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}.

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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}