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Little bit confuse on vector space 
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#1
Apr2309, 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?



#2
Apr2309, 07:59 AM

Sci Advisor
PF Gold
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. 


#3
Apr2309, 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 


#4
Apr2309, 08:33 AM

Sci Advisor
PF Gold
P: 1,776

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 "eigenvector" of the D operator with eigenvalue r. The set of all such function forms an "eigenbasis" so any solution must be a linear combination of exponential functions and you can find the r's algebraically. 


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