Results 1 to 5 of 5

Math Help - Method of Characteristics

  1. #1
    Senior Member bugatti79's Avatar
    Joined
    Jul 2010
    Posts
    461

    Method of Characteristics

    Hi All,

    Can any one confirm that \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
    \frac{\partial u}{\partial x}-\frac{\partial u}{\partial y}=\frac{u}{x+y}

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,591
    Thanks
    1445
    Is y a function of x, or are we treating it as constant?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,368
    Thanks
    43
    What you have is correct but not applicable here. Your method od charactistics are

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

    From the first pair

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

    Then from the first and third

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

    or

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

    Your general solution is c_2 = f(c_1) or u = e^{\frac{x}{x+y}}f(x+y)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member bugatti79's Avatar
    Joined
    Jul 2010
    Posts
    461
    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 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
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,368
    Thanks
    43
    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 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 x + y = k. You must use this. Thus, \dfrac{du}{dx} = \dfrac{u}{k}.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Method of Characteristics
    Posted in the Calculus Forum
    Replies: 0
    Last Post: November 11th 2011, 07:16 AM
  2. Cauchy problem, method of characteristics
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: September 4th 2011, 09:30 AM
  3. [SOLVED] Method of Characteristics: Prove the u and v is a solution
    Posted in the Differential Equations Forum
    Replies: 17
    Last Post: October 25th 2010, 10:04 AM
  4. 1st order PDE through Method of Characteristics
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: June 30th 2010, 05:51 PM
  5. Method of characteristics
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: February 7th 2010, 12:06 AM

/mathhelpforum @mathhelpforum