Difference Equation and Dirac Delta

[Larrytsai]

Homework Statement

y[n] - (2/3)y[n-1] = x[n]

what is y[n] if x[n] = diracdelta[n]

The Attempt at a Solution

for some reason, i argued that y[n-1] = diracdelta[n-1]

so

y[n] = diracdelta[n] + (2/3)diracdelta[n-1]

Im pretty sure this is wrong, anybody can help?

 

[the_amateur]

The answer is y[n] = $(\frac{2}{3})^{n}$U[n]

 

[Larrytsai]

You mind explaining how you got to what y[n] is?