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Math Help - Second order ODE's, proof of solution method

  1. #1
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    Second order ODE's, proof of solution method

    When solving second order ODE's, we always assume that all solutions will be proportional to e^rt.

    How hard is it to prove that this has to be the case?

    If there's a simple proof, please don't just state it. In stead give me a hint of where to begin, so I can try to do it myself.

    If the proof is long and complicated, I don't need to know all the details (because I wouldn't understand them anyway). But it still would be nice to get a brief outline of what it involves.
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  2. #2
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    Re: Second order ODE's, proof of solution method

    Here's one way but I don't know if it's a proof or not. Consider the ODE

    y'' -(a+b)y' + aby = 0,\;\;\; a\ne b   \;\;(1)

    where a and b are constant. I'm assuming that you meant constant coefficients otherwise the solutions are not usually in the form e^{rt}. Now we will factor this equation using operators. We can re-write (1) as

    \left(\dfrac{d}{dt} - a\right)\left(\dfrac{d}{dt} - b\right) y = 0.  \;\; (2).

    If we let

    u = \left(\dfrac{d}{dt} - b\right) y
    then (2) becomes

    \left(\dfrac{d}{dt} - a\right)u = 0,   (2).

    or

    u' - au = 0.

    We solve (it's separable) giving

    u = c_1 e^{at}

    Now

    \left(\dfrac{d}{dt} - b\right) y = u = c_1 e^{at}

    or

    y' - by = c_1 e^{at}

    This is linear which we integrate giving

    y = c_1 e^{at} + c_2 e^{bt}

    noting that I have absorbed some constants into c_1.
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