Wednesday, 10 April 2019
10:33 AM
Given that the general solution of the homogeneous recurrence
relation is …−6_(n−2)=0%20is%20hn=A(−2)%5en+B(3)%5en…_files/image001.png){kind=small}
Find the general solution of the following recurrence relations
−6_(n−2)=0%20is%20hn=A(−2)%5en+B(3)%5en…_files/image002.png){kind=small}
Do NOT use −6_(n−2)=0%20is%20hn=A(−2)%5en+B(3)%5en…_files/image003.png){kind=small}
This will give −6_(n−2)=0%20is%20hn=A(−2)%5en+B(3)%5en…_files/image004.png){kind=small}
−6_(n−2)=0%20is%20hn=A(−2)%5en+B(3)%5en…_files/image005.png){kind=small}
−6_(n−2)=0%20is%20hn=A(−2)%5en+B(3)%5en…_files/image006.png){kind=small}
… That didn't help!
Use −6_(n−2)=0%20is%20hn=A(−2)%5en+B(3)%5en…_files/image007.png){kind=small}
−6_(n−2)=0%20is%20hn=A(−2)%5en+B(3)%5en…_files/image008.png){kind=small}
−6_(n−2)=0%20is%20hn=A(−2)%5en+B(3)%5en…_files/image009.png){kind=small}
−6_(n−2)=0%20is%20hn=A(−2)%5en+B(3)%5en…_files/image010.png){kind=small}
Thus we obtain −6_(n−2)=0%20is%20hn=A(−2)%5en+B(3)%5en…_files/image011.png){kind=small}
General solution is then −6_(n−2)=0%20is%20hn=A(−2)%5en+B(3)%5en…_files/image012.png){kind=small}
−6_(n−2)=0%20is%20hn=A(−2)%5en+B(3)%5en…_files/image013.png){kind=small}
Try
Cancel out the factor to get−6_(n−2)=0%20is%20hn=A(−2)%5en+B(3)%5en…_files/image014.png)
−6_(n−2)=0%20is%20hn=A(−2)%5en+B(3)%5en…_files/image015.png){kind=small}
Thus −6_(n−2)=0%20is%20hn=A(−2)%5en+B(3)%5en…_files/image016.png){kind=small}
General solution is now −6_(n−2)=0%20is%20hn=A(−2)%5en+B(3)%5en…_files/image017.png){kind=small}
−6_(n−2)=0%20is%20hn=A(−2)%5en+B(3)%5en…_files/image018.png){kind=small}
Try −6_(n−2)=0%20is%20hn=A(−2)%5en+B(3)%5en…_files/image019.png){kind=small}
Since shares a term with the solution of the homgeneous equation, we use −6_(n−2)=0%20is%20hn=A(−2)%5en+B(3)%5en…_files/image020.png){kind=small}
This gives −6_(n−2)=0%20is%20hn=A(−2)%5en+B(3)%5en…_files/image021.png){kind=small}
Cancel out the factor −6_(n−2)=0%20is%20hn=A(−2)%5en+B(3)%5en…_files/image022.png){kind=small}
−6_(n−2)=0%20is%20hn=A(−2)%5en+B(3)%5en…_files/image023.png){kind=small}
Thus −6_(n−2)=0%20is%20hn=A(−2)%5en+B(3)%5en…_files/image024.png){kind=small}
−6_(n−2)=0%20is%20hn=A(−2)%5en+B(3)%5en…_files/image025.png){kind=small}
−6_(n−2)=0%20is%20hn=A(−2)%5en+B(3)%5en…_files/image026.png){kind=small}
won't work, as it shares the same
term of the homogeneous equation−6_(n−2)=0%20is%20hn=A(−2)%5en+B(3)%5en…_files/image027.png){kind=small}
So use, −6_(n−2)=0%20is%20hn=A(−2)%5en+B(3)%5en…_files/image028.png){kind=small}
−6_(n−2)=0%20is%20hn=A(−2)%5en+B(3)%5en…_files/image029.png){kind=small}
Cancel the factor from both sides−6_(n−2)=0%20is%20hn=A(−2)%5en+B(3)%5en…_files/image030.png){kind=small}
−6_(n−2)=0%20is%20hn=A(−2)%5en+B(3)%5en…_files/image031.png){kind=small}
−6_(n−2)=0%20is%20hn=A(−2)%5en+B(3)%5en…_files/image032.png){kind=small}