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Two-parameter family of solutions of the second-order DE

  1. Sep 21, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex]y = c_{1}e^{x} + c_{2}e^{-x}[/tex] is a two-parameter family of solutions of the second-order DE [tex]y^{''} - y = 0[/tex]. Find a solution of the second-order initial-value problem consisting of this differential equation and the given initial conditions.

    2. Relevant equations

    [tex]y(1) = 0[/tex]
    [tex]y^{'}(1) = e[/tex]

    [tex]y^{''} - y = 0[/tex]
    [tex]y = c_{1}e^{x} + c_{2}e^{-x}[/tex]

    3. The attempt at a solution

    I am not sure if I found the solution correctly.

    First derivative of the family of solutions:

    [tex]y^{'} = c_{1}e^{x} - c_{2}e^{-x}[/tex]

    Solving for [tex]c_{1}[/tex]:

    [tex]0 = c_{1}e^{1} + c_{2}e^{-1}[/tex]
    [tex]c_{1} = -\left(\frac{c_{2}}{e^{2}}\right)[/tex]

    Solving for [tex]c_{2}[/tex]:

    [tex]e^{1} = \left(-c_{2}e^{-2}\right)e^{1} - c_{2}e^{-1}[/tex]
    [tex]e^{1} = \left(-c_{2}e^{-1}\right) - c_{2}e^{-1}[/tex]
    [tex]e^{1} = \left(-2c_{2}e^{-1}\right)[/tex]
    [tex]c_{2} = -\left(\frac{e^{1}}{2e^{-1}}\right) = -\left(\frac{e^{2}}{2}\right)[/tex]

    [tex]c_{1} = -\left(\frac{c_{2}}{e^{2}}\right) = \left(\frac{e^{2}}{2e^{2}}\right)[/tex]

    Therefore [tex]y = \left(\frac{e^{2}}{2e^{2}}\right)e^{x} - \left(\frac{e^{2}}{2}\right)e^{-x}[/tex]
     
  2. jcsd
  3. Sep 21, 2009 #2
    Looks good. c_1 simplifies to 1/2.
     
  4. Sep 21, 2009 #3
    Great thanks.
     
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