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## Main Question or Discussion Point

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?

- Thread starter zhfs
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- #1

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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

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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

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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

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[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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