Question

diff eq

Answer 1

\[\frac{ d(\frac{ mv }{ \sqrt{1-v^2/c^2} }) }{ dt }= F\]

Answer 2

solve for v(t)

Answer 3

\[\frac{ F }{ m}= \frac{ d }{ dt }(\frac{ v}{\sqrt{1-v^2/c^2} })\]

Answer 4

\[\frac{ Ft }{ m }= \frac{ v }{ \sqrt{1-v^2/c^2 }}\]

Answer 5

stuck here. @Michele_Laino

Answer 6

is the force \(F\) constant with respect to time?

Answer 7

yes

Answer 8

we can try to take the square of both sides

Answer 9

I got this:
\[\Large {v^2}\left( {1 + {{\left( {\frac{{Ft}}{{mc}}} \right)}^2}} \right) = {\left( {\frac{{Ft}}{m}} \right)^2}\]

Answer 10

now, we can solve for \(v(t)\)

Answer 11

got it thanks

Answer 12

:)

Answer 13

please wait, after integration, we have to add the arbitrary constant, so we get:
\[\Large \frac{{Ft}}{m} + k = \frac{v}{{\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} }},\quad k \in \mathbb{R}\]

Answer 14

got it!

Answer 15

ok! :)