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Sorry for this dumb question, but I need help finding an error in my proof.

Assume you have a difference equation $$u_{n+1} + au_n + bu_{n-1} = 0.$$ We assume the solution is $$A \lambda^{n+1} + aA\lambda^n + bA\lambda^{n-1}.$$ Then the auxiliary equation is $$\lambda^2 + a\lambda + b = 0.$$ Assume this has repeated roots $$\lambda = \lambda_1 = \lambda_2.$$ Then the textbook tells me that the solution is of the form $$u_n = (A+Bn)\lambda^n.$$ I tried proving this but I got stuck:

\begin{align} (A + B(n+1))\lambda^{n+1} + a(A+Bn)\lambda^n + b(A+B(n-1))\lambda^{n-1} &= 0 \\ (A + Bn + B)\lambda^{n+1} + a(A + Bn)\lambda^n + b(A + Bn - B)\lambda^{n-1} &= 0 \\ (A + Bn)\lambda^{n+1} + B\lambda^{n+1} + a(A + Bn)\lambda^n + b(A + Bn)\lambda^{n-1} - bB\lambda^{n-1}&= 0 \\ \lambda^{n-1}(A+Bn)(\lambda^2 + a\lambda + b) + B\lambda^{n+1} -bB\lambda^{n-1} &= 0 \\ B\lambda^{n+1} -bB\lambda^{n-1} &= 0 \\ \lambda^2 &= b \\ \end{align}

This is clearly not correct. I'm feeling really dumb but I'm just stuck and I don't see what I did wrong up there ...

Hinton
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    If $\lambda_1 = \lambda_2$ then $a^2-4b= 0$ and the solution is on the form $$u_n=(-\sqrt b)^n C_1 + n (-\sqrt b)^n C_2$$ – Cesareo Jul 07 '19 at 15:16
  • How did you get to $a^2 - 4b = 0$? – Hinton Jul 08 '19 at 07:52
  • If $\lambda^2 + a\lambda + b = 0$ has equal roots then $a^2-4b=0$ – Cesareo Jul 08 '19 at 08:25
  • omg hahaha just by setting the root of the quadratic solution equation to zero ... helppp me haha. This is what happens when you set a CS major loose on some "introduction to probability" lecture notes – Hinton Jul 08 '19 at 14:54

1 Answers1

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You haven’t done anything wrong. You simply haven’t applied the fact that the auxiliary equation has a repeated root. It is therefore a perfect square, so factors as $(\lambda\pm\sqrt b)^2$, from which $\lambda^2=b$. (Note that this also means that $b\ge0$ for the equation to have a repeated root.)

amd
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