 expeditiupc

2022-01-21

Closed form solutions of $\stackrel{¨}{x}\left(t\right)-x{\left(t\right)}^{n}=0$ Paul Mitchell

Expert

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

Expert

This is the answer given by Mathematica: $x=x\left(t\right)$ is implicitly given by
$\left(n+1\right)x{\left(t\right)}^{2}{\left({c}_{1}n+{c}_{1}+2x{\left(t\right)}^{n+1}\right)}^{2}$
$\frac{2{F}_{1}{\left(\frac{1}{2},\frac{1}{n+1};1+\frac{1}{n+1};-\frac{2x{\left(t\right)}^{n+1}}{n{c}_{1}+{c}_{1}}\right)}^{2}}{\left({c}_{1}n+{c}_{1}\right){\left({c}_{1}\left(n+1\right)+2x{\left(t\right)}^{n+1}\right)}^{2}}={\left({c}_{2}+t\right)}^{2}$.
In general, I do not believe that y may be written down in terms of elementary or special functions.

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