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1st order diff. equation problem

  • #1

Homework Statement


Problem asks what 'A' and 'a' represent,(which I think represent air resistance and acceleration), and also asks if this equation accurately describes the descent of a falling object taking air resistance into account.

The question can be found here:
http ://edisk. fandm. edu/annalisa.crannell/writing_in_math/projects/ Grapefruit. html
(have to get rid of the spacings)


Homework Equations


v=A(e^(at-1))/(e^(at+1))


The Attempt at a Solution


Tried starting from the beginning using integrating factors like e^(at-1) or just (at-1), but I don't understand how you can end up with an answer that has an e^(x)/e^(x). The book would have me get the equation in the form y'+xy=C and use x as the integrating factor, but what x could I use to get the above equation? Any help would be much appreciated.
 

Answers and Replies

  • #2
HallsofIvy
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I am told that that website address does not exist. Not knowing what the question is, we can't answer.
 
  • #3
tiny-tim
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Welcome to PF!

v=A(e^(at-1))/(e^(at+1))
Hi Sprinkle159! Welcome to PF! :smile:

I don't think it's (e^(at-1))/(e^(at+1)).

I think it's (e^(at) - 1)/(e^(at) + 1), which is the same as tanh(at/2).

Does that help? :smile:
 
  • #4
Thank you for the reply you were correct about it being (e^(at)-1)/(e^(at)+1), which looks much more like an equation that takes into account air resistance. I still have a couple questions though.

Problem can be found here: ****(PLEASE remove the spaces between h t t p)****
h t t p://edisk.fandm.edu/annalisa.crannell/writing_in_math/projects/Grapefruit.html


So my problem now is how to undo the differential equation to get the initial problem. From what I understand from my book you start with a problem like :
xy' + zy=K.
From here it has you divide the entire equation by x to get y' alone, giving:
y' + (z/x)y=K/x.
From here you use (z/x) as the integrating factor, and take:
e^(integral of z/x) which gives you: e^(integral of z/x)y'=k/x.
From here you'd get the integral of k/x:
e^(integral of z/x)y=(integral)k/x.
Then I'd divide the left side to isolate y.
y=(integral k/x)/(e^integral of z/x).

So when comparing this to the problem I'm given, which is y=A(tanh(at/2)) Im a bit confused at how to find the initial x and z values.
 
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  • #5
On second though: e^(integral of x)=(e^(x)+e^(-x))/2
integral(x)= ln(e^(x)+e^(-x))/2)
x=(derivative) ln(e^(x)+e^(-x))/2)
using chain rule gives: (2/(e^(x)-e^(-x)) * (2/(e^(x)-e^(-x))=
4/((e^(x)-e^(-x))^2)
does this seem reasonable?
 
  • #6
tiny-tim
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So my problem now is how to undo the differential equation to get the initial problem. From what I understand from my book you start with a problem like :
xy' + zy=K.
From here it has you divide the entire equation by x to get y' alone, giving:
y' + (z/x)y=K/x.
From here you use (z/x) as the integrating factor, and take:
e^(integral of z/x) which gives you: e^(integral of z/x)y'=k/x.
From here you'd get the integral of k/x:
e^(integral of z/x)y=(integral)k/x.
Then I'd divide the left side to isolate y.
y=(integral k/x)/(e^integral of z/x).

So when comparing this to the problem I'm given, which is y=A(tanh(at/2)) Im a bit confused at how to find the initial x and z values.
Hi Sprinkle159! :smile:

I'm sorry … I don't understand any of this.

The given solution is for vertical motion, so what are x and z supposed to be?
On second though: e^(integral of x)=(e^(x)+e^(-x))/2
integral(x)= ln(e^(x)+e^(-x))/2)
x=(derivative) ln(e^(x)+e^(-x))/2)
using chain rule gives: (2/(e^(x)-e^(-x)) * (2/(e^(x)-e^(-x))=
4/((e^(x)-e^(-x))^2)
does this seem reasonable?
Yes … d/dx (log coshx) = sinhx/coshx = tanhx …

but how does that help? :confused:
 
  • #7
Oh no I just had x and z there to represent equations. I should have explained that the problem description says: "concept: differential equations; using initial values to find exact solutions"
So I went to the section in my calculus book talking about first order linear differential equations, which explained that it is an equation of the form: (dy/dx) + P(X)y=Q(X)
The problem as I saw it was to take equation he gave me for velocity which was
V=(A)tanh(at/2) and put it back into the original form of (dy/dx) + P(X)y=Q(X).
I'm trying to verify the v=(A)tanh(at/2) is the correct formula for velocity taking air resistance into account, and since the problem hint mentioned differential equations I thought I should start there and see what happens when I get the original problem. Does this make sense or am I moving in the wrong direction? (I've been assuming a=acceleration due to gravity and I'm not quite sure what to make of the big A; the tanh(at/2) seems to mimic very well what you'd expect air resistance to be as v increases)

as an ex from the book:
y' + (1/x)y=2
integrating factor = x
xy' +y=(xy)'
(xy)'=2x
y=x+ (C/x)

And so what I was thinking with (4/((e^(x)-e^(-x))^2)) was that it was the orginal P(x) term in the orginal equation of the form
(dy/dx) + P(X)y=Q(X). Although looking at it I'm not sure how this will help me prove that v=(A)tanh(at/2) is the correct formula for velocity.
 
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