a -4 p \right )}{\left (2 a^{2} p^{3}+3 a^{2} p -2 p^{3}+\sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a -2 p \right ) \left (a^{2} p^{2}-p^{2}-1\right ) \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}}\\ q(p) &=0 \end {align*}
Hence the ode is \begin {align*} \frac {d}{d p}x \left (p \right )-\frac {x \left (p \right ) a \left (5 a^{4} p^{3}+3 p^{2} \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a^{3}-a^{2} p^{3}-3 p^{2} \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a -4 p^{3}+13 a^{2} p +3 \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a -4 p \right )}{\left (2 a^{2} p^{3}+3 a^{2} p -2 p^{3}+\sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a -2 p \right ) \left (a^{2} p^{2}-p^{2}-1\right ) \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}} = 0 \end {align*}
The integrating factor \(\mu \) is \[ \mu = {\mathrm e}^{\int -\frac {a \left (5 a^{4} p^{3}+3 p^{2} \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a^{3}-a^{2} p^{3}-3 p^{2} \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a -4 p^{3}+13 a^{2} p +3 \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a -4 p \right )}{\left (2 a^{2} p^{3}+3 a^{2} p -2 p^{3}+\sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a -2 p \right ) \left (a^{2} p^{2}-p^{2}-1\right ) \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}}d p} \] The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}p}} \mu x &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}p}} \left ({\mathrm e}^{\int -\frac {a \left (5 a^{4} p^{3}+3 p^{2} \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a^{3}-a^{2} p^{3}-3 p^{2} \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a -4 p^{3}+13 a^{2} p +3 \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a -4 p \right )}{\left (2 a^{2} p^{3}+3 a^{2} p -2 p^{3}+\sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a -2 p \right ) \left (a^{2} p^{2}-p^{2}-1\right ) \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}}d p} x\right ) &= 0 \end {align*}
Integrating gives \begin {align*} {\mathrm e}^{\int -\frac {a \left (5 a^{4} p^{3}+3 p^{2} \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a^{3}-a^{2} p^{3}-3 p^{2} \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a -4 p^{3}+13 a^{2} p +3 \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a -4 p \right )}{\left (2 a^{2} p^{3}+3 a^{2} p -2 p^{3}+\sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a -2 p \right ) \left (a^{2} p^{2}-p^{2}-1\right ) \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}}d p} x &= c_{4} \end {align*}
