# Finding zero input response with only one initial condition

I have to find the Zero input response of the following system if $$y[-2]= 0.5$$

$$x[n] - 3y[n-1] - 4y[n-2] = 0$$ I am not sure if this is the correct way since every other example I have done had two initial conditions given.

My solution:

$$-3y[n-1] - 4[n-2] = 0$$ ( Homogeneous response equal to zero)

Let's set $$y(n) = Ar^n$$. We have:

$$-r^{n-2}(3r+4)=0 \leftrightarrow -r^{n-2} = 0$$ gives $$r_1 =0$$

$$3r+4=0$$ gives $$r_2= -4/3$$

Finding coefficient $$A$$

$$y[-2] = A(-4/3)^{-2} =0.5 \leftrightarrow A(9/16) = 0.5 \leftrightarrow A= 8/9$$

Zero input response $$y[n] = \frac{8}{9}\left(-\frac{4}{3}\right)^n$$

• Yes, your solution looks correct to me. Next time, please make an effort to correctly format your question!
– Jdip
Nov 23, 2023 at 0:19

Your solution is correct, but it's good to understand the reason why you only need one initial condition in this case: there are only two output values in the difference equation, and they are only one sample instant apart. So for $$x[n]=0$$ (zero input), you simply have
$$y[n-1]=-\frac43 y[n-2]\tag{1}$$
\begin{align*} y[-1] &= \frac12\cdot\left(-\frac43\right) \\ y[0] &= \frac12\cdot\left(-\frac43\right)^2 \\ y[1] &= \frac12\cdot\left(-\frac43\right)^3 \\\vdots \end{align*}
$$y[n]=\frac12\cdot\left(-\frac43\right)^{n+2}=\frac89\cdot\left(-\frac43\right)^{n}$$