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Calc I - Why is it possible to combine these constants of integration?

  1. Oct 25, 2005 #1

    I have the following equation

    [tex]\int \!v{dv}=-{\it GM}\,\int \!{y}^{-2}{dy}\]}[/tex]

    Integrating I get...

    [tex]\[\frac{1}{2}v ^{2}+C _{1}=-GM\essdot\left(-\frac{1}{y }+C _{2}\right)\]}

    Now how do I go from what I have above to this

    [tex]v ^{2}=2{\it GM} \left(\frac{1}{y }\right)+C \]}[/tex]

    How it is possible to combine the separate constants of integration into a single constant of integration? Do these two equations follow in logical steps or are there steps missing which would help explain? Any web sites which could help me understand why we can do this?

    I'm also having problems previewing my latex? Why does it always say reload?

    Also whenever after I go back to the editing box after previewing I lose all of my line breaks.
    Is there any way to prevent that? I always have to copy, preview, go back and paste, then edit... Firefox problem?

    Last edited: Oct 25, 2005
  2. jcsd
  3. Oct 25, 2005 #2


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    a + C1 = b + C2, where a and b are variables, C1 and C2 constants of integration.


    a + C1 = b + C2
    a = b + (C2 - C1)
    Let C = C2 - C1
    a = b + C
  4. Oct 25, 2005 #3
    Your explanation makes sense but I don't see how to apply it to the above equations.
  5. Oct 25, 2005 #4
    [tex]\frac{1}{2}v ^{2}=-GM\left(-\frac{1}{y }+C _{2}\right)-C _{1}$}

    Simplifying the right side of the equation gives us

    [tex]\[\frac{1}{2}v ^{2}=\frac{{\it GM} }{y }-GMC _{2}-C _{1}\]}[/tex]

    I'm trying to step through this algebraically to help me understand but I don't think it is helping. How can C1 and C2 be combined as a single constant from the above equation?
    Last edited: Oct 25, 2005
  6. Oct 25, 2005 #5
    C1 and C2 are just real numbers. You just add like terms in an equation:

    \int vdv + GM\int y^{-2}dy = (\frac{1}{2}v + C_1) + (GM\cdot (-\frac{1}{y} + C_2)) = \frac{1}{2}v - \frac{GM}{y} + C_1 + GMC_2

    Because [itex]C_1[/itex] and [itex]GMC_2[/itex] are real constants, you just add them up to get

    \frac{1}{2}v - \frac{GM}{y} + C

    - Kamataat
    Last edited: Oct 25, 2005
  7. Oct 25, 2005 #6
    What makes GMC_2 a constant?

    In this problem it is a constant because G is the gravitational constant and M is the mass of the Earth.
    Is that why you say the above is a constant, because you recognized the above equation?
    But what if GM was lets say Z which was a variable? Is ZC_2 still a constant? Why?
    Last edited: Oct 25, 2005
  8. Oct 25, 2005 #7
    I sorta guessed G and M are constants. Of course, I could have been wrong. My apologies for that, then.

    Of course, if GM is not a constant, but is some non-constant function of a variable, then you can't add them like that.

    edit: PS: You can add them b/c the derivative of a constant is zero, so it won't make a difference if there's one or two (or more) of them.

    - Kamataat
    Last edited: Oct 25, 2005
  9. Oct 25, 2005 #8
    Ok, I understand now. Thanks for taking the time to explain this problem to me :)
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