= Solution
Adding the steady and transient parts gives
$$
\boxed{u(x,t)=\frac{G}{\nu n^2\pi^2}
\left(1-e^{-\nu n^2\pi^2t}\right)\sin(n\pi x)}.
$$
Integrating the <sine> across the channel gives $\int_0^1\sin(n\pi x)\,dx=[1-(-1)^n]/(n\pi)$. Thus
$$
\boxed{Q(t)=\frac{G[1-(-1)^n]}{\nu n^3\pi^3}
\left(1-e^{-\nu n^2\pi^2t}\right)}.
$$
For even $n$ the signed flow rate is zero, because positive and negative velocity lobes cancel. For odd $n$ the bracket is two. Both expressions vanish initially and approach their steady values exponentially; $G=0$ gives the trivial zero solution throughout.
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