I have a system given by $$y[n] - \frac{1}{4} y[n-1] - \frac{1}{8} y[n-2] =3x[n] $$

I want to solve for $y[n]$ for $x[n]=(\frac{1}{2})^nu[n]$.

The complementary solution evaluates to $[k_1(\frac{1}{2})^n+k_2 (\frac{-1}{4})^n]u[n]$.

But when I evaluate the particular solution consider $y_p[n]=k(\frac{1}{2})^n u[n]$ I get an absurd answer as $0=3$. And boundary conditions are not given either. What initial conditions am I to consider? I want to solve this difference equation without using Z-transform or Fourier transform.

Edit: I took the particular solution of the form $$y_p[n]=kn(\frac{1}{2})^nu[n]$$ since the input is of the same form as the roots of the characteristic equation.

and get k as equal to 2.

and evaluate the coefficients of complementary solution as $$y[n]=\frac{8}{3} (\frac{1}{2})^n u[n]+\frac{1}{3} (\frac{-1}{4})^n u[n]+2n (\frac{1}{2})^n u[n]$$

but the solution given is $$\frac{1}{3} (\frac{1}{4})^n u[n] + 4(n+1)(\frac{1}{2})^{(n+1)} u[n+1] + \frac{2}{3} (\frac{1}{2})^n u[n]$$

I don't where the anomaly in my evaluation is? Or is the given solution wrong? I'm new to the concept of difference equations.

I just want to know if my understanding is right.

I wish to solve this difference equation without using z transform.

  • $\begingroup$ How did you come up with that particular solution? $\endgroup$
    – Matt L.
    Commented Jan 26, 2021 at 16:34
  • $\begingroup$ Isn't the particular solution suppose to be of the same form as the input. The particular solution for the input $a^n u[n]$ is of the form $ka^n u[n]$....or so was told to me by my professor....or does this take a different form because the input is the same as one of the roots of the characteristic equation...I'm confused.. $\endgroup$
    – Orpheus
    Commented Jan 26, 2021 at 17:05
  • $\begingroup$ Yes, the problem is that the input has the same form as a characteristic mode of the system. $\endgroup$
    – Matt L.
    Commented Jan 26, 2021 at 17:22
  • $\begingroup$ So how do I resolve it?...I can't seem to find anything online $\endgroup$
    – Orpheus
    Commented Jan 26, 2021 at 17:24

1 Answer 1


The solution you came up with is the correct homogeneous solution (i.e., when $x[n]=0$). Thing is, this is a non-homogeneous difference equation, and its solutions are of the form $$y[n]=y_h[n]+y_p[n]=k_1\Big(\frac{1}{2}\Big)^nu[n]+ k_2\Big(\frac{-1}{4}\Big)^nu[n]+y_p[n]$$

where $y_p[n]$ is the particular solution that solves for the $x[n]$ term, and you are missing the corresponding particular solution in your evaluation. One way to find $y_p[n]$ is to guess the form of $y_p[n]$ with unknown coefficients given the form of $x[n]$, and use the method of undetermined coefficients to find those coefficients.

For $x[n]=r^nu[n]$, you would try $y_p[n]=ar^nu[n]$ at first. But, in this case $x[n]$ corresponds to a mode of the system. Therefore, the particular solution cannot be of the form $a(\tfrac{1}{2})^nu[n]$, but it has to be of the form $an(\tfrac{1}{2})^nu[n]$. Therefore, the solution of the equation is

$$y[n]=k_1\Big(\frac{1}{2}\Big)^nu[n]+ k_2\Big(\frac{-1}{4}\Big)^nu[n]+an\Big(\frac{1}{2}\Big)^nu[n],$$

and you find $a$ by evaluating this solution in the equation.


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