Could you help me solve first order diff eq?

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

http://spiro.fisica.unipd.it/~mimo/altro/sol.JPG
 
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hi hangainlover! :smile:

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

learn your list of standard integrals from the pf library :wink:
 
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.
 
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".
 
There is the following linear Volterra equation of the second kind $$ y(x)+\int_{0}^{x} K(x-s) y(s)\,{\rm d}s = 1 $$ with kernel $$ K(x-s) = 1 - 4 \sum_{n=1}^{\infty} \dfrac{1}{\lambda_n^2} e^{-\beta \lambda_n^2 (x-s)} $$ where $y(0)=1$, $\beta>0$ and $\lambda_n$ is the $n$-th positive root of the equation $J_0(x)=0$ (here $n$ is a natural number that numbers these positive roots in the order of increasing their values), $J_0(x)$ is the Bessel function of the first kind of zero order. I...
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