1st order diff. equation problem

• Sprinkle159
So I'm not sure where to go from here.In summary, the conversation discusses finding the equation for the descent of a falling object taking air resistance into account. The given equation is v=(A)tanh(at/2). The individual asking the question is trying to verify if this equation is correct and is using the concept of first order linear differential equations to do so. However, there is confusion about the variables x and z and how they relate to the given equation. The conversation also mentions that the given equation accurately represents air resistance as velocity increases.

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.

I am told that that website address does not exist. Not knowing what the question is, we can't answer.

Welcome to PF!

Sprinkle159 said:
v=A(e^(at-1))/(e^(at+1))

Hi Sprinkle159! Welcome to PF!

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?

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)) I am a bit confused at how to find the initial x and z values.

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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?

Sprinkle159 said:
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)) I am a bit confused at how to find the initial x and z values.

Hi Sprinkle159!

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?
Sprinkle159 said:
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?

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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