Solution to $ y_{t+2} −4y_{t+1} + 3y_{t} = t^2 + 2t$

Question

I have the following second order difference equation to solve: $y_{t+2} −4y_{t+1} + 3y_{t} = t^2 + 2t; y_{0} = 1; y_{1} = 2$. I have found the solution for the homogeneous part but i have difficulty in solving the non-homogeneous part. Can anyone advise me on what should i do in order to find the particular solution? Thanks a lot.

Answer 1

SInce you have a second degree on the rhs, you can assume a third degree for the particular solution.

So, let $$y_t=a+b t + c t^2+d t^3$$ replace in the lhs and group terms to get $$(4 d-2 b)-4 c t-6 d t^2=2t+t^2$$ So, comparing the coefficient for each power, $d=-\frac 16$, $c=-\frac 12$, $b=-\frac 13$.

Combining all results $$y_t=c_1+c_2\, 3^t +a-\frac{t^3}{6}-\frac{t^2}{2}-\frac{t}{3}=c_3+c_2 \,3^t -\frac{t^3}{6}-\frac{t^2}{2}-\frac{t}{3} $$ Use the conditions to get $c_2$ and $c_3$.

Answer 2

$$(\mathbb E-1)(\mathbb E-3) y_t = t^2+2t$$

Since the $3^{rd}$ difference of a quadratic polynomial is zero, we have

$$(\mathbb E-1)^3 (t^2+2t) = 0$$

Therefore $$(\mathbb E-1)^4(\mathbb E-3) y_t = 0$$ is homogeneous. The roots of the characteristic equations are $1$ (with multiplicity $4$) and $3$ (with multiplicity $1$). Now you can solve.

Here $\mathbb E$ is the forward shift operator (i.e., $\mathbb E^i y_t = y_{t+i}, \forall i \in \mathbb N$).