I'm stuck at the last step of this exercise:
b) Use the Banach fixed point theorem to show that there is a unique function $f \in C[0,1]$ for which the equation $$f(t) + \int_0^1e^{\tau+t-3}f(\tau)d\tau = 1 \quad \forall t \in [0,1]$$ holds. Determine $f$. Use the fact that $(X,\|\cdot \|_{\infty})$ is a Banach space.
I already proved existence of $f$ with the fixed point theorem. However, I am stuck determinating $f$. Can you help me?
Notice $$ \int_0^1 e^{\tau + t - 3} f(\tau) \, d\tau = e^t \int_0^1 e^{\tau - 3} f(\tau) \, d\tau .$$ So the equation can be rewritten as $$ f(t) = 1 - C e^t ,\tag 1$$ where $$ C = \int_0^1 e^{\tau - 3} f(\tau) \, d\tau .\tag 2$$ Now substitute equation (1) into equation (2), and solve for $C$.