How can we find the expected value of $u(t)$ in terms of the following information: $$u(t)=\int_{0}^{t}(f(s)+(T-s)Y)(t-s)X(s)ds$$ where:
$X(s)$ is a wide sense stationary process with known probability density function, statistical moments, and (higher order) spectral density function(s).
$f(t)$ is a known real function.
$T>0$ is a constant and $0<t<T$.
$Y$ is an unknown random variable which satisfies $u(T)=0$.
Note that $$ Y=-\frac1Z\int_{0}^Tf(s)(T-s)X(s)ds,\qquad Z=\int_{0}^{T}(T-s)^2X(s)ds, $$ hence $$ u(t)=\int_{0}^{t}f(s)(t-s)X(s)ds-\frac1Z\int_{0}^Tf(u)(T-u)X(u)du\int_0^t(T-s)(t-s)X(s)ds. $$ The expectation of the first integral is $$ E[X(0)]\int_{0}^{t}f(s)(t-s)ds, $$ but the denominator $Z$ in $Y$ makes unlikely that any simple expression exists for the expectations $$E\left[\frac{X(u)X(s)}Z\right]$$ that are needed to compute the expectation of the second term.