Arrangements of roots in recurrence relations

Question

I came across the recurrence relation $T(k)-7T(k-2)+6T(k-3)=0$ where $T(0)=8,T(1)=6 $and $T(2)=22$.
I found the roots of the characteristic equation to be $1,-3,2$ and the constants to be $1,2,5$. So, the solution $T(k)=5(1)^k+2(-3)^k+1(2)^k$. But my book states the solution to be $ T(k)=5+2(2)^k+(-3)^k$. I can notice that this has been obtained by just assigning different constants to different roots. So, my question is how do we determine which constants are to be assigned to which root?

Answer 1

You properly found the roots of the characteristic equation. So $$T_k=c_1 (1)^k+c_2 (2)^k+c_3 (-3)^k=c_1 +c_2 (2)^k+c_3 (-3)^k$$ Now apply the conditions $$8=c_1+c_2+c_3$$ $$6=c_1+2c_2-3c_3$$ $$22=c_1+4c_2+9c_3$$

This seems to be easy to solve.