1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Absolute value in a differential equation

  1. Dec 27, 2015 #1
    1. The problem statement, all variables and given/known data
    ##\displaystyle (x+3)\frac{dy}{dx} = y - 2##, where x is not 3 and y is not 2.

    2. Relevant equations


    3. The attempt at a solution

    ##\displaystyle (x+3)\frac{dy}{dx} = y - 2##
    ##\displaystyle \frac{dy}{y-2} = \frac{dx}{x+3}##
    ##\displaystyle \int \frac{dy}{y-2} = \int \frac{dx}{x+3}##
    ##\ln|y-2| = \ln|x+3| + c##
    ##\displaystyle |y-2| = e^{\ln |x+3|+c}##
    ##\displaystyle |y-2| = ke^{\ln |x+3|}##, where ##k=e^c##
    ##\displaystyle |y-2| = k|x+3|##

    This is as far as I can get. I am not sure how I can get rid of the absolute value signs in order to obtain a single function.
     
  2. jcsd
  3. Dec 27, 2015 #2

    andrewkirk

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You can't, because there are two functions, not one, so you have to choose one of them - sometimes called 'choosing a branch'.
    To find out what the two functions are, write the right-hand side, which is a well-defined function, as f(x), and then explicitly write out the two possibilities created by the absolute value function on the left hand side.

    Rearranging, you will get two different expressions for y in terms of f(x), corresponding to the two different branches of the absolute value.
     
  4. Dec 27, 2015 #3
    So are the two possible functions

    ##y = 2 + k|x+3|##
    and
    ##y = 2-k|x+3|##
    ?

    So I have to choose one of these as a solution?
     
  5. Dec 27, 2015 #4

    Mark44

    Staff: Mentor

    Without an initial condition you can't know which is the correct solution. In that case, you can leave your solution as ## y = 2 \pm |x + 3|##, with k being a positive constant.
     
  6. Dec 28, 2015 #5
    Given that the initial condition is (x = -1, y = 1), how do you go from ##y = 2 \pm k|x+3|## to a specific solution?
     
  7. Dec 28, 2015 #6

    Samy_A

    User Avatar
    Science Advisor
    Homework Helper

    Plug in the initial condition in each possible solution and see what you get.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Absolute value in a differential equation
Loading...