Internal problem ID [4657]
Internal file name [OUTPUT/4150_Sunday_June_05_2022_12_29_25_PM_80642096/index.tex]
Book: Ordinary Differential Equations, By Tenenbaum and Pollard. Dover, NY 1963
Section: Chapter 8. Special second order equations. Lesson 35. Independent variable x
absent
Problem number: Exercise 35.7, page 504.
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]]
\[ \boxed {r^{\prime \prime }+\frac {k}{r^{2}}=0} \]
This is missing independent variable second order ode. Solved by reduction of order by using substitution which makes the dependent variable \(r\) an independent variable. Using \begin {align*} r' &= p(r) \end {align*}
Then \begin {align*} r'' &= \frac {dp}{dt}\\ &= \frac {dr}{dt} \frac {dp}{dr}\\ &= p \frac {dp}{dr} \end {align*}
Hence the ode becomes \begin {align*} p \left (r \right ) \left (\frac {d}{d r}p \left (r \right )\right ) r^{2} = -k \end {align*}
Which is now solved as first order ode for \(p(r)\). In canonical form the ODE is \begin {align*} p' &= F(r,p)\\ &= f( r) g(p)\\ &= -\frac {k}{p \,r^{2}} \end {align*}
Where \(f(r)=-\frac {k}{r^{2}}\) and \(g(p)=\frac {1}{p}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{p}} \,dp &= -\frac {k}{r^{2}} \,d r \\ \int { \frac {1}{\frac {1}{p}} \,dp} &= \int {-\frac {k}{r^{2}} \,d r} \\ \frac {p^{2}}{2}&=\frac {k}{r}+c_{1} \\ \end{align*} The solution is \[ \frac {p \left (r \right )^{2}}{2}-\frac {k}{r}-c_{1} = 0 \] For solution (1) found earlier, since \(p=r^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} \frac {{r^{\prime }}^{2}}{2}-\frac {k}{r}-c_{1} = 0 \end {align*}
Solving the given ode for \(r^{\prime }\) results in \(2\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} r^{\prime }&=\frac {\sqrt {2}\, \sqrt {r \left (c_{1} r+k \right )}}{r} \tag {1} \\ r^{\prime }&=-\frac {\sqrt {2}\, \sqrt {r \left (c_{1} r+k \right )}}{r} \tag {2} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin{align*} \int \frac {r \sqrt {2}}{2 \sqrt {r \left (c_{1} r +k \right )}}d r &= \int d t \\ -\frac {\sqrt {2}\, \left (\operatorname {arctanh}\left (\frac {\sqrt {r \left (c_{1} r+k \right )}}{r \sqrt {c_{1}}}\right ) k -\sqrt {r \left (c_{1} r+k \right )}\, \sqrt {c_{1}}\right )}{2 c_{1}^{\frac {3}{2}}}&=t +c_{2} \\ \end{align*} Solving equation (2)
Integrating both sides gives \begin{align*} \int -\frac {r \sqrt {2}}{2 \sqrt {r \left (c_{1} r +k \right )}}d r &= \int d t \\ \frac {\sqrt {2}\, \left (\operatorname {arctanh}\left (\frac {\sqrt {r \left (c_{1} r+k \right )}}{r \sqrt {c_{1}}}\right ) k -\sqrt {r \left (c_{1} r+k \right )}\, \sqrt {c_{1}}\right )}{2 c_{1}^{\frac {3}{2}}}&=t +c_{3} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} -\frac {\sqrt {2}\, \left (\operatorname {arctanh}\left (\frac {\sqrt {r \left (c_{1} r+k \right )}}{r \sqrt {c_{1}}}\right ) k -\sqrt {r \left (c_{1} r+k \right )}\, \sqrt {c_{1}}\right )}{2 c_{1}^{\frac {3}{2}}} &= t +c_{2} \\ \tag{2} \frac {\sqrt {2}\, \left (\operatorname {arctanh}\left (\frac {\sqrt {r \left (c_{1} r+k \right )}}{r \sqrt {c_{1}}}\right ) k -\sqrt {r \left (c_{1} r+k \right )}\, \sqrt {c_{1}}\right )}{2 c_{1}^{\frac {3}{2}}} &= t +c_{3} \\ \end{align*}
Verification of solutions
\[ -\frac {\sqrt {2}\, \left (\operatorname {arctanh}\left (\frac {\sqrt {r \left (c_{1} r+k \right )}}{r \sqrt {c_{1}}}\right ) k -\sqrt {r \left (c_{1} r+k \right )}\, \sqrt {c_{1}}\right )}{2 c_{1}^{\frac {3}{2}}} = t +c_{2} \] Verified OK.
