Closed form solutions of $\ddot x(t)-x(t)^n=0$ Given the ODE:
$$\ddot x(t)-x(t)=0$$
the solution is:
$$x(t)=C_1\exp(-t)+C_2\exp(t)$$
If we square the $x(t)$ we have:
$$\ddot x(t)-x(t)^2=0$$
and the solution is given by:
$$x(t)=6\wp(t+C_1;0,C_2)$$
and so for $x(t)^3$ which gives:
$$x(t)=\operatorname{sn}\left(\left(\frac{1}{2}i\sqrt t +C_1\right)C_2,i\right)$$
More generally, is it possible to find closed form solutions for the equation:
$$\ddot x(t)-x(t)^n=0$$
?
Thanks.
 A: $$\begin{align*}
y''-y^n&=0\\
y'y''&=y'y^n\\
\int y'y'' dx &=\int y'y^n dx\\
\frac{(y')^2}{2} &=\frac{y^{n+1}}{n+1} +c_1\\
y' &=\sqrt {\frac{2y^{n+1}}{n+1} +2c_1 }\\
\int \frac{dy}{\sqrt {\frac{2y^{n+1}}{n+1} +2c_1 }} &=\int dx =x+a\\
\frac{1}{(2c_1)^{1/2}}\int \frac{dy}{ \sqrt{1+\frac{y^{n+1}}{c_1(n+1)}}} &=x+a\\
\int \frac{dy}{ \sqrt{1+\frac{y^{n+1}}{c_1(n+1)}}} &=(2c_1)^{1/2}x+a(2c_1)^{1/2}=(2c_1)^{1/2}x+c_2
\end{align*}$$
After here you can change variable and use the binomial expansion to evaluate integral.
$u=\frac{y^{n+1}}{c_1(n+1)}$
$${(1+u)}^{\alpha}= \sum_{n=0}^\infty \binom{\alpha}{n} u^n=1+ \alpha \frac{u}{1!}+\alpha (\alpha-1) \frac{u^2}{2!}+\dots$$
I avoided doing many calculations, and I preferred to use the quick way: ask Wolfram Alpha. Here is a link to the solution.
$k=\frac{1}{c_1(n+1)}$
$\int \frac{dy}{ \sqrt{1+ky^{n+1}}} =y. _2F_1 (\frac{1}{2},\frac{1}{n+1};\frac{n+2}{n+1}; -ky^{n+1}) = y. _2F_1 (\frac{1}{2},\frac{1}{n+1};\frac{n+2}{n+1}; -\frac{y^{n+1}}{c_1(n+1)})=(2c_1)^{1/2}x+c_2 $
Solution in closed form as Hypergeometric function
$$_2F_1 \left(\frac{1}{2},\frac{1}{n+1};\frac{n+2}{n+1}; -\frac{y^{n+1}}{c_1(n+1)}\right)=\frac{1}{y}\left(\sqrt{2c_1}x+c_2\right)$$
A: As an addendum of sorts to Mathlover's answer: the differential equation
$$(y^\prime)^2=2\left(\frac{y^{n+1}}{n+1}+c_1\right)$$
is in fact the very sort of equation that is solved by Abelian (hyperelliptic) functions, in the sense that the integral
$$\int\frac{\mathrm dt}{\sqrt{t^{n+1}+c}}$$
is a hyperelliptic (Abelian) integral (which, as already noted, can be represented in terms of the Gaussian hypergeometric function), and the original differential equation is solved by the inversion of this integral; i.e., with Abelian functions. Since Mathematica and Wolfram Alpha know nothing about Abelian functions, they are unable to proceed further, and they just leave an implicit equation as a result.
As expected, when $n=2$ or $3$, the hyperelliptic integral becomes an elliptic integral, and thus the differential equation is expected to have elliptic function solutions. I'll carry out the inversion explicitly for those two cases.

For $n=2$, we have, after absorption of arbitrary constants, the expression
$$\int\frac{\mathrm dy}{\sqrt{y^3+C_1}}=\sqrt{\frac23}x+C_2$$
To make the integral on the left a bit more recognizable, we multiply by a constant on both sides:
$$\frac12\int\frac{\mathrm dy}{\sqrt{y^3+C_1}}=\frac12\left(\sqrt{\frac23}x+C_2\right)$$
which turns into
$$\int\frac{\mathrm dy}{\sqrt{4y^3+4C_1}}=\frac{x}{\sqrt 6}+\frac{C_2}{2}$$
and we now recognize the Weierstrass elliptic integral corresponding to the cubic $4y^3-g_2 y-g_3$ on the left side, with the invariants $g_2=0$ and $g_3=-4C_1$. Inversion yields
$$y=\wp\left(\frac{x}{\sqrt 6}+\frac{C_2}{2};0,-4C_1\right)$$
or, after absorption of arbitrary constants,
$$y=\wp\left(\frac{x}{\sqrt 6}+C_2;0,C_1\right)=6\wp\left(x+C_2;0,C_1\right)$$
where the homogeneity relation for $\wp$ was used to obtain the final expression.

