General solution $y_{t+2}+y_{t+1}-2y_t=3$

Question

I would like to ask if there is a way to get a General solution for $y_{t+2}+y_{t+1}-2y_t=3$ without using the characteristic polynomial approach (similar to solving Fibonacci Equation) ? That would help me a lot.

Answer 1

As one can rewrite the left side as $$ (y_{t+2}+2y_{t+1})-(y_{t+1}+2y_{t})=3 \\~\\ \iff (y_{t+2}+2y_{t+1}-3(t+1))-(y_{t+1}+2y_{t}-3t)=0 $$ one gets as first reduction $$ y_{t+1}+2y_{t}=3t+(y_{1}+2y_{0}) $$ which now is a first order recursion.

Set now $u_t=(-\frac12)^t·y_t$ to find $$ u_{t+1}=u_t+(-\frac12)^{t+1}·(3t+y_{1}+2y_{0}) $$ which now amounts to the summation of geometric and derived sequences.