by frostpad
Last Updated August 02, 2018 19:20 PM

I apologize if this question is far too obvious, but I'm a bit confused (and new to this subject). I have to solve the recurrence relation:

$a_n=6a_{n-1}-9a_{n-2}$

$n=2,3,..$

Using Euler substitution, I get

$x^2-6x-9=0$

Solving this equation gives me $x_1=x_2=3$

The solution seemed to be obvious (since there weren't any initial conditions, I cannot determine $\lambda$ coefficients): $a_n=\lambda_1*3^n$

But the official one from my book is: $a_n=\lambda_1*3^n+\lambda_2*n*3^n$

The only explanation I'm left with is *"Along with $a_n=\lambda_1*3^n$, it's obvious that $\lambda_2*n*3^n$ is another solution as well, which gives us $a_n=\lambda_1*3^n+\lambda_2*n*3^n$.*

If anyone understands how they got the second solution, I would really appreciate an explanation. I assume it's something really obvious, but I can't see it. If the question isn't clear, please point me to it since I'm not sure I've translated it properly.

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