# DE question involving a falling object

1. May 4, 2009

### Emethyst

1. The problem statement, all variables and given/known data
Note that this is a first year, Calculus 12 DE question:
The velocity,v, of an object falling through the earth's atmosphere obeys the DE dv/dt=g-kv, where k is the called the drag constant and g is the gravitational constant. This equation states that the acceleration of the object is g reduced by an amount that represents air resistance (kv). Air resistance is proportional to the velocity of the object.
a) Find the function v(t) assuming that v(0)=0.

2. Relevant equations
All first year calculus methods used to find DE's.

3. The attempt at a solution
I just seem to be lost on how to convert the derivative given into an equation that contains the variables given by dv/dt. I have no problems finding antiderivatives, it just seems to be the introduction of variables instead of numbers here that is throwing me off. If anyone can point me along in the right direction it would be greatly appreciated, thanks in advance.

2. May 4, 2009

### FedEx

Well if you take kv on the other side than you could see that it is a linear diff equation

3. May 5, 2009

### Emethyst

That's what I originally tried to do, but I am not sure if I can then integrate both sides, as from what I did:

dv/dt + kv = g
dv + (dt)(kv) = (g)(dt)

I don't know if I am going wrong somewhere here, as I thought for this to be integrated I would have to get it into the form (kv)(dv) = (g)(dt), and from what I did it doesn't look like it will get into that form easily :tongue:

4. May 5, 2009

### Staff: Mentor

You need to look at this as two separate equations:
1) v' + kv = 0 (the homogeneous linear DE)
2) v' + kv = g (the nonhomogeneous linear DE)

Can you find the set of solutions vh for the first equation? Hint: There will be a bunch of them, and they are multiples of the same exponential function.

Now can you find a particular solution vp to the second equation?

Your general solution vg will be vh + vp.

Be sure to use your initial condition, v(0) = 0.

5. May 5, 2009

### FedEx

Havent you been taught to solve linear diff equation with the help of integration factor?

6. May 5, 2009

### Cyosis

There is another way of doing this, however depending on whether you're a maths or physics student this may or may not be allowed. So I would ask you to check up on this method with your professor since this method is usually frowned upon by mathematicians. Physicists however use it all the time.

Divide by g-kv on both sides and multiple both sides with dt gives

$$\frac{dv}{g-kv}=dt$$

This is easily integrated, don't forget the constant!

7. May 5, 2009

### Staff: Mentor

This is a better, and simpler approach than the one I suggested. I was thinking that the equation was separable, but for some reason this very simple step didn't come to me.

8. May 7, 2009

### Emethyst

Thanks for all of that help, I managed to find the equation, but now i'm having problems with b) of this question, which states: Let vT=g/k. When v=vT, the acceleration, dv/dt, is zero and thus the object maintains a constant terminal velocity. Rewrite v(t) in terms of vT, and sketch the graph of v(t).

Graphing is no problem for me, I am just not sure how to find the new equation in terms of vT. So far I have the equation written as v(t)=vT(1-e^-kt), but don't know how to continue on with the question. Any help pointing me in the right direction here would be greatly appreciated, thanks.

9. May 7, 2009

### Cyosis

You've done what they asked of you, rewritten the equation in terms of v_T.