Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2019 / Paper 2 / 15 / b / iii

The first initial condition gives $A=1$. The original first equation gives $u^{\prime }(0)=-3u(0)+v(0)=-2$, whereas the general solution gives $u^{\prime }(0)=B-A$. Thus $B=-1$ andThe corresponding $v=e^{-t}(\cos t-3\sin t)$ has $v(0)=1$, checking the other initial condition.