\[ \frac {\sqrt {2}\, \left (\operatorname {arctanh}\left (\frac {\sqrt {r \left (c_{1} r+k \right )}}{r \sqrt {c_{1}}}\right ) k -\sqrt {r \left (c_{1} r+k \right )}\, \sqrt {c_{1}}\right )}{2 c_{1}^{\frac {3}{2}}} = t +c_{3} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d t}r^{\prime }\right ) r^{2}=-k \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d t}r^{\prime } \\ \bullet & {} & \textrm {Define new dependent variable}\hspace {3pt} u \\ {} & {} & u \left (t \right )=r^{\prime } \\ \bullet & {} & \textrm {Compute}\hspace {3pt} \frac {d}{d t}r^{\prime } \\ {} & {} & u^{\prime }\left (t \right )=r^{\prime \prime } \\ \bullet & {} & \textrm {Use chain rule on the lhs}\hspace {3pt} \\ {} & {} & r^{\prime } \left (\frac {d}{d r}u \left (r \right )\right )=r^{\prime \prime } \\ \bullet & {} & \textrm {Substitute in the definition of}\hspace {3pt} u \\ {} & {} & u \left (r \right ) \left (\frac {d}{d r}u \left (r \right )\right )=r^{\prime \prime } \\ \bullet & {} & \textrm {Make substitutions}\hspace {3pt} r^{\prime }=u \left (r \right ),\frac {d}{d t}r^{\prime }=u \left (r \right ) \left (\frac {d}{d r}u \left (r \right )\right )\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & u \left (r \right ) \left (\frac {d}{d r}u \left (r \right )\right ) r^{2}=-k \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d r}u \left (r \right )=-\frac {k}{u \left (r \right ) r^{2}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & u \left (r \right ) \left (\frac {d}{d r}u \left (r \right )\right )=-\frac {k}{r^{2}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} r \\ {} & {} & \int u \left (r \right ) \left (\frac {d}{d r}u \left (r \right )\right )d r =\int -\frac {k}{r^{2}}d r +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {u \left (r \right )^{2}}{2}=\frac {k}{r}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (r \right ) \\ {} & {} & \left \{u \left (r \right )=\frac {\sqrt {2}\, \sqrt {r \left (c_{1} r +k \right )}}{r}, u \left (r \right )=-\frac {\sqrt {2}\, \sqrt {r \left (c_{1} r +k \right )}}{r}\right \} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (r \right ) \\ {} & {} & u \left (r \right )=\frac {\sqrt {2}\, \sqrt {r \left (c_{1} r +k \right )}}{r} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (r \right )=r^{\prime },r =r \\ {} & {} & r^{\prime }=\frac {\sqrt {2}\, \sqrt {r \left (c_{1} r+k \right )}}{r} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & r^{\prime }=\frac {\sqrt {2}\, \sqrt {r \left (c_{1} r+k \right )}}{r} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {r^{\prime } r}{\sqrt {r \left (c_{1} r+k \right )}}=\sqrt {2} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {r^{\prime } r}{\sqrt {r \left (c_{1} r+k \right )}}d t =\int \sqrt {2}d t +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\sqrt {r^{2} c_{1} +r k}}{c_{1}}-\frac {k \ln \left (\frac {\frac {k}{2}+c_{1} r}{\sqrt {c_{1}}}+\sqrt {r^{2} c_{1} +r k}\right )}{2 c_{1}^{\frac {3}{2}}}=\sqrt {2}\, t +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} r \\ {} & {} & \left \{-\frac {4 \sqrt {c_{1}}\, k \,{\mathrm e}^{\mathit {RootOf}\left (64 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} \sqrt {2}\, c_{1}^{\frac {5}{2}} \textit {\_Z} k t +64 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{1}^{\frac {5}{2}} c_{2} \textit {\_Z} k +128 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} \sqrt {2}\, c_{1}^{4} c_{2} t +64 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{1}^{4} c_{2}^{2}+128 