Internal problem ID [4292]
Internal file name [OUTPUT/3785_Sunday_June_05_2022_10_56_27_AM_58818420/index.tex]
Book: Ordinary differential equations and their solutions. By George Moseley Murphy.
1960
Section: Various 36
Problem number: 1073.
ODE order: 1.
ODE degree: 3.
The type(s) of ODE detected by this program : "dAlembert"
Maple gives the following as the ode type
[[_1st_order, _with_linear_symmetries], _dAlembert]
\[ \boxed {2 y {y^{\prime }}^{3}-3 x y^{\prime }+2 y=0} \]
Let \(p=y^{\prime }\) the ode becomes \begin {align*} 2 y \,p^{3}-3 x p +2 y = 0 \end {align*}
Solving for \(y\) from the above results in \begin {align*} y &= \frac {3 x p}{2 \left (p^{3}+1\right )}\tag {1A} \end {align*}
This has the form \begin {align*} y=xf(p)+g(p)\tag {*} \end {align*}
Where \(f,g\) are functions of \(p=y'(x)\). The above ode is dAlembert ode which is now solved. 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 {3 p}{2 p^{3}+2}\\ g &= 0 \end {align*}
Hence (2) becomes \begin {align*} p -\frac {3 p}{2 p^{3}+2} = x \left (\frac {3}{2 p^{3}+2}-\frac {18 p^{3}}{\left (2 p^{3}+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 {3 p}{2 p^{3}+2} = 0 \end {align*}
Solving for \(p\) from the above gives \begin {align*} p&=0\\ p&=\frac {2^{\frac {2}{3}}}{2}\\ p&=-\frac {2^{\frac {2}{3}}}{4}+\frac {i \sqrt {3}\, 2^{\frac {2}{3}}}{4}\\ p&=-\frac {2^{\frac {2}{3}}}{4}-\frac {i \sqrt {3}\, 2^{\frac {2}{3}}}{4} \end {align*}
Substituting these in (1A) gives \begin {align*} y&=0\\ y&=\frac {2^{\frac {2}{3}} x}{2}\\ y&=-\frac {i 2^{\frac {2}{3}} \sqrt {3}\, x}{4}-\frac {2^{\frac {2}{3}} x}{4}\\ y&=\frac {i 2^{\frac {2}{3}} \sqrt {3}\, x}{4}-\frac {2^{\frac {2}{3}} x}{4} \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 {3 p \left (x \right )}{2 p \left (x \right )^{3}+2}}{x \left (\frac {3}{2 p \left (x \right )^{3}+2}-\frac {18 p \left (x \right )^{3}}{\left (2 p \left (x \right )^{3}+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 {3}{2 p^{3}+2}-\frac {18 p^{3}}{\left (2 p^{3}+2\right )^{2}}\right )}{p -\frac {3 p}{2 p^{3}+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 {3}{p \left (p^{3}+1\right )}\\ q(p) &=0 \end {align*}
Hence the ode is \begin {align*} \frac {d}{d p}x \left (p \right )+\frac {3 x \left (p \right )}{p \left (p^{3}+1\right )} = 0 \end {align*}
The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int \frac {3}{p \left (p^{3}+1\right )}d p} \\ &= {\mathrm e}^{-\ln \left (p^{2}-p +1\right )+3 \ln \left (p \right )-\ln \left (p +1\right )} \\ \end{align*} Which simplifies to \[ \mu = \frac {p^{3}}{\left (p^{2}-p +1\right ) \left (p +1\right )} \] 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 (\frac {p^{3} x}{\left (p^{2}-p +1\right ) \left (p +1\right )}\right ) &= 0 \end {align*}
Integrating gives \begin {align*} \frac {p^{3} x}{\left (p^{2}-p +1\right ) \left (p +1\right )} &= c_{3} \end {align*}
Dividing both sides by the integrating factor \(\mu =\frac {p^{3}}{\left (p^{2}-p +1\right ) \left (p +1\right )}\) results in \begin {align*} x \left (p \right ) &= \frac {c_{3} \left (p^{2}-p +1\right ) \left (p +1\right )}{p^{3}} \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 {{\left (\left (-4 y+2 \sqrt {-\frac {2 \left (-2 y^{3}+x^{3}\right )}{y}}\right ) y^{2}\right )}^{\frac {1}{3}}}{2 y}+\frac {x}{{\left (\left (-4 y+2 \sqrt {-\frac {2 \left (-2 y^{3}+x^{3}\right )}{y}}\right ) y^{2}\right )}^{\frac {1}{3}}}\\ p&=-\frac {{\left (\left (-4 y+2 \sqrt {-\frac {2 \left (-2 y^{3}+x^{3}\right )}{y}}\right ) y^{2}\right )}^{\frac {1}{3}}}{4 y}-\frac {x}{2 {\left (\left (-4 y+2 \sqrt {-\frac {2 \left (-2 y^{3}+x^{3}\right )}{y}}\right ) y^{2}\right )}^{\frac {1}{3}}}+\frac {i \sqrt {3}\, \left (\frac {{\left (\left (-4 y+2 \sqrt {-\frac {2 \left (-2 y^{3}+x^{3}\right )}{y}}\right ) y^{2}\right )}^{\frac {1}{3}}}{2 y}-\frac {x}{{\left (\left (-4 y+2 \sqrt {-\frac {2 \left (-2 y^{3}+x^{3}\right )}{y}}\right ) y^{2}\right )}^{\frac {1}{3}}}\right )}{2}\\ p&=-\frac {{\left (\left (-4 y+2 \sqrt {-\frac {2 \left (-2 y^{3}+x^{3}\right )}{y}}\right ) y^{2}\right )}^{\frac {1}{3}}}{4 y}-\frac {x}{2 {\left (\left (-4 y+2 \sqrt {-\frac {2 \left (-2 y^{3}+x^{3}\right )}{y}}\right ) y^{2}\right )}^{\frac {1}{3}}}-\frac {i \sqrt {3}\, \left (\frac {{\left (\left (-4 y+2 \sqrt {-\frac {2 \left (-2 y^{3}+x^{3}\right )}{y}}\right ) y^{2}\right )}^{\frac {1}{3}}}{2 y}-\frac {x}{{\left (\left (-4 y+2 \sqrt {-\frac {2 \left (-2 y^{3}+x^{3}\right )}{y}}\right ) y^{2}\right )}^{\frac {1}{3}}}\right )}{2} \end {align*}
Substituting the above in the solution for \(x\) found above gives \begin{align*} x&=\frac {\left (\left (2 x +2 y\right ) \left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {2}{3}}+y \left (-2^{\frac {2}{3}} \left (x +y-\frac {\sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}}{2}\right ) \left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {1}{3}}+2^{\frac {1}{3}} \left (x^{2}+2 y^{2}-y \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )\right )\right ) y \left (2^{\frac {1}{3}} \left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {2}{3}}+y \left (2^{\frac {2}{3}} x +2 \left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {1}{3}}\right )\right ) c_{3}}{\left (2^{\frac {1}{3}} x y+\left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {2}{3}}\right )^{3}} \\ x&=\frac {2 y \left (\left (4 x +4 y\right ) \left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {2}{3}}+y \left (2^{\frac {2}{3}} \left (1+i \sqrt {3}\right ) \left (x +y-\frac {\sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}}{2}\right ) \left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {1}{3}}+\left (i \sqrt {3}-1\right ) 2^{\frac {1}{3}} \left (x^{2}+2 y^{2}-y \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )\right )\right ) \left (\left (1-i \sqrt {3}\right ) 2^{\frac {1}{3}} \left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {2}{3}}+y \left (-4 \left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {1}{3}}+\left (1+i \sqrt {3}\right ) 2^{\frac {2}{3}} x \right )\right ) c_{3}}{{\left (\left (1-i \sqrt {3}\right ) \left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {2}{3}}+y x \left (1+i \sqrt {3}\right ) 2^{\frac {1}{3}}\right )}^{3}} \\ x&=-\frac {2 \left (\left (-i \sqrt {3}-1\right ) 2^{\frac {1}{3}} \left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {2}{3}}+y \left (4 \left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {1}{3}}+\left (i \sqrt {3}-1\right ) 2^{\frac {2}{3}} x \right )\right ) y \left (\left (-4 x -4 y\right ) \left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {2}{3}}+\left (\left (i \sqrt {3}-1\right ) 2^{\frac {2}{3}} \left (x +y-\frac {\sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}}{2}\right ) \left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {1}{3}}+\left (1+i \sqrt {3}\right ) 2^{\frac {1}{3}} \left (x^{2}+2 y^{2}-y \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )\right ) y\right ) c_{3}}{{\left (\left (-i \sqrt {3}-1\right ) \left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {2}{3}}+\left (i \sqrt {3}-1\right ) y x 2^{\frac {1}{3}}\right )}^{3}} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= 0 \\ \tag{2} y &= \frac {2^{\frac {2}{3}} x}{2} \\ \tag{3} y &= -\frac {i 2^{\frac {2}{3}} \sqrt {3}\, x}{4}-\frac {2^{\frac {2}{3}} x}{4} \\ \tag{4} y &= \frac {i 2^{\frac {2}{3}} \sqrt {3}\, x}{4}-\frac {2^{\frac {2}{3}} x}{4} \\ \tag{5} x &= \frac {\left (\left (2 x +2 y\right ) \left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {2}{3}}+y \left (-2^{\frac {2}{3}} \left (x +y-\frac {\sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}}{2}\right ) \left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {1}{3}}+2^{\frac {1}{3}} \left (x^{2}+2 y^{2}-y \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )\right )\right ) y \left (2^{\frac {1}{3}} \left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {2}{3}}+y \left (2^{\frac {2}{3}} x +2 \left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {1}{3}}\right )\right ) c_{3}}{\left (2^{\frac {1}{3}} x y+\left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {2}{3}}\right )^{3}} \\ \tag{6} x &= \frac {2 y \left (\left (4 x +4 y\right ) \left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {2}{3}}+y \left (2^{\frac {2}{3}} \left (1+i \sqrt {3}\right ) \left (x +y-\frac {\sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}}{2}\right ) \left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {1}{3}}+\left (i \sqrt {3}-1\right ) 2^{\frac {1}{3}} \left (x^{2}+2 y^{2}-y \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )\right )\right ) \left (\left (1-i \sqrt {3}\right ) 2^{\frac {1}{3}} \left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {2}{3}}+y \left (-4 \left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {1}{3}}+\left (1+i \sqrt {3}\right ) 2^{\frac {2}{3}} x \right )\right ) c_{3}}{{\left (\left (1-i \sqrt {3}\right ) \left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {2}{3}}+y x \left (1+i \sqrt {3}\right ) 2^{\frac {1}{3}}\right )}^{3}} \\ \tag{7} x &= -\frac {2 \left (\left (-i \sqrt {3}-1\right ) 2^{\frac {1}{3}} \left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {2}{3}}+y \left (4 \left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {1}{3}}+\left (i \sqrt {3}-1\right ) 2^{\frac {2}{3}} x \right )\right ) y \left (\left (-4 x -4 y\right ) \left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {2}{3}}+\left (\left (i \sqrt {3}-1\right ) 2^{\frac {2}{3}} \left (x +y-\frac {\sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}}{2}\right ) \left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {1}{3}}+\left (1+i \sqrt {3}\right ) 2^{\frac {1}{3}} \left (x^{2}+2 y^{2}-y \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )\right ) y\right ) c_{3}}{{\left (\left (-i \sqrt {3}-1\right ) \left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {2}{3}}+\left (i \sqrt {3}-1\right ) y x 2^{\frac {1}{3}}\right )}^{3}} \\ \end{align*}
Verification of solutions
\[ y = 0 \] Verified OK.
\[ y = \frac {2^{\frac {2}{3}} x}{2} \] Verified OK.
\[ y = -\frac {i 2^{\frac {2}{3}} \sqrt {3}\, x}{4}-\frac {2^{\frac {2}{3}} x}{4} \] Verified OK.
\[ y = \frac {i 2^{\frac {2}{3}} \sqrt {3}\, x}{4}-\frac {2^{\frac {2}{3}} x}{4} \] Verified OK.