Dividing both sides by the integrating factor \(\mu ={\mathrm e}^{\int -\frac {a \left (5 a^{4} p^{3}+3 p^{2} \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a^{3}-a^{2} p^{3}-3 p^{2} \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a -4 p^{3}+13 a^{2} p +3 \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a -4 p \right )}{\left (2 a^{2} p^{3}+3 a^{2} p -2 p^{3}+\sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a -2 p \right ) \left (a^{2} p^{2}-p^{2}-1\right ) \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}}d p}\) results in \begin {align*} x \left (p \right ) &= c_{4} {\mathrm e}^{a \left (\int \frac {\left (\left (3 a^{3}-3 a \right ) p^{2}+3 a \right ) \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}+\left (5 a^{4}-a^{2}-4\right ) p^{3}+\left (13 a^{2}-4\right ) p}{\left (2 a^{2} p^{3}+3 a^{2} p -2 p^{3}+\sqrt {4+\left (5 a^{2}+4\right ) p^{2}}\, a -2 p \right ) \left (a^{2} p^{2}-p^{2}-1\right ) \sqrt {4+\left (5 a^{2}+4\right ) p^{2}}}d p \right )} \end {align*}
Since the solution \(x \left (p \right )\) has unresolved integral, unable to continue.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {-3 a^{2} x \sqrt {-\frac {2 \left (3 a^{2}+\sqrt {5 a^{4}-2 a^{2}+1}-1\right )}{a^{2}-1}}-a x \sqrt {-\frac {2 \left (15 a^{4}+5 \sqrt {5 a^{4}-2 a^{2}+1}\, a^{2}-a^{2}+4 \sqrt {5 a^{4}-2 a^{2}+1}+4\right )}{a^{2}-1}}}{6 a^{2}+2 \sqrt {5 a^{4}-2 a^{2}+1}+2} \\ \tag{2} y &= \frac {3 a^{2} x \sqrt {-\frac {2 \left (3 a^{2}+\sqrt {5 a^{4}-2 a^{2}+1}-1\right )}{a^{2}-1}}-a x \sqrt {-\frac {2 \left (15 a^{4}+5 \sqrt {5 a^{4}-2 a^{2}+1}\, a^{2}-a^{2}+4 \sqrt {5 a^{4}-2 a^{2}+1}+4\right )}{a^{2}-1}}}{6 a^{2}+2 \sqrt {5 a^{4}-2 a^{2}+1}+2} \\ \tag{3} y &= \frac {3 \sqrt {2}\, a^{2} x \sqrt {\frac {-3 a^{2}+1+\sqrt {5 a^{4}-2 a^{2}+1}}{a^{2}-1}}+\sqrt {2}\, a x \sqrt {\frac {-15 a^{4}+5 \sqrt {5 a^{4}-2 a^{2}+1}\, a^{2}+a^{2}+4 \sqrt {5 a^{4}-2 a^{2}+1}-4}{a^{2}-1}}}{-6 a^{2}+2 \sqrt {5 a^{4}-2 a^{2}+1}-2} \\ \tag{4} y &= \frac {-3 \sqrt {2}\, a^{2} x \sqrt {\frac {-3 a^{2}+1+\sqrt {5 a^{4}-2 a^{2}+1}}{a^{2}-1}}+\sqrt {2}\, a x \sqrt {\frac {-15 a^{4}+5 \sqrt {5 a^{4}-2 a^{2}+1}\, a^{2}+a^{2}+4 \sqrt {5 a^{4}-2 a^{2}+1}-4}{a^{2}-1}}}{-6 a^{2}+2 \sqrt {5 a^{4}-2 a^{2}+1}-2} \\ \tag{5} y &= \frac {3 a^{2} x \sqrt {-\frac {2 \left (3 a^{2}+\sqrt {5 a^{4}-2 a^{2}+1}-1\right )}{a^{2}-1}}+a x \sqrt {-\frac {2 \left (15 a^{4}+5 \sqrt {5 a^{4}-2 a^{2}+1}\, a^{2}-a^{2}+4 \sqrt {5 a^{4}-2 a^{2}+1}+4\right )}{a^{2}-1}}}{6 a^{2}+2 \sqrt {5 a^{4}-2 a^{2}+1}+2} \\ \tag{6} y &= \frac {3 \sqrt {2}\, a^{2} x \sqrt {\frac {-3 a^{2}+1+\sqrt {5 a^{4}-2 a^{2}+1}}{a^{2}-1}}-\sqrt {2}\, a x \sqrt {\frac {-15 a^{4}+5 \sqrt {5 a^{4}-2 a^{2}+1}\, a^{2}+a^{2}+4 \sqrt {5 a^{4}-2 a^{2}+1}-4}{a^{2}-1}}}{-6 a^{2}+2 \sqrt {5 a^{4}-2 a^{2}+1}-2} \\ \tag{7} y &= \frac {-3 \sqrt {2}\, a^{2} x \sqrt {\frac {-3 a^{2}+1+\sqrt {5 a^{4}-2 a^{2}+1}}{a^{2}-1}}-\sqrt {2}\, a x \sqrt {\frac {-15 a^{4}+5 \sqrt {5 a^{4}-2 a^{2}+1}\, a^{2}+a^{2}+4 \sqrt {5 a^{4}-2 a^{2}+1}-4}{a^{2}-1}}}{-6 a^{2}+2 \sqrt {5 a^{4}-2 a^{2}+1}-2} \\ \end{align*}
Verification of solutions
\[ y = \frac {-3 a^{2} x \sqrt {-\frac {2 \left (3 a^{2}+\sqrt {5 a^{4}-2 a^{2}+1}-1\right )}{a^{2}-1}}-a x \sqrt {-\frac {2 \left (15 a^{4}+5 \sqrt {5 a^{4}-2 a^{2}+1}\, a^{2}-a^{2}+4 \sqrt {5 a^{4}-2 a^{2}+1}+4\right )}{a^{2}-1}}}{6 a^{2}+2 \sqrt {5 a^{4}-2 a^{2}+1}+2} \] Verified OK.
