# General solution to homogeneous difference equation

With a given example

$$a_{n-1} = ca_{n-2}$$

general solution:

$$a_{n} = c . c . a_{n-2}$$
$$= c . c . a_{n-3}$$
$$= c^n a_0$$

Question: Find the general solution for the homogeneous equation $$a_{n} = 5a_{n-1}$$

General solution:

$$a_{n} = 5^n a_{0}$$

Is my general solution correct based on the given example?

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I believe you have a typo in the first equation, should it be $a_n = c \cdot a_{n-1}$? –  Nicolas Villanueva May 5 '11 at 7:59
Yes. There is a typo in the first displayed formula, you meant $a_n=c\cdot a_{n-1}$, and in the third displayed line, you meant $a_n=c\cdot c\cdot c\cdot a_{n-3}$. And I would prefer to go forwards than backwards, $a_1=5a_0$, $a_2=5a_1=5^2a_0$, and so on. But backwards is fine. –  André Nicolas May 5 '11 at 8:00
liangteh, I notice you have a 0% accept rate, but you have now asked 10 questions. If you find an answer helpful, please accept it by clicking the check mark to the left of the answer. More people may be inclined to answer you in the future if you do so, and it's generally considered the polite thing to do here. –  yunone May 5 '11 at 8:03
@Nicolas Villanueva: I have corrected the typo. @yunone: I am new to mathse. Thanks for the info. I will accept the answers –  optimus May 5 '11 at 8:10
@liangneh: The correction of the first typo should have been to $a_n=c\cdot a_{n-1}$. The correction you made is correct, but not as helpful. The next typo, on the third displayed line, remains uncorrected as of now. –  André Nicolas May 5 '11 at 8:37
Even if that first equation is a typo, that is the correct general solution. If a proof is necessary, I suggest using Induction on $n$.