Past exam of the mathematics course of the University of Cambridge / 2002 / ia / Paper 2 / 1D (Differential Equations)

The characteristic equation of the homogeneous linear differential equation is $m^2+m-2=(m-1)(m+2)=0$, giving the homogeneous solution $A{e}^t+B{e}^{-2t}$. For the first forcing, a particular solution of the form $C{e}^{-t}$ gives $-2C=1$, soThe initial conditions impose $A+B=1/2$ and $A-2B=-1/2$, whence $A=1/6$, $B=1/3$. Thus the first solution isFor the second forcing, $e^t$ is already a homogeneous solution, so the usual exponential trial must be multiplied by $t$. Substitution of the particular solution $Ct{e}^t$ gives $3C{e}^t=e^t$, hence $C=1/3$. Now $y=A{e}^t+B{e}^{-2t}+t{e}^t/3$, and the initial conditions give $A+B=0$, $A-2B=-1/3$. Therefore the resonant solution isThe additional factor $t$ reflects resonance in a differential equation.