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Math Help - 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:

    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)
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  2. #2
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    I am told I can use the substitution 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:

    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
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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!

    \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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  7. #7
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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 \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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    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 $$A and $$B such that:

    \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 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
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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,

    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 ?
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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,

    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
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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 \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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