Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2012 / Paper 2 / 13

The characteristic equation is $r^2+r-2=(r-1)(r+2)=0$, giving $y=A{e}^x+B{e}^{-2x}$. The initial conditions give $A+B=1$ and $A-2B=-1$. Solving yields $A=1/3$, $B=2/3$, henceThe two distinct characteristic roots supply all solutions of this second-order linear ordinary differential equation.

The characteristic equation has the repeated root $r=-3$. Its homogeneous solution is therefore $(C_1+C_{2}x)e^{-3x}$. For a particular solution, try $y_p=Ax+B$. Substitution gives $9Ax+6A+9B=3x+1$, so $A=1/3$ and $B=-1/9$. Thus the general solution isThe extra factor $x$ in the second homogeneous term is required by the repeated root.

The characteristic equation factors as $(r+1)(r+2)=0$, giving $y_h=C_1{e}^{-x}+C_2{e}^{-2x}$. The superposition principle lets us find particular solutions for the two forcing terms separately. A trial $Ax+B$ gives $2A=1$ and $3A+2B=0$, so it is $x/2-3/4$. A trial $C{e}^{3x}$ gives $(9+9+2)C=1$, so $C=1/20$. Therefore