Internal problem ID [10100]
Internal file name [OUTPUT/9047_Monday_June_06_2022_06_16_56_AM_76475572/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 6, non-linear second order
Problem number: 1779 (book 6.188).
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_ode_missing_x"
Maple gives the following as the ode type
[[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1]]
\[ \boxed {y^{2} y^{\prime \prime }=a} \]
This is missing independent variable second order ode. Solved by reduction of order by using substitution which makes the dependent variable \(y\) an independent variable. Using \begin {align*} y' &= p(y) \end {align*}
Then \begin {align*} y'' &= \frac {dp}{dx}\\ &= \frac {dy}{dx} \frac {dp}{dy}\\ &= p \frac {dp}{dy} \end {align*}
Hence the ode becomes \begin {align*} y^{2} p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right ) = a \end {align*}
Which is now solved as first order ode for \(p(y)\). In canonical form the ODE is \begin {align*} p' &= F(y,p)\\ &= f( y) g(p)\\ &= \frac {a}{y^{2} p} \end {align*}
Where \(f(y)=\frac {a}{y^{2}}\) and \(g(p)=\frac {1}{p}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{p}} \,dp &= \frac {a}{y^{2}} \,d y \\ \int { \frac {1}{\frac {1}{p}} \,dp} &= \int {\frac {a}{y^{2}} \,d y} \\ \frac {p^{2}}{2}&=-\frac {a}{y}+c_{1} \\ \end{align*} The solution is \[ \frac {p \left (y \right )^{2}}{2}+\frac {a}{y}-c_{1} = 0 \] For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} \frac {{y^{\prime }}^{2}}{2}+\frac {a}{y}-c_{1} = 0 \end {align*}
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 {\sqrt {2}\, \sqrt {y \left (c_{1} y-a \right )}}{y} \tag {1} \\ y^{\prime }&=-\frac {\sqrt {2}\, \sqrt {y \left (c_{1} y-a \right )}}{y} \tag {2} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin{align*} \int \frac {y \sqrt {2}}{2 \sqrt {y \left (c_{1} y -a \right )}}d y &= \int d x \\ \frac {\sqrt {2}\, \left (\operatorname {arctanh}\left (\frac {\sqrt {-y \left (-c_{1} y+a \right )}}{y \sqrt {c_{1}}}\right ) a +\sqrt {-y \left (-c_{1} y+a \right )}\, \sqrt {c_{1}}\right )}{2 c_{1}^{\frac {3}{2}}}&=x +c_{2} \\ \end{align*} Solving equation (2)
Integrating both sides gives \begin{align*} \int -\frac {y \sqrt {2}}{2 \sqrt {y \left (c_{1} y -a \right )}}d y &= \int d x \\ -\frac {\sqrt {2}\, \left (\operatorname {arctanh}\left (\frac {\sqrt {-y \left (-c_{1} y+a \right )}}{y \sqrt {c_{1}}}\right ) a +\sqrt {-y \left (-c_{1} y+a \right )}\, \sqrt {c_{1}}\right )}{2 c_{1}^{\frac {3}{2}}}&=x +c_{3} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} \frac {\sqrt {2}\, \left (\operatorname {arctanh}\left (\frac {\sqrt {-y \left (-c_{1} y+a \right )}}{y \sqrt {c_{1}}}\right ) a +\sqrt {-y \left (-c_{1} y+a \right )}\, \sqrt {c_{1}}\right )}{2 c_{1}^{\frac {3}{2}}} &= x +c_{2} \\ \tag{2} -\frac {\sqrt {2}\, \left (\operatorname {arctanh}\left (\frac {\sqrt {-y \left (-c_{1} y+a \right )}}{y \sqrt {c_{1}}}\right ) a +\sqrt {-y \left (-c_{1} y+a \right )}\, \sqrt {c_{1}}\right )}{2 c_{1}^{\frac {3}{2}}} &= x +c_{3} \\ \end{align*}
Verification of solutions
\[ \frac {\sqrt {2}\, \left (\operatorname {arctanh}\left (\frac {\sqrt {-y \left (-c_{1} y+a \right )}}{y \sqrt {c_{1}}}\right ) a +\sqrt {-y \left (-c_{1} y+a \right )}\, \sqrt {c_{1}}\right )}{2 c_{1}^{\frac {3}{2}}} = x +c_{2} \] Verified OK.
