= Solution
Vary the quadratic <Einstein-Hilbert action> within the transverse trace-free tensor sector. Integration by parts gives
$$
\delta S^{(2)}=-M_{\rm Pl}^2\int d\eta\,d^3x\,
\left[(a^2E_{ij}^{T\prime})'-a^2\nabla^2E_{ij}^T\right]\delta E^{Tij},
$$
with endpoint terms set to zero. Hence the tensor equation is
$$
\boxed{E_{ij}^{T\prime\prime}+2\mathcal H E_{ij}^{T\prime}-\nabla^2E_{ij}^T=0}.
$$
A Fourier mode on <super-Hubble scales> has $k^2\ll\mathcal H^2$, so its leading equation is $(a^2E_{ij}^{T\prime})'=0$. Its solutions are
$$
E_{ij}^T=C_{ij}+D_{ij}\int^\eta\frac{d\eta'}{a(\eta')^2}
=C_{ij}+D_{ij}\int^t\frac{dt'}{a(t')^3}.
$$
There is thus a \b[constant tensor mode]. During inflation the second mode decays; in exact de Sitter expansion it is proportional to $a^{-3}$ after shifting the constant of integration. Constancy is the leading super-Hubble behaviour, exactly realized for the zero-gradient solution.
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