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Manipulation of Arbitrary Constants in Differential Equations

  1. Mar 6, 2013 #1
    1. The problem statement, all variables and given/known data

    yy''+(y')2 = 0

    2. Relevant equations

    yv(dv/dy)+v2=0

    3. The attempt at a solution

    Variable separable when solving for the first step the result is:

    - ln |v| = ln |y| + C1

    Now, I've looked at the remainder of the solution with a few other sources and the cause of my mistake results in the constant.

    After turning the equation into: ln |vy| + C1 = 0;

    I raise everything to the e so that I can solve for v.

    It seems all the solutions do that as well but yet they don't raise the constant to the e.

    Is it because eC1 will still be a constant and therefore completely arbitrary?
     
  2. jcsd
  3. Mar 6, 2013 #2

    SammyS

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    [itex]\displaystyle e^{C_{\,1}}\ [/itex] is positive, otherwise it's as arbitrary as using a redefined C1 .
     
  4. Mar 6, 2013 #3

    HallsofIvy

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    You could, at this point, take the exponential of both sides:
    [itex]e^{-ln|v|}= e^{ln|y|+ C_1}[/itex]
    [itex]\frac{1}{|v|}= e^{C_1}|y|[/itex]
    Now, as you say, since C1 is an arbitrary constant, so is [itex]e^{C_1}[/itex] so let's just call it "C2". And if allow C2 to be either positive or negative, it we can drop the absolute values: [itex]\frac{1}{v}= C_2y[/itex] which is, of course, the same as [itex]yv= \frac{1}{C_2}[/itex] or [itex]yv= C_3[/itex] where [itex]C_3= 1/C_2[/itex].
    Some texts just don't bother to label the different constants differently.

     
  5. Mar 6, 2013 #4
    That makes sense, thank you.

    Seems like they just don't want things to get messy so they reuse the same constants.
     
  6. Mar 6, 2013 #5

    HallsofIvy

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    A "conservation of constants" law!!!
     
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