# solve an difference equation's particular solution with unit step input

I have a question like this $y[n] - \frac 2 5y[n-1] = 2x[n]$ and $x[n] = 2u[n]$, the manual say that $y^p[n] = k u[n]$ and subtitute it into the equation will become $k - \frac 25k = 4$. But I think that it should be $ku[n] - \frac 25ku[n-1] = 4u[n]$, and I don't know how to solve it.

My question is why the u[n] disappear and how to solve it? If I change the input to delta, the method is the same?

x[n] is input signal and y[n] is output signal, u[n] is unit step function that u[n] = 1, n>=0, u[n] = 0, otherwise.

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Your problem statement is nearly incomprehensible to me. What is $u$ and $x$? maybe a little detail on what you did might help. – Ron Gordon Apr 9 '13 at 18:36
I have add it in my problem in the bottom – Liang-Yu Pan Apr 9 '13 at 19:01

This all seems to depend on what your initial condition is. For example, if you are concerned with just $n \ge 0$, then you will need to specify what $y_0$ is. Otherwise, the equation would deal with $n > 0$ and the unit step is irrelevant - just the constant $4$. You would then be solving

$$y_n - \frac{2}{5} y_{n-1} = 4$$

for $n \ge 1$, given $y_0$. The homogeneous solution is

$$y_n^{(H)} = A \left (\frac{2}{5} \right )^n$$

The particular solution is a constant $B$ which is found from

$$B - \frac{2}{5} B = 4 \implies B = \frac{20}{3}$$

The generic solution is then

$$y_n = A \left (\frac{2}{5} \right )^n + \frac{20}{3}$$

We find $A$ form the initial condition:

$$A + \frac{20}{3} = y_0$$

Then the full solution is

$$y_n = \left ( y_0 - \frac{20}{3} \right ) \left (\frac{2}{5} \right )^n + \frac{20}{3}$$

That said, this may not be what your problem is asking. Maybe you have all $n \in \{\infty,\ldots,\infty\}$; in that case, the step function may be significant.

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