Internal problem ID [4202]
Internal file name [OUTPUT/3695_Sunday_June_05_2022_10_17_11_AM_58503920/index.tex
]
Book: Ordinary differential equations and their solutions. By George Moseley Murphy.
1960
Section: Various 33
Problem number: 969.
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`], _dAlembert]
\[ \boxed {x \left (x -2 y\right ) {y^{\prime }}^{2}+6 y y^{\prime } x -2 y x +y^{2}=0} \]
Let \(p=y^{\prime }\) the ode becomes \begin {align*} x \left (x -2 y \right ) p^{2}+6 y p x -2 x y +y^{2} = 0 \end {align*}
Solving for \(y\) from the above results in \begin {align*} y &= \left (p^{2}-3 p +1+\sqrt {p^{4}-6 p^{3}+10 p^{2}-6 p +1}\right ) x\tag {1A}\\ y &= \left (p^{2}-3 p +1-\sqrt {p^{4}-6 p^{3}+10 p^{2}-6 p +1}\right ) x\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 &= p^{2}-3 p +1+\sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}}\\ g &= 0 \end {align*}
Hence (2) becomes \begin {align*} 4 p -p^{2}-1-\sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}} = x \left (2 p -3+\frac {\left (2 p -4\right ) \left (p -1\right )^{2}+2 \left (p^{2}-4 p +1\right ) \left (p -1\right )}{2 \sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\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*} 4 p -p^{2}-1-\sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}} = 0 \end {align*}
Solving for \(p\) from the above gives \begin {align*} p&=2+\sqrt {3}\\ p&=2-\sqrt {3} \end {align*}
Substituting these in (1A) gives \begin {align*} y&=-x \sqrt {3}+2 x\\ y&=x \sqrt {3}+2 x \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 {4 p \left (x \right )-p \left (x \right )^{2}-1-\sqrt {\left (p \left (x \right )^{2}-4 p \left (x \right )+1\right ) \left (p \left (x \right )-1\right )^{2}}}{x \left (2 p \left (x \right )-3+\frac {\left (2 p \left (x \right )-4\right ) \left (p \left (x \right )-1\right )^{2}+2 \left (p \left (x \right )^{2}-4 p \left (x \right )+1\right ) \left (p \left (x \right )-1\right )}{2 \sqrt {\left (p \left (x \right )^{2}-4 p \left (x \right )+1\right ) \left (p \left (x \right )-1\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 (2 p -3+\frac {\left (2 p -4\right ) \left (p -1\right )^{2}+2 \left (p^{2}-4 p +1\right ) \left (p -1\right )}{2 \sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}}}\right )}{4 p -p^{2}-1-\sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{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 {2 p^{3}+2 p \sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}}-9 p^{2}-3 \sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}}+10 p -3}{\sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}}\, \left (-4 p +p^{2}+1+\sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}}\right )}\\ q(p) &=0 \end {align*}
Hence the ode is \begin {align*} \frac {d}{d p}x \left (p \right )+\frac {x \left (p \right ) \left (2 p^{3}+2 p \sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}}-9 p^{2}-3 \sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}}+10 p -3\right )}{\sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}}\, \left (-4 p +p^{2}+1+\sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}}\right )} = 0 \end {align*}
The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int \frac {2 p^{3}+2 p \sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}}-9 p^{2}-3 \sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}}+10 p -3}{\sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}}\, \left (-4 p +p^{2}+1+\sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}}\right )}d p} \\ &= {\mathrm e}^{\frac {\sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}}\, \ln \left (p -2+\sqrt {p^{2}-4 p +1}\right )}{\left (p -1\right ) \sqrt {p^{2}-4 p +1}}+\frac {\ln \left (p^{2}-4 p +1\right )}{2}} \\ \end{align*} Which simplifies to \[ \mu = \sqrt {p^{2}-4 p +1}\, \left (p -2+\sqrt {p^{2}-4 p +1}\right )^{\frac {\sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}}}{\left (p -1\right ) \sqrt {p^{2}-4 p +1}}} \] 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 (\sqrt {p^{2}-4 p +1}\, \left (p -2+\sqrt {p^{2}-4 p +1}\right )^{\frac {\sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}}}{\left (p -1\right ) \sqrt {p^{2}-4 p +1}}} x\right ) &= 0 \end {align*}
