# Thread: differencial equation

1. ## differencial equation

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

I have:

$
ln|v + 1| = - gx + C
$

2. Originally Posted by billym
$
v \frac{dv}{dx} = -\frac{g}{b^2}(v^2 + b^2)
$

I have:

$
ln|v + 1| = - gx + C
$
1. you have no dependence on $b$ so its unlikeley to be right.

2. $\frac{v}{v^2+b^2}~\frac{dv}{dx}=-\frac{g}{b^2}$

so (assuming $g$ and $b$ are constants):

$\log (v^2+b^2)=-\frac{2g}{b^2}x+C$

RonL

3. Originally Posted by billym
$
v \frac{dv}{dx} = -\frac{g}{b^2}(v^2 + b^2)
$

I have:

$
ln|v + 1| = - gx + C
$
$\int \frac{v}{v^2+b^2}dv=\frac{1}{2}\ln|v^2+b^2|$

4. Originally Posted by CaptainBlack
1. you have no dependence on $b$ so its unlikeley to be right.

2. $\frac{v}{v^2+b^2}~\frac{dv}{dx}=-\frac{g}{b^2}$

so (assuming $g$ and $b$ are constants):

$\log (v^2+b^2)=-\frac{2g}{b^2}x+C$

RonL
Alternatively, reply #6 in this thread would provide the answer.