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Math Help - another differentiation question ( PDE )

  1. #1
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    another differentiation question ( PDE )

    hi sorry for the insurgent of questions ..
    anyways we have :

    du/dx + x*du/dy = 0

    with boundary conditions
    i) u = sinh(y) on x=0
    ii) u = cosh(x) on y=0

    question: solve pde above and show solution is single valued throughout (x,y) plane.

    heres what i have so far:
    I get 2 integrals I = u and J = y-(x^2)/2 so general solution to pde is
    u = f(y-(x^2)/2)

    but not sure where to go from here : /
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  2. #2
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    Quote Originally Posted by Richyie View Post
    hi sorry for the insurgent of questions ..
    anyways we have :

    du/dx + x*du/dy = 0

    with boundary conditions
    i) u = sinh(y) on x=0
    ii) u = cosh(x) on y=0

    question: solve pde above and show solution is single valued throughout (x,y) plane.

    heres what i have so far:
    I get 2 integrals I = u and J = y-(x^2)/2 so general solution to pde is
    u = f(y-(x^2)/2)

    but not sure where to go from here : /
    Look for a seperable solution: u = f(x) g(y).

    Substitute into the PDE. After a modest amount of algebra you get:

    \frac{1}{x f} \frac{df}{dx} = - \frac{1}{g} \frac{dg}{dy} ......
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  3. #3
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    according to the solutions i have for
    the condition u(0,y) = sinh(y)

    u = sinh(y-x^2/2) is a solution

    i dont think i get the logic of how he got that tho,

    i guess u = sinh(y-x^2/2) would satisfy the general solution u(x,y) = f(y-x^2/2) and the condition u(0,y) = sinh(y), but then could i say
    u = sinh(y-x^2/2)+ b(x) is a solution aswell ? ? ( b(x) being an arbitary function of x)
    one of the things i am confused about here is what the f means in u(x,y) = f(y-x^2/2)

    for example if u(x,y) = f(x+y) .. since f is an arbitary function. .. . cant this imply: u(x,y) = f(x+y) => u(x,y) = (x+y)^2 => u(x,y) = f(x^2+y^2+2xy) . ..


    im not sure i understand your approach my fantastic : /
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