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Clarification regarding the proper use of constant C.

  1. Sep 24, 2011 #1
    Please forgive the rather simple nature of my question. I am rather confused about it. I have just started a class on differential equations and I am confused on the proper use of the constant C after integrating. There are occasions in which I end up with something like
    y = et + c. If this is the same as etec, then is ec essentially the same as c? Can anyone enlighten me as to the proper use of c in similar situations such as e6t + 6c, etc.?
  2. jcsd
  3. Sep 24, 2011 #2
    c is usually a variable you find so that your solution satisfies given initial conditions. Typically c is in front of the exponent. Ie c1 exp^2t.
  4. Sep 25, 2011 #3


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    Remember that c is an arbitrary constant. There are often different ways to write the "same" constant value.

    As a simple example, if you integrate [itex]\int 2x\, dx[/itex], it doesn't really matter if you write the answer is [itex]x^2 + c[/itex] or [itex]x^2 + c + 1[/itex] or [itex]x^2 - c[/itex], or whatever. The expressions give different values for a particular value of c (for example c = 0) but the complete set of expressions for every possible value of c is the same in each case.

    But in that example, there isn't any obvious reason to write anything apart from [itex]x^2 + c[/itex], so that is what you will find in a textbook as "the answer".

    In your example, the same idea applies, but the alternative ways to write the answer look "more different" than the previous example.

    If [itex]y = e^{t+c}[/itex] then you are right that [itex]y = e^t e^c[/itex]. You would normally let [itex]e^c = a[/itex] where [itex]a[/itex] is another constant, and write [itex] y = ae^t[/itex] rather than [itex]y = e^t e^c[/itex].

    If this was part of a bigger problem, sometimes [itex]y = e^{t+c}[/itex] is easier to work with, and sometimes [itex] y = ae^t[/itex]. You have to get used to the fact that in math, there are often different ways of writing "the same thing".
  5. Sep 25, 2011 #4
    Thank you so much. Great explanation. It answered my question.
  6. Sep 25, 2011 #5


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    Sometimes a particular form of the constant will cause one to gain or lose solutions.
    y1=C exp(t)
    -exp(t) may of may or not be a solution y1 tends to make us think of it as one while y2 may make us forget about it (if working wiht real numbers).
    we might forget to consider y=1 as a possible solution while some other form may remind us of that possiblity.
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