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Math Help - Integrating the integral of ydx

  1. #1
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    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?
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  2. #2
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    Quote Originally Posted by SteveBasnof View Post
    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*
    Attached Thumbnails Attached Thumbnails Integrating the integral of ydx-picture15.gif  
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  3. #3
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    Joined
    Mar 2007
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    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.
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