What are the restrictions such that $f'(x) = f(x)/x$?

Question

Let $C(y) \geq 0$ denote some cost function. Let $MC(y) = C'(y)$. Let $AC(y) = \frac{C(y)}{y}$. I am considering the economic case where $$MC (y)= AC(y)$$ This boils down to a simple math problem as follows:

Given $x> 0$ and $f(x) \geq 0$, what are the necessary restrictions on $f$ such that $$f'(x) = \frac{f(x)}{x}$$? I want to know this to understand what kinds of functions will work as cost functions and still have this property.

Answer 1

Case $x\neq0$ and $f(x)=0$ then $f'(x)=0$ . Hense $f(x)=0$ is a solution.

Case $x\neq0$ and $f(x)\neq0$ $$\frac{f'(x)}{f(x)}=\frac{1}{x}$$ After integration : $$\ln|f(x)|=\ln|x|+c$$ $c$ is a constant. $$e^{\ln|f(x)|}=e^{\ln|x|}e^c$$ $$|f(x)|=e^c|x|$$ With $C=\pm e^c$ $$f(x)=C\:x$$ any constant $C$ including the case $C=0$ which corresponds to the the case $f(x)=0$ seen at first.

Answer 2

Multiply both sides by $\frac{1}{x}$ and rearrange to get

$$\frac{1}{x}f'(x)-\frac{1}{x^2}f(x)=0$$

Then note that you have

$$\frac{d}{dx}\left(\frac{1}{x}f(x)\right)=0$$

$$\frac{1}{x}f(x)=c$$

$$f(x)=cx$$

Answer 3

We'd like to solve the differential equation $f'(x) = \frac{f(x)}{x}, x>0$ (and we'd like to restrict ourselves to cases where $f(x) \geq 0$ for $x>0$, but we'll get to this after we found all solutions).

There are some funtions which obviously solve the equation, namely linear functions, as for some $c \in \mathbb R$ and $f(x)=c x$ we get $f'(x)=c= f(x)/x$.

But are these functions the only solutions for this differential equation? It turns out:

Yes, they are.

The differential equation is a linear first order ODE which has a one-dimentisonal solution space (of which the one-dimesional space of linear function must be a subset, as we checkt above).

If we want to have $f(x)=cx\geq 0$ for $x>0,$ we need to impose the condition $c\geq 0$.

Answer 4

Suppose we don't want to solve the ODE $$f'(x)= \frac{f(x)}{x} \mbox{ for all } x >0 ,$$

but we are interested in conditions on $f$ such that we can conclude

$$f'(x)= \frac{f(x)}{x} \mbox{ for some } x >0 .$$

It turns out that this is almost no restriction to $f$.

So let $g$ be any function which is continuously differentiable on $[0,\infty).$ Let $f(x) = g(x) + g'(1) - g(1),$ which implies $g'(1)=f(x) - g(x) + g(1) - g(1) = \frac{f(1)}{1}$.

Then, since $f' =g'$ we have $f'(1) = g'(1) = f(1) - g(1) +g(x) = \frac{f(1)}{1},$ i.e. we have found some $x$ (namely $1$) such that the equation is fufilled.