The questions asks for the $y(t)$ and gives:
$$4\int_0^ty(u)\,du+y'(t)=\int_0^1y(u)\cos(t-u)\,du\,\,\,\,\text{if }y(0)=1$$
I've been dealing with equations of the type:
$$y(t)=f(t)+\int_0^ty(s)g(t-s)\,ds$$
Which I can solve by Laplace Transform, setting $$Y=\mathcal{L}(y(t)),F=\mathcal{L}(f(t)),G=\mathcal{L}(g(t))$$
Then isolate $Y$ and find $y(t)=\mathcal{L}^{-1}(Y)$.
I could write the first as $\frac{4Y}{s}+sY-1$, and to the same to the second if it weren't for that '1' instead of $t$. Could this be a typo in the list of exercises? Thanks in advance.
Edit: Could I write it as $\lim_{t\to 1}\int_0^1y(u)\cos(t-u)\,du$?