From Spivak
Find all continuous functions such that $$f(x)^2 = \int_0^x f(\xi)\frac{\xi}{1+\xi^2}d\xi.$$
$f(x)^2$ is clearly differentiable. I'd like to write $\dfrac{d}{dx}f(x)^2=2f(x)f'(x).$ In order to justify this, I want to say that $$\sqrt{\left| \int_0^x f(\xi)\frac{\xi}{1+\xi^2}d\xi \right|},$$ which is equal to $|f|,$ is differentiable everywhere except where the value of the integral is zero. So at every point at which $f(x)\neq 0$, $f$ is differentiable, and $$2f(x)f'(x) = f(x) \frac{x}{1+x^2} \implies \\ 4f'(x) = \frac{2x}{1+x^2} \implies \\ f(x) = \frac14 (\log (1+x^2) + C).$$
Now, any solution must have the value zero at $x=0$. The behavior of the solution to the left of zero and the right of zero should be looked at separately since neither affects the other. Further, the analysis that follows is symmetric on both sides (note $\dfrac{\xi}{1+\xi^2}$ is odd), and so we can determine the possible solutions on $x\geq 0$ and then all the same solutions are equally good (backwards) on $x\leq 0$ (and we can cut and paste any two from either side together).
So let's consider $x\geq 0$. The "first thing" $f$ does after being zero for some period of time cannot be to become negative, since then $f(x)^2 =\int_0^x f(\xi)\dfrac{\xi}{1+\xi^2}d\xi$ would be negative for some $x$, a contradiction. Therefore the "first thing" it does must be to become positive. But then it must follow $\log$ on an upward trajectory forever, and so its behavior is determined for all time.
To put this previous paragraph in more rigorous language, note that $\log(1+x^2) + C$, once it has positive derivative at some point, must have positive derivative for all following points. Since $f(x)^2 =\int_0^x f(\xi)\dfrac{\xi}{1+\xi^2}d\xi$, it cannot be that $f(x)$ takes only negative values in any interval $[0,a]$, so if it takes a nonzero value, we can use the mean value theorem to find a point where it has positive derivative, and then its behavior is determined for all time after that point. So the only question is when (if ever) it becomes positive. This is equivalent to choosing a solution from the family
$$\begin{cases} 0 & x\leq \sqrt{e^{-C}-1} \\[12pt] \dfrac{1}{4}(\log(1+x^2) + C) & x > \sqrt{e^{-C}-1} \end{cases}$$ where $C\leq 0$.
Pick a function from the above family for $x>0$ and (the reflection of) one for $x<0$ (or take the zero function for either), and you have a solution.
Now we can check that $$\int_0^x f(\xi)\frac{\xi}{1+\xi^2}d\xi = \frac1{16}(\log (1+x^2) - C)^2 = f(x)^2,\, \,\forall x \text{ s.t. } f>0,$$ so all solutions of the above form do indeed satisfy the diffEQ.
Is my analysis correct? Can anyone see a slicker way to it?
The solution is okay, but a bit messy about the neighborhood of $x=0$ ("first thing", etc). I would begin with: suppose there is $x_0>0$ such that $f(x_0)\ne 0$. Let $a=\sup \{x<x_0 : f(x)=0\}$ and observe that $a\ge 0$ because $f(0)=0$. Also, let $b=\inf\{x>x_0:f(x)=0\}$; this is understood as $b=\infty$ if the set is empty.
On the interval $(a,b)$ we have $f'(x) = \dfrac12 \dfrac{x}{x^2+1}$, hence $f(x) = \dfrac14\ln(x^2+1)+C$. Since this function is increasing and $f(a)=0$, it follows that
This completes the picture on $(0,\infty)$: the function is either identically $0$ there, or is of the form $\dfrac14\left(\ln(x^2+1)+C\right)^+$.
The same description applies to $\tilde f(x)=f(-x)$, which also satisfies the given integral equation.