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Math Help - [SOLVED] separable ode

  1. #1
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    [SOLVED] separable ode

    A parachutist falls under gravity and the drag on the chute is known to have an air resistance which is proportional to the square of the velocity v and given by kv2 per unit mass. According to Newton’s Law of Motion the velocity of the parachutist satisfies the differential equation

    dv/dt = g - kv^2

    where g is a gravitational constant. Find the velocity of the particle at any time t subject to the initial condition that v=0 at t=0. Hence confirm that the velocity of the particle tends to the terminal velocity.

    Im finding it tricky to differentiate the dv/g-kv^2 bit
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  2. #2
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    Quote Originally Posted by ben.mahoney@tesco.net View Post
    A parachutist falls under gravity and the drag on the chute is known to have an air resistance which is proportional to the square of the velocity v and given by kv2 per unit mass. According to Newton’s Law of Motion the velocity of the parachutist satisfies the differential equation

    dv/dt = g - kv^2

    where g is a gravitational constant. Find the velocity of the particle at any time t subject to the initial condition that v=0 at t=0. Hence confirm that the velocity of the particle tends to the terminal velocity.

    Im finding it tricky to differentiate the dv/g-kv^2 bit
    I hope what you are trying to do is integrate dv/(g- kv^2), not differentiate it! Since that is a rational function, you can integrate using "partial fractions". If you think of the denominator as a "difference of squares" it's easy to factor: g- kv^2= (\sqrt{g}+ \sqrt{k}v)(\sqrt{g}- \sqrt{k}v)
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  3. #3
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    ok i can get the simplified logs:
    1 *log(sqrt(g)+sqrt(k)*v) = t +C
    sqrt(k)*(sqrt(g)+sqrt(k)) -1+v

    the final answer is given as

    v = sqrt(g) * (e^(2*sqrt(kg)*t) -1)
    (k) (e^(2*sqrt(kg)*t) +1)

    Any ideas on how it simplifies???
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  4. #4
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    It's hard to read what you have written there. I recommend you look at one of the threads on "latex". You can see the code for any Latex on this board by clicking it.

    Yes, you can write \frac{1}{g- kv^2}= \frac{1}{\sqrt{g}+\sqrt{k}v}+ \frac{1}{\sqrt{g}-\sqrt{k}v} so, integrating,
    \frac{1}{\sqrt{k}}(ln(\sqrt{g}+\sqrt{k}v)- ln(\sqrt{g}-\sqrt{k}v)
    = \frac{1}{\sqrt{k}}ln\left(\frac{\sqrt{g}+ \sqrt{k}v}{\sqrt{g}-\sqrt{k}v}\right)= t+ c
    Now multiply both sides by \sqrt{k} and take the exponential of each side to get
    \frac{\sqrt{g}+ \sqrt{k}v}{\sqrt{g}-\sqrt{k}v}= Ce^{\sqrt{k}t}
    Multiply both sides by \sqrt{g}-\sqrt{k}v and multiply on the right so you have
    \sqrt{g}+ \sqrt{k}v= Ce^{\sqrt{k}t}\sqrt{g}- \sqrt{k}vCe^{\sqrt{k}t}
    and isolate the "v" terms
    \sqrt{k}v+ \sqrt{k}vCe^{\sqrt{k}t}= \sqrt{k}(1+ Ce^{\sqrt{k}t})v
    = Ce^{\sqrt{k}t}\sqrt{g}- \sqrt{g}= \sqrt{g}(Ce^{\sqrt{k}t}- 1)
    so
    v= \sqrt{\frac{g}{k}}\frac{Ce^{\sqrt{k}t}- 1}{Ce^{\sqrt{k}t}+ 1}
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