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Rate of change

  1. Jun 3, 2012 #1
    1. The problem statement, all variables and given/known data

    This actually part of the Friedman equation but it's the math part I'm having trouble with. I'm simplifying down to the nitty-gritty.

    a'/a = x

    a' = ax

    a = e^x

    And the lecturer said that when a rate of change is proportional to thing itself then you multiply the thing itself by exp. I don't recall learning about that rule. What is it's name so that I can go back and practice it.
     
    Last edited: Jun 3, 2012
  2. jcsd
  3. Jun 3, 2012 #2

    LCKurtz

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    If ##a## is a function of ##x## so that ##a' =\frac {da}{dx}##, the statement that the rate of change is proportional to itself means ##\frac {da}{dx} = ka##, where ##k## is a constant. That is not the type of equation you have written.

    But when you do have ##\frac {da}{dx} = ka##, the solution is ##a = Ce^{kx}## for any constant ##C##, as you can verify by plugging it in.
     
  4. Jun 3, 2012 #3
    but what's the name of this operation so that i can go back and look it up and practice it. the lecturer did have a constant written in the equation but I didn't write it for unknown reasons.
     
  5. Jun 3, 2012 #4

    HallsofIvy

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    The operation being used is integration and I feel sure you have heard of that before!
    [tex]a'= \frac{da}{dx}= ax[/tex]
    so that, in terms of differentials,
    [tex]\frac{da}{a}= xdx[/tex]

    Now integrate both sides: [itex]ln(a)= (1/2)x^2+ C[/itex] so that [itex]a= e^{(1/2)x^2+ C}[/itex] or [itex]a(x)= C_1e^{(1/2)x^2}[/itex] where [itex]C_1= e^C[/itex].

    That is NOT the result LCKurtz gave because, as he said, the equation he was looking at was not the one you posted. You said "the lecturer said that when a rate of change is proportional to thing itself then you multiply the thing itself by exp." which would be the case with a'= ka, which is what LCKurtz solved, NOT with your a'= xa, which is what I solved.

    a'= da/dx= ka becomes da/a= k dx which, integrated, gives ln(a)= kx+ C so that [/itex]a= C'e^{kx}[/itex] as LCKurtz said.
     
    Last edited: Jun 3, 2012
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