What kind of equation is this and how could I solve it?

[lonewolf219]

Homework Statement

I was wondering if anyone can tell me about this equation:

f(x)=f(y)+cx

How can I select points for this equation and graph it? What values are constant and what values are changing? Thanks

Homework Equations

f(x)=f(y)+cx

The Attempt at a Solution

Well... maybe choose a fixed value for y and for c?

 

[LCKurtz]

Homework Statement

I was wondering if anyone can tell me about this equation:

f(x)=f(y)+cx

How can I select points for this equation and graph it? What values are constant and what values are changing? Thanks

Homework Equations

f(x)=f(y)+cx

The Attempt at a Solution

Well... maybe choose a fixed value for y and for c?

What happens if you put y = 0?

[Edit-Added]Are you sure you have stated the problem completely and correctly? Putting y=0 gives f(x) = f(0) + cx. Putting y = t gives f(x) = f(t) + cx. Subtracting those says f(0) = f(t) so f is constant. But that doesn't work unless c = 0.

 

[lonewolf219]

OK, that's a possibility, thanks LCKurtz

 

[LCKurtz]

OK, that's a possibility, thanks LCKurtz

Make sure you notice my edited comment.

 

[lonewolf219]

Oh thanks for that message, I did not see your edited response...You are right, there is more! The equation is \alpha^(-1)(t)=\alpha^(-1)(m)+(1/2\pi)(β)(t). This is for the running of the couplings, haven't done any of this stuff until this summer... Hope you can shed some light!

 

[haruspex]

The equation is \alpha^(-1)(t)=\alpha^(-1)(m)+(1/2\pi)(β)(t).

I assume you want to arrive at y as a function of x, y=y(x) (or in the revised version m = m(t)).
Please clarify what alpha is here. Judging from the original post it is an arbitrary unknown function. If so, I see no hope of saying much about m(t). If you assume it's invertible then you can show m(0) = 0.