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Determing Whether A Function Is A Solution To A Differential Equation

  1. Aug 28, 2013 #1
    The differential equation: y' -2ty = 1.

    The possible solution: [itex]y=e^{t^2} \int^6_0e^{-s^2}ds + e^{t^2}[/itex].

    For the integral, I employed integration by parts:

    Let [itex]u=e^{-s^2} \rightarrow du = -s2e^{-s^2}ds[/itex]

    and

    Let [itex]dv = ds \rightarrow v=s[/itex].

    This lead to:

    [itex][se^{-s^2}|^t_0 - \int_0^t -2s^2e^{-s^2}ds[/itex]

    My first thought was to perform another integration by parts; however, after having run through the process in my mind, this would seem of no avail. What am I missing?
     
  2. jcsd
  3. Aug 28, 2013 #2
    Okay, so the point is you can't put that integral into closed form. Just leave it as it is. What you _can_ do is differentiate is. So what you want to do is simply differentiate your possible solution and put that back into the original differential equation.
     
  4. Aug 28, 2013 #3
    How would you define "closed form?"
     
  5. Aug 28, 2013 #4

    HallsofIvy

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    That function, [itex]e^{-x^2}[/itex] is well known NOT to have any elementary integrand. The "error function", erf(x), used in statistics and probability, is defined to be its integral and tables of the normal distribution are created using numerical integration.

    In any case, there is no reason to actually do the integration- since it is a definite integral it is simply a constant. The problem is actually asking you to show that [tex]y(t)= Ce^{t^2}+ e^{t^2}= De^{t^2}[/tex]. where D= C+ 1, satisfies that equation. Differentiate that to find y', put it into the equation and see if the equation is satisfied.
    (Note that the problem says "possible" solution.)
     
  6. Aug 28, 2013 #5
    I don't think so, Halls. I think he just mis-wrote the first integral. It's supposed to be erf(t) not erf(6). Just look at his later calculations.

    Anyway, what Halls said is basically what I meant. 'Closed form' means that you can write the integral in terms of elementary functions.
     
  7. Aug 28, 2013 #6

    HallsofIvy

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    Okay, if it is in fact supposed to be [itex]y= e^{t^2}\int_0^t e^{-s^2}ds+ e^{t^2}[/itex], then use the "Fundamental Theorem of Calculus to differentiate:
    [itex]y'= 2te^{t^2}\int_0^t e^{-s^2}ds+ e^{t^2}(e^{-t^2})+ 2te^{t^2}[/itex]
    But "[itex]2t(e^{t^2}\int_0^t e^{-s^2}ds+ e^{t^2})[/itex]" is just [itex]2ty[/itex] and [itex]e^{t^2}e^{-t^2}= 1[/itex].
     
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