For $n=3$, we have (changing the form of one of the arbitrary constants for convenience)
$$\int\frac{\mathrm dy}{\sqrt{y^4+C_1^4}}=\frac{x}{\sqrt 2}+C_2$$
We can try a Weierstrass substitution $y=C_1 \cot\frac{t}{2}$ here:
$$\frac1{2C_1}\int\frac{\mathrm dt}{\sqrt{1-\frac12\sin^2 t}}=\frac{x}{\sqrt 2}+C_2$$
and we recognize the incomplete elliptic integral of the first kind at this point (and absorbing arbitrary constants while we're at it):
$$\frac1{C_1}F\left(2\mathrm{arccot}\frac{y}{C_1}\mid\frac12\right)=\frac{x}{\sqrt 2}+C_2$$
We solve for $y$ in stages:
$$\begin{align*}
F\left(2\mathrm{arccot}\frac{y}{C_1}\mid\frac12\right)&=C_1 x+C_2\\
2\mathrm{arccot}\frac{y}{C_1}&=\mathrm{am}\left(C_1 x+C_2\mid\frac12\right)\\
y&=C_1\cot\left(\frac12\mathrm{am}\left(C_1 x+C_2\mid\frac12\right)\right)\\
y&=C_1\frac{\sin\left(\mathrm{am}\left(C_1 x+C_2\mid\frac12\right)\right)}{1-\cos\left(\mathrm{am}\left(C_1 x+C_2\mid\frac12\right)\right)}\\
y&=C_1\frac{\mathrm{sn}\left(C_1 x+C_2\mid\frac12\right)}{1-\mathrm{cn}\left(C_1 x+C_2\mid\frac12\right)}
\end{align*}$$
and that's how Jacobian elliptic functions turn up in the solution.

As it turns out, the hyperelliptic functions for $n$ an odd integer can theoretically be expressed in terms of elliptic functions, but the expressions look rather unwieldy. See Byrd and Friedman for details (p. 252 onwards).
A: This is the answer given by Mathematica: $x=x(t)$ is implicitly given by
$$\frac{(n+1) x(t)^2 \left(c_1 n+c_1+2 x(t)^{n+1}\right)^2
   \, _2F_1\left(\frac{1}{2},\frac{1}{n+1};1+\frac{1}{n+1};-\frac{2
   x(t)^{n+1}}{n c_1+c_1}\right){}^2}{\left(c_1 n+c_1\right) \left(c_1
   (n+1)+2 x(t)^{n+1}\right){}^2}=\left(c_2+t\right)^2.$$
In general, I do not believe that $y$ may be written down in terms of elementary or special functions.
Remark: this was already suggested by Willie Wong in a comment. I wrote it just to improve readability. Moreover, I guess that a rigorous answer (i.e. a proof that it is impossible to find a closed form) is really hard to build.
A: Assuming mathematica special functions are closed forms, here is such a solution to the equation using Inverse Beta Regularized $\text I^{-1}_s(a,b)$ for $a,b>0$ and $0\le s\le1$
Following from both of the first two answers and simplifying into Incomplete Beta function $\text B_z(a,b)$ form:
$$y\,_2\text F_1\left(\frac12,\frac1{n+1};\frac1{n+1}+1;-\frac{y^{n+1}}{c_1(n+1)}\right)=\frac{\left(-\frac1{c_1(n+1)}\right)^{-\frac1{n+1}}}{n+1}\text B_{-\frac{y^{n+1}}{c_1(n+1)}}\left(\frac1{n+1},\frac12\right)=\sqrt{2c_1}x+c_2$$
Inverting,
$$\begin{align}y’’=y^n\implies y=\sqrt[n+1]{-c_1(n+1)\text I^{-1}_{\frac{\sqrt[n+1]{c_1}(n+1)^{\frac1{n+1}+1}\cdot\left(\sqrt{2c_1}x+c_2\right)}{\text B\left(\frac1{n+1},\frac12\right)}}\left(\frac1{n+1},\frac12\right)},n>-1,c_{1,2}\in\Bbb C,0\le \frac{\sqrt[n+1]{c_1}(n+1)^{\frac1{n+1}+1}\cdot\left(\sqrt{2c_1}x+c_2\right)}{\text B\left(\frac1{n+1},\frac12\right)} \le 1\end{align}$$
which is true. Please see the “alternate form assuming…” section.
The solution also has a similar $\text I^{-1}_{•}\left(\frac1{n+1},\frac1{n+1}\right)$ form which may have multiple branches for $\sqrt[n+1]{\quad}$.
The solution represents one period of a hyperelliptic function, so please transform the solution accordingly for another period. The function also represents simple cases of Jacobi and Weierstrass $\wp$ elliptic functions.
This answer is also a “draft” version, so please bring up any mistakes.
