Does this PDE only have the trivial solution? Let $(M,g)$ be a closed Einstein manifold of dimension $m>2$ and
$$
\mathrm{Ricc}(g)=\lambda g,
$$
$h$ a symmetric $2$-covariant tensor, $\Delta=\nabla^*\nabla$ the Laplacian on functions as well as the rough or connection Laplacian on symmetric $2$-covariant tensors, and $\delta_g$ the divergence on symmetric $2$-covariant tensors as well as on $1$-forms, so $\delta_g\delta_gh$ is the double-divergence and is a function.  Write $\mathrm{tr}h$ for the metric trace of $h$ wrt $g$.  Consider the following PDE:
$$
\Delta(\mathrm{tr}h)=\lambda\mathrm{tr}h+\frac{m-2}{2}\delta_g\delta_gh.
$$
Questions:  Does this PDE have only the trivial solution $h=0$?  Are there conditions under which it has only the trivial solution?  I'm interested in the case that $\lambda<0$.
 A: There are lots of nontrivial solutions for any negative $\lambda$.  Here's how to construct them all.
First, we can decompose an arbitrary symmetric $2$-tensor $h$ as $h=fg+u$, where $f$ is a scalar function and $u$ is trace-free. It follows that $\operatorname{tr} h = mf$, and $\delta_g\delta_g h = -\Delta f + \delta_g\delta_g u$. After some simplification, therefore, your equation becomes
$$
\left(\Delta  - \frac{2m\lambda }{3m - 2}\right) f = \frac{m-2}{3m-2} \delta_g\delta_g u.\tag{$*$}
$$
Since the eigenvalues of $\Delta$ are all nonnegative, the fact that $\lambda<0$ ensures that the operator in parentheses on the left-hand side is invertible on $C^\infty(M)$. Thus we can choose $u$ to be an arbitrary trace-free symmetric $2$-tensor field, and then let $f$ be the unique solution to ($*$).
Notice that this has nothing to do with the manifold being Einstein. This argument works on any compact Riemannian manifold for any negative constant $\lambda$. 
The case $\lambda\ge 0$ is a bit more complicated. In that case, if $2m\lambda /(3m-2)$ isn't an eigenvalue of $\Delta$ the same argument works; but if it happens to be an eigenvalue, then there will be solutions only for those $u$ for which $\delta_g\delta_g u$ is $L^2$-orthogonal to the kernel of the operator on the left-hand side. 