\[ y = \frac {3 a^{2} x \sqrt {-\frac {2 \left (3 a^{2}+\sqrt {5 a^{4}-2 a^{2}+1}-1\right )}{a^{2}-1}}-a x \sqrt {-\frac {2 \left (15 a^{4}+5 \sqrt {5 a^{4}-2 a^{2}+1}\, a^{2}-a^{2}+4 \sqrt {5 a^{4}-2 a^{2}+1}+4\right )}{a^{2}-1}}}{6 a^{2}+2 \sqrt {5 a^{4}-2 a^{2}+1}+2} \] Verified OK.
\[ y = \frac {3 \sqrt {2}\, a^{2} x \sqrt {\frac {-3 a^{2}+1+\sqrt {5 a^{4}-2 a^{2}+1}}{a^{2}-1}}+\sqrt {2}\, a x \sqrt {\frac {-15 a^{4}+5 \sqrt {5 a^{4}-2 a^{2}+1}\, a^{2}+a^{2}+4 \sqrt {5 a^{4}-2 a^{2}+1}-4}{a^{2}-1}}}{-6 a^{2}+2 \sqrt {5 a^{4}-2 a^{2}+1}-2} \] Verified OK.
\[ y = \frac {-3 \sqrt {2}\, a^{2} x \sqrt {\frac {-3 a^{2}+1+\sqrt {5 a^{4}-2 a^{2}+1}}{a^{2}-1}}+\sqrt {2}\, a x \sqrt {\frac {-15 a^{4}+5 \sqrt {5 a^{4}-2 a^{2}+1}\, a^{2}+a^{2}+4 \sqrt {5 a^{4}-2 a^{2}+1}-4}{a^{2}-1}}}{-6 a^{2}+2 \sqrt {5 a^{4}-2 a^{2}+1}-2} \] Verified OK.
\[ y = \frac {3 a^{2} x \sqrt {-\frac {2 \left (3 a^{2}+\sqrt {5 a^{4}-2 a^{2}+1}-1\right )}{a^{2}-1}}+a x \sqrt {-\frac {2 \left (15 a^{4}+5 \sqrt {5 a^{4}-2 a^{2}+1}\, a^{2}-a^{2}+4 \sqrt {5 a^{4}-2 a^{2}+1}+4\right )}{a^{2}-1}}}{6 a^{2}+2 \sqrt {5 a^{4}-2 a^{2}+1}+2} \] Verified OK.
\[ y = \frac {3 \sqrt {2}\, a^{2} x \sqrt {\frac {-3 a^{2}+1+\sqrt {5 a^{4}-2 a^{2}+1}}{a^{2}-1}}-\sqrt {2}\, a x \sqrt {\frac {-15 a^{4}+5 \sqrt {5 a^{4}-2 a^{2}+1}\, a^{2}+a^{2}+4 \sqrt {5 a^{4}-2 a^{2}+1}-4}{a^{2}-1}}}{-6 a^{2}+2 \sqrt {5 a^{4}-2 a^{2}+1}-2} \] Verified OK.
