You may also like

problem icon

Weekly Challenge 43: A Close Match

Can you massage the parameters of these curves to make them match as closely as possible?

problem icon

Weekly Challenge 44: Prime Counter

A weekly challenge concerning prime numbers.

problem icon

Weekly Challenge 28: the Right Volume

Can you rotate a curve to make a volume of 1?

Weekly Challenge 29: Integral Equation

Stage: 5 Short Challenge Level: Challenge Level:1


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 $$f(x) = {-k\over 3} e^{x/3}.$$

It was also noted by Nat that the problem can be solved rapidly using the university-level technique of Laplace transforms. He says

You could have it done with the Laplace transform (which I think is more ideal if you really want to solve integro-diff equations). So let $s$ be the frequential variable and $F(s)$ the Laplace transform of $f(x)$

Then $$\frac{F(s)}{s} =3F(s) + \frac{k}{s}$$

$$F(s) = - \frac{k}{3}\cdot \frac{1}{s-\frac{1}{3}}$$

Then $$f(x) = - \frac{k}{3}\cdot e^{\frac{x}{3}}$$