Solve the difference equation y_{n+1}=-0.9y_{n} in terms of the initial value y_{o}.
If you start with y_{0},
y_{1} = -0.9y_{0}
y_{2} = -0.9y_{1} = (-0.9)(-0.9y_{0}) = (-0.9)^{2}y_{0}
You can see that y_{n} = (-0.9)^{n}y_{0}
Solve the difference equation y_{n+1}=-0.9y_{n} in terms of the initial value y_{o}.
If you start with y_{0},
y_{1} = -0.9y_{0}
y_{2} = -0.9y_{1} = (-0.9)(-0.9y_{0}) = (-0.9)^{2}y_{0}
You can see that y_{n} = (-0.9)^{n}y_{0}
Comments
But that's not the answer. It's yn=(-1)^{n}(0.9)^{n}y_{o}.
On the contrary!
(-1)^{n}(0.9)^{n}y_{o} is the same as (-0.9)^{n}y_{0}
How is it the same?
Because -0.9 = (-1) x 0.9. Then you have to use the rule of exponents:
(ab)^{n} =a^{n}b^{n}
Thus
(- 0.9)^{n} = [(-1) x (0.9)]^{n } = (-1)^{n} (0.9)^{n}