# Thread: more inegration

1. ## more inegration

I need to solve for v... not a clue...

$
\frac{dv}{dt} = -\frac{g}{b^2}(v^2 + b^2)
$

2. Originally Posted by billym
I need to solve for v... not a clue...

$
\frac{dv}{dt} = -\frac{g}{b^2}(v^2 + b^2)
$
This is a seperable equation

$
\frac{dv}{v^2+b^2} = -\frac{g}{b^2}dt
$

Now we integrate both sides

$
\int \frac{dv}{v^2+b^2} = \int -\frac{g}{b^2}dt \iff \frac{1}{b^2} \int \frac{dv}{1+(v/b)^2}= - \frac{g}{b^2} \int dt \ =$

$\int \frac{dv}{1+(v/b)^2} = -g \int dt
$

finally we take the integral and we get

$b \tan^{-1}(\frac{v}{b})=-gt+c$

Now all you need to do is solve for v.

Good luck