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Math Help - Partial Differential Quotation - Maple 15

  1. #1
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    Partial Differential Quotation - Maple 15

    Hello!

    u(x,y)

    du/dx +2*x*u=0

    du/dx is the partial derivate.

    How do you do this in Maple?

    I have tried this:
    PDE:=du/dx u(x,y)+2*x*(u(x,y)=0
    ans(PDE)

    but get wrong answer

    Kjell
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  2. #2
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    Re: Partial Differential Quotation - Maple 15

    u may well be a function of two variables, say, x and y, but since y does not appear in the equation, it is can be treated as an ordinary equation. I also cannot imagine why one would want to use 'Maple'. The equation du/dx= -2xu can be separated to du/u= -2xdx and then, integrating both sides, ln(|u|)= -x^2+ C. Now, if u is a function of x and y (or more variables) the other variables are treated as constants in differentiating with respect to x so the "constant", C may be functions of those variables. If we are given that u is a function of x and y only, then u(x,y)= e^{-x^2+ f(y)}= e^{f(y)}e^{-x^2}. Of course, e to the power of an unknown function of y is itself an unknown function of y so we could also have used the solution u(x)= Ce^{-x^2} to get u(x,y)= f(y)e^{-x^2} where f(y) is an arbitrary function of y.

    What answer did Maple give?
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  3. #3
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    Re: Partial Differential Quotation - Maple 15

    Hello!
    Thank you very much for you answer.I understand now.
    Maple gave now the same: f(x,y)=F(y)*e^-x^2

    Kjell
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