I need to find x.
any help would be great:
$\displaystyle
v \frac{dv}{dx} = -\frac{g}{b^2}(v^2 + b^2)
$
first of all you can always check if your answer is correct by plugging it into the ODE. (your answer is wrong)
let us make the following substitution:
$\displaystyle
\begin{gathered}
z = \frac{1}
{2}v^2 \hfill \\
\frac{{dz}}
{{dx}} = v\frac{{dv}}
{{dx}} \hfill \\
\end{gathered}
$
thus we get the following equation:
$\displaystyle
z' + \frac{{2g}}
{{b^2 }}z = - g
$
This is a first order ODE which can be easily solved...