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Thread: Method of Characteristics

  1. #1
    Senior Member bugatti79's Avatar
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    Method of Characteristics

    Hi All,

    Can any one confirm that $\displaystyle \int\frac{1}{x+y}dx=ln(x+y)+C$ since the top is the derivative of the bottom? It is part of my bigger question for solving for u using method of characteristics
    $\displaystyle \frac{\partial u}{\partial x}-\frac{\partial u}{\partial y}=\frac{u}{x+y}$

    Thanks
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  2. #2
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    Is $\displaystyle y$ a function of $\displaystyle x$, or are we treating it as constant?
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  3. #3
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    What you have is correct but not applicable here. Your method od charactistics are

    $\displaystyle \dfrac{dx}{1} = \dfrac{dy}{-1} = \dfrac{(x+y)du}{u}$

    From the first pair

    $\displaystyle \dfrac{dx}{1} = \dfrac{dy}{-1} \;\; \Rightarrow\;\; x + y = c_1$

    Then from the first and third

    $\displaystyle \dfrac{dx}{1} = \dfrac{(x+y)du}{u}$

    or

    $\displaystyle \dfrac{dx}{c_1} = \dfrac{du}{u}\;\; \Rightarrow\;\; u e^{-\frac{x}{x+y} = c_2}$

    Your general solution is $\displaystyle c_2 = f(c_1)$ or $\displaystyle u = e^{\frac{x}{x+y}}f(x+y)$
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  4. #4
    Senior Member bugatti79's Avatar
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    Danny and Prove it
    Not sure I understand how you got the last term on the 2nd last line...? Can you clarify. FYI, my procedure is

    dy/dx = -1, therefore y=-x+k, du/dx=u/(x+y) therefore $\displaystyle ln u=\int\frac{1}{x+y} dx = ln (x+y) +f(k)$.
    Not sure what I have done wrong so far...
    prove it, the above integration is based on assuming y is constant....not sure of this right

    Thanks
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  5. #5
    MHF Contributor
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    Quote Originally Posted by bugatti79 View Post
    Danny and Prove it
    Not sure I understand how you got the last term on the 2nd last line...? Can you clarify. FYI, my procedure is

    dy/dx = -1, therefore y=-x+k, du/dx=u/(x+y) therefore $\displaystyle ln u=\int\frac{1}{x+y} dx = ln (x+y) +f(k)$.
    Not sure what I have done wrong so far...
    prove it, the above integration is based on assuming y is constant....not sure of this right

    Thanks
    But from what you have $\displaystyle x + y = k$. You must use this. Thus, $\displaystyle \dfrac{du}{dx} = \dfrac{u}{k}$.
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