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How do I get d^2 y/dx^2 for a Cauchy-Euler, differential equation?

  1. Feb 23, 2014 #1

    s3a

    User Avatar

    1. The problem statement, all variables and given/known data
    How do I get d^2 y/dx^2 for a Cauchy-Euler, differential equation?

    Basically, how do I derive d^2 y/dx^2, as given in the following link (since I don't want to just memorize that equation)?:
    http://www.sosmath.com/diffeq/second/euler/euler.html

    2. Relevant equations
    * Cauchy-Euler, differential Equation.
    * x = e^t
    * Differentiating.
    * Chain rule

    3. The attempt at a solution
    I do get how to derive dy/dx.

    When I try to derive d^2 y/dx^2, I get a very close result, but it's different.

    Could someone please tell me where my mistake is, or if I'm doing something completely wrong, tell me what it is I am doing wrong?

    Here is my work.:
    x = e^t

    dy/dx = 1/x dy/dt

    d/dx (dy/dx) = d/dx(1/x dy/dt)

    d^2 x/dx^2 = d/dx (1/x) dy/dt + (1/x) d/dx (dy/dt)

    d^2 x/dx^2 = d/dx (1/x) dy/dt + (1/x) d/dt (dy/dx)

    d^2 x/dx^2 = d/dx (1/x) dy/dt + (1/x) d/dt (1/x dy/dt)

    d^2 x/dx^2 = d/dx (1/x) dy/dt + (1/x) d/dt (e^(-t) dy/dt)

    d^2 x/dx^2 = (-1/x^2) dy/dt + [e^(-t)] d/dt (e^(-t) dy/dt)

    d^2 x/dx^2 = [-e^(-2t)] dy/dt + [e^(-t)] (-[e^(-t)] [dy/dt] + [e^(-t)] [d^2 y/dt^2])

    d^2 x/dx^2 = -e^(-2t) dy/dt – e^(-2t) dy/dt + e^(-2t) d^2 y/dt^2

    d^2 x/dx^2 = -2e^(-2t) dy/dt + e^(-2t) d^2 y/dt^2

    d^2 x/dx^2 = e^(-2t) [d^2 y/dt^2 - 2 dy/dt]


    Any help in deriving d^2 y/dx^2 would be GREATLY appreciated!
     
  2. jcsd
  3. Feb 24, 2014 #2

    pasmith

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    Homework Helper

    I assume you mean [itex]d^2 y/dx^2[/itex] on the left, but your error is in assuming
    [tex]
    \frac{d}{dx}\left(\frac{dy}{dt}\right) = \frac{d}{dt}\left(\frac{dy}{dx}\right)
    [/tex]
    This doesn't work here:
    [tex]
    \frac{d}{dx}\left(\frac{dy}{dt}\right) = \frac1x \frac{d^2 y}{dt^2}
    [/tex]
    but
    [tex]
    \frac{d}{dt}\left(\frac{dy}{dx}\right)
    = \frac{d}{dt}\left(e^{-t} \frac{dy}{dt}\right)
    = e^{-t}\frac{d^2 y}{dt^2} - e^{-t} \frac{dy}{dt}
    = \frac1x \frac{d^2 y}{dt^2} - \frac1x \frac{dy}{dt}
    [/tex]

    I would start with the chain rule in the form
    [tex]
    \frac{d}{dt} = x \frac{d}{dx}
    [/tex]
    and then
    [tex]
    \frac{d^2 y}{dt^2} = x\frac{d}{dx}\left( x \frac{dy}{dx}\right)
    = x \frac{dy}{dx} + x^2 \frac{d^2 y}{dx^2}
    = \frac{dy}{dt} + x^2 \frac{d^2 y}{dx^2}
    [/tex]
    so that
    [tex]
    \frac{d^2 y}{dx^2} =
    \frac1{x^2} \left(\frac{d^2 y}{dt^2} - \frac{dy}{dt}\right) \\
    = e^{-2t} \left(\frac{d^2 y}{dt^2} - \frac{dy}{dt}\right)
    [/tex]
     
  4. Feb 25, 2014 #3

    s3a

    User Avatar

    Thank you very much!
     
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