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Math Help - Help verifying solution for Differential equation

  1. #1
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    Help verifying solution for Differential equation

    I am having problems verifying a solution for a differential equation. The equation is

     y\prime -2ty =1

    The solution to the equation is

       e^{t^2} \int\limits_{0}^{t} e^{-s^2}\, ds + e^{t^2}

    My problem is calculating the integral that is contained on the solution. I thought there was no way to calculate an integral of

       e^{t^2}

    at least that is what a tutor told me at the tutoring service the university provides.
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  2. #2
    Super Member Random Variable's Avatar
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     \int^{t}_{0} e^{-s^{2}} ds cannot be expressed in terms of a finite number of elementary functions. But the value of the integral can be approximated for different values of t.
    Last edited by Random Variable; August 31st 2009 at 06:30 PM.
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  3. #3
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    Quote Originally Posted by Random Variable View Post
     \int^{t}_{0} e^{-s^{2}} dt cannot be expressed in terms of a finite number of elementary functions. But the value of the integral can be approximated for different values of t.
    how would I go in approximating it for different values?
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  4. #4
    Super Member Random Variable's Avatar
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    Quote Originally Posted by fishguts View Post
    how would I go in approximating it for different values?
    There are several numerical integration techniques (Newton-Cotes, Gaussian Quadrature, etc.).

    But it would be best to let a computer program do it and plot the result.
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  5. #5
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    Shouldn't there be a constant of integration in there? When I solve for the integrating factor and integrate I get:

    \mathop\int\limits_{\substack{t=t_0 \\ y(t_0)}}^{\substack{t=t \\ y(t)}} d\Big[ye^{-t^2}\Big]=\int_{t_0}^{t} e^{-t^2}

    y(t)=e^{t^2}\int_{t_0}^{t} e^{-s^2}ds+y(t_0)e^{t^2}

    Now differentiate y(t):

    \frac{d}{dt} y(t)=e^{t^2} e^{-t^2}+2te^{t^2}\int_{t_0}^{t} e^{-s^2}ds+y(t_0)2te^{t^2}

    Now you can back-substitute that into the differential equation to verify the solution. Also, just for fun, you can work with it in Mathematica: define y(t), y(0)=a, differentiate it, substitute it back into the DE:

    Code:
    In[73]:=
    y[t_] := Exp[t^2]*Integrate[Exp[-s^2], 
         {s, 0, t}] + a*Exp[t^2]
    FullSimplify[D[y[t], t] - 2*t*y[t]]
    
    Out[74]= 1
    Last edited by shawsend; September 1st 2009 at 08:03 AM.
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  6. #6
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    Quote Originally Posted by shawsend View Post
    Shouldn't there be a constant of integration in there?
    I copied the problem as the book has it and there was no constant of integration in there.
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  7. #7
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    Quote Originally Posted by fishguts View Post
    I copied the problem as the book has it and there was no constant of integration in there.
    The differential equation in the OP is a linear one. The standard solution to this is given in any textbook on differential equations. The resulting solution I got for this equation is:

    y(t)=e^{t^2}\cdot \left[K+\int e^{-t^2}dt \right]

    So, the integration constant is K and the derivative of the integral can easily be found to become the DE again. I do not see why one should have a definite integral in the solution.

    coomast
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