1. ## Where to start with this one:

I have to following differential equation. I need to solve it. I have gotten so far but got stuck... I have tried solving the equation using seperation of variables, which I think is where I've gone wrong. This is the equation:

$v\frac{dv}{dx}=-\frac{g}{10}(2+3v)$

Is seperation of variables the way to go, i.e, should I start by dividing through by:

$(2+3v)$

2. I am told I can use the substitution $u=(2+3v)$

3. Originally Posted by MaverickUK82
I have to following differential equation. I need to solve it. I have gotten so far but got stuck... I have tried solving the equation using seperation of variables, which I think is where I've gone wrong. This is the equation:

$v\frac{dv}{dx}=-\frac{g}{10}(2+3v)$

Is seperation of variables the way to go, i.e, should I start by dividing through by:

$(2+3v)$
Yes, you should end with:

$\displaystyle \int \frac{v}{2+3v} dv=\int-\frac{g}{10}dx$

CB

4. O.k, thanks!

I have got that, was struggling with integration of the left hand side. The right hand side is easier.

i'll struggle on a bit more - glad I have the beginning of the method correct...

Thanks for your help Captain Black!

5. Originally Posted by MaverickUK82
O.k, thanks!

I have got that, was struggling with integration of the left hand side. The right hand side is easier.

i'll struggle on a bit more - glad I have the beginning of the method correct...

Thanks for your help Captain Black!

$\dfrac{v}{3v+2}=\dfrac{1}{3}-\dfrac{2}{3(3v+2)}$

CB

6. I understand I need to get the v out of the numerator, but am struggling to do this today also, can anyone point me to a reference when I can get a pointer on how to do this again?

7. I dont understand how you've done that CB...

8. Originally Posted by MaverickUK82
I dont understand how you've done that CB...
Use polynomial long division to divide 3v + 2 into v. Alternatively, note that

$\displaystyle \frac{v}{3v + 2} = \frac{1}{3} \left( \frac{3v}{3v + 2}\right) = \frac{1}{3} \left( \frac{(3v + 2) - 2}{3v + 2}\right) = \frac{1}{3} \left( 1 - \frac{2}{3v + 2}\right) = ....$

9. Originally Posted by MaverickUK82
I dont understand how you've done that CB...
As Mr F says use polynomial division. Alternatively think of this as a degenerate form of partial fractions, we seek $A$ and $B$ such that:

$\dfrac{v}{3v+2}=A+\dfrac{B}{3v+2}$

which can be done in a number of ways.

CB

10. Originally Posted by MaverickUK82
I am told I can use the substitution $u=(2+3v)$
Or if you like:
$\displaystyle \int \frac{v}{2+3v} dv = \int \frac{\frac{1}{3}(u - 2)}{u} \frac{1}{3}~du$

$\displaystyle = \frac{1}{9} \int \left ( 1 - \frac{2}{u} \right ) ~du$

-Dan

11. thanks Dan for the explanation. Your method it the method I am supposed to be using, although, no doubt CB and Mr Fantastic's methods are equally correct.

However, could you explain what you have done there because I don't truly understand how you've achieved it...

12. I'm sorry, that was a bit vague, how does what you have done fit in with the integration by substitution rule, i.e,

$S f(g(x))g'(x) dx=Sf(u) du,$where $u=g(x)$

Hope you don;t mind me using S's as integral signs, don't know how to do them in latex and double clicking yours brings up an error ?

13. Originally Posted by MaverickUK82
I'm sorry, that was a bit vague, how does what you have done fit in with the integration by substitution rule, i.e,

$S f(g(x))g'(x) dx=Sf(u) du,$where $u=g(x)$

Hope you don;t mind me using S's as integral signs, don't know how to do them in latex and double clicking yours brings up an error ?
Let $u = 2 + 3v$. That means that $\displaystyle v = \frac{1}{3} \cdot (u - 2)$. This also means that $\displaystyle du = 3 dv \implies dv = \frac{1}{3} du$. Now just sub in for v and dv in the integral.

-Dan

14. I'm sorry topsquark. I'm just not getting it. Why do I need to do that. In order to use the substitution rule, doesn't one have to be a derivative of the other? I'm really sorry if I appear dense. I probably am.

15. Originally Posted by MaverickUK82
I'm sorry topsquark. I'm just not getting it. Why do I need to do that. In order to use the substitution rule, doesn't one have to be a derivative of the other? I'm really sorry if I appear dense. I probably am.
You don't necessarily need to do it, it just makes the form of the integrand simpler. (When the concept works, that is.) In this case integrating $\displaystyle \int \frac{v}{2 + 3v}dv$ is harder to integrate than $\displaystyle \int \frac{1}{u}du$. This case is almost trivial because it is not hard to see how to integrate the expression in v, but your integrand might be more complicated, such as $\displaystyle \int \frac{x}{\sqrt{4 - x^2}}$ for which the substitution of $u = 4 - x^2$ makes the problem much simpler.

-Dan

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