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DEQ/Calculus (FTC) Question Simplifying Nonintegrable Function

  1. Jul 22, 2011 #1
    1. The problem statement, all variables and given/known data
    Greetings,

    I am having trouble understanding the solution to this DEQ equation. I am able to solve the equation using the variation of parameters technique, however, when it comes to simplifying the integral I am lost. It appears they are using the Fundamental Theorem of Calculus to simplify the integral, but I am confused as to how this is done?


    2. Relevant equations
    1) Original DEQ equation:

    [itex]
    y''-5y'+6y=g(t)
    [/itex]

    2) Solving the characteristic equation of the homogenous equation yields:

    [itex]
    y_{H}=c_{1}e^{2t}+c_{2}e^{3t}
    [/itex]

    3) Solving for [itex]y_{1}, y'_{1}, y_{2}, y'_{2}[/itex], and the Wronskian, and plugging into the theorem of variation of parameters yields:

    [itex]
    Y(t)=-e^{2t}\int{\frac{e^{3t}g(t)}{e^{5t}}dt}+e^{3t}\int{\frac{e^{2t}g(t)}{e^{5t}}dt}
    [/itex]

    4) Simplified:

    [itex]
    Y(t)=-e^{2t}\int{e^{-2t}g(t)dt}+e^{3t}\int{e^{-3t}g(t)dt}
    [/itex]

    5) BUT, Texbook Solution:

    [itex]
    Y(t)=\int{[e^{3(t-s)}-e^{2(t-s)}]g(s)ds}
    [/itex]

    3. The attempt at a solution
    So, my problem is how to get from 4) to 5)? I realize they probably applied the FTC, but I am not clear on how to use this to derive the solution in the textbook? The farthest I can get is as below:

    [itex]
    Y(t)=-e^{2t}\int_{0}^{s}{e^{-2t}g(t)dt}+e^{3t}\int_{0}^{s}{e^{-3t}g(t)dt}
    [/itex]

    Any help is greatly appreciated.
     
  2. jcsd
  3. Jul 22, 2011 #2
    Just throwing out a guess, but perhaps the book used Laplace Transforms since it introduced s into the solution and was able to make the answer look a little neater.
     
  4. Jul 22, 2011 #3
    While that is possible, I don't think they used a LaPlace transform since that topic is not introduced until section 6 of the textbook and this problem is from section 3.

    Therefore, I'm still under the assumption that they are using the fundamental theorem of calculus to somehow simplify the integral, but I don't know how to do this?
     
    Last edited: Jul 22, 2011
  5. Jul 22, 2011 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi Fish Sauce! Fishy welcome to PF! :smile:
    No no no no noooo :cry:

    you can't have the same variable (t) inside the integral and outside it …

    the inside variable is a "dummy" variable, and disappears on the integration …

    use s inside this integral instead of t, and put t (and 0) as the limit :wink:
     
  6. Jul 22, 2011 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    No, Laplace transforms are not relevant here. But be careful about the "variable of integration", also called the "dummy variable" is important!

    If you have the variable "t" for functions outside the integration, and want the result of the integral to be in terms of "t", it is a very bad idea to use "t" for the "dummy" variable of integration. If you write
    [tex]-e^{-2t}\int \frac{e^{2t}g(t)}{e^{5t}}dt[/tex]
    there is a grave danger that you will (mistakenly) take the "[itex]e^{-2t}[/itex] inside the integral to get
    [tex]-\int \frac{g(t)}{e^{5t}}dt[/tex]
    which is, of course, incorrect.

    Instead, use t outside the integral and, say, "s" for the "dummy" variable:
    [tex]-e^{-2t}\int \frac{e^{2s}g(x)}{e^{5s}}ds[/tex]
    to make the difference explicit. To make it clear that you want the integral to be written as a function of t, you can make the upper limite "t":
    [tex]-e^{-2t}\int_a^t \frac{e^{2s}g(x)}{e^{5s}}ds[/tex]
    where the lower limit, a, can be any thing you want- that only affects the choice of "constant of integration". Here you only want a single solution to the entire equation so you can choose "a" to be anything you want. In this example, they chose it to be 0.
     
  7. Jul 22, 2011 #6
    Re: Welcome to PF!

    Hi tiny-tim and HallsofIvy,

    Thank you for the help.

    Following the advice, my solution now looks like this?

    [itex]
    Y(t)=-e^{2t}\int_{0}^{t}{e^{-2s}g(s)ds}+e^{3t}\int_{0}^{t}{e^{-3s}g(s)ds}
    [/itex]

    Assuming the answer above is correct, how does this simplify to this form?

    [itex]
    Y(t)=\int{[e^{3(t-s)}-e^{2(t-s)}]g(s)ds}
    [/itex]
     
  8. Jul 22, 2011 #7

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi Fish Sauce! :smile:

    Just shove that t stuff through the ∫ …

    it's a constant (relative to s), so you can do that. :wink:
     
  9. Jul 22, 2011 #8
    Thank you tiny-tim.

    I did not consider t as a constant relative to the s integral. But since t is considered a constant (since it is not a function of s) then I can see how it can be moved inside the integral.

    Thank you all for your help!!!
     
  10. Jul 22, 2011 #9
    Re: Welcome to PF!

    That's an interesting method I never saw before; very nice!


    I did want to mention one thing...
    I've always seen functions of t placed directly in front of integrals like that when using variation of parameters. The general formula usually has y instead of something explicitly with the variable t, but I know in diff eq class, we put the actual function y in front of the integral in parentheses...
     
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