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Math Help - How To Integrate Partial Derivative Functions

  1. #1
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    Question How To Integrate Partial Derivative Functions

    Hi Suppose We have a partial equation like below :


    u_{xx} + u_{x} - 2u =0

    I am wondering how to solve that ? I have misunderstanding for partial functions...
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  2. #2
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    Re: How To Integrate Partial Derivative Functions

    Since the only derivatives are with respect to x, for this particular example, you can (almost) treat this as an ordinary differential equation. u_{xx}- u_x+ 2u= 0 is a linear equation with constant coefficients. Its characteristic equation is r^2- r+ 2= 0. The only change for the partial differential equation is that the "constants" in the general solution may be functions of whatever the other variables are.
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    Re: How To Integrate Partial Derivative Functions

    Quote Originally Posted by HallsofIvy View Post
    Since the only derivatives are with respect to x, for this particular example, you can (almost) treat this as an ordinary differential equation. u_{xx}- u_x+ 2u= 0 is a linear equation with constant coefficients. Its characteristic equation is r^2- r+ 2= 0. The only change for the partial differential equation is that the "constants" in the general solution may be functions of whatever the other variables are.
    Thanks for your quick reply ,so what is the answer of Integral ? what is U ?
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    Re: How To Integrate Partial Derivative Functions

    I have solved the problem by it's index equation and I have found the index equation roots (x1,x2 ) and i.e it has answer in form of : e^{x1,x2 : roots of the equation}
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  5. #5
    MHF Contributor chisigma's Avatar
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    Re: How To Integrate Partial Derivative Functions

    Quote Originally Posted by SilverRenderer View Post
    I have solved the problem by it's index equation and I have found the index equation roots (x1,x2 ) and i.e it has answer in form of : e^{x1,x2 : roots of the equation}
    The discriminant of the charactestic equation is <0, so that the roots are complex. If \alpha \pm i\ \beta are the roots then the solution is...

    u= e^{\alpha\ x} (c_{1}\ \cos \beta x + c_{2}\ \sin \beta x) (1)

    ... where c_{1} and c_{2}, as HalsofIvy has precised, are independent from x...



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