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

## 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:

$∫f(x)dx+f'(x)+rg'(r)-g(r)=0$

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

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.

$\frac{dy}{dx}=g(x)f(y)$

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

$∫f(x)dx+f'(x)+rg'(r)-g(r)=A \cos{x}$

gives

$f(x)=0.5Ax \cos{x}+K \cos{x} + L \cos{x}$

and

$g(r)=Hr$

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

$∫f(x)dx+f'(x)+rg'(r)-g(r)=0$

to

$∫f(x)dx+f'(x)=-rg'(r)+g(r)$

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

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

I am seriously getting the feeling that I am missing something extremely obvious.

Last edited:

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PeroK
Homework Helper
Gold Member
Is there some relationship between r and x?

ohw wait, I mistyped, theta=x, ill fix it. There is no relation between r and x, it is a polar coordinate system.

PeroK
Homework Helper
Gold Member
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?

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.

PeroK
Homework Helper
Gold Member
But where does g(r) go when finding the particular solution for f(x)?
Can you perhaps state that both $∫f(x)dx+f'(x)$ and $rg'(r)-g(r)$ 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.

Ray Vickson
Homework Helper
Dearly Missed

## 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:

$∫f(x)dx+f'(x)+rg'(r)-g(r)=0$

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

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.

$\frac{dy}{dx}=g(x)f(y)$

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

$∫f(x)dx+f'(x)+rg'(r)-g(r)=A \cos{x}$

gives

$f(x)=0.5Ax \cos{x}+K \cos{x} + L \cos{x}$

and

$g(r)=Hr$

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

$∫f(x)dx+f'(x)+rg'(r)-g(r)=0$

to

$∫f(x)dx+f'(x)=-rg'(r)+g(r)$

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

But where does g(r) go when finding the particular solution for f(x)?
Can you perhaps state that both $∫f(x)dx+f'(x)$ and $rg'(r)-g(r)$ 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
$$L(x) = R(r),\\ L(x) = \int f(x) \, dx + f'(x) \\ R(r) = g(r) - r g'(r)$$
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".