homoonous solution(discret time signal)

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TAKYMOUNIR

Joined Jun 23, 2008
352
y[n]=y[n-1]+u[n]
where u[n] is step function
y[n]=yp[n]+yh[n]
wher yp[n] is the particular solution
and yh[n] is the homogenous solution
in this example yh[n]=k^n so why ? how this come from
i know in homogenous solution we put input to be zero so u[n]=zero but how yh[n]=k^n
thanks
 
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