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

  • Thread starter Baashaas
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Homework Statement



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.

Homework Equations



none given.


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.
 
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Answers and Replies

  • #2
PeroK
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Is there some relationship between r and x?
 
  • #3
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ohw wait, I mistyped, theta=x, ill fix it. There is no relation between r and x, it is a polar coordinate system.
 
  • #4
PeroK
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ohw wait, I mistyped, theta=x, ill fix it. There is no relation between r and x, it is a polar coordinate system.
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?
 
  • #5
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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?
Yeah, I am sorry, x should be theta in all equations. That would make things a lot clearer.
 
  • #6
PeroK
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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.
If x and r (or θ and r) are independent variables, then this implies that both the above are constant.
 
  • #7
Ray Vickson
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Homework Statement



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.

Homework Equations



none given.


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.
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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