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Math Help - First Order, Linear, PDE

  1. #1
    Junior Member
    Joined
    Mar 2009
    Posts
    31

    First Order, Linear, PDE

     u_x + u_y + u = 1
     u = sin(x) on y = x + x^2, x>0
    Let x = s be the parameter along the initial line  y = x + x^2

    Solution:

    Characteristic Equation:
     \frac{dx}{dt} = 1, x = t + s
     \frac{dy}{dt} = 1, y = t + s +s^2
     \frac{du}{dt} = 1 - u, u(0) = sin(s)

    Hence:

    y = s + x^2
    u = 1 + (sin(s) - 1)e^{-t}

    and the solution continues on...
    ------------------------------------------------------

    What I don't understand is how to get u = 1 + (sin(s) - 1)e^{-t}.

    I sort of understand the  \frac{dx}{dt} = 1, x = t + s part, but can someone give me an explanation as well?

    Thanks =]
    Last edited by Creebe; October 9th 2010 at 09:43 AM.
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  2. #2
    Senior Member
    Joined
    Mar 2010
    Posts
    280
    <br />
\displaystyle{\frac{du}{1-u}=dt}<br />

    <br />
\displaystyle{\frac{d(1-u)}{1-u}=-dt}<br />

    <br />
ln(1-u)=-t+C<br />

    <br />
1-u=Ce^{-t}<br />

    <br />
**(1)** \;  \: \:  u=1-Ce^{-t}<br />

    <br />
u(0)=sin(s)<br />

    <br />
u(0)=sin(s)=1-C<br />

    <br />
C=1-sin(s)<br />

    <br />
Inserting \; to \; (1)<br />

    <br />
u=1-(1-sin(s))e^{-t}<br />
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