# Solve the difference equation?

1. Jul 12, 2013

### Success

Solve the difference equation yn+1=-0.9yn in terms of the initial value y0.

y1=-0.9y0
y2=-0.9y1=(-0.9)2y0
yn=(-0.9)ny0
Is this the answer? Because the answer in the textbook says yn=(-1)n(0.9)ny0. Please help.

2. Jul 12, 2013

### krome

Your solution and the solution in your textbook are the same.

3. Jul 14, 2013

### HallsofIvy

Staff Emeritus
$$(ab)^n= a^n b^n$$

4. Jul 14, 2013

### Staff: Mentor

The textbook wrote it that way to make it clear that even terms are positive and odd terms are negative.

5. Jul 15, 2013

### HallsofIvy

Staff Emeritus
And, as I pointed out in your previous thread, you are not finished until you have shown that your solution does satisfy the equation. If yn= (-0.9)ny0 then -0.9yn= -0.9((-0.9)ny0)= (-0.9)n+1y0= yn+1.

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