# Non-homogeneous equation is also a difference equation

The problem is this:

The impulse response of a system is the output from this system when excited by an input signal δ(k) that is zero everywhere, except at k = 0, where it is equal to 1. Using this definition and the general form of the solution of a difference equation, write the output of a linear system described by:

y(k) – 3y(k – 1) – 4y(k – 2) = δ(k) + 2δ(k – 1)

The initial conditions are: y(–2) = y(–1) = 0.

My question is: How can the particular solution be found using the method of undetermined coefficients if the non-homogeneous equation is also a difference equation?

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It looks you've copied from a textbook. You need be more specific; what do you call a general form of the solution? Anyway, if you can find the impulse response, name it $h(k)$, and it is easy to see that the solution you are looking for is $h(k) + 2h(k-1)$ why? (remember your system is LTI!) – fmoura2005 Oct 22 '12 at 0:42
Sorry, the last part is my actual question. A general solution is the sum of the characteristic solution to the homogeneous equation and the particular solution to the non-homogeneous equation. I know the characteristic solution is link. – photon Oct 22 '12 at 4:17
Can we equal homogenous thing with non homogeneous things ? – user50841 Nov 27 '12 at 5:26

OK. So, I suppose you want to solve your difference equation for $k\geq 0$. If you set $k=0$ we easily get $y(0)=1$ and you could find $y(k)$ recursively for $k>0$, right? that's the point. To obtain the impulse response, you can see it as a response of an homogeneous equation with initial conditions different from zero. Then, check it out that the impulse response $h(k)$ to your problem is the same of this one: $$y(k) - 3y(k-1) - 4y(k-2)=0, \quad y(-2)=1/4, y(-1)=0$$ Why? simply because they have the same solution for $k\geq 0$. Now its easy to find $h(k)$ right? The final solution is then $h(k)+2h(k-1)$.