Solution of a 1st order homogeneous difference equation
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Solution of a 1st order homogeneous difference equation

[From: ] [author: ] [Date: 11-07-08] [Hit: ]
Thus, a(n) = c1 * 5^n is the general solution.Since a(0) = 2 we see that a(0) = 2 = c1 * 5^0 implies c1 = 2.Thus, a(n) = 2 * 5^n.-First,......
Given an+1 - 5 an = 0 and a0=2
what is the particular solution?
i know its an= 2*5^n but how? :S

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a(n+1) - 5a(n) = 0

Characteristic equation: r - 5 = 0. Then r = 5. Thus, a(n) = c1 * 5^n is the general solution.

Since a(0) = 2 we see that a(0) = 2 = c1 * 5^0 implies c1 = 2. Thus, a(n) = 2 * 5^n.

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First, rewrite the equation (parentheses added for clarity):
a(n+1) - 5a(n) = 0
a(n+1) = 5a(n)

Now, think about it: for each new value of a(n+1), you just multiply the old a(n) by 5. Suppose a(0) is k: then the sequence is k, 5k, 25k, 125k, ..., (5^n)k, in other words the general solution is a(n)= (5^n)k. You are given that k=2, so the particular solution is a(n) = (5^n)2, which is what you wanted.
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keywords: equation,st,difference,order,Solution,homogeneous,of,Solution of a 1st order homogeneous difference equation
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