# Integrating the integral of ydx

• Mar 8th 2007, 06:14 PM
SteveBasnof
Integrating the integral of ydx
Ok well I'm doing a problem involving throwing a ball up in the air and figuring out it's position function with consideration of air resistance.
The short in sweet is that I'm trying to integrate the integral Vdt, where V is velocity and is dependent on t, time. My calculator says the integral ydx = x*y, but I believe the calculator thinks y is just a generic variable not dependant on x. Here's the details.

I've come to the following formula.

m(dV/dt) = pV - mg

where m=mass, V=velocity, t=time, g=gravity, p=some constant.

I need to get this equation into closed form solving for V, but I've come across a problem dealing with the equation.

If the equation was:

m(dV/dt) = pV it would be simple enough, just make it: m(dV)/V = p(dt)
and do simple implicit differentiation.

However since I have the added "- mg" in the equation I can't seperate out V and t to different sides of the equals.

So I did this:
m(dV/dt) = pV - mg
becomes

dV = ((p/m)V -g)dt

If I integrate both sides:

V = (p/m)*integral(Vdt) - gt + C.

But I still can't figure out the correct way to integrate integral(Vdt). I did the rest of the problem assuming the calculator was right and integral(Vdt) = Vt, but when I plotted the function I found when solving for C against the slope field in winplot it didn't line up quite right. Though it was close.

I tried integration by parts with: int(udv) = uv - int(vdu), and came across this.

int(Vdt) = Vt - int(t(dV/dt))

I end up with the same mess, instead of Vdt, I am stuck at t(dV/dt).

If I use integration by parts for int(t(dV/dt) and plug it into the above equation I end up with: int(Vdt) = int(Vdt) :/ so that doesn't really help me at all.

Anyone know what it is I'm not seeing here?
• Mar 8th 2007, 06:55 PM
ThePerfectHacker
Quote:

Originally Posted by SteveBasnof
Ok well I'm doing a problem involving throwing a ball up in the air and figuring out it's position function with consideration of air resistance.
The short in sweet is that I'm trying to integrate the integral Vdt, where V is velocity and is dependent on t, time. My calculator says the integral ydx = x*y, but I believe the calculator thinks y is just a generic variable not dependant on x. Here's the details.

I've come to the following formula.

m(dV/dt) = pV - mg

where m=mass, V=velocity, t=time, g=gravity, p=some constant.

I need to get this equation into closed form solving for V, but I've come across a problem dealing with the equation.

If the equation was:

m(dV/dt) = pV it would be simple enough, just make it: m(dV)/V = p(dt)
and do simple implicit differentiation.

However since I have the added "- mg" in the equation I can't seperate out V and t to different sides of the equals.

So I did this:
m(dV/dt) = pV - mg

I am not sure what you are trying to do? Apparently you are trying to solve a differencial equation. It seems that you are trying to find an integrating factor. But when I look upon thy work it seems like something else.
You know there exists a formula.

Here is a general formula I give below.
*Bring everything to the form in the formual to use it properly*
• Mar 8th 2007, 07:18 PM
SteveBasnof
Hmmz, well that's a helpful formula but since I don't know how it is derived I can't use it in this particular way.

However I think I found my error.

I was trying to get the following form of an equation into closed form.

(dy/dx) = y - k where k is a constant.

If it was: (dy/dx) = y
I could change it to: (dy)/y = dx, then integrate to get:
ln(y) = x --> y = exp(x).

With the k constant though I was thinking that I couldn't do this, but after looking over your post and thinking, I realized that I can do

(dy/dx) = y - k
(dy)/(y - k) = dx

then integrate from there.

I was just making the problem more difficult than it was. Thanks for your help.