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I've taken these steps:

The defining equation is $r^{n+1} + 2r^n = 0$ Which we can then rewrite as $r + 2 = 0$

And we get that $r = -2$

Assuming that $x_n = x_n^h + x_n^p$

Then the homogenous solutions $x_n^h$ are $A(-2)^n$

And to find $x_n^p$ we assume that it can be written as something multiplied with $3^n$. Or $c \cdot 3^n$

If we put $c \cdot 3^n$ into the equation we get that $c3^{n+1} + 2c3^n = 3^n$

And then dividing by $3^n$ we get $c3^1 + 2c = 1$.

Therefore $5c = 1$ and $c = \frac{1}{5}$

We then put this into $x_n = x_n^h + x_n^p$.

And get that $x_n = A(-2)^n + \frac{1}{5}3^n$

And with $x_0 = 0$ we get that $x_n = A(-2)^0 + \frac{1}{5}3^0 = 0$ Which can be written as $A + \frac{1}{5} = 0$

We then get that $A = -\frac{4}{5}$

And that the final equation is $x_n = -\frac{4}{5}(-2)^n + \frac{1}{5}3^n$

I'm unsure if this is the right answer, please give me feedback.

Thanks.

Dr. Richard Klitzing
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1 Answers1

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This is almost correct. The only detail that you are missing is that $A+\frac15=0\implies A=-\frac15$. When I have time, I will edit this answer to prove that your solution satisfies the recurrence.

Kamal Saleh
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    If you use questionable steps along the way, and you get an answer that satisfies the recurrence, you are done. If that happens, then there is no need to justify those questionable steps. – GEdgar Dec 12 '23 at 16:35
  • @GEdgar thanks for your comment. – Kamal Saleh Dec 12 '23 at 16:36
  • I agree that if the final computation solves the recurrence then the computation must be right. However, hopefully, the given answer will also educate the OP (i.e. original poster) as to what steps are intended by the problem composer. – user2661923 Dec 12 '23 at 17:07