Emily is correct.

Yaodong: there is a notion of linear operator; $L(\alpha u+\beta v)= \alpha L(u)+\beta L(v)$ but talking about equations or systems and not necessarily differential, may be algebraic we have

$Lu=0$ is linear homogeneous

$Lu=f$ with $f$ (which does not depend on $u$!) is linear inhomogeneous (linear algebra!!)

So linearity test applies only to $L$.

Remark: $Lu=0$ describes a *linear subspace*, $Lu=f$ (if has solution) describes *affine subspace* (former linear space but shifted, so we have no longer a linear space!)