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Skydiver differential equation

  • Thread starter darryw
  • Start date
  • #1
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Homework Statement


the velocity of a skydiver falling to the ground is governed by:
m dv/dt = mg - kv^2

solve for v(t) with initial conditions v(0) = 0

i rearranged equation into standard form:

v' + (k/m)v^2 = g

so i got an integrating factor, mu(x) = e^integ(k/m), but does the fact that the v is squared prohibit me from using an integrating factor?

thanks

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The Attempt at a Solution


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The Attempt at a Solution

 

Answers and Replies

  • #2
Cyosis
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Strictly a non linear differential equation doesn't prohibit you from using an integrating factor. It does however prohibit you from using the integrating factor in the way you did. You used a 'formula' for an integrating factor which was derived for linear ODEs, which does not apply in this case.

Hint: Note that the equation is separable.
 
  • #3
1,796
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No, you can't do that. Need y'+ay=b(x) for that. Surely you can take the expression:

[itex]a\frac{dv}{dt}=G(v)[/itex]

and separate variables to get [itex]a\frac{dv}{G(v)}=dt[/itex].

Try that.
 
  • #4
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how do you know just from looking at an equation if it is separable or not? What do you look for specifically? What is a giveaway that tells you that, nope this equation cant be separable also.
thanks
 
  • #5
Cyosis
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how do you know just from looking at an equation if it is separable or not? What do you look for specifically? What is a giveaway that tells you that, nope this equation cant be separable also.
thanks
You look whether or not you can put all v-s on one side of the equation and all t-s on the other side.

And yes this equation is separable.
 
  • #6
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m dv/dt = mg - kv^2

dv/dt + kv^2 / m = g

dv + v^2 = mg/k dt

its separated but it looks sort of weird to me??
also, can u please clarify this: that k,g and m are all constants, right? thanks
 
  • #7
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but if this is correct i will proceed, thank u.
 
  • #8
Cyosis
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No that is not correct. Compare your equation to the one in post #3.
 
  • #9
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i glossed over post 3 because i wasnt sure where the a comes from, and why the caps G.
Is the a acceleration?
thanks
 
  • #10
Cyosis
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'a' is just a constant and with G(v) he means a function that depends on v. He might as well have called it f(v) or g(v) etc.
 
  • #11
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53
Sorry you're having problems with my notation. You have got to review Separation of Variables today in some elementary DE text book. Tons in the library. You need to get the DE in the form [itex]A(x)dx+B(y)dy=0[/itex] to separate the variables. You have:

[itex]mdv=(mg-kv^2)dt[/itex]

now I bet you can put that into the form [itex]A(v)dv=B(t)dt[/itex] and thus separated. Wouldn't that just be [itex]\frac{m}{mg-kv^2}dv=dt[/itex]
 
  • #12
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actually i do understand what you are doing, but i think all the constants are confusing me.

so, just to clarify: m, g and k are all constants right?

Also I was trying to get all the constants to go with "dt", but am i right in saying that it doesnt matter what side of the equation i put the constants?

(i was trying to get it so that equations was v^2 dv = "something" dt and it was really difficult.

So i now have:

m/mg-kv^2 dv = (1)dt

integrate both sides...

so, because m,g and k are all constants i can take them out of the integral right?

m/(mg-k) integ 1/(v^2) = (1)dt

m/(mg-k) * -(1/v) = t+c

m/(mg-k)v = t+c

v = (t/m) (mg-k) + (c/m) (mg-k)

is this correct so far? thanks
 
  • #13
Cyosis
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You really need to brush up on basic algebraic manipulations.

so, because m,g and k are all constants i can take them out of the integral right?

m/(mg-k) integ 1/(v^2) = (1)dt
What you're claiming here is that [itex]mg-kv^2=(mg-k)v^2[/itex]. I suggest you work out the brackets so you can see it's wrong.
 
  • #14
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i have the following equality, that i need to now integrate (correctly)

integ: m/(mg-kv^2) dv = integ :(1)dt

heres how i integrated LHS...

u = mg-kv^2
du = -2kv

dv = m
v = v

integ: m/(mg-kv^2) dv = mg-kv^2(v) - integ (v)(-2kv)

................................= mg-kv^2(v) - (-2)((1/3)(v^3)

simplify..

................................ = mgv-kv^3 +(2v^3/3)

then add the RHS integration

mgv-kv^3 +(2v^3/3) = t + c

is this correct so far?
 
  • #15
Cyosis
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u = mg-kv^2
du = -2kv
Wrong if [itex]u=mg-kv^2[/itex] then [itex]du=-2kvdv[/itex]

dv = m
v = v
What are you doing here?

integ: m/(mg-kv^2) dv = mg-kv^2(v) - integ (v)(-2kv)
It seems that you have tried to do something like partial integration. Notice how [itex]mg-v^2[/itex] is in the denominator on the left side and suddenly in the numerator on the right side.. I have no idea how you managed to do that, but I can tell you it is wrong.

As a result all other steps are wrong as well.

integ: m/(mg-kv^2) dv = integ :(1)dt
This is the only correct part of your post.

I suggest you use [itex]1-v^2=(1-v)(1+v)[/itex] in combination with partial fractions.
 
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  • #16
1,796
53
May I offer a modern approach which I believe may help Darry but is controversial. I do realize I'm new here and don't wish to cause trouble. Let me know.

You have:

[itex]\frac{dv}{mg-kv^2}=dt[/itex]

and you know the right side is just t+c and you want to integrate the left side. Here comes the controversial part:

First enter the equation in Wolfram Alpha:

Wolfram Alpha: Integrate[m/(m g-k v^2),v]

Now that gives you an answer in terms of Arctanh. So immediately, you go to your Calculus text and find "Integrals yielding Inverse Hyperbolic Functions" and I find the one that says:

[itex]\int \frac{du}{a^2-u^2}=\begin{cases} 1/a \tanh^{-1} u/a+c &|u|<a \\
1/a \coth^{-1} u/a+c & |u|>a
\end{cases}
[/itex]

Now, it's not difficult to put your integral into that a^2-u^2 form which you can then integrate using the formula above. Try this approach in general when you have integration problems but always try to do it yourself first and not come to rely on a CAS to do your work for you.
 
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