1. Limited time only! Sign up for a free 30min personal 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!

Solving ∫f(x)dx+f'(x)+rg'(r)-g(r)=0 for f and g

  1. Sep 7, 2014 #1
    1. The problem statement, all variables and given/known data

    This is a part of a much larger engineering problem I am working out, the functions f and g are integration functions appearing when a strain is integrated to determine a displacement.

    I am stuck with the following problem:

    determine f(x) and g(r) from the following equation:


    A hint in the problem description says you have to apply separation of variables.

    2. Relevant equations

    none given.

    3. The attempt at a solution

    I am getting stuck at this equation because it doesn't seem to be writeable in the standard form for separating variables.

    [itex] \frac{dy}{dx}=g(x)f(y) [/itex]

    An example (without any elaboration) of a non homogeneous equation of the same form states that the solution of:

    [itex]∫f(x)dx+f'(x)+rg'(r)-g(r)=A \cos{x}[/itex]


    [itex]f(x)=0.5Ax \cos{x}+K \cos{x} + L \cos{x}[/itex]



    Now I can see that this is a valid solution, but I have no idea how you separate the 2 functions so that that f is only a function of x and g is only a function of r.
    I can of course write




    and than find a homogeneous solution of the form [itex]f(x)=K \cos{x} + L \cos{x}[/itex]

    But where does g(r) go when finding the particular solution for f(x)?
    Can you perhaps state that both [itex]∫f(x)dx+f'(x)[/itex] and [itex]rg'(r)-g(r)[/itex] are constant?

    I am seriously getting the feeling that I am missing something extremely obvious.
    Last edited: Sep 7, 2014
  2. jcsd
  3. Sep 7, 2014 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Is there some relationship between r and x?
  4. Sep 7, 2014 #3
    ohw wait, I mistyped, theta=x, ill fix it. There is no relation between r and x, it is a polar coordinate system.
  5. Sep 7, 2014 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    So, I guess you don't count ##r^2 = x^2 + y^2## as a relationship? Or, is x something other than the x-coordinate?
  6. Sep 7, 2014 #5
    Yeah, I am sorry, x should be theta in all equations. That would make things a lot clearer.
  7. Sep 7, 2014 #6


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    If x and r (or θ and r) are independent variables, then this implies that both the above are constant.
  8. Sep 7, 2014 #7

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    You are missing the whole point about "separation of variables" Write your equation as
    [tex] L(x) = R(r),\\
    L(x) = \int f(x) \, dx + f'(x) \\
    R(r) = g(r) - r g'(r) [/tex]
    If you fix ##r## but vary ##x##, the equation ##L(x) = R(r)## requires that ##L(x)## remain unchanged; that is, ##L(x) = c## for some constant ##c##. Of course, that also requires that ##R(r) = c##, so you have enough to determine both ##x## and ##r## up to some"boundary coonditions".
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted