I was hoping for some help with this integral relation. I would prefer hints to answers. I'm trying to find any solution to this relation:
$$ f(x) = \int_0^x f(s) \, \mathrm ds $$
I feel like it should be some variation of the exponential function that is $0$ for $x \le 0$, but I'm not sure.
Any help would be appreciated!
Take the derivative of both sides
$$\frac{d}{dx}f(x)=\frac{d}{dx}\int_{0}^{x}f(s)\space ds$$
Note that $\int_{0}^{x} f(s) ds=F(x)-F(0)$. The derivative of $F(x)$ is $f(x)$ and the derivative of $F(0)=0$. Hence
$$f'(x)=f(x)$$
There are two possiblities.
Once is $f(x)=0$, another is $f(x)=ce^x$
But if we substitute $f(x)=ce^x$ into the first equation the left hand side is
$$\frac{d}{dx}ce^x=ce^x$$
And the right hand side is
$$\int_{0}^{x}ce^s\space ds=ce^x-c$$
If set both side to equal eachother we get
$$0=-c$$
Hence $f(x)=0$ is the only solution.