Finding particular solution to recurrence relation

[vincentvance]

Hi,

I have a question about how to find the particular solutions when trying to solve recurrence relations. For example, trying to solve

a_n+2 = -4a_n + 8n2ⁿ ,

I begin with finding the roots in the characteristic polynomial associated with the homogeneous equation, so r₁ = 2i and r₂ = -2i.

Then, because the roots are complex, the general solution is

a_n^(h) = 2ⁿ*(αcos(πn/2) + βsin(πn/2)).Now, my textbook suggests trying a function of the form

(An+B)2ⁿ

when trying to find the particual solution. I don't understand why and I have come across a couple of other examples which have made me equally confused as I am this time. Could anyone shed some light on the matter?

 

[chisigma]

Hi,

I have a question about how to find the particular solutions when trying to solve recurrence relations. For example, trying to solve

a_n+2 = -4a_n + 8n2ⁿ ,

I begin with finding the roots in the characteristic polynomial associated with the homogeneous equation, so r₁ = 2i and r₂ = -2i.

Then, because the roots are complex, the general solution is

a_n^(h) = 2ⁿ*(αcos(πn/2) + βsin(πn/2)).Now, my textbook suggests trying a function of the form

(An+B)2ⁿ

when trying to find the particural solution. I don't understand why and I have come across a couple of other examples which have made me equally confused as I am this time. Could anyone shed some light on the matter?

The procedure to 'attack' this type of problem is illustrated in...

http://mathhelpboards.com/discrete-mathematics-set-theory-logic-15/difference-equation-tutorial-draft-part-ii-860.html#post4671

Using the 'useful table' You arrive to the following relation...

$\displaystyle \gamma_{n}= 8\ n\ 2^{n} \implies w_{n} = 8\ 2^{n}\ (\chi_{0} + \chi_{1}\ n)\ (1)$

Kind regards

$\chi$ $\sigma$

 

[vincentvance]

The procedure to 'attack' this type of problem is illustrated in...

http://mathhelpboards.com/discrete-mathematics-set-theory-logic-15/difference-equation-tutorial-draft-part-ii-860.html#post4671

Using the 'useful table' You arrive to the following relation...

$\displaystyle \gamma_{n}= 8\ n\ 2^{n} \implies w_{n} = 8\ 2^{n}\ (\chi_{0} + \chi_{1}\ n)\ (1)$

Kind regards

$\chi$ $\sigma$

Thank you!

In the post you linked it says "Some caution is to be adopted using the table 2.1 because the $w_n$ are valid only if the sequences themselves aren’t solution of the homogeneous DE and if that is the case a different procedure must be adopted."

Isn't the sequence a solution to the homogeneous equation in this example because the general solution is a product with $2^n$

 

[chisigma]

Thank you!

In the post you linked it says "Some caution is to be adopted using the table 2.1 because the $w_n$ are valid only if the sequences themselves aren’t solution of the homogeneous DE and if that is the case a different procedure must be adopted."

Isn't the sequence a solution to the homogeneous equation in this example because the general solution is a product with $2^n$

The homogeneous equation is...

$\displaystyle a_{n+2} + 4\ a_{n} = 0\ (1)$

... and its solution is...

$\displaystyle a_{n} = 2^{n}\ \{ c_{0}\ i^{n} + c_{1}\ (- i)^{n}\}\ (2)$

The particular solution we have found is...

$\displaystyle w_{n} = 8\ 2^{n}\ (\chi_{0} + \chi_{1}\ n)\ (3)$

... and (3) is not a solution of (1)...

Kind regards

$\chi$ $\sigma$

 

[vincentvance]

The homogeneous equation is...

$\displaystyle a_{n+2} + 4\ a_{n} = 0\ (1)$

... and its solution is...

$\displaystyle a_{n} = 2^{n}\ \{ c_{0}\ i^{n} + c_{1}\ (- i)^{n}\}\ (2)$

The particular solution we have found is...

$\displaystyle w_{n} = 8\ 2^{n}\ (\chi_{0} + \chi_{1}\ n)\ (3)$

... and (3) is not a solution of (1)...

Kind regards

$\chi$ $\sigma$

Oh, I actually think I get it now. Thank you for your patience!