Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2023 / Paper 2 / 15 / b

Differentiate the first ordinary differential equation and substitute the second:The original first equation also gives $bv=u^{\prime }-au$, soTherefore the requested coefficients areThey are the trace and determinant of the coefficient matrix. This derivation never divides by $b$ and is valid even if $b=0$. In that degenerate case the scalar second-order equation is a necessary consequence, while the original constraint $u^{\prime }=au$ must also be retained when reconstructing the coupled system.

Here $\alpha =4$ and $\beta =4$, so the reduced ordinary differential equation is $(D-2{)}^{2}u=0$. Its repeated characteristic root gives $u=(A+Bt)e^{2t}$. The first initial condition sets $A=0$. The original equation $u^{\prime }=3u-v$ gives $u^{\prime }(0)=-1$, hence $B=-1$.

Recover $v=3u-u^{\prime }$ to obtainDirect differentiation verifies $v^{\prime }=u+v$ and both initial conditions.