Solving dv/dt = g - (k/m)v²

[hangainlover]

dv/dt=g-(k/m)*v^2
i thought about all sorts of diff eq solving techniques
but nothing comes up to my mind

dv/(g-(k/m)*v^2)=dt
but it doesn't really help ...

 

[MaxPlank]

I think it's quite correct, but there could be some errors in calculation..anyway:

http://spiro.fisica.unipd.it/~mimo/altro/sol.JPG

 

[tiny-tim]

hi hangainlover!

(try using the X² tag just above the Reply box )

learn your list of standard integrals from the pf library

 

[Delta2]

Separation of variables seems to work. Just try to transform the integral $$\int \frac{1}{g-\frac{k}{m}v^2}dv$$ to the case xxvi of the standard integrals link provided by tiny-tim.

 

[HallsofIvy]

Specifically,
$$g- \frac{k}{m}v^2$$
, being a "difference of squares", factors as
$$(\sqrt{g}- \sqrt{k/m}v)(\sqrt{g}+ \sqrt{k/m}v)$$
and so that function can be integrated by "partial fractions".