$y[n]-3y[n-1]-4y[n-2]=x[n]$
i solved it as follows :
1.first calculated natural-response(ZIR i.e,zero input response)
by using characteristic equation it came out as
$y_n[n]=[c_1(-1)^{n}+c_2(4)^{n}]$ here i couldn't be able to calculate $c_1$ and $c_2$ because there were no initial conditions given in the question.
2.And,Then calculated forced-response(ZSR i.e,with all initial conditions zero) and ,since we know it consists of two parts
homogeneous solution and particular solution i.e,($ y_h[n] +y_p[n]$) i calculated both of parts separately as follows: first $ y_p[n]=0 $ because input is $\delta [n] $ and secondly $y_h[n] $ =$[C_1(-1)^{n}+C_2(4)^{n}$ (here i was able to calculate values of $C_1$ and $C_2$ as $\dfrac{-1}{5}$ and $\dfrac{6}{5}$ respectively by using zero initial conditions i.e, $y[-1]=y[-2]=0$,$x(0)=1$ ,as per definition of forced response)
3.and finally i added all solutions to get total solution i.e, $y[n]= y_n[n]+y_h[n]+y_p[n]$ and got answer as $ y_n[n]=[c_1(-1)^{n}+c_2(4)^{n}-0.2(-1)^{n}+1.2(4)^{n}+0] $ but answer in the book is given as $y [n]=[-0.2(-1)^{n}+1.2(4)^{n}]$ so my problem is why didn't they incorporated natural response term in final answer . thus, my questions are
1.does "finding impulse response in numerical" means finding only "forced response" and leaving natural response.terms.unlike we do in other input cases(like $x_\left[n\right]=\left[\dfrac{1}{2}\right]^n$) where we also calculate natural responses(where of course initial conditions are provided so that we can calculate $c_1$ and $c_2$ as well)...... is impulse input an exceptional case of inputs?
2.same they have done for questions regarding 'step response'
i haven't read z-transform and generating function method to solve difference equations so please do not give answer by using those techniques.
