Find all smooth functions such that $f(x)f(y)=\int_{x-y}^{x+y}f(t)dt$ for all $x,y \in \mathbb{R}$ From Art of Problem Solving:

Find all continuous, differentiable functions f with domain $\mathbb{R}$ such that
  $$f(x)f(y)=\int_{x-y}^{x+y}f(t)dt$$
  for all $x,y \in \mathbb{R}$.

Find all 


*

*$Polynomial Functions$

*$Trigonometric Functions$

*$Functions$
Then prove all functions found are the only functions that satisfy the equation.
As some hints provided by the book, the following functions work:


*

*$2x$

*$c$

*$\frac{2}{c}\sin(cx)$

*It is also noted that there is another family of functions that satisfies the equation.


The problem should be able to be solved without any multivariable methods.
As of now I have only been able to conclude types of functions that satisfy the equation when $x=0$ or $y=0$. 
 A: Let $f\colon \Bbb R\to\Bbb R$ be a continuous function with 
$$\tag1 f(x)f(y)=\int_{x-y}^{x+y}f(t)\,\mathrm dt$$
for all $x,y\in\Bbb R$.
We shall see that $f$ must belong to the four solutions/solution families (named trivial, parabolic, elliptic, hyperbolic case for some deeper reasons) listed below.

Plugging $x=y=0$ into $(1)$ we get
$$\tag2f(0)=0.$$
Lemma 1.
The function $f$ is odd.
Proof. By the Fundamental Theorem,
$$f(y)+f(-y)= \frac{\mathrm d}{\mathrm dy}\int_{-y}^{y}f(t)\,\mathrm dt = f(0)f(y)=0 $$
holds for all $y$. $\square$
Solution 1. [Trivial case] We have the trivial solution $f(x)=0$. $\square$
Assume from now on that there exists $a\in\Bbb R$ with $b:=f(a)\ne 0$.
 Then 
$$f(x)=\frac 1b\int_{x-a}^{x+a}f(t)\,\mathrm dt $$
and by the Fundamental Theorem the right hand side has derivative $\frac{f(x+a)-f(x-a)}{b}$, showing that $f$ is continuously differentiable even without assuming it beforehand: 
$$\tag3f'(x)=\frac{f(x+a)-f(x-a)}{b}.$$ Specifically, using lemma 1
$$\tag4f'(0)=2.$$
Knowing that $f'$ exists, we dare to apply $\frac{\mathrm d}{\mathrm dx}$ to both sides of $(1)$ (viewing $y$ as constant) to obtain
$$\tag5 f'(x)f(y)=f(x+y)-f(x-y)$$
for all $x,y\in\Bbb R$.
Differentiating $(3)$ and afterwards using $(3)$ again to get rid of $f'$ on the right hand side we obtain
$$f''(x)=\frac{f(x+2a)-2f(x)+f(x-2a)}{b^2} $$
and by repeating the procedure
$$\tag6f'''(x)=\frac{f(x+3a)-3f(x+a)+3f(x-a)-f(x-3a)}{b^3}.$$
But then using $(5)$ with $y=3a$ and with $y=a$ we find
$$f'''(x)=\frac{f(3a)-3b}{b^3}f'(x), $$
i.e., $f'$ is a solution to the differential equation $$\tag{7} y''+cy=0,$$ for somereal number $c$.
The general solutions to $(7)$  are well-known:
Letting $\gamma=\sqrt{|c|}$, there are $\alpha,\beta\in\Bbb R$ such that 


*

*If $c=0$, $y=\alpha x+\beta$

*If $c>0$, $y=\alpha\sin(\gamma x)+\beta \cos(\gamma x)$

*If $c<0$, $y=\alpha\sinh(\gamma x)+\beta\cosh(\gamma x)$.


From $(4)$ we find $\beta=2$ in all three cases.
Also, by lemma 1, $f'$ is even so that we conclude $\alpha=0$.
Now we obtain $f$ by integration, observing that the integration constant is determined by $(2)$.
This gives us the following solutions (actually just candidate solutions, but one directly verifies that $(1)$ holds for them):
Solution 2. [Parabolic case] We have the solution $f(x)=2x$. $\square$ 
Solution (family) 3. [Elliptic case] For arbitrary $\gamma>0$ we have the solution  $f(x)=\frac 2\gamma\sin(\gamma x)$. $\square$
Solution (family) 4. [Hyperbolic case] For arbitrary $\gamma>0$ we have the solution  $f(x)=\frac 2\gamma\sinh(\gamma x)$. $\square$
