lets T(n) =r^n
so its corresponding recurrence eqn will be r^n =2r^(n-1) + r^(n-2) {not considering c as it will affect the complexity}
=r^2=2r+1
=r^2-2r-1=0
r=1+root(2) and 1-root(2)
so, T(n) = a[1+root(2)]^n + b*(1-root(2))^n
since t(0)=t(1)=1
putting these value in the eqn we will get
t(0) = a+b=1
t(1) = a(1+root(2)) +b(1-root(2))=1
solving these 2 eqn gives
a=1/2 and b=1/2
so , t(n)=((1+root(2))^n + (1-root(2))^n)/2
= 0((1+root(2)^n)) {1+root(2)is the leading tern in the eqn}
0(2.4n)