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Proving 2nd Order Differential Eqns

  1. Mar 26, 2007 #1
    1. The problem statement, all variables and given/known data
    If y = (3x)/e^2x, find the value of x when d^2y/dx^2 = 0


    2. Relevant equations
    I'll just abbv d^2y/dx^2 as d2ydx2

    3. The attempt at a solution
    I kept getting stuck at:
    d2ydx2 = 6/e^2
    When d2ydx2 = 0,

    6/e^2 = 0
    6 = e^2

    Then where's my x?? :confused: :confused:

    P.S: i have serious issues with differentiation =.=
     
  2. jcsd
  3. Mar 26, 2007 #2
    y = 3x*e^(-2x)
    y' = 3e^(-2x) - 6xe^(-2x)
    y'' = -6e^(-2x) - 6e^(-2x) + 12xe^(-2x) = 0
    -12e^(-2x) + 12xe^(-2x) = 0
    12xe^(-2x) = 12e^(-2x)
    x = 1

    Differentiated using the chain rule (for exponentials) and product rule.
     
  4. Mar 28, 2007 #3
    thanx! ^^
    i thought that (e^2x)squared = (e^4x) lol :biggrin:
     
  5. Mar 28, 2007 #4
    It is.

    Filler Filler Filler.
     
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