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Second order linear differential equation with constant coefficients

  1. Jan 25, 2012 #1
    1. The problem statement, all variables and given/known data

    For the differential equation y'' - 4y' + y = 0,
    (a) Show that if we let x = y' (i.e. x(t) = y'(t)), then this leads to the system:
    x' = 4x -y
    y' = x

    (b) Conversely, show that the system in (a) leads to y'' - 4y' + y = 0 (and x'' -4x' + x = 0 also).

    2. Relevant equations

    None.

    3. The attempt at a solution

    part (a) seems easy enough, let x = y', then x' must equal y'', then substitute:

    y'' - 4y' + y = 0
    x' - 4x + y = 0
    x' = 4x - y as expected (and then we have x = y' initially given), so the system is correct.

    part (b) seems extremely obvious, which makes me wonder if I'm missing something. Anyways,, I begin with

    x' = 4x - y and y' = x. y'' = x' follows.

    by substituting back in, we have
    y'' = 4y' - y
    y'' - 4y' + y = 0 as required.

    To show find x'' - 4x' + x = 0, I don't know. I tried finding dy/dx:

    dy/dx = y'/x' = x/(4x - y) which is not separable:
    dy(4x - y) = xdx

    so my questions are: is part (b) as simple as it seems? Or is my method not correct? And how do I show x'' - 4x' + x = 0? Thanks in advance!
     
  2. jcsd
  3. Jan 25, 2012 #2

    vela

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    Yes, it's as simple as it seems.

    You find the differential equation for x similarly. Start by differentiating the equation for x'.
     
  4. Jan 25, 2012 #3
    Thank you for your help. I end up with x'' - 4x' + x = 0 as required.
     
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