Past exam of the mathematics course of the University of Cambridge / 2010 / ia / Paper 2 / 1A (Differential Equations)

Substitution of $y_n=r^n$ gives the characteristic equation of a linear recurrenceThe repeated characteristic root of a linear recurrence $1$ supplies the solutions $1$ and $n$, while the simple root $-2$ supplies $(-2{)}^n$. They are linearly independent: their values at three consecutive indices have nonzero determinant. A third-order linear recurrence relation is uniquely determined by three consecutive initial values, so these three solutions span its solution space. The general solution is

Let $L[y{]}_n=y_{n+3}-3y_{n+1}+2y_n$. For the exponential trial sequence $D{2}^n$,Thus $D=1/4$ gives a particular solution. The difference of any two solutions of the inhomogeneous linear recurrence satisfies the homogeneous linear recurrence relation in part (i). Therefore

The forcing base $-2$ is a root of the characteristic equation of a linear recurrence, so a constant multiple of $(-2{)}^n$ cannot be a particular solution. For $P(r)=r^3-3r+2$, direct substitution givesFor the resonant case of exponential forcing in a linear recurrence, at $r=-2$, $P(-2)=0$ and $(-2)P^{\prime }(-2)=-18$. Hence $-(n/18)(-2{)}^n$ is a particular solution of the inhomogeneous linear recurrence. The general solution is

The superposition principle applies because the recurrence operator is linear. Add the particular solutions from parts (ii) and (iii), then add the general homogeneous linear recurrence relation solution from part (i). This gives