Closed form solutions of x¨(t)−x(t)n=0

${\displaystyle y{}^{″}-y^n=0}$
${\displaystyle y^{\prime }y{}^{″}=y^{\prime }{y}^n}$
${\displaystyle \int y^{\prime }y{}^{″}dx=\int y^{\prime }{y}^{n}dx}$
${\displaystyle \frac{{\left(y^{\prime }\right)}^2}{2}=\frac{y^{n+1}}{n+1}+c_1}$
${\displaystyle y^{\prime }=\sqrt{\frac{2y^{n+1}}{n+1}}+2c_1}$
${\displaystyle \int \frac{dy}{\sqrt{\frac{2y^{n+1}}{n+1}}+2c_1}=\int dx=x+a}$
${\displaystyle \frac{1}{{\left(2c_1\right)}^{\frac{1}{2}}}\int \frac{dy}{\sqrt{1+\frac{y^{n+1}}{c_1(n+1)}}}=x+a}$
${\displaystyle \int \frac{dy}{\sqrt{1+\frac{y^{n+1}}{c_1(n+1)}}}={\left(2c_1\right)}^{\frac{1}{2}}x+a{\left(2c_1\right)}^{\frac{1}{2}}={\left(2c_1\right)}^{\frac{1}{2}}x+c_2}$
After here you can change variable and use the binomial expansion to evaluate integral.
${\displaystyle u=\frac{y^{n+1}}{c_1(n+1)}}$
$(1+u{)}^{\alpha }=\sum _{n=0}(\begin{array}{c}\alpha \\ n\end{array})u^n=1+\alpha \frac{u}{1!}+\alpha (\alpha -1)\frac{u^2}{2!}+...$
I avoided doing many calculations, and I preferred to use the quick way: ask Wolfram Alpha.
${\displaystyle k=\frac{1}{c_1(n+1)}}$