Integrating gives \begin {align*} \sqrt {p^{2}-4 p +1}\, \left (p -2+\sqrt {p^{2}-4 p +1}\right )^{\frac {\sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}}}{\left (p -1\right ) \sqrt {p^{2}-4 p +1}}} x &= c_{2} \end {align*}
Dividing both sides by the integrating factor \(\mu =\sqrt {p^{2}-4 p +1}\, \left (p -2+\sqrt {p^{2}-4 p +1}\right )^{\frac {\sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}}}{\left (p -1\right ) \sqrt {p^{2}-4 p +1}}}\) results in \begin {align*} x \left (p \right ) &= \frac {c_{2} \left (p -2+\sqrt {p^{2}-4 p +1}\right )^{-\frac {\sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}}}{\left (p -1\right ) \sqrt {p^{2}-4 p +1}}}}{\sqrt {p^{2}-4 p +1}} \end {align*}
Now we need to eliminate \(p\) between the above and (1A). One way to do this is by solving (1) for \(p\). This results in \begin {align*} p&=\frac {-3 y x +\sqrt {2 x y^{3}+4 y^{2} x^{2}+2 x^{3} y}}{x \left (x -2 y\right )}\\ p&=-\frac {3 y x +\sqrt {2 x y^{3}+4 y^{2} x^{2}+2 x^{3} y}}{x \left (x -2 y\right )} \end {align*}
Substituting the above in the solution for \(x\) found above gives \begin{align*} x&=\frac {c_{2} {\left (\frac {x \left (x -2 y\right ) \sqrt {\frac {-4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}}{x \left (x -2 y\right )^{2}}}-2 x^{2}+y x +\sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}}{x \left (x -2 y\right )}\right )}^{\frac {\sqrt {\frac {\left (-\sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x \left (y+x \right )\right )^{2} \left (-4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}\right )}{x^{3} \left (x -2 y\right )^{4}}}\, x \left (x -2 y\right )}{\sqrt {\frac {-4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}}{x \left (x -2 y\right )^{2}}}\, \left (-\sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x \left (y+x \right )\right )}}}{\sqrt {\frac {-4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}}{x \left (x -2 y\right )^{2}}}} \\ x&=\frac {c_{2} {\left (\frac {x \left (x -2 y\right ) \sqrt {\frac {4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}}{x \left (x -2 y\right )^{2}}}-2 x^{2}+y x -\sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}}{x \left (x -2 y\right )}\right )}^{\frac {\sqrt {\frac {\left (\sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x \left (y+x \right )\right )^{2} \left (4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}\right )}{x^{3} \left (x -2 y\right )^{4}}}\, x \left (x -2 y\right )}{\left (\sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{2}+y x \right ) \sqrt {\frac {4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}}{x \left (x -2 y\right )^{2}}}}}}{\sqrt {\frac {4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}}{x \left (x -2 y\right )^{2}}}} \\ \end{align*} 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 &= p^{2}-3 p +1-\sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}}\\ g &= 0 \end {align*}
Hence (2) becomes \begin {align*} 4 p -p^{2}-1+\sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}} = x \left (2 p -3-\frac {\left (2 p -4\right ) \left (p -1\right )^{2}+2 \left (p^{2}-4 p +1\right ) \left (p -1\right )}{2 \sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\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*} 4 p -p^{2}-1+\sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}} = 0 \end {align*}
