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Stuck on: PDE with boundry conditions OR series expantion of x^c

  1. Nov 19, 2007 #1
    PDE with boundary conditions

    Full question
    A function u(x,y) has two independent variables x and y and satisfies the 1st order PDE
    [tex]x \frac{du}{dx} - \frac{y}{2}\frac{du}{dy}= 0[/tex]
    by first looking for a separable solution u(x,y)=X(x)Y(y), find the general solution of the equation.
    determine the u(x,y) that satisfies the boundary condition u(1,y) = 1 + sin y

    a. Relevant equations
    [tex]x \frac{du}{dx} = \frac{y}{2} \frac{du}{dy}[/tex]

    a. The attempt at a solution
    i think ive got this right:
    so i sub in X(x)Y(y) and get
    [tex]x Y(y) \frac{dX(x)}{dx} = X(x) \frac{y}{2}\frac{dY(y)}{dy}[/tex]

    [tex]\frac{x}{X(x)}\frac{dX(x)}{dx} = \frac{y}{2Y(y)}\frac{dY(y)}{dy} = c[/tex]
    this gives
    [tex]\int\frac{1}{X(x)} dX = \int\frac{c}{x} dx[/tex]

    [tex]ln[\frac{X(x)}{A_{1}}] = c ln[x][/tex]
    [tex]X = A_{1} x^{c}[/tex]
    doing the same for y gives
    [tex]Y = A_{2} y^{2c}[/tex]

    [tex]u(x,y) = X(x)Y(y) = A_{1}A_{2}x^{c}y^{2c} = A (x y^{2})^{c} [/tex]

    or with [tex] b=2c[/tex] [tex] u=A(y \sqrt{x})^{b}[/tex]

    b. relevant equations
    however when i apply the boundary condition i get
    [tex]1 + Sin[y] = A y^{b}[/tex]

    b. attempts and ideas
    i tried a few e^-iy routes e.g [tex] \frac{e^{-iy}-e^{iy}}{2i} = Sin[y] [/tex] but i dont think that helps

    the only thing i can think is that i have to relate a power series of a^b to 1 + Sin[a] but i cant remember what a^b is is it related to e^b
    which is [tex]\sum\frac{x^{n}}{n!}[/tex]

    or have a i gone wrong already?
    Last edited: Nov 20, 2007
  2. jcsd
  3. Nov 20, 2007 #2
    is there anyway to Change the title after thinking about it im pretty sure it should just be
    "PDE with boundary condition"
    but i think im stuck with the current tittle
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