\[ -\frac {\sqrt {2}\, \left (\operatorname {arctanh}\left (\frac {\sqrt {-y \left (-c_{1} y+a \right )}}{y \sqrt {c_{1}}}\right ) a +\sqrt {-y \left (-c_{1} y+a \right )}\, \sqrt {c_{1}}\right )}{2 c_{1}^{\frac {3}{2}}} = x +c_{3} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{2} \left (\frac {d}{d x}y^{\prime }\right )=a \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Define new dependent variable}\hspace {3pt} u \\ {} & {} & u \left (x \right )=y^{\prime } \\ \bullet & {} & \textrm {Compute}\hspace {3pt} \frac {d}{d x}y^{\prime } \\ {} & {} & u^{\prime }\left (x \right )=y^{\prime \prime } \\ \bullet & {} & \textrm {Use chain rule on the lhs}\hspace {3pt} \\ {} & {} & y^{\prime } \left (\frac {d}{d y}u \left (y \right )\right )=y^{\prime \prime } \\ \bullet & {} & \textrm {Substitute in the definition of}\hspace {3pt} u \\ {} & {} & u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )=y^{\prime \prime } \\ \bullet & {} & \textrm {Make substitutions}\hspace {3pt} y^{\prime }=u \left (y \right ),\frac {d}{d x}y^{\prime }=u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & y^{2} u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )=a \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )=\frac {a}{y^{2} u \left (y \right )} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )=\frac {a}{y^{2}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} y \\ {} & {} & \int u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )d y =\int \frac {a}{y^{2}}d y +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {u \left (y \right )^{2}}{2}=-\frac {a}{y}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & \left \{u \left (y \right )=\frac {\sqrt {2}\, \sqrt {y \left (c_{1} y -a \right )}}{y}, u \left (y \right )=-\frac {\sqrt {2}\, \sqrt {y \left (c_{1} y -a \right )}}{y}\right \} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=\frac {\sqrt {2}\, \sqrt {y \left (c_{1} y -a \right )}}{y} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=\frac {\sqrt {2}\, \sqrt {y \left (c_{1} y-a \right )}}{y} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\sqrt {2}\, \sqrt {y \left (c_{1} y-a \right )}}{y} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime } y}{\sqrt {y \left (c_{1} y-a \right )}}=\sqrt {2} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime } y}{\sqrt {y \left (c_{1} y-a \right )}}d x =\int \sqrt {2}d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\sqrt {-y a +c_{1} y^{2}}}{c_{1}}+\frac {a \ln \left (\frac {-\frac {a}{2}+c_{1} y}{\sqrt {c_{1}}}+\sqrt {-y a +c_{1} y^{2}}\right )}{2 c_{1}^{\frac {3}{2}}}=\sqrt {2}\, x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{\frac {4 c_{1} \left ({\mathrm e}^{\mathit {RootOf}\left (-64 \sqrt {2}\, c_{1}^{\frac {5}{2}} \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} \textit {\_Z} a x -64 c_{1}^{\frac {5}{2}} \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{2} \textit {\_Z} a +128 \sqrt {2}\, c_{1}^{4} \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{2} x +64 c_{1}^{4} \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{2}^{2}+128 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{1}^{4} x^{2}+16 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{1} \textit {\_Z}^{2} a^{2}-16 c_{1}^{2} \left ({\mathrm e}^{\textit {\_Z}}\right )^{4}+8 c_{1} a^{2} \left ({\mathrm e}^{\textit {\_Z}}\right )^{2}-a^{4}\right )}\right )^{2}+4 \sqrt {c_{1}}\, a \,{\mathrm e}^{\mathit {RootOf}\left (-64 \sqrt {2}\, c_{1}^{\frac {5}{2}} \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} \textit {\_Z} a x -64 c_{1}^{\frac {5}{2}} \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{2} \textit {\_Z} a +128 \sqrt {2}\, c_{1}^{4} \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{2} x +64 c_{1}^{4} \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{2}^{2}+128 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{1}^{4} x^{2}+16 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{1} \textit {\_Z}^{2} a^{2}-16 c_{1}^{2} \left ({\mathrm e}^{\textit {\_Z}}\right )^{4}+8 c_{1} a^{2} \left ({\mathrm e}^{\textit {\_Z}}\right )^{2}-a^{4}\right )}+a^{2}}{8 \,{\mathrm e}^{\mathit {RootOf}\left (-64 \sqrt {2}\, c_{1}^{\frac {5}{2}} \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} \textit {\_Z} a x -64 c_{1}^{\frac {5}{2}} \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{2} \textit {\_Z} a +128 \sqrt {2}\, c_{1}^{4} \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{2} x +64 c_{1}^{4} \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{2}^{2}+128 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{1}^{4} x^{2}+16 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{1} \textit {\_Z}^{2} a^{2}-16 c_{1}^{2} \left ({\mathrm e}^{\textit {\_Z}}\right )^{4}+8 c_{1} a^{2} \left ({\mathrm e}^{\textit {\_Z}}\right )^{2}-a^{4}\right )} c_{1}^{\frac {3}{2}}}\right \} \\ \bullet & {} & \textrm {Solve 2nd ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=-\frac {\sqrt {2}\, \sqrt {y \left (c_{1} y -a \right )}}{y} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=-\frac {\sqrt {2}\, \sqrt {y \left (c_{1} y-a \right )}}{y} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {\sqrt {2}\, \sqrt {y \left (c_{1} y-a \right )}}{y} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime } y}{\sqrt {y \left (c_{1} y-a \right )}}=-\sqrt {2} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime } y}{\sqrt {y \left (c_{1} y-a \right )}}d x =\int -\sqrt {2}d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\sqrt {-y a +c_{1} y^{2}}}{c_{1}}+\frac {a \ln \left (\frac {-\frac {a}{2}+c_{1} y}{\sqrt {c_{1}}}+\sqrt {-y a +c_{1} y^{2}}\right )}{2 c_{1}^{\frac {3}{2}}}=-\sqrt {2}\, x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{\frac {4 c_{1} \left ({\mathrm e}^{\mathit {RootOf}\left (-64 \sqrt {2}\, c_{1}^{\frac {5}{2}} \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} \textit {\_Z} a x +64 c_{1}^{\frac {5}{2}} \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{2} \textit {\_Z} a +128 \sqrt {2}\, c_{1}^{4} \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{2} x -64 c_{1}^{4} \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{2}^{2}-128 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{1}^{4} x^{2}-16 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{1} \textit {\_Z}^{2} a^{2}+16 c_{1}^{2} \left ({\mathrm e}^{\textit {\_Z}}\right )^{4}-8 c_{1} a^{2} \left ({\mathrm e}^{\textit {\_Z}}\right )^{2}+a^{4}\right )}\right )^{2}+4 \sqrt {c_{1}}\, a \,{\mathrm e}^{\mathit {RootOf}\left (-64 \sqrt {2}\, c_{1}^{\frac {5}{2}} \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} \textit {\_Z} a x +64 c_{1}^{\frac {5}{2}} \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{2} \textit {\_Z} a +128 \sqrt {2}\, c_{1}^{4} \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{2} x -64 c_{1}^{4} \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{2}^{2}-128 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{1}^{4} x^{2}-16 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{1} \textit {\_Z}^{2} a^{2}+16 c_{1}^{2} \left ({\mathrm e}^{\textit {\_Z}}\right )^{4}-8 c_{1} a^{2} \left ({\mathrm e}^{\textit {\_Z}}\right )^{2}+a^{4}\right )}+a^{2}}{8 \,{\mathrm e}^{\mathit {RootOf}\left (-64 \sqrt {2}\, c_{1}^{\frac {5}{2}} \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} \textit {\_Z} a x +64 c_{1}^{\frac {5}{2}} \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{2} \textit {\_Z} a +128 \sqrt {2}\, c_{1}^{4} \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{2} x -64 c_{1}^{4} \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{2}^{2}-128 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{1}^{4} x^{2}-16 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{1} \textit {\_Z}^{2} a^{2}+16 c_{1}^{2} \left ({\mathrm e}^{\textit {\_Z}}\right )^{4}-8 c_{1} a^{2} \left ({\mathrm e}^{\textit {\_Z}}\right )^{2}+a^{4}\right )} c_{1}^{\frac {3}{2}}}\right \} \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying 2nd order Liouville trying 2nd order WeierstrassP trying 2nd order JacobiSN differential order: 2; trying a linearization to 3rd order trying 2nd order ODE linearizable_by_differentiation trying 2nd order, 2 integrating factors of the form mu(x,y) trying differential order: 2; missing variables `, `-> Computing symmetries using: way = 3 -> Calling odsolve with the ODE`, (diff(_b(_a), _a))*_b(_a)-a/_a^2 = 0, _b(_a), HINT = [[_a, -(1/2)*_b]]` *** Sublevel 2 *** symmetry methods on request `, `1st order, trying reduction of order with given symmetries:`[_a, -1/2*_b]
✓ Solution by Maple
Time used: 0.047 (sec). Leaf size: 369
dsolve(y(x)^2*diff(diff(y(x),x),x)-a=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \frac {c_{1} \left (c_{1}^{2} a^{2}+2 a c_{1} {\mathrm e}^{\operatorname {RootOf}\left (\operatorname {csgn}\left (\frac {1}{c_{1}}\right ) c_{1}^{4} a^{2}-2 \textit {\_Z} \,c_{1}^{3} a \,{\mathrm e}^{\textit {\_Z}}-\operatorname {csgn}\left (\frac {1}{c_{1}}\right ) {\mathrm e}^{2 \textit {\_Z}} c_{1}^{2}-2 \,\operatorname {csgn}\left (\frac {1}{c_{1}}\right ) {\mathrm e}^{\textit {\_Z}} c_{2} -2 \,\operatorname {csgn}\left (\frac {1}{c_{1}}\right ) {\mathrm e}^{\textit {\_Z}} x \right )}+{\mathrm e}^{2 \operatorname {RootOf}\left (\operatorname {csgn}\left (\frac {1}{c_{1}}\right ) c_{1}^{4} a^{2}-2 \textit {\_Z} \,c_{1}^{3} a \,{\mathrm e}^{\textit {\_Z}}-\operatorname {csgn}\left (\frac {1}{c_{1}}\right ) {\mathrm e}^{2 \textit {\_Z}} c_{1}^{2}-2 \,\operatorname {csgn}\left (\frac {1}{c_{1}}\right ) {\mathrm e}^{\textit {\_Z}} c_{2} -2 \,\operatorname {csgn}\left (\frac {1}{c_{1}}\right ) {\mathrm e}^{\textit {\_Z}} x \right )}\right ) {\mathrm e}^{-\operatorname {RootOf}\left (\operatorname {csgn}\left (\frac {1}{c_{1}}\right ) c_{1}^{4} a^{2}-2 \textit {\_Z} \,c_{1}^{3} a \,{\mathrm e}^{\textit {\_Z}}-\operatorname {csgn}\left (\frac {1}{c_{1}}\right ) {\mathrm e}^{2 \textit {\_Z}} c_{1}^{2}-2 \,\operatorname {csgn}\left (\frac {1}{c_{1}}\right ) {\mathrm e}^{\textit {\_Z}} c_{2} -2 \,\operatorname {csgn}\left (\frac {1}{c_{1}}\right ) {\mathrm e}^{\textit {\_Z}} x \right )}}{2} \\ y \left (x \right ) &= \frac {c_{1} \left (c_{1}^{2} a^{2}+2 a c_{1} {\mathrm e}^{\operatorname {RootOf}\left (\operatorname {csgn}\left (\frac {1}{c_{1}}\right ) c_{1}^{4} a^{2}-2 \textit {\_Z} \,c_{1}^{3} a \,{\mathrm e}^{\textit {\_Z}}-\operatorname {csgn}\left (\frac {1}{c_{1}}\right ) {\mathrm e}^{2 \textit {\_Z}} c_{1}^{2}+2 \,\operatorname {csgn}\left (\frac {1}{c_{1}}\right ) {\mathrm e}^{\textit {\_Z}} c_{2} +2 \,\operatorname {csgn}\left (\frac {1}{c_{1}}\right ) {\mathrm e}^{\textit {\_Z}} x \right )}+{\mathrm e}^{2 \operatorname {RootOf}\left (\operatorname {csgn}\left (\frac {1}{c_{1}}\right ) c_{1}^{4} a^{2}-2 \textit {\_Z} \,c_{1}^{3} a \,{\mathrm e}^{\textit {\_Z}}-\operatorname {csgn}\left (\frac {1}{c_{1}}\right ) {\mathrm e}^{2 \textit {\_Z}} c_{1}^{2}+2 \,\operatorname {csgn}\left (\frac {1}{c_{1}}\right ) {\mathrm e}^{\textit {\_Z}} c_{2} +2 \,\operatorname {csgn}\left (\frac {1}{c_{1}}\right ) {\mathrm e}^{\textit {\_Z}} x \right )}\right ) {\mathrm e}^{-\operatorname {RootOf}\left (\operatorname {csgn}\left (\frac {1}{c_{1}}\right ) c_{1}^{4} a^{2}-2 \textit {\_Z} \,c_{1}^{3} a \,{\mathrm e}^{\textit {\_Z}}-\operatorname {csgn}\left (\frac {1}{c_{1}}\right ) {\mathrm e}^{2 \textit {\_Z}} c_{1}^{2}+2 \,\operatorname {csgn}\left (\frac {1}{c_{1}}\right ) {\mathrm e}^{\textit {\_Z}} c_{2} +2 \,\operatorname {csgn}\left (\frac {1}{c_{1}}\right ) {\mathrm e}^{\textit {\_Z}} x \right )}}{2} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.267 (sec). Leaf size: 65
DSolve[-a + y[x]^2*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [\left (\frac {2 a \text {arctanh}\left (\frac {\sqrt {-\frac {2 a}{y(x)}+c_1}}{\sqrt {c_1}}\right )}{c_1{}^{3/2}}+\frac {y(x) \sqrt {-\frac {2 a}{y(x)}+c_1}}{c_1}\right ){}^2=(x+c_2){}^2,y(x)\right ] \]