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Recurrence relation $$a_n = 2 a_{n-1} - a_{n-2} + 3$$ ?

The initial conditions are $a_0$ = 2 and $a_1$ = 5.

I know the characteristic equation is: $$t^2-2t+1$$ The roots are 1 and 1.

The homogeneous part is: $$( c_1 + c_2 * n ) * 1^n$$

How do you solve the particular part?

I guess the particular part to be: A

I substituted in recurrence relation: A = 2A - A + 3 but I get A = A+3 --> 0 = 3.

What did I do wrong?

Malcolm Scott
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1 Answers1

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Because this is a "failure case" you have to try $c_3n^2$ for the particular solution, since both $c_1$ and $c_2n$ are already "taken".

On substitution you will find that $c_3=\frac32$, and hopefully you can continue from there...

David Quinn
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