can you please explain this example to me

i can't get the last step (Initial Conditions)

Example-1: tn - 3th-1 = 4n to = 0, t=4 Let's apply Theorem 3: 4n of the form bp(n), b = 4 & degree d of p(n) = 0 characteristic equation (ao rk + ajrk-1 + ... + a)(r - b)d+1 = 0, that is (r - 3) (r - 4) = = 0 Replace n with n-1 in the original recurrence: tn-1 - 3th -2 = 4n-1 (1) (1/4)tn-(7/4)n-1 + 3tn-2=0 Divide the original recurrence by 4: (1/4)t, (3/4)),-1 = 41-1 Multiply by 4: we obtain a homogeneous linear recurrence In - 7th-1 + 12tn - 2 = 0 Characteristic equation: r2 - 7r + 12 = (r - 3) (r - 4) = 0, roots are r= 3 and r=4 General Solution: tn = c 31 + c24n Use Initial Conditions: th= 4n+1 - 4(34)