Homogeneous first and second order difference equations

117 Views Asked by At

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?

1

There are 1 best solutions below

1
On

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.