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Solve for initial Value

  1. Nov 2, 2012 #1
    1. The problem statement, all variables and given/known data
    Solve for initial Value

    The attempt at a solution

    http://store2.up-00.com/Sep12/RjY55784.jpg [Broken]
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Nov 2, 2012 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    The way you have written this is wrong and very confusing! You have the initial value problem
    [tex]\frac{d^2y}{dx^2}= 2- 6x[/tex]
    y'(0)= 4, y(0)= 1.

    Since the right side is a function of x only this really just an interation problem.
    Integrating once,
    [tex]\frac{dy}{dx}= 2x- 3x^2+ C[/tex]
    You have "[itex]d^2y/dx^2[/itex]" still on the left side which is "wrong and confusing".
    Because dy/dx= 4 when x= 0, C= 4 so
    [tex]\frac{dy}{dx}= 2x- 3x^2+ 4[/tex]
    which is what you have on your sixth line although now you have it equal to "y", rather than dy/dx.

    Integrating again, [itex]y= x^2- x^3+ 4x+ C[/itex] and, since y(0)= 1, C= 1 (in your hand written derivation, that "1" looks an awful lot like a "2") so your final answer, [itex]y(x)= x^2- x^3+ 4x+ 1[/itex] is correct.
     
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