Question

solve the recarency relation ar-7ar-1+10ar-2=2r,a0=0,and a1=6

Answer 1

SOLUTION

TO SOLVE

The recurrence relation

$\sf{a_r - 7a_{r - 1} + 10a_{r - 2} = 0}$

With initial conditions

$\sf{a_0 = 0 \: \: \: and \: \:a_1 = 6 }$

EVALUATION

Here the given recurrence relation is

$\sf{a_r - 7a_{r - 1} + 10a_{r - 2} = 0}$

This is a second order homogeneous difference equation with constant coefficients

The corresponding auxiliary equation is

$\sf{ {x}^{2} - 7x + 10 =0 }$

$\implies \sf{ \sf{ (x - 2)(x - 5) = 0 }}$

$\implies \sf{ \sf{x = 2 \:, \: 5}}$

The roots of the auxiliary equation are real and distinct

Hence there are two constants b and c such that

$\sf{a_r = b .\: {2}^{r} +c. \: {5}^{r} \: \: \: \: \: \: \: for \: all \: r \geqslant 0 }$

Now the initial conditions are

$\sf{a_0 = 0 \: \: \: and \: \:a_1 = 6 }$

Now

$\sf{a_0 = 0 \: \: gives \: \: b + c = 0 }$

$\sf{a_1 = 6 \: \: \: gives \: \: \:2b + 5c = 6 }$

Above gives

$\sf{ - 2c + 5c = 6 }$

$\implies \sf{ 3c = 6 }$

$\implies \sf{ c = 2 }$

$\implies \sf{ b = - c = - 2 }$

Now

$\sf{a_r = b .\: {2}^{r} +c. \: {5}^{r} \: \: \: \: \: \: gives\: }$

$\sf{a_r = - 2.\: {2}^{r} +2. \: {5}^{r}}$

$\sf{ \implies \: a_r = - \: {2}^{r + 1} +2. \: {5}^{r}}$

Hence the required solution is given by

$\sf{ \: a_r = - \: {2}^{r + 1} +2. \: {5}^{r}} \: \: \: \: \: \: \: \: \: \: for \: all \: r \geqslant 0$

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