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First order Linear DE

  1. Jul 30, 2006 #1
    Hello all,

    I'm trying to prove to myself that the following solution to the DE shown works. I can't start using it until i prove to my self it works (it's this psycological thing i have were i can't use anything unless i know where it comes from). :rofl:

    Here is the Equation and it's solution
    http://img142.imageshack.us/img142/3437/defa7.png

    and here is me trying to prove to my self it works...
    http://img137.imageshack.us/img137/1831/desolhv6.png


    Am I doing anything wrong? Or can anyone please show me a proof which shows that this is a solution to the Differential Equation?

    Thanks in advance :redface:


    edit: I don't think this is a homework question, as you know, im just trying to prove it to my self.
     
  2. jcsd
  3. Jul 30, 2006 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Let me see if I can put into Latex what you have so others won't have to wait for those to load:

    Your first reference asserts that the solution to the first order, linear, differential equation
    [tex]\frac{dx}{dt}= ax(t)+ f(t)[/tex]
    with x(0) and f(t) given is
    [tex]x(t)= e^{at}x(0)+ \int_0^t e^{a(t-s)}ds[/itex]

    2) You method of solution is: an integrating factor for the problem is e-at so
    [tex]e^{-at}\frac{dx}{dt}= ae^{-at}x(t)+ e^{-at}f(t)[/tex]
    [tex]e^{-at}\frac{dx}{dt}- ae^{-at}x(t)= e^{-at}f(t)[/tex]
    [tex]\frac{de^{-at}x}{dt}= e^{-at}f(t)[/tex]

    Yes, so far this is completely correct. You then integrate to get
    [tex]e^{-at}x(t)= \int e^{-at}f(t)dt+ C[/tex]
    so
    [tex]x(t)= e^{at}\int e^{-at}f(t)dt+ Ce^{at}[/tex]
    and try to determine C by setting t= 0
    [tex]x(0)= e^{a0}\int e^{-a(0)}f(0)dt+ Ce^{a0}[/tex]
    That's your mistake! You are treating the "t" inside the integral as if it were the same as the "t" outside. It's not- it's a "dummy" variable.
    Remember that [itex]\int_0^1 t^2dt[/itex]= 3. You can't "set" t equal to 0 and declare that [itex]\int_0^1 0^2 dt= 3[/itex]!

    Go back and use a different variable in your integral:
    [tex]e^{-at}x(t)= \int^t e^{-as}f(s)ds[/tex]
    Notice the single "t" as a limit on the integral. That tells people we mean for the final result of the integral to be in the variable t. Also notice there is no "C". Strictly speaking, that is included in the indefinite integral. A better technique, which you should learn, is to write that indefinite integral as a definite integral with a variable limit:
    [tex]e^{-at}x(t)= \int_{0}^t e^{-as}f(s)ds+ C[/tex]
    I now have "+ C" because choosing a lower limit is the same as choosing a specific constant for the indefinite integral which we don't want to do yet.
    I took the lower limit as 0 because we know x(0). The upper limit is the variable t. Of course, if t= 0, that integral is from 0 to 0 and so is 0 no matter what is being integrated:
    [tex]e^{-a(0)}x(0)= x(0)= \int_0^0 e^{-as}f(s)ds+ C= C[/tex]
    so
    [tex]e^{-at}x(t)= \int_0^t e^{-as}f(s)ds+ x(0)[/tex]
    Now multiply by eat to get
    [tex]x(t)= e^{at}\int_0^t e^{-as}f(s)ds+ x(0)e^{at}[/tex]
    [tex]x(t)= \int_0^t e^{a(t-s)}f(s)ds+ x(0)e^{at}[/tex]
    as claimed. (Of course we can take that eat inside the integral as if it were a constant because it does not depend on the variable of integration, s.)

    (You don't think this is a homework question? Don't you know for sure?:rofl:
     
  4. Jul 31, 2006 #3
    HallsofIvy,

    Thank you very much for that. What a disgrace, this is even a fundemental part of first year calculas!. I kind of knew something was wrong in that line - hence those red question marks. :redface:

    As for the homework part, :tongue2: What i meant to say was that i don't think this should be in the homework section (I wasn't sure what constitutes as "homework" in this forum). And Latex looks powerful, I think i'd better learn it.
     
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