Find explicit solutions to the recurrences: an + 1 = 2an + 11, n 0, with a0 = 3 bn + 1 = 4bn - 3, n 0, with b0 = 1 Solution the general solution of the type of the equation yn+1 = ryn + b is of the form : yk = crk + y ; where y is the equilibrium solution! a) for very large values n, we will have an+1 = an s0, a = 2a + 11 => a = -11 is an equilibrium solution : now, the General solution is : ak = c2k - 11 given , a0 = 3 => c - 11 = 3 => c = 14 so, the explicit solution is : ak = 14.2k -11 b) for very large values n, we will have bn+1 =bn s0, b = 4b -3 => b = 1 is an equilibrium soluiton now, the General solution is : bk= c.4k + 1 given , b0 = 1 => c + 1 =1 => c =0 so, the explicit solution is bk = 1 , for all values of k..

1. Find explicit solutions to the recurrences: an + 1 = 2an + 11, n 0, with a0 = 3 bn + 1 = 4bn - 3,
n 0, with b0 = 1
Solution
the general solution of the type of the equation yn+1 = ryn + b is of the form : yk = crk + y ;
where y is the equilibrium solution!
a)
for very large values n, we will have an+1 = an
s0, a = 2a + 11
=> a = -11 is an equilibrium solution :
now, the General solution is : ak = c2k - 11
given , a0 = 3
=> c - 11 = 3
=> c = 14
so, the explicit solution is : ak = 14.2k -11
b)
for very large values n, we will have bn+1 =bn
s0, b = 4b -3
=> b = 1 is an equilibrium soluiton
now, the General solution is : bk= c.4k + 1
given , b0 = 1
=> c + 1 =1
=> c =0
so, the explicit solution is bk = 1 , for all values of k.