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{1}^{4} t^{2}+16 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{1} \textit {\_Z}^{2} k^{2}-16 c_{1}^{2} \left ({\mathrm e}^{\textit {\_Z}}\right )^{4}+8 c_{1} k^{2} \left ({\mathrm e}^{\textit {\_Z}}\right )^{2}-k^{4}\right )}-4 c_{1} \left ({\mathrm e}^{\mathit {RootOf}\left (64 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} \sqrt {2}\, c_{1}^{\frac {5}{2}} \textit {\_Z} k t +64 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{1}^{\frac {5}{2}} c_{2} \textit {\_Z} k +128 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} \sqrt {2}\, c_{1}^{4} c_{2} t +64 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{1}^{4} c_{2}^{2}+128 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{1}^{4} t^{2}+16 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{1} \textit {\_Z}^{2} k^{2}-16 c_{1}^{2} \left ({\mathrm e}^{\textit {\_Z}}\right )^{4}+8 c_{1} k^{2} \left ({\mathrm e}^{\textit {\_Z}}\right )^{2}-k^{4}\right )}\right )^{2}-k^{2}}{8 \,{\mathrm e}^{\mathit {RootOf}\left (64 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} \sqrt {2}\, c_{1}^{\frac {5}{2}} \textit {\_Z} k t +64 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{1}^{\frac {5}{2}} c_{2} \textit {\_Z} k +128 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} \sqrt {2}\, c_{1}^{4} c_{2} t +64 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{1}^{4} c_{2}^{2}+128 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{1}^{4} t^{2}+16 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{1} \textit {\_Z}^{2} k^{2}-16 c_{1}^{2} \left ({\mathrm e}^{\textit {\_Z}}\right )^{4}+8 c_{1} k^{2} \left ({\mathrm e}^{\textit {\_Z}}\right )^{2}-k^{4}\right )} c_{1}^{\frac {3}{2}}}\right \} \\ \bullet & {} & \textrm {Solve 2nd ODE for}\hspace {3pt} u \left (r \right ) \\ {} & {} & u \left (r \right )=-\frac {\sqrt {2}\, \sqrt {r \left (c_{1} r +k \right )}}{r} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (r \right )=r^{\prime },r =r \\ {} & {} & r^{\prime }=-\frac {\sqrt {2}\, \sqrt {r \left (c_{1} r+k \right )}}{r} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & r^{\prime }=-\frac {\sqrt {2}\, \sqrt {r \left (c_{1} r+k \right )}}{r} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {r^{\prime } r}{\sqrt {r \left (c_{1} r+k \right )}}=-\sqrt {2} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \frac {r^{\prime } r}{\sqrt {r \left (c_{1} r+k \right )}}d t =\int -\sqrt {2}d t +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\sqrt {r^{2} c_{1} +r k}}{c_{1}}-\frac {k \ln \left (\frac {\frac {k}{2}+c_{1} r}{\sqrt {c_{1}}}+\sqrt {r^{2} c_{1} +r k}\right )}{2 c_{1}^{\frac {3}{2}}}=-\sqrt {2}\, t +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} r \\ {} & {} & \left \{-\frac {4 \sqrt {c_{1}}\, k \,{\mathrm e}^{\mathit {RootOf}\left (64 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} \sqrt {2}\, c_{1}^{\frac {5}{2}} \textit {\_Z} k t -64 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{1}^{\frac {5}{2}} c_{2} \textit {\_Z} k +128 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} \sqrt {2}\, c_{1}^{4} c_{2} t -64 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{1}^{4} c_{2}^{2}-128 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{1}^{4} t^{2}-16 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{1} \textit {\_Z}^{2} k^{2}+16 c_{1}^{2} \left ({\mathrm e}^{\textit {\_Z}}\right )^{4}-8 c_{1} k^{2} \left ({\mathrm