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Name: Closed form solutions of x¨(t)−x(t)n=0
Uploaded: 2022-02-23T17:22:57+00:00
Description: Closed form solutions of x¨(t)−x(t)n=0

expeditiupc

Answered question

2022-01-21

Closed form solutions of

\ddot{x} (t) - x {(t)}^{n} = 0

Answer & Explanation

Paul Mitchell

Beginner2022-01-21Added 40 answers

$y^{″} - y^{n} = 0$
$y^{'} y^{″} = y^{'} y^{n}$
$\int y^{'} y^{″} d x = \int y^{'} y^{n} d x$
$\frac{{(y^{'})}^{2}}{2} = \frac{y^{n + 1}}{n + 1} + c_{1}$
$y^{'} = \sqrt{\frac{2 y^{n + 1}}{n + 1}} + 2 c_{1}$
$\int \frac{d y}{\sqrt{\frac{2 y^{n + 1}}{n + 1}} + 2 c_{1}} = \int d x = x + a$
$\frac{1}{{(2 c_{1})}^{\frac{1}{2}}} \int \frac{d y}{\sqrt{1 + \frac{y^{n + 1}}{c_{1} (n + 1)}}} = x + a$
$\int \frac{d y}{\sqrt{1 + \frac{y^{n + 1}}{c_{1} (n + 1)}}} = {(2 c_{1})}^{\frac{1}{2}} x + a {(2 c_{1})}^{\frac{1}{2}} = {(2 c_{1})}^{\frac{1}{2}} x + c_{2}$
After here you can change variable and use the binomial expansion to evaluate integral.
$u = \frac{y^{n + 1}}{c_{1} (n + 1)}$
$(1 + u)^{α} = \sum_{n = 0} (\begin{matrix} α \\ n \end{matrix}) u^{n} = 1 + α \frac{u}{1!} + α (α - 1) \frac{u^{2}}{2!} + . . .$
I avoided doing many calculations, and I preferred to use the quick way: ask Wolfram Alpha.
$k = \frac{1}{c_{1} (n + 1)}$

kaluitagf

Beginner2022-01-22Added 38 answers

This is the answer given by Mathematica:

x = x (t)

is implicitly given by

(n + 1) x {(t)}^{2} {(c_{1} n + c_{1} + 2 x {(t)}^{n + 1})}^{2}

\frac{2 F_{1} {(\frac{1}{2}, \frac{1}{n + 1}; 1 + \frac{1}{n + 1}; - \frac{2 x {(t)}^{n + 1}}{n c_{1} + c_{1}})}^{2}}{(c_{1} n + c_{1}) {(c_{1} (n + 1) + 2 x {(t)}^{n + 1})}^{2}} = {(c_{2} + t)}^{2}

.
In general, I do not believe that y may be written down in terms of elementary or special functions.

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