$\text{Solve the following recurrence relation:} \\ a_n-5a_{n-1}+6a_{n-2}=0 \\ a_n-5a_{n-1}+6a_{n-2}=2^\pi \text{with initial conditions}a_0 = -1 \ \ and \ \ a

Welcome back.

and 3 others joined a min ago.

2.5kviews

$\text{Solve the following recurrence relation:} \\ a_n-5a_{n-1}+6a_{n-2}=0 \\ a_n-5a_{n-1}+6a_{n-2}=2^\pi \text{with initial conditions}a_0 = -1 \ \ and \ \ a_1 =1$

written 8.2 years ago by

teamques10 ★ 69k

modified 3.1 years ago by

pedsangini276 • 4.8k

Mumbai University > Computer Engineering > Sem 3 > Discrete Structures

Marks: 7 Marks

Year: Dec 2015

discrete mathematics

ADD COMMENT EDIT

1 Answer

58views

written 8.2 years ago by

teamques10 ★ 69k

The characteristic equation of the recurrence relation is $r^2-5r+6=0$

Its roots are $r_1=3, r_2=2$ . Hence the sequence { $a_n$ } is a solution to the recurrence relation if and only if

$a_n=α_1*3^n+α_2*2^n$

For some constant $α_1$ and $α_2$ .

From the initial condition, it follows that

$a_0=-1=α_1 + α_2 \\ a_1= 1 = 3α_1 + 2α_2$

Solving the equations, we get $α_1=3, α_2=-4$

Hence the solution is the sequence { $a_n$ } with

$a_n =3*3^n-4*2^n$

ADD COMMENT EDIT

Community

Content

Company