Solving I. y[n+2]-(1/3)y[n+1]=sin(n) and II. y[n+2]+3y[n+1]-4y[n]=n-1 difference equations

Question

I have two difference equations, which I just can't solve. I hardly even get the method, so if you could help me with the steps, I would be grateful.

1. $y_{n+2}-\frac{1}{3}y_{n+1}=\sin(n)$

2. $y_{n+2}+3y_{n+1}-4y_{n}=n-1$

Answer 1

at first solve the equation $y_{n+2}+3y_{n+1}-4y_n=0$ with the ansatz $y_n=q^n$