I am new to DSP and was going through different responses of a system subjected to an input. My understanding of zero input response is: it is the response/output of the system when the input signal is set to zero. In other words if a system is described by a linear constant coefficient difference equation the zero input response is the homogeneous solution. 

However if the $\mathcal Z$-transform of the input is a rational function $X(z)=N(z)/Q(z)$ and that of the LTI system function is $H(z)=B(z)/A(z)$ and **the system is initially relaxed**, then $Y(z)= H(z)X(z) = N(z)B(z)/A(z)Q(z)$. Assuming distinct zeros(real  only) and poles(real only) of $X(z)$ and $H(z)$ then 

$$Y(z) = \sum_{k=1}^N \frac{A_k}{1-p_kz^{-1}} + \sum_{k=1}^L \frac{Q_k}{1-q_kz^{-1}}$$ 

which gives 

$$y(n) = \sum_{k=1}^N A_k(p_k)^{n}u(n) + \sum_{k=1}^L Q_k(q_k)^{n}u(n)$$ 

where $p_k$ and $q_k$ are the poles of system $H(z)$ and input signal $X(z)$ respectively and $u(n)$ is the unit step function. Now the first term is referred to as the natural response of the system $H(z)$. It's very confusing to grasp the difference between zero input and natural response. 

Edit: The reference of the question is to book DSP : Principles , Algorithms and Applications by John Proakis and D Manolakis 
[pdf of the book is here][1]
Page no 203 and 204. The two paragraphs after formula 3.6.4 explains the difference between zero input response and natural response


 


  [1]: http://itl7.elte.hu/~zsolt/Oktatas/editable_Digital_Signal_Processing_Principles_Algorithms_and_Applications_Third_Edition.pdf