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    Hi,

    I am stuck on this differential equation question. It concerns a particle projected vertically upwards at a speed of U subject to air resistance such that if it were in freefall its terminal velocity would be V. In the first part of the question I am supposed to show that

     \frac{dv}{dt} = -(1 + \frac{v}{V})g Where v is the velocity at time t.

    The next part of the question has stumped me however. It has told me to use that  \frac {dv}{dt} = v \frac{dv}{dx} to find an expression for the maximum height H. I'm not sure how to do this however. I have tried rewriting the earlier differential equation as:

     \frac{dv}{dx} = -(\frac{1}{v} + \frac{1}{V})g But when I separate variables the integral with respect to velocity is just nasty and I cannot seem to get to the correct answer. Am I missing something or is this how it is supposed to be?
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    (Original post by 16Characters....)
    Hi,

    I am stuck on this differential equation question. It concerns a particle projected vertically upwards at a speed of U subject to air resistance such that if it were in freefall its terminal velocity would be V. In the first part of the question I am supposed to show that

     \frac{dv}{dt} = -(1 + \frac{v}{V})g Where v is the velocity at time t.

    The next part of the question has stumped me however. It has told me to use that  \frac {dv}{dt} = v \frac{dv}{dx} to find an expression for the maximum height H. I'm not sure how to do this however. I have tried rewriting the earlier differential equation as:

     \frac{dv}{dx} = -(\frac{1}{v} + \frac{1}{V})g But when I separate variables the integral with respect to velocity is just nasty and I cannot seem to get to the correct answer. Am I missing something or is this how it is supposed to be?

    you can still split the fraction by division or manipulation or the alternative is to leave the constant in the RHS, find a solution by separating variables, then seek a PI for the constant term
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    (Original post by 16Characters....)
     \frac{dv}{dx} = -(\frac{1}{v} + \frac{1}{V})g But when I separate variables the integral with respect to velocity is just nasty and I cannot seem to get to the correct answer. Am I missing something or is this how it is supposed to be?
    Presumably you're integrating \dfrac{vV}{v+V} on the left.

    V in just a constant. You can use polynomial division (in v rather than x)
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    (Original post by TeeEm)
    you can still split the fraction by division or manipulation or the alternative is to leave the constant in the RHS, find a solution by separating variables, then seek a PI for the constant term
    (Original post by ghostwalker)
    Presumably you're integrating \dfrac{vV}{v+V} on the left.

    V in just a constant. You can use polynomial division (in v rather than x)
    Aaah  \frac{vV}{v + V} = V(1 - \frac{V}{v + V}) How did I not notice that! Thanks for your help.
 
 
 
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