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Solution to first order ODE

  1. Mar 4, 2014 #1
    1. The problem statement, all variables and given/known data

    Can anyone point out where I have gone wrong with this?

    Verify that the given function is a solution of the differential equation.

    [tex]y' -2ty =1[/tex] [tex]y= e^{t^2}\int^t_0 e^{-s^2}ds+e^{t^2}[/tex]






    3. The attempt at a solution

    The steps I have taken are the following:

    i) Evaluate integral in expression for y
    ii) Differentiate the expression for y
    iii)Substitute these values into [tex]\frac{dy}{dt}-2ty=1[/tex]



    i)Evaluate the integral

    [tex]\int^t_0e^{-s^2}ds[/tex]

    by substitution:

    [tex]u=s^2[/tex]; [tex]ds=\frac{1}{-2s}du[/tex]

    [tex]\frac{1}{-2s}\int^t_0e^{u}du = \left[\frac{e^{-s^2}}{-2s}\right]^t_0 =\frac{e^{-t^2}}{-2t}[/tex]



    ii)Differentiate expression for y

    [tex]y=e^{t^2}\cdot \frac{e^{-t^2}}{-2t}+e^{t^2}[/tex]

    [tex] y=e^{t^2}-\frac{1}{2}t^{-1}[/tex]

    [tex]y'=2te^{t^2}+\frac{1}{2}t^{-2}[/tex]



    iii) Substitute these into [tex]y'-2ty=1

    [tex]2te^{t^2}+\frac{1}{2}t^{-2} -2t(2te^t{2}+\frac{1}{2}t^{-2})[/tex]

    [tex]2te^{t^2}+\frac{1}{2}t^{-2}-4t^2e^{t^2}-t^{-1}[/tex]


    Which is not what is required.
     
  2. jcsd
  3. Mar 4, 2014 #2
    Hi jellicorse!

    The quoted step is incorrect. There exists no antiderivative for ##e^{-x^2}##. While evaluating the integral, you took ##s## outside the integral but ##s## is a function of ##u## so what you did is not valid.

    For the given problem, you have to differentiate the given function using Fundamental Theorem of Calculus. Can you find the derivative wrt t for:
    $$\int_0^t e^{-s^2}\,ds$$
    ?
     
  4. Mar 4, 2014 #3
    Thanks a lot Pranav!

    I need to go and revise the Fundamental Theorem of Calculus again to check...
     
  5. Mar 4, 2014 #4
    Oh, I forgot. Welcome to PF! :smile:
     
  6. Mar 4, 2014 #5
    Thanks, Pranav!

    I see that comes to [tex]e^{-t^2}[/tex] according to the Fundamental Theorem of Calculus... I am getting a bit rusty on that so need to revise it a bit!
     
  7. Mar 4, 2014 #6
    Your welcome. :)

    Yes. :approve:
     
  8. Mar 4, 2014 #7

    hilbert2

    User Avatar
    Science Advisor
    Gold Member

    It sure does exist, and it's called the error function, but you can't represent it in terms of elementary functions.

    http://en.wikipedia.org/wiki/Error_function
     
  9. Mar 4, 2014 #8
    Yes, that's what I meant, sorry for the confusion. :redface:
     
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