# Thread: Integration... can't do this one either...

1. ## Integration... can't do this one either...

I need to find x.
any help would be great:

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

2. Is this right?

$
ln(v + 1) = - gx + C
$

or this:

$
\frac{1}{2} ln(b^2 + v^2) = - \frac{g}{b^2} + C
$

3. Originally Posted by billym
Is this right?

$
ln(v + 1) = - gx + C
$

let us make the following substitution:

$
\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:

$
z' + \frac{{2g}}
{{b^2 }}z = - g
$

This is a first order ODE which can be easily solved...

4. Originally Posted by billym
I need to find x.
any help would be great:

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

Make the substitution $u = v^2 + b$.

5. um...

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

6. Originally Posted by billym
um...

$
x = -\frac{b^2}{2g}ln(v^2 + b)+C
$
No.

$x = - \frac{b^2}{2g} \ln |v^2 + b| + C$.

Note that $\int \frac{1}{u} \, du = \ln |u| + C$ ....

7. Originally Posted by mr fantastic
$\Rightarrow \frac{dx}{dv} = -\frac{b^2}{g} \int \frac{v}{v^2 + b} \, dv$.
(chuckles) Are you sure about that? I think the LHS is just $\int dx$.

-Dan

8. Originally Posted by topsquark
(chuckles) Are you sure about that? I think the LHS is just $\int dx$.

-Dan

9. I think it is all a bit confusing to the original poster. The equation is one that can be separated and thus written as:

$\int\frac{v \cdot dv}{v^2 + b^2}=-\frac{g}{b^2}\int dx$

This should get the original poster a step further.