# Homogeneous first and second order difference equations

I don't really get what I am meant to be doing here, ALSO (...)- ANYTHING inside brackets are subscripts since I don't know how to do that here. Solve the following first order difference eqn : $$2\,x_{n+1} + 3\,x_n = 0$$ I know the answer is X(n)=A[-3/2]^n (sorry gave you the wrong answer must have confused you my bad!

I have worked it out $$2\,x+3=0\quad X_n=A\left(-\frac 32\right)^n$$ $$2\,x=-3$$ $$x=-\frac 32$$ and the formula $x_n=C.[a^n]$ but I actually don't really know or understand what I am doing :(
this will probably seem very easy to most of you mathematicians out there but could you please explain every single step?

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I have edited your question. Could you verify the result (I am not sure of the $A(-3/2)^n$ part since a closing parenthesis was missing...). If $A=x_0$ please specify it. – Raymond Manzoni May 16 '13 at 13:43
(If I understood correctly... $a, A$ should be specified) You may rewrite your equation as $\;x_{n+1}=-\frac 32\;x_n\;$ and get indeed $x_n=\left(-\frac 32\right)^n\;x_0$. If $x_0=a$ you may substitute it. But the meaning of $x$ in your question is unclear... – Raymond Manzoni May 16 '13 at 14:00
Thankyou sooo much your comment is very help ful and sorry since I didnt actually know what I was doing I came to a conclusion that x=-3/2 out of confusion my bad, thanyou your comment is very helpful! – Unistudent9 May 16 '13 at 14:17
Glad it helped @Unistudent9 ! – Raymond Manzoni May 16 '13 at 14:25

The basic idea is to assume that any solution has the form $$x_n = n^kz^n$$ for some $z$. For a particular choice of $k$, that turns the (homogenous) equation into a polynomial in $z$, and the roots of the polynomial thus correspond to the solutions of your difference equation.
You always start with $k=0$, i.e. $x_n = z^n$, and only needs to consider higher values of $k$ if that doesn't yield "enough" solutions.