HELP!
Show that if $$\lambda _1=\lambda _2=\lambda$$, then the general solution to $$x_{n+2}+bx_{n+1}+cx_n=0$$
is given by $$x_n=C_1\lambda ^n+C_2n\lambda ^n$$

Question

HELP!
Show that if $$\lambda _1=\lambda _2=\lambda$$, then the general solution to $$x_{n+2}+bx_{n+1}+cx_n=0$$
is given by $$x_n=C_1\lambda ^n+C_2n\lambda ^n$$

loser66 · Answer

Do you have any initial condition? like $X_0=??$ and $X_1=??$

loser66 · Answer

Characteristic equation: $\lambda^2 +b\lambda +c =0$
hence if $\lambda_1=\lambda_2=\lambda$, then, the solution of recursive equation is
$x_n = C_1\lambda^n+C_2n \lambda^n$, that is the formula. But if you need prove, let see..

anonymous · Answer

There are no initial conditions.. I need a prove for that but i don't know how to go with it

loser66 · Answer

Let me look up at my book,

loser66 · Answer

@SithsAndGiggles

loser66 · Answer

I tried consider $X_n $ in both derivative and quadratic, but get the definition. How to prove the definition is... hehhe... out of my head.
I was taught that consider it as ODE to get y" +by'+cy =0 but at the end up, still have to use definition. :)

anonymous · Answer

I tried going the same route as i was taught in DE but i didnt get anywhere

loser66 · Answer

For recursive part like this, I looked up my discrete math, but all problems have initial condition. That's why I looked back to ODE

loser66 · Answer

I am sorry for being helpless. Hopefully someone can help you so that I learn more. :)

anonymous · Answer

@dan815, Any help on this please

dan815 · Answer

ya cuz its a 2 by 2 matrix you only get 1 lin dep eigen vector  classs

anonymous · Answer

How do i prove this because am confused

dan815 · Answer

see wat happens with eigenvectors when u got repeate eigen value

dan815 · Answer

ok to show its true show that AXn  = A(c1...+c2...)

anonymous · Answer

So how would the matrix look like. Am really confused.

dan815 · Answer

ok lets start by wrting this in matrix form

dan815 · Answer

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