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Math Help - PDE

  1. #1
    Member Altair's Avatar
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    PDE

    u_y + 2yu = 0

    I have to solve this as ODE...How can it be done ? u is a function of x and y but I do not have any idea how to transform this equation into derivative of one variable only
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  2. #2
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    mr fantastic's Avatar
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    Quote Originally Posted by Altair View Post
    u_y + 2yu = 0

    I have to solve this as ODE...How can it be done ? u is a function of x and y but I do not have any idea how to transform this equation into derivative of one variable only
    Assume a seperable solution: U = X(x) Y(y).

    This leads to \frac{dY}{dy} + 2yY = 0 which is seperable.

    You end up with U = e^{-y^2} X(x).
    Last edited by mr fantastic; May 9th 2009 at 02:34 PM. Reason: Fixed a typo
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  3. #3
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    Quote Originally Posted by Altair View Post
    u_y + 2yu = 0

    I have to solve this as ODE...How can it be done ? u is a function of x and y but I do not have any idea how to transform this equation into derivative of one variable only
    Or, since x appears neither explictely nor as a variable of differentiation, ignore the fact that it is a partial differential equation!
    The problem \frac{du}{dy}+ 2yu= 0 can be written as \frac{du}{dy}= -2yu and is a separable equation: \frac{du}{u}= -2ydy. Integrating, ln(u)= -y^2+ C so, taking the exponential of both sides, u= e^{-y^2+ C}= e^Ce^{-y^2}= C'e^{-y^2}.

    Finally, use the fact that we have been ignoring x- that is, treating it like a constant: the constant C' might be any function of x. If we call that "F(x)" then we have u(x,y)= F(x)e^{-y^2} where F(x) is an arbitrary function of x, exactly the solution mr. fantastic got.
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