Solving for \(p\) from the above gives \begin {align*} p&=0\\ p&=2+\sqrt {3}\\ p&=2-\sqrt {3} \end {align*}
Substituting these in (1A) gives \begin {align*} y&=0\\ y&=-x \sqrt {3}+2 x\\ y&=x \sqrt {3}+2 x \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 {4 p \left (x \right )-p \left (x \right )^{2}-1+\sqrt {\left (p \left (x \right )^{2}-4 p \left (x \right )+1\right ) \left (p \left (x \right )-1\right )^{2}}}{x \left (2 p \left (x \right )-3-\frac {\left (2 p \left (x \right )-4\right ) \left (p \left (x \right )-1\right )^{2}+2 \left (p \left (x \right )^{2}-4 p \left (x \right )+1\right ) \left (p \left (x \right )-1\right )}{2 \sqrt {\left (p \left (x \right )^{2}-4 p \left (x \right )+1\right ) \left (p \left (x \right )-1\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 (2 p -3-\frac {\left (2 p -4\right ) \left (p -1\right )^{2}+2 \left (p^{2}-4 p +1\right ) \left (p -1\right )}{2 \sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}}}\right )}{4 p -p^{2}-1+\sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{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 {-2 p^{3}+2 p \sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}}+9 p^{2}-3 \sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}}-10 p +3}{\sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}}\, \left (4 p -p^{2}-1+\sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}}\right )}\\ q(p) &=0 \end {align*}
Hence the ode is \begin {align*} \frac {d}{d p}x \left (p \right )-\frac {x \left (p \right ) \left (-2 p^{3}+2 p \sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}}+9 p^{2}-3 \sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}}-10 p +3\right )}{\sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}}\, \left (4 p -p^{2}-1+\sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}}\right )} = 0 \end {align*}
The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int -\frac {-2 p^{3}+2 p \sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}}+9 p^{2}-3 \sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}}-10 p +3}{\sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}}\, \left (4 p -p^{2}-1+\sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}}\right )}d p} \\ &= {\mathrm e}^{-\frac {\sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}}\, \ln \left (p -2+\sqrt {p^{2}-4 p +1}\right )}{\left (p -1\right ) \sqrt {p^{2}-4 p +1}}+\frac {\ln \left (p^{2}-4 p +1\right )}{2}} \\ \end{align*} Which simplifies to \[ \mu = \sqrt {p^{2}-4 p +1}\, \left (p -2+\sqrt {p^{2}-4 p +1}\right )^{-\frac {\sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}}}{\left (p -1\right ) \sqrt {p^{2}-4 p +1}}} \] 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 (\sqrt {p^{2}-4 p +1}\, \left (p -2+\sqrt {p^{2}-4 p +1}\right )^{-\frac {\sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}}}{\left (p -1\right ) \sqrt {p^{2}-4 p +1}}} x\right ) &= 0 \end {align*}
Integrating gives \begin {align*} \sqrt {p^{2}-4 p +1}\, \left (p -2+\sqrt {p^{2}-4 p +1}\right )^{-\frac {\sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}}}{\left (p -1\right ) \sqrt {p^{2}-4 p +1}}} x &= c_{4} \end {align*}
Dividing both sides by the integrating factor \(\mu =\sqrt {p^{2}-4 p +1}\, \left (p -2+\sqrt {p^{2}-4 p +1}\right )^{-\frac {\sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}}}{\left (p -1\right ) \sqrt {p^{2}-4 p +1}}}\) results in \begin {align*} x \left (p \right ) &= \frac {c_{4} \left (p -2+\sqrt {p^{2}-4 p +1}\right )^{\frac {\sqrt {\left (p^{2}-4 p +1\right ) \left (p -1\right )^{2}}}{\left (p -1\right ) \sqrt {p^{2}-4 p +1}}}}{\sqrt {p^{2}-4 p +1}} \end {align*}
