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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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