Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2017 / Paper 2 / 17 / b

Substitute $\mathbf{y}=\mathbf{v}{e}^{-\gamma t}$ into the unforced system. It gives $(K-\gamma I)\mathbf{v}=0$, so a nonzero single-exponential solution must use one of the eigendirections in part (a):These are all nonzero solutions having one common decay constant. Their arbitrary linear combination is the general unforced solution; all decay because the three eigenvalues are positive.