Solve the following recurrences using the characteristics equation

T(n) = 4T(n-1) – 3T(n-2)

T(1) = 1

T(2) = 2

T(n) – 4T(n-1) + 3T (n-2) = 0

r^n r^n-1 r^n-2

r^n – 4r^n-1 + 3r^n-2= 0

r^n-2 (r^2 -4r + 3) = 0

r = 3, r= 1

T(n) = C1 3n + C2 1n

For T(1) =1

C1 3^1 + C2 1^1 = 1

For T(2)=2

C1 3^2 + C2 1^2 = 2

9C1 + C2 = 2

Solving C1

C2 = 1 - 3C1

9C1 + (1 - 3C1 ) = 2

6C1 + 1 = 2

C1 = 1/6

Solving for C2

3*(1/6) + C2 = 1

(½) + C2 = 1

C2 = ½

for T(2) = 2

C1 3^2 + C2 1^2 = 1 for T(2) = 2

Therefore: T(n) = (1/6) 3n + (½) 1n

Did I do this correctly?