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 / ii

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.