Question

Solve the recurrence relation s(k)-10s(k-1)+9s(k-2)=0 with n
s(0)=3,s(1)=11.

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO SOLVE

The recurrence relation given by

$\sf{S_k - 10 S_{k - 1} + 9S_{k - 2} = 0 \: }$

With initial conditions

$\sf{S_0 = 3 \: \: \: and \: \: \: S_{ 1} = 11\: }$

CALCULATION

The given difference equation is

$\sf{S_k - 10 S_{k - 1} + 9S_{k - 2} = 0 \: }$

This is a second order homogeneous difference equation with constant coefficients.

The corresponding auxiliary equation is

$\sf{ {x}^{2} - 10x + 9 = 0 \: \: }$

$\implies \: \sf{ {x}^{2} - 9x - x + 9 = 0 \: \: }$

$\implies \: \sf{ x(x - 9)- 1(x - 9) = 0 \: \: }$

$\implies \: \sf{ (x - 9)(x - 1) = 0 \: \: }$

$\implies \: \sf{ x = 1 \: , \: \: 9 \: \: }$

Hence 1 & 9 are the roots of the auxiliary equation and the roots are distinct.

$\sf{So \: there \: exists \: two \: constants \: t_1 \: and \: t_2 \: such \: that }$

$\sf{S_k = t_1 \times {(1)}^{k} + t_2 \times {(9)}^{k} } \: \: \: for \: all \: k \geqslant 0$

$\implies \: \sf{S_k = t_1 + t_2 \times {(9)}^{k} } \: \: \: ........(1)$

$\sf{ Now\: \: S_0 =3 \: \: \: gives \: \: }$

$\: \sf{ t_1 + t_2 \times {(9)}^{0} = 3}$

$\implies \: \sf{ t_1 + t_2 = 3 } \: \: \: ........(2)$

$\sf{ Now\: \: S_1 =11 \: \: \: gives \: \: }$

$\: \sf{ t_1 + t_2 \times {(9)}^{1} = 11}$

$\implies \: \sf{ t_1 + 9 t_2 = 11 } \: \: \: ........(3)$

Now Equation (2) - Equation (2) gives

$\: \sf{ 8 t_2 = 8 } \: \: \:$

$\implies \: \sf{ t_2 = 1 } \: \: \:$

From Equation (2)

$\implies \: \sf{ t_1 = 3 - 1 } \: \: \:$

$\implies \: \sf{ t_1 = 2 } \: \: \:$

Hence from Equation (1)

$\sf{S_k = 2 + {(9)}^{k} } \: \: \:$

RESULT

The solution of the given recurrence relation is

$\boxed{\sf{ \: \: \: S_k = 2 + {(9)}^{k} } \: \: \: }$

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LEARN MORE FROM BRAINLY

A sequence {an} is defined recursively, with a1 = -1, and, for n > 1, an = an-1 + (-1)^n.

Find the first five terms

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