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First-order linear Diff Eqs

  1. Dec 9, 2012 #1
    http://img22.imageshack.us/img22/9595/capturecv.png [Broken]

    This image is from Thomas's calculus, 10th edition, chapter 9 section 2. It is solving an example problem about first-order linear differential equations.

    To solve the equation, one needs to multiply the standard form of it by [itex] v = e^{\int{Pdx}} [/itex].

    I can't seem to understand the part highlighted in blue. Why is the constant of integration equal to 0 (when v is being computed)? Are you always allowed to do this?

    BiP
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Dec 9, 2012 #2

    micromass

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    In order to calculate v, you need to compute the integral

    [tex]\int P(x)dx[/tex]

    But those integrals always have infinitely many solutions. For example,

    [tex]\int \frac{-3}{x}dx = -3 ln|x| + C.[/tex]

    So if we let C vary, then we get other solutions.

    Now, every possible solution will give you a valid integrating factor (why?). For example, the functions

    [tex]e^{-3 ln |x|},~~ e^{-3 ln|x| +1},~~ e^{-3 ln|x|+100^{100^\pi}}[/tex]

    are all valid integrating factors and they will all help you solve the equation.

    However, the use of an integrating factor is to multiply them with the ODE. So we don't need full generality. If we just find an integrating factor, then we can already solve the ODE. Nobody cares about finding all possible integrating factors, we just need one.

    If you're going to choose an integrating factor, then it makes sense to choose it as simple *** possible. The simplest one will clearly be the one where C=0.
     
  4. Dec 9, 2012 #3
    Thank you micro, but I am afraid I don't compltely understand.

    So are you saying that every value of C yields the same solution, so the choice of C=0 simplifies things?

    Or are you saying that we choose C=0 obtaining a particular solution? But in this case, why are we only aiming for a particular solution if we want the set of general solutions to the ODE?

    Thanks again.

    BiP
     
  5. Dec 9, 2012 #4

    micromass

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    Yes, this is right.

    Try it for yourself. Take the most general integrating factor (so the one that still has the C). Multiply that with your ODE and solve the ODE. Do you get the same solutions as when you took the particular case C=0?
     
  6. Dec 13, 2012 #5
    Try using this integrating factor approach with C = ln k, for example. Multiply the entire differential equation by the integrating factor. You will see that a factor of k can be factored out of all terms on both sides of the equation, with no loss of generality.
    Chet
     
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