Finding a solution for Relativistic Acceleration

I tried all my mathcad software such as Maple and I can't seem to find a solution to this time based differential equation.

$f'(t) = \frac{F}{m}\left ( 1 - \frac{\left [ f(t) \right ]^{2}}{c^{2}} \right )^{\frac{3}{2}}$
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 Quote by Michio Cuckoo I tried all my mathcad software such as Maple and I can't seem to find a solution to this time based differential equation. $f'(t) = \frac{F}{m}\left ( 1 - \frac{\left [ f(t) \right ]^{2}}{c^{2}} \right )^{\frac{3}{2}}$
Hi !

Very surprising ! They should solve this EDO without any difficulty, since it is an EDO of the "separable variables" kind. This can be handly carried out.

 Quote by Michio Cuckoo I tried all my mathcad software such as Maple and I can't seem to find a solution to this time based differential equation. $f'(t) = \frac{F}{m}\left ( 1 - \frac{\left [ f(t) \right ]^{2}}{c^{2}} \right )^{\frac{3}{2}}$
Whenever you have a messy-looking equation, it's sometimes helpful to at least initially, get rid of all the fluff by canonicalizing the equation. In the case above, write it simply as:

$$\frac{dy}{dt}=k(1-y^2/a)^{3/2}$$

See, just that lil' bit is good progress. Now, you can't separate variables and integrate? Telll you what, can you integrate just:

$$\int \frac{dy}{(1-y^2/a)^{3/2}}$$

I haven't tried so I don't know. You try it.

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Finding a solution for Relativistic Acceleration

This might be a two stage process, first try a substitution of $y=\sqrt{a}\sin\theta$ and see what you get.