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Little bit confuse on vector space

by zhfs
Tags: confuse, space, vector
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zhfs
#1
Apr23-09, 07:44 AM
P: 10
how to proof if the solution set of a second order diffential equation af''+bf'+cf=0 is a real vector space w.r.t. the usual opeations?
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jambaugh
#2
Apr23-09, 07:59 AM
Sci Advisor
PF Gold
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P: 1,776
Since the set of differentiable functions is itself a vector space the solutions would form a subspace. It thus is sufficient to show that the set is closed under the operations of addition and scalar multiplication.

Given any subset of a vector space you already have all the properties of associativity, distribution under scalar multiplication and vector addition, etc. The only issue is closure under the basic operations.
zhfs
#3
Apr23-09, 08:11 AM
P: 10
james,

first of all thank you very much for your explaining, that helps me a lot.

but still i have some question to ask you, can you please help me out as well?

i just got no idea what is the set of the soluiton of those d.e.
do i need to use y=a^ex to solve them or i need to reduce them into first order matrix system?

but if i reduce into first order matrix system, how can i proof it is closeure under addition and scalar multiplication?

many thanks!!

regards,
tony

jambaugh
#4
Apr23-09, 08:33 AM
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Little bit confuse on vector space

Either method works for finding solutions but solving the system directly is the most straightforward (presuming a,b, and c are constants). The way to look at this equation is in terms of the derivative as an operator D:

[aD^2 +bD + c1] f = 0
The exponential function f=e^(rx) is an "eigen-vector" of the D operator with eigen-value r. The set of all such function forms an "eigen-basis" so any solution must be a linear combination of exponential functions and you can find the r's algebraically.


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