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:

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

2. Is this right?

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

or this:

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

3. Originally Posted by billym
Is this right?

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

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...

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

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

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

5. um...

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

6. Originally Posted by billym
um...

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

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

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

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

-Dan

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

-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:

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

This should get the original poster a step further.