Solving for $g(y)$ in the following integral equation

Question

Given the following integral equation where $f, h$ are known:

$$ f(t) = \int_{t-1}^t{g(y)\cdot h(y)\ \textrm{d}y}$$

Is it possible to solve this for $g(t)$, i.e. get an equation of the form $g(t) = \dots$ (supposing $g, h$ are suitably nice)? I have no idea where to start. I acknowledge that an analytical solution may not even exist in the general case, but I remain hopeful!

Answer 1

If $ g $ and $ h $ are continuous at $ \Bbb R $ , then $ f $ will be differentiable at $ \Bbb R $ and, by chain rule,

$$f'(t)=g(t)h(t)-g(t-1)h(t-1)$$

So,

$$g(t)=g(t-1)\frac{h(t-1)}{h(t)}+\frac{f'(t)}{h(t)}$$

which is a kind of recursive relation.

Answer 2

ok lets suppose we have some function: $z(y)=g(y)h(y)$ then we can say: $$f(t)=\int_{t-1}^tz(y)dy$$ now suppose that $z$ satisfies the equation $Z'(y)=z(y)$ so we have: $$f(t)=Z(t)-Z(t-1)$$ now if we differentiate both sides we get: $$f'(t)=Z'(t)-Z'(t-1)=z(t)-z(t-1)$$ now substituting back in we get: $$f'(t)=g(t)h(t)-g(t-1)h(t-1)$$ how if we isolate out $g(t)$ we get: $$g(t)=\frac{f'(t)+g(t-1)h(t-1)}{h(t)}$$ and so: $$g(t)=\frac{h(t-1)}{h(t)}\color{red}{g(t-1)}+\frac{f'(t)}{h(t)}$$ so whilst we cannot get a pure solution for $g(t)$ we can get this recurrance relation style which I believe you could get a solution from using generating functions depending on what the other functions are. Hope this helps