\[ x = \frac {\left (\left (2 x +2 y\right ) \left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {2}{3}}+y \left (-2^{\frac {2}{3}} \left (x +y-\frac {\sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}}{2}\right ) \left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {1}{3}}+2^{\frac {1}{3}} \left (x^{2}+2 y^{2}-y \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )\right )\right ) y \left (2^{\frac {1}{3}} \left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {2}{3}}+y \left (2^{\frac {2}{3}} x +2 \left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {1}{3}}\right )\right ) c_{3}}{\left (2^{\frac {1}{3}} x y+\left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {2}{3}}\right )^{3}} \] Warning, solution could not be verified
\[ x = \frac {2 y \left (\left (4 x +4 y\right ) \left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {2}{3}}+y \left (2^{\frac {2}{3}} \left (1+i \sqrt {3}\right ) \left (x +y-\frac {\sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}}{2}\right ) \left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {1}{3}}+\left (i \sqrt {3}-1\right ) 2^{\frac {1}{3}} \left (x^{2}+2 y^{2}-y \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )\right )\right ) \left (\left (1-i \sqrt {3}\right ) 2^{\frac {1}{3}} \left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {2}{3}}+y \left (-4 \left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {1}{3}}+\left (1+i \sqrt {3}\right ) 2^{\frac {2}{3}} x \right )\right ) c_{3}}{{\left (\left (1-i \sqrt {3}\right ) \left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {2}{3}}+y x \left (1+i \sqrt {3}\right ) 2^{\frac {1}{3}}\right )}^{3}} \] Warning, solution could not be verified
\[ x = -\frac {2 \left (\left (-i \sqrt {3}-1\right ) 2^{\frac {1}{3}} \left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {2}{3}}+y \left (4 \left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {1}{3}}+\left (i \sqrt {3}-1\right ) 2^{\frac {2}{3}} x \right )\right ) y \left (\left (-4 x -4 y\right ) \left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {2}{3}}+\left (\left (i \sqrt {3}-1\right ) 2^{\frac {2}{3}} \left (x +y-\frac {\sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}}{2}\right ) \left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {1}{3}}+\left (1+i \sqrt {3}\right ) 2^{\frac {1}{3}} \left (x^{2}+2 y^{2}-y \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )\right ) y\right ) c_{3}}{{\left (\left (-i \sqrt {3}-1\right ) \left (-2 y^{3}+y^{2} \sqrt {\frac {-2 x^{3}+4 y^{3}}{y}}\right )^{\frac {2}{3}}+\left (i \sqrt {3}-1\right ) y x 2^{\frac {1}{3}}\right )}^{3}} \] Warning, solution could not be verified
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 y {y^{\prime }}^{3}-3 x y^{\prime }+2 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 {{\left (\left (2 \sqrt {2}\, \sqrt {\frac {2 y^{3}-x^{3}}{y}}-4 y\right ) y^{2}\right )}^{\frac {1}{3}}}{2 y}+\frac {x}{{\left (\left (2 \sqrt {2}\, \sqrt {\frac {2 y^{3}-x^{3}}{y}}-4 y\right ) y^{2}\right )}^{\frac {1}{3}}}, y^{\prime }=-\frac {{\left (\left (2 \sqrt {2}\, \sqrt {\frac {2 y^{3}-x^{3}}{y}}-4 y\right ) y^{2}\right )}^{\frac {1}{3}}}{4 y}-\frac {x}{2 {\left (\left (2 \sqrt {2}\, \sqrt {\frac {2 y^{3}-x^{3}}{y}}-4 y\right ) y^{2}\right )}^{\frac {1}{3}}}-\frac {\mathrm {I} \sqrt {3}\, \left (\frac {{\left (\left (2 \sqrt {2}\, \sqrt {\frac {2 y^{3}-x^{3}}{y}}-4 y\right ) y^{2}\right )}^{\frac {1}{3}}}{2 y}-\frac {x}{{\left (\left (2 \sqrt {2}\, \sqrt {\frac {2 y^{3}-x^{3}}{y}}-4 y\right ) y^{2}\right )}^{\frac {1}{3}}}\right )}{2}, y^{\prime }=-\frac {{\left (\left (2 \sqrt {2}\, \sqrt {\frac {2 y^{3}-x^{3}}{y}}-4 y\right ) y^{2}\right )}^{\frac {1}{3}}}{4 y}-\frac {x}{2 {\left (\left (2 \sqrt {2}\, \sqrt {\frac {2 y^{3}-x^{3}}{y}}-4 y\right ) y^{2}\right )}^{\frac {1}{3}}}+\frac {\mathrm {I} \sqrt {3}\, \left (\frac {{\left (\left (2 \sqrt {2}\, \sqrt {\frac {2 y^{3}-x^{3}}{y}}-4 y\right ) y^{2}\right )}^{\frac {1}{3}}}{2 y}-\frac {x}{{\left (\left (2 \sqrt {2}\, \sqrt {\frac {2 y^{3}-x^{3}}{y}}-4 y\right ) y^{2}\right )}^{\frac {1}{3}}}\right )}{2}\right ] \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {{\left (\left (2 \sqrt {2}\, \sqrt {\frac {2 y^{3}-x^{3}}{y}}-4 y\right ) y^{2}\right )}^{\frac {1}{3}}}{2 y}+\frac {x}{{\left (\left (2 \sqrt {2}\, \sqrt {\frac {2 y^{3}-x^{3}}{y}}-4 y\right ) y^{2}\right )}^{\frac {1}{3}}} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {{\left (\left (2 \sqrt {2}\, \sqrt {\frac {2 y^{3}-x^{3}}{y}}-4 y\right ) y^{2}\right )}^{\frac {1}{3}}}{4 y}-\frac {x}{2 {\left (\left (2 \sqrt {2}\, \sqrt {\frac {2 y^{3}-x^{3}}{y}}-4 y\right ) y^{2}\right )}^{\frac {1}{3}}}-\frac {\mathrm {I} \sqrt {3}\, \left (\frac {{\left (\left (2 \sqrt {2}\, \sqrt {\frac {2 y^{3}-x^{3}}{y}}-4 y\right ) y^{2}\right )}^{\frac {1}{3}}}{2 y}-\frac {x}{{\left (\left (2 \sqrt {2}\, \sqrt {\frac {2 y^{3}-x^{3}}{y}}-4 y\right ) y^{2}\right )}^{\frac {1}{3}}}\right )}{2} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {{\left (\left (2 \sqrt {2}\, \sqrt {\frac {2 y^{3}-x^{3}}{y}}-4 y\right ) y^{2}\right )}^{\frac {1}{3}}}{4 y}-\frac {x}{2 {\left (\left (2 \sqrt {2}\, \sqrt {\frac {2 y^{3}-x^{3}}{y}}-4 y\right ) y^{2}\right )}^{\frac {1}{3}}}+\frac {\mathrm {I} \sqrt {3}\, \left (\frac {{\left (\left (2 \sqrt {2}\, \sqrt {\frac {2 y^{3}-x^{3}}{y}}-4 y\right ) y^{2}\right )}^{\frac {1}{3}}}{2 y}-\frac {x}{{\left (\left (2 \sqrt {2}\, \sqrt {\frac {2 y^{3}-x^{3}}{y}}-4 y\right ) y^{2}\right )}^{\frac {1}{3}}}\right )}{2} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\mathit {workingODE} , \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: 0.078 (sec). Leaf size: 734
dsolve(2*y(x)*diff(y(x),x)^3-3*x*diff(y(x),x)+2*y(x) = 0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \frac {2^{\frac {2}{3}} x}{2} \\ y \left (x \right ) &= -\frac {2^{\frac {2}{3}} \left (1+i \sqrt {3}\right ) x}{4} \\ y \left (x \right ) &= \frac {2^{\frac {2}{3}} \left (-1+i \sqrt {3}\right ) x}{4} \\ y \left (x \right ) &= 0 \\ y \left (x \right ) &= \operatorname {RootOf}\left (-\ln \left (x \right )+\int _{}^{\textit {\_Z}}-\frac {2 {\left (\left (\sqrt {2}\, \sqrt {\frac {1}{\textit {\_a} \left (2 \textit {\_a}^{3}-1\right )}}\, \textit {\_a}^{2}+1\right ) \left (2 \textit {\_a}^{3}-1\right )^{2}\right )}^{\frac {1}{3}} \textit {\_a}^{3}+2 \textit {\_a}^{3}-{\left (\left (\sqrt {2}\, \sqrt {\frac {1}{\textit {\_a} \left (2 \textit {\_a}^{3}-1\right )}}\, \textit {\_a}^{2}+1\right ) \left (2 \textit {\_a}^{3}-1\right )^{2}\right )}^{\frac {2}{3}}-{\left (\left (\sqrt {2}\, \sqrt {\frac {1}{\textit {\_a} \left (2 \textit {\_a}^{3}-1\right )}}\, \textit {\_a}^{2}+1\right ) \left (2 \textit {\_a}^{3}-1\right )^{2}\right )}^{\frac {1}{3}}-1}{\textit {\_a} \left (2 \textit {\_a}^{3}-1\right ) {\left (\left (\sqrt {2}\, \sqrt {\frac {1}{\textit {\_a} \left (2 \textit {\_a}^{3}-1\right )}}\, \textit {\_a}^{2}+1\right ) \left (2 \textit {\_a}^{3}-1\right )^{2}\right )}^{\frac {1}{3}}}d \textit {\_a} +c_{1} \right ) x \\ y \left (x \right ) &= \operatorname {RootOf}\left (-2 \ln \left (x \right )+\int _{}^{\textit {\_Z}}\frac {2 i \sqrt {3}\, \textit {\_a}^{3}+i \sqrt {3}\, {\left (\left (\sqrt {2}\, \sqrt {\frac {1}{\textit {\_a} \left (2 \textit {\_a}^{3}-1\right )}}\, \textit {\_a}^{2}+1\right ) \left (2 \textit {\_a}^{3}-1\right )^{2}\right )}^{\frac {2}{3}}-4 {\left (\left (\sqrt {2}\, \sqrt {\frac {1}{\textit {\_a} \left (2 \textit {\_a}^{3}-1\right )}}\, \textit {\_a}^{2}+1\right ) \left (2 \textit {\_a}^{3}-1\right )^{2}\right )}^{\frac {1}{3}} \textit {\_a}^{3}+2 \textit {\_a}^{3}-i \sqrt {3}-{\left (\left (\sqrt {2}\, \sqrt {\frac {1}{\textit {\_a} \left (2 \textit {\_a}^{3}-1\right )}}\, \textit {\_a}^{2}+1\right ) \left (2 \textit {\_a}^{3}-1\right )^{2}\right )}^{\frac {2}{3}}+2 {\left (\left (\sqrt {2}\, \sqrt {\frac {1}{\textit {\_a} \left (2 \textit {\_a}^{3}-1\right )}}\, \textit {\_a}^{2}+1\right ) \left (2 \textit {\_a}^{3}-1\right )^{2}\right )}^{\frac {1}{3}}-1}{{\left (\left (\sqrt {2}\, \sqrt {\frac {1}{\textit {\_a} \left (2 \textit {\_a}^{3}-1\right )}}\, \textit {\_a}^{2}+1\right ) \left (2 \textit {\_a}^{3}-1\right )^{2}\right )}^{\frac {1}{3}} \textit {\_a} \left (2 \textit {\_a}^{3}-1\right )}d \textit {\_a} +2 c_{1} \right ) x \\ y \left (x \right ) &= \operatorname {RootOf}\left (-2 \ln \left (x \right )-\left (\int _{}^{\textit {\_Z}}\frac {2 i \sqrt {3}\, \textit {\_a}^{3}+i \sqrt {3}\, {\left (\left (\sqrt {2}\, \sqrt {\frac {1}{\textit {\_a} \left (2 \textit {\_a}^{3}-1\right )}}\, \textit {\_a}^{2}+1\right ) \left (2 \textit {\_a}^{3}-1\right )^{2}\right )}^{\frac {2}{3}}+4 {\left (\left (\sqrt {2}\, \sqrt {\frac {1}{\textit {\_a} \left (2 \textit {\_a}^{3}-1\right )}}\, \textit {\_a}^{2}+1\right ) \left (2 \textit {\_a}^{3}-1\right )^{2}\right )}^{\frac {1}{3}} \textit {\_a}^{3}-2 \textit {\_a}^{3}-i \sqrt {3}+{\left (\left (\sqrt {2}\, \sqrt {\frac {1}{\textit {\_a} \left (2 \textit {\_a}^{3}-1\right )}}\, \textit {\_a}^{2}+1\right ) \left (2 \textit {\_a}^{3}-1\right )^{2}\right )}^{\frac {2}{3}}-2 {\left (\left (\sqrt {2}\, \sqrt {\frac {1}{\textit {\_a} \left (2 \textit {\_a}^{3}-1\right )}}\, \textit {\_a}^{2}+1\right ) \left (2 \textit {\_a}^{3}-1\right )^{2}\right )}^{\frac {1}{3}}+1}{\textit {\_a} \left (2 \textit {\_a}^{3}-1\right ) {\left (\left (\sqrt {2}\, \sqrt {\frac {1}{\textit {\_a} \left (2 \textit {\_a}^{3}-1\right )}}\, \textit {\_a}^{2}+1\right ) \left (2 \textit {\_a}^{3}-1\right )^{2}\right )}^{\frac {1}{3}}}d \textit {\_a} \right )+2 c_{1} \right ) x \\ \end{align*}
✓ Solution by Mathematica
Time used: 172.826 (sec). Leaf size: 10331
DSolve[2 y[x] (y'[x])^3 -3 x y'[x]+2 y[x]==0,y[x],x,IncludeSingularSolutions -> True]
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