Weekly Challenge 29: Integral Equation
The integral equation is: $$\int_0^x f(t)\,dt = 3f(x)+k,\quad
\quad(\star)$$ where $k$ is a constant. Differentiating both sides
of $(\star)$ gives $$f(x) = 3f'(x)$$ If there is a solution of
$(\star)$ it must be of the form $$f(x) = Ae^{x/3},$$ for some
constant $A$. We check to see whether or not this is a solution.
For $f(x)=Ae^{x/3}$ we have $$\int_0^x Ae^{t/3}\,dt =
\Big[3Ae^{t/3}\Big]_0^x = 3Ae^{x/3}-3A.$$ Thus $f(x)=Ae^{x/3}$ is a
solution if and only if $A=-k/3$. The unique solution is $$f(x) =
{-k\over 3} e^{x/3}.$$