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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:

    [itex]∫f(x)dx+f'(x)+rg'(r)-g(r)=0[/itex]

    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]

    gives

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

    and

    [itex]g(r)=Hr[/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

    [itex]∫f(x)dx+f'(x)+rg'(r)-g(r)=0[/itex]

    to

    [itex]∫f(x)dx+f'(x)=-rg'(r)+g(r)[/itex]

    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

    PeroK

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    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

    PeroK

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    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

    PeroK

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    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

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    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".
     
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