Now we need to eliminate \(p\) between the above and (1A). One way to do this is by solving (1) for \(p\). This results in \begin {align*} p&=\frac {-3 y x +\sqrt {2 x y^{3}+4 y^{2} x^{2}+2 x^{3} y}}{x \left (x -2 y\right )}\\ p&=-\frac {3 y x +\sqrt {2 x y^{3}+4 y^{2} x^{2}+2 x^{3} y}}{x \left (x -2 y\right )} \end {align*}
Substituting the above in the solution for \(x\) found above gives \begin{align*} x&=\frac {c_{4} {\left (\frac {x \left (x -2 y\right ) \sqrt {\frac {-4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}}{x \left (x -2 y\right )^{2}}}-2 x^{2}+y x +\sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}}{x \left (x -2 y\right )}\right )}^{-\frac {\sqrt {\frac {\left (-\sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x \left (y+x \right )\right )^{2} \left (-4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}\right )}{x^{3} \left (x -2 y\right )^{4}}}\, x \left (x -2 y\right )}{\sqrt {\frac {-4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}}{x \left (x -2 y\right )^{2}}}\, \left (-\sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x \left (y+x \right )\right )}}}{\sqrt {\frac {-4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}}{x \left (x -2 y\right )^{2}}}} \\ x&=\frac {c_{4} {\left (\frac {x \left (x -2 y\right ) \sqrt {\frac {4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}}{x \left (x -2 y\right )^{2}}}-2 x^{2}+y x -\sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}}{x \left (x -2 y\right )}\right )}^{-\frac {\sqrt {\frac {\left (\sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x \left (y+x \right )\right )^{2} \left (4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}\right )}{x^{3} \left (x -2 y\right )^{4}}}\, x \left (x -2 y\right )}{\left (\sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{2}+y x \right ) \sqrt {\frac {4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}}{x \left (x -2 y\right )^{2}}}}}}{\sqrt {\frac {4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}}{x \left (x -2 y\right )^{2}}}} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= -x \sqrt {3}+2 x \\ \tag{2} y &= x \sqrt {3}+2 x \\ \tag{3} x &= \frac {c_{2} {\left (\frac {x \left (x -2 y\right ) \sqrt {\frac {-4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}}{x \left (x -2 y\right )^{2}}}-2 x^{2}+y x +\sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}}{x \left (x -2 y\right )}\right )}^{\frac {\sqrt {\frac {\left (-\sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x \left (y+x \right )\right )^{2} \left (-4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}\right )}{x^{3} \left (x -2 y\right )^{4}}}\, x \left (x -2 y\right )}{\sqrt {\frac {-4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}}{x \left (x -2 y\right )^{2}}}\, \left (-\sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x \left (y+x \right )\right )}}}{\sqrt {\frac {-4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}}{x \left (x -2 y\right )^{2}}}} \\ \tag{4} x &= \frac {c_{2} {\left (\frac {x \left (x -2 y\right ) \sqrt {\frac {4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}}{x \left (x -2 y\right )^{2}}}-2 x^{2}+y x -\sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}}{x \left (x -2 y\right )}\right )}^{\frac {\sqrt {\frac {\left (\sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x \left (y+x \right )\right )^{2} \left (4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}\right )}{x^{3} \left (x -2 y\right )^{4}}}\, x \left (x -2 y\right )}{\left (\sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{2}+y x \right ) \sqrt {\frac {4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}}{x \left (x -2 y\right )^{2}}}}}}{\sqrt {\frac {4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}}{x \left (x -2 y\right )^{2}}}} \\ \tag{5} y &= 0 \\ \tag{6} y &= -x \sqrt {3}+2 x \\ \tag{7} y &= x \sqrt {3}+2 x \\ \tag{8} x &= \frac {c_{4} {\left (\frac {x \left (x -2 y\right ) \sqrt {\frac {-4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}}{x \left (x -2 y\right )^{2}}}-2 x^{2}+y x +\sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}}{x \left (x -2 y\right )}\right )}^{-\frac {\sqrt {\frac {\left (-\sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x \left (y+x \right )\right )^{2} \left (-4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}\right )}{x^{3} \left (x -2 y\right )^{4}}}\, x \left (x -2 y\right )}{\sqrt {\frac {-4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}}{x \left (x -2 y\right )^{2}}}\, \left (-\sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x \left (y+x \right )\right )}}}{\sqrt {\frac {-4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}}{x \left (x -2 y\right )^{2}}}} \\ \tag{9} x &= \frac {c_{4} {\left (\frac {x \left (x -2 y\right ) \sqrt {\frac {4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}}{x \left (x -2 y\right )^{2}}}-2 x^{2}+y x -\sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}}{x \left (x -2 y\right )}\right )}^{-\frac {\sqrt {\frac {\left (\sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x \left (y+x \right )\right )^{2} \left (4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}\right )}{x^{3} \left (x -2 y\right )^{4}}}\, x \left (x -2 y\right )}{\left (\sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{2}+y x \right ) \sqrt {\frac {4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}}{x \left (x -2 y\right )^{2}}}}}}{\sqrt {\frac {4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}}{x \left (x -2 y\right )^{2}}}} \\ \end{align*}
Verification of solutions
\[ y = -x \sqrt {3}+2 x \] Verified OK.
\[ y = x \sqrt {3}+2 x \] Verified OK.
\[ x = \frac {c_{2} {\left (\frac {x \left (x -2 y\right ) \sqrt {\frac {-4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}}{x \left (x -2 y\right )^{2}}}-2 x^{2}+y x +\sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}}{x \left (x -2 y\right )}\right )}^{\frac {\sqrt {\frac {\left (-\sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x \left (y+x \right )\right )^{2} \left (-4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}\right )}{x^{3} \left (x -2 y\right )^{4}}}\, x \left (x -2 y\right )}{\sqrt {\frac {-4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}}{x \left (x -2 y\right )^{2}}}\, \left (-\sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x \left (y+x \right )\right )}}}{\sqrt {\frac {-4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}}{x \left (x -2 y\right )^{2}}}} \] Warning, solution could not be verified
\[ x = \frac {c_{2} {\left (\frac {x \left (x -2 y\right ) \sqrt {\frac {4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}}{x \left (x -2 y\right )^{2}}}-2 x^{2}+y x -\sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}}{x \left (x -2 y\right )}\right )}^{\frac {\sqrt {\frac {\left (\sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x \left (y+x \right )\right )^{2} \left (4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}\right )}{x^{3} \left (x -2 y\right )^{4}}}\, x \left (x -2 y\right )}{\left (\sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{2}+y x \right ) \sqrt {\frac {4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}}{x \left (x -2 y\right )^{2}}}}}}{\sqrt {\frac {4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}}{x \left (x -2 y\right )^{2}}}} \] Warning, solution could not be verified
\[ y = 0 \] Verified OK.
\[ y = -x \sqrt {3}+2 x \] Verified OK.
\[ y = x \sqrt {3}+2 x \] Verified OK.
