# little bit confuse on vector space

by zhfs
Tags: confuse, space, vector
 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?
 Sci Advisor PF Gold P: 1,767 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.
 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
PF Gold
P: 1,767

## 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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