# Can function be derived from this relation?

I'm trying to figure out what function $f(x)$ is from the information below. I don't think the following relation is enough to determine exactly what function $f(x)$ is, but I would appreciate some advice as to what this relation tells us about $f(x)$. Does it narrow it down to a particular "class" of functions or something like that?

$$f(x)=x\frac{df}{dx}$$

Thanks

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What you have is a first order linear differential equation. –  user38268 Jan 3 '12 at 4:28

Let us call $f(x)$ for the moment $y$. Then rewriting what you have, we get

$$x \cdot \frac{dy}{dx} - y = 0.$$

This is a first order total differential equation (total because no partial derivatives are involved, and first order because the highest degree of the derivative(s) involved is one).

It is linear because the coefficient of $\frac{dy}{dx}$ is a function of $x$.

If you know the method of separation of variables, you can solve your differential equation like this. Write it as

$$x \cdot \frac{dy}{dx} = y.$$

Then by seperation of variables, we have that

$$\frac{1}{y} dy = \frac{1}{x} dx,$$

hence integrating both sides we get that

$$\ln|y| = \ln|x| + C,$$ where $C$ is some constant. Taking the exponential of both sides gives

$$y = e^Cx.$$

Hence the general class of functions that satisfy your relation in the question are the straight lines through the origin of positive slope (because $e^C$ is always positive).

This post may be of interest to you, the answers given here why the method of separation of variables is justified.

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Thanks! That clears it up for me. –  ben Jan 3 '12 at 6:11
@ben $e^{{\ln x} + C} = e^{\ln x} \cdot e^C = e^Cx$. –  user38268 Jan 3 '12 at 6:13
You may also want to take a look at the post below by André. If you already know existence uniqueness theorems for differential equations, then you can apply them to your differential equation to know that the solution(s) above are the only ones. –  user38268 Jan 3 '12 at 6:15

There are general procedures for solving differential equations related to the one you mention. But it may be useful to look at your problem in isolation, and see what can be done by careful application of standard tools. It is clear that for any constant $A$, $f(x)=Ax$ is a solution of your equation. We show that there are no other solutions.

Suppose that $f(x)$ satisfies your differential equation. Let $y=f(x)$ and let $$z=\frac{y}{x}.$$ Differentiate with respect to $x$. By the Quotient Rule, $$\frac{dz}{dx}=\frac{x\frac{dy}{dx}-y}{x^2}.\qquad\qquad(\ast)$$ We are given that $y=x\frac{dy}{dx}$. It follows that the numerator in $(\ast)$ is $0$.

So we know that $\frac{dz}{dx}=0$ on the interval $(0,\infty)$ and also on the interval $(-\infty,0)$. (There is potential trouble at $x=0$, since $z$ is not defined there.)

A function whose derivative is $0$ on an interval is constant on that interval. So on $(0,\infty)$, $f(x)=Ax$ for some constant $A$. Also, on $(-\infty,0)$, $f(x)=Bx$ for some constant $B$. But since $f(x)$ is differentiable at $x=0$, we have $A=B$.

We conclude that $f(x)=Ax$ for some constant $A$.

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