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Second order differential

  1. Apr 5, 2008 #1
    y'' - 8y' + 16 = 0 ; y(0) = 4 ; y'(0) = 9


    this should be not too bad but im stuck in the same place.
    m^2 - 8m + 16
    m1 = m2 = 4

    y = C1e^(mx) + C2(e^(mx))
    sub in y(0) = 4
    4 = C1 + C2
    C1 = 4-C2

    y' = mC1e^(mx) + mC2e^(mx)
    sub in y'(0) = 9
    9 = mC1 + mC2
    C1 = 4 - C2 (from above), m=4
    9 = 4(4-C2) + 4C2
    9 = 16 - 4C2 + 4C2
    9 = 16
    In this case would there just be no solution or am i missing something?!!?
    Thank you!!!!
     
  2. jcsd
  3. Apr 5, 2008 #2

    nicksauce

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    Science Advisor
    Homework Helper

    If I recall correctly, the general solution in the case of a repeated root for a second order constant coefficients ODE is
    [tex]c_1e^{m_1x} + c_2xe^{m_1x}[/tex]
     
  4. Apr 5, 2008 #3
    Hey
    The formula you are using for y is only valid if you have to different solutions m, i.e.
    [tex]m_{1}\neq{m_{2}} [/tex]

    In your case y is given by

    [tex]y=(C_{1}+C_{2}x)e^{mx}[/tex]
     
  5. Apr 5, 2008 #4
    thanks!!!!!
     
  6. Apr 5, 2008 #5

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Well done, guys.
     
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