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Thread: Where to start with this one:

  1. #1
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    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:

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

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

    $\displaystyle (2+3v)$
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    I am told I can use the substitution $\displaystyle u=(2+3v)$
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  3. #3
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    Quote Originally Posted by MaverickUK82 View Post
    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:

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

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

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

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

    CB
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    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!
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    Quote Originally Posted by MaverickUK82 View Post
    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!

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

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

    $\displaystyle \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) = .... $
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  9. #9
    Grand Panjandrum
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    Quote Originally Posted by MaverickUK82 View Post
    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 $\displaystyle $$A$ and $\displaystyle $$B$ such that:

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

    which can be done in a number of ways.

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

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

    -Dan
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  11. #11
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    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...
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    I'm sorry, that was a bit vague, how does what you have done fit in with the integration by substitution rule, i.e,

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

    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 ?
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  13. #13
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by MaverickUK82 View Post
    I'm sorry, that was a bit vague, how does what you have done fit in with the integration by substitution rule, i.e,

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

    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 $\displaystyle u = 2 + 3v$. That means that $\displaystyle \displaystyle v = \frac{1}{3} \cdot (u - 2)$. This also means that $\displaystyle \displaystyle du = 3 dv \implies dv = \frac{1}{3} du$. Now just sub in for v and dv in the integral.

    -Dan
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  14. #14
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    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.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by MaverickUK82 View Post
    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 $\displaystyle \displaystyle \int \frac{v}{2 + 3v}dv$ is harder to integrate than $\displaystyle \displaystyle \int \frac{1}{u}du$. 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 $\displaystyle \displaystyle \int \frac{x}{\sqrt{4 - x^2}}$ for which the substitution of $\displaystyle u = 4 - x^2$ makes the problem much simpler.

    -Dan
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