How was this ODE solution found? Doesn't seem to be the normal solution.

[InvisibleMan1]

Homework Statement

The solution for the differential equation on this page http://electron9.phys.utk.edu/phys135d/modules/m5/Friction.htm#Drag checks out, but I can't figure out how they found it. Both my solution and theirs check out. A couple people I asked for help reached the same solution I did.

The problem appears to be with the constant of integration. I don't see a way of getting from mine to theirs though.
What I am guessing theirs is: C = -((kv(0)/g + 1)
Mine: C = (kv(0))/g - 1

I'm starting to suspect that their solution is a very lucky typo. "Lucky" since it is actually a valid solution, unless I checked it wrong.

Homework Equations

The differential equation: dv/dt = g - (b/m)v

Their solution: v = (g/k) - [(kv(0) + g)/k]e^(-kt) Where k = b/m

The solution I found: v = (g/k) + [(kv(0) - g)/k]e^(-kt) Where k = b/m
The integrating factor I found: u = e^(kt)
The constant of integration I found: C = (kv(0))/g - 1

The Attempt at a Solution

Write the ODE in standard linear form, and solve it as a first-order linear ODE using an integrating factor u(x)=e^(int(p(x)dx). The solution I and a couple other people reached is given above.


On a side-note, I also can't figure out why they multiplied (kv(0))/g in C by k/k. I did the same thing in my solution just because I couldn't see a reason why not (and it makes it easier to compare the two solutions). I suspect the way they found their solution may have had something to do with k/k, but it is also possible they just did that to make it easier to do/show something.

 

[fzero]

For their solution

$$v(0) = -v_0.$$

They probably have a sign wrong. The only other possibility is that they're trying to choose a convention where the velocity is negative for a falling object, but that doesn't seem to be compatible with the other terms in their solution.

 

[InvisibleMan1]

Is my theory about it being a lucky typo correct then?

 

[fzero]

Is my theory about it being a lucky typo correct then?

I obtain the same solution as you did. Since you wrote v(0) rather than v₀ there's no room for confusion.

 

[InvisibleMan1]

Erm, sorry, but I don't see how that answers my question.

 

[fzero]

Erm, sorry, but I don't see how that answers my question.

Their solution is a solution to the DE, but the notation doesn't correspond well to the initial value condition, since

$$v_0 = - v(0).$$

The minus sign could very well be a typo, the only time they refer to v₀ is to say that it's the speed at t=0 and the sign didn't matter for that.

You got the right solution and handled the initial value condition the right way.

 

[InvisibleMan1]

Alright, thanks for the help.