\[ x = \frac {c_{4} {\left (\frac {x \left (x -2 y\right ) \sqrt {\frac {-4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}}{x \left (x -2 y\right )^{2}}}-2 x^{2}+y x +\sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}}{x \left (x -2 y\right )}\right )}^{-\frac {\sqrt {\frac {\left (-\sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x \left (y+x \right )\right )^{2} \left (-4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}\right )}{x^{3} \left (x -2 y\right )^{4}}}\, x \left (x -2 y\right )}{\sqrt {\frac {-4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}}{x \left (x -2 y\right )^{2}}}\, \left (-\sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x \left (y+x \right )\right )}}}{\sqrt {\frac {-4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}}{x \left (x -2 y\right )^{2}}}} \] Warning, solution could not be verified
\[ x = \frac {c_{4} {\left (\frac {x \left (x -2 y\right ) \sqrt {\frac {4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}}{x \left (x -2 y\right )^{2}}}-2 x^{2}+y x -\sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}}{x \left (x -2 y\right )}\right )}^{-\frac {\sqrt {\frac {\left (\sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x \left (y+x \right )\right )^{2} \left (4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}\right )}{x^{3} \left (x -2 y\right )^{4}}}\, x \left (x -2 y\right )}{\left (\sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{2}+y x \right ) \sqrt {\frac {4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}}{x \left (x -2 y\right )^{2}}}}}}{\sqrt {\frac {4 \left (x -\frac {y}{2}\right ) \sqrt {2}\, \sqrt {x y \left (y+x \right )^{2}}+x^{3}+10 y x^{2}-7 x y^{2}+2 y^{3}}{x \left (x -2 y\right )^{2}}}} \] Warning, solution could not be verified
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (x -2 y\right ) {y^{\prime }}^{2}+6 y y^{\prime } x -2 y x +y^{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=\frac {3 y x -\sqrt {2 x y^{3}+4 y^{2} x^{2}+2 x^{3} y}}{x \left (-x +2 y\right )}, y^{\prime }=\frac {3 y x +\sqrt {2 x y^{3}+4 y^{2} x^{2}+2 x^{3} y}}{x \left (-x +2 y\right )}\right ] \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {3 y x -\sqrt {2 x y^{3}+4 y^{2} x^{2}+2 x^{3} y}}{x \left (-x +2 y\right )} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {3 y x +\sqrt {2 x y^{3}+4 y^{2} x^{2}+2 x^{3} y}}{x \left (-x +2 y\right )} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\mathit {workingODE} , \mathit {workingODE}\right \} \end {array} \]
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 <- symmetries for implicit equations successful`
✓ Solution by Maple
Time used: 0.797 (sec). Leaf size: 115
dsolve(x*(x-2*y(x))*diff(y(x),x)^2+6*x*y(x)*diff(y(x),x)-2*x*y(x)+y(x)^2 = 0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= 0 \\ y \left (x \right ) &= \operatorname {RootOf}\left (-2 \ln \left (x \right )-\left (\int _{}^{\textit {\_Z}}\frac {2 \textit {\_a}^{2}+\sqrt {2}\, \sqrt {\textit {\_a} \left (\textit {\_a} +1\right )^{2}}-4 \textit {\_a}}{\textit {\_a} \left (\textit {\_a}^{2}-4 \textit {\_a} +1\right )}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 {\sqrt {2}\, \sqrt {\textit {\_a} \left (\textit {\_a} +1\right )^{2}}-2 \textit {\_a}^{2}+4 \textit {\_a}}{\textit {\_a} \left (\textit {\_a}^{2}-4 \textit {\_a} +1\right )}d \textit {\_a} +2 c_{1} \right ) x \\ \end{align*}
✓ Solution by Mathematica
Time used: 7.579 (sec). Leaf size: 196
DSolve[x(x-2 y[x]) (y'[x])^2+6 x y[x] y'[x]-2 x y[x]+y[x]^2==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to 2 x-\sqrt {x \left (3 x-2 e^{\frac {c_1}{2}}\right )}-e^{\frac {c_1}{2}} \\ y(x)\to 2 x+\sqrt {x \left (3 x-2 e^{\frac {c_1}{2}}\right )}-e^{\frac {c_1}{2}} \\ y(x)\to 2 x-\sqrt {x \left (3 x+2 e^{\frac {c_1}{2}}\right )}+e^{\frac {c_1}{2}} \\ y(x)\to 2 x+\sqrt {x \left (3 x+2 e^{\frac {c_1}{2}}\right )}+e^{\frac {c_1}{2}} \\ y(x)\to 2 x-\sqrt {3} \sqrt {x^2} \\ y(x)\to \sqrt {3} \sqrt {x^2}+2 x \\ \end{align*}