Concept of finding forced response from difference equation?

Question

I know how the process works, first from difference equation make a homogeneous equation and then find natural response , then find a particular response (in the form of input) substitute it back in the equation and find constants. And we get a forced response. I have a doubt in finding constants.

This example is Drill problem 2.12 in "Signals and Systems" by Simon Haykin:

For the equation: $$y[n] - (0.25)*y[n-2] = 2*x[n]+x[n-1]$$

find a forced response for $x[n]=u[n].$

I have found the natural response and for particular response. Say, $y[n] = k*u[n]$

Now the equation is:

$$k*u[n] - 0.25*k*u[n-2] = 2*u[n] + u[n-1]$$

The answer according to the book is $k=4$, but that is true only for $n>=2$. Can somebody explain what happens if $2>n>0$; $k=4$ does not seem correct in this case, as $u[n-2]=0$ for $n<2$ and $u[n-1]=0$ for $n<1$.

Why can we just substitute all $u[n-c] = 1$, in the difference equation for finding value of constants in particular solution?

Answer 1

considering your equation,

$$k\times u[n] - 0.25\times k\times u[n-2] = 2\times u[n] + u[n-1]$$

the left half of the equation will result in a series as shown below:

$$\begin{array}{cccccccc} [k & k & \frac{3}{4}k & \frac{3}{4}k & \frac{3}{4}k & \frac{3}{4}k & \ldots &]\end{array}$$

the right hand of the equation is $$\begin{array}{cccccccc} [2 & 3 & 3 & 3 & 3 & 3 & \ldots & ]\end{array}$$

for these two series to be equal, $$\begin{array}{cccccccc} k = 2 \\ k=3\\ \frac{3}{4}k=3 \\ \frac{3}{4}k=3 \\ \vdots & \end{array}$$

It is not possible to equate these two sequences without removing first two points. In that case, $k = 4$. The sequences are equal or the equation holds for $k = 4$ and $n \ge 2$.

What we are getting here is total solution, so it will have the natural response and forced response. The output get stabilized and reach the stable value only after $n \ge 2$. This 'transients' before $n < 2$ are due to the natural response.