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let us suppose we have following recurrence relation

$T(n)=T(n-1)+1$ where $T_{0}=1$

we need to find homogeneous solution and particular solution , for homogeneous solution, we have

$T(n)-T(n-1)=0$ so in index form $a_n-a_{n-1}=0$ therefore $r-1=0$ , from where we have $r=1$ and general solution will be $a_n=\alpha*r^n$ , because $r=1$ we have $a_n=\alpha$ now for particular solution we have $f(n)=1$ , that why simple take particular solution $T_p=A$ if we put get

$A=A+1$ but how to solve this equation? please help me

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  • $\begingroup$ any help guys?of course i can use iterated version but i need to know how to solve using characteristic equation $\endgroup$ – dato datuashvili Oct 7 '18 at 12:40
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The solution to $T(n) = T(n-1) + 1$ is $T(n) = n + T(0)$.
The solution to $T(n) = T(n-1)$ is $T(n) = T(0)$.

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  • $\begingroup$ i know but can i find it using characteristic equation? $\endgroup$ – dato datuashvili Oct 7 '18 at 20:27
  • $\begingroup$ i think that there was a misunderstanding ,, i have asked about if it is possible to solve this system using characteristic equation , homogeneous and non homogeneous techniques $\endgroup$ – dato datuashvili Oct 7 '18 at 20:53

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