\[ y = \frac {-3 \sqrt {2}\, a^{2} x \sqrt {\frac {-3 a^{2}+1+\sqrt {5 a^{4}-2 a^{2}+1}}{a^{2}-1}}-\sqrt {2}\, a x \sqrt {\frac {-15 a^{4}+5 \sqrt {5 a^{4}-2 a^{2}+1}\, a^{2}+a^{2}+4 \sqrt {5 a^{4}-2 a^{2}+1}-4}{a^{2}-1}}}{-6 a^{2}+2 \sqrt {5 a^{4}-2 a^{2}+1}-2} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (-a^{2}+1\right ) y^{2} {y^{\prime }}^{2}-3 a^{2} x y y^{\prime }+y^{2}=a^{2} x^{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=-\frac {3 a^{2} x +\sqrt {5 a^{4} x^{2}+4 a^{2} y^{2}+4 a^{2} x^{2}-4 y^{2}}}{2 \left (a^{2}-1\right ) y}, y^{\prime }=\frac {-3 a^{2} x +\sqrt {5 a^{4} x^{2}+4 a^{2} y^{2}+4 a^{2} x^{2}-4 y^{2}}}{2 \left (a^{2}-1\right ) y}\right ] \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {3 a^{2} x +\sqrt {5 a^{4} x^{2}+4 a^{2} y^{2}+4 a^{2} x^{2}-4 y^{2}}}{2 \left (a^{2}-1\right ) y} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {-3 a^{2} x +\sqrt {5 a^{4} x^{2}+4 a^{2} y^{2}+4 a^{2} x^{2}-4 y^{2}}}{2 \left (a^{2}-1\right ) y} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\mathit {workingODE} , \mathit {workingODE}\right \} \end {array} \]
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 <- symmetries for implicit equations successful`
✓ Solution by Maple
Time used: 0.062 (sec). Leaf size: 198
dsolve((-a^2+1)*y(x)^2*diff(y(x),x)^2-3*a^2*x*y(x)*diff(y(x),x)-a^2*x^2+y(x)^2 = 0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \operatorname {RootOf}\left (-2 \ln \left (x \right )-\left (\int _{}^{\textit {\_Z}}\frac {\left (2 \textit {\_a}^{2} a^{2}-2 \textit {\_a}^{2}+3 a^{2}+\sqrt {4 \textit {\_a}^{2} a^{2}+5 a^{4}-4 \textit {\_a}^{2}+4 a^{2}}\right ) \textit {\_a}}{a^{2} \textit {\_a}^{4}-\textit {\_a}^{4}+3 \textit {\_a}^{2} a^{2}-\textit {\_a}^{2}+a^{2}}d \textit {\_a} \right )+2 c_{1} \right ) x \\ y \left (x \right ) &= \operatorname {RootOf}\left (-2 \ln \left (x \right )+\int _{}^{\textit {\_Z}}-\frac {\left (2 \textit {\_a}^{2} a^{2}-2 \textit {\_a}^{2}+3 a^{2}-\sqrt {4 \textit {\_a}^{2} a^{2}+5 a^{4}-4 \textit {\_a}^{2}+4 a^{2}}\right ) \textit {\_a}}{a^{2} \textit {\_a}^{4}-\textit {\_a}^{4}+3 \textit {\_a}^{2} a^{2}-\textit {\_a}^{2}+a^{2}}d \textit {\_a} +2 c_{1} \right ) x \\ \end{align*}
✓ Solution by Mathematica
Time used: 1.255 (sec). Leaf size: 342
DSolve[(1-a^2)y[x]^2 (y'[x])^2 -2 a^2 x y[x] y'[x]-a^2 x y[x] y'[x]-a^2 x^2+y[x]^2==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} \text {Solve}\left [\frac {\log \left (-\left (a^2 \left (\frac {2 y(x)^2}{x^2}+3\right )\right )+\sqrt {5 a^4+4 a^2 \left (\frac {y(x)^2}{x^2}+1\right )-\frac {4 y(x)^2}{x^2}}+\frac {2 y(x)^2}{x^2}\right )-\frac {2 \arctan \left (\frac {1-\sqrt {5 a^4+4 a^2 \left (\frac {y(x)^2}{x^2}+1\right )-\frac {4 y(x)^2}{x^2}}}{\sqrt {-5 a^4+2 a^2-1}}\right )}{\sqrt {-5 a^4+2 a^2-1}}}{4 a^2-4}&=\frac {\log \left (-2 \left (a^2-1\right ) x\right )}{2-2 a^2}+c_1,y(x)\right ] \\ \text {Solve}\left [\frac {\log \left (a^2 \left (\frac {2 y(x)^2}{x^2}+3\right )+\sqrt {5 a^4+4 a^2 \left (\frac {y(x)^2}{x^2}+1\right )-\frac {4 y(x)^2}{x^2}}-\frac {2 y(x)^2}{x^2}\right )-\frac {2 \arctan \left (\frac {\sqrt {5 a^4+4 a^2 \left (\frac {y(x)^2}{x^2}+1\right )-\frac {4 y(x)^2}{x^2}}+1}{\sqrt {-5 a^4+2 a^2-1}}\right )}{\sqrt {-5 a^4+2 a^2-1}}}{4 a^2-4}&=\frac {\log \left (-2 \left (a^2-1\right ) x\right )}{2-2 a^2}+c_1,y(x)\right ] \\ \end{align*}