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?