1. ## [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

2. Originally Posted by ben.mahoney@tesco.net
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 $\displaystyle dv/(g- kv^2)$, 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: $\displaystyle g- kv^2= (\sqrt{g}+ \sqrt{k}v)(\sqrt{g}- \sqrt{k}v)$

3. 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???

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

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