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

First divide each of the first two equations by two. ThenThe equal-component and opposite-component directions diagonalize the upper block, while the third coordinate is independent. Direct multiplication gives the eigenvalues and unit eigenvectorsThey form an orthonormal eigenbasis because $K$ is a real symmetric matrix.

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.

At a steady state, $K{\mathbf{y}}_0={\mathbf{f}}_0$. All eigenvalues of $K$ are nonzero, so it is invertible andFor the specified constant source this givesIndeed the first two rows of $K{\mathbf{y}}_0$ each give one, and the third gives $4(1/4)=1$.

Take the switch-off time to be $t=0$. The state remains continuous through the source step, so $\mathbf{y}(0)=(1,1,1/4{)}^T$. Its eigendirection decomposition isThe antisymmetric mode has zero amplitude because the first two initial components are equal. Applying the decay factors from part (b),The first two responses decay at rate one and the third at rate four, approaching the new zero-source steady state.