e}^{\textit {\_Z}}\right )^{2}+k^{4}\right )}-4 c_{1} \left ({\mathrm e}^{\mathit {RootOf}\left (64 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} \sqrt {2}\, c_{1}^{\frac {5}{2}} \textit {\_Z} k t -64 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{1}^{\frac {5}{2}} c_{2} \textit {\_Z} k +128 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} \sqrt {2}\, c_{1}^{4} c_{2} t -64 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{1}^{4} c_{2}^{2}-128 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{1}^{4} t^{2}-16 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{1} \textit {\_Z}^{2} k^{2}+16 c_{1}^{2} \left ({\mathrm e}^{\textit {\_Z}}\right )^{4}-8 c_{1} k^{2} \left ({\mathrm e}^{\textit {\_Z}}\right )^{2}+k^{4}\right )}\right )^{2}-k^{2}}{8 \,{\mathrm e}^{\mathit {RootOf}\left (64 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} \sqrt {2}\, c_{1}^{\frac {5}{2}} \textit {\_Z} k t -64 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{1}^{\frac {5}{2}} c_{2} \textit {\_Z} k +128 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} \sqrt {2}\, c_{1}^{4} c_{2} t -64 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{1}^{4} c_{2}^{2}-128 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{1}^{4} t^{2}-16 \left ({\mathrm e}^{\textit {\_Z}}\right )^{2} c_{1} \textit {\_Z}^{2} k^{2}+16 c_{1}^{2} \left ({\mathrm e}^{\textit {\_Z}}\right )^{4}-8 c_{1} k^{2} \left ({\mathrm e}^{\textit {\_Z}}\right )^{2}+k^{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)+k/_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: 0.062 (sec). Leaf size: 369
dsolve(diff(r(t),t$2)=-k/(r(t)^2),r(t), singsol=all)
\begin{align*} r \left (t \right ) &= \frac {c_{1} \left (c_{1}^{2} k^{2}-2 k c_{1} {\mathrm e}^{\operatorname {RootOf}\left (\operatorname {csgn}\left (\frac {1}{c_{1}}\right ) c_{1}^{4} k^{2}+2 \textit {\_Z} \,c_{1}^{3} k \,{\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}} t \right )}+{\mathrm e}^{2 \operatorname {RootOf}\left (\operatorname {csgn}\left (\frac {1}{c_{1}}\right ) c_{1}^{4} k^{2}+2 \textit {\_Z} \,c_{1}^{3} k \,{\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}} t \right )}\right ) {\mathrm e}^{-\operatorname {RootOf}\left (\operatorname {csgn}\left (\frac {1}{c_{1}}\right ) c_{1}^{4} k^{2}+2 \textit {\_Z} \,c_{1}^{3} k \,{\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}} t \right )}}{2} \\ r \left (t \right ) &= \frac {c_{1} \left (c_{1}^{2} k^{2}-2 k c_{1} {\mathrm e}^{\operatorname {RootOf}\left (\operatorname {csgn}\left (\frac {1}{c_{1}}\right ) c_{1}^{4} k^{2}+2 \textit {\_Z} \,c_{1}^{3} k \,{\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}} t \right )}+{\mathrm e}^{2 \operatorname {RootOf}\left (\operatorname {csgn}\left (\frac {1}{c_{1}}\right ) c_{1}^{4} k^{2}+2 \textit {\_Z} \,c_{1}^{3} k \,{\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}} t \right )}\right ) {\mathrm e}^{-\operatorname {RootOf}\left (\operatorname {csgn}\left (\frac {1}{c_{1}}\right ) c_{1}^{4} k^{2}+2 \textit {\_Z} \,c_{1}^{3} k \,{\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}} t \right )}}{2} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.169 (sec). Leaf size: 65
DSolve[r''[t]==-k/(r[t]^2),r[t],t,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [\left (\frac {r(t) \sqrt {\frac {2 k}{r(t)}+c_1}}{c_1}-\frac {2 k \text {arctanh}\left (\frac {\sqrt {\frac {2 k}{r(t)}+c_1}}{\sqrt {c_1}}\right )}{c_1{}^{3/2}}\right ){}^2=(t+c_2){}^2,r(t)\right ] \]