1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Initial value nonhomogeneous DE

  1. Nov 28, 2009 #1
    1. The problem statement, all variables and given/known data

    (D^3 - D^2 + D - 1)y = e^x

    and y(0) = 0, y'(0) = 1, y"(0) = 0

    2. Relevant equations

    D = d/dx

    3. The attempt at a solution

    Factoring the left side of the equation gives :
    (D^2 + 1)(D - 1)y = e^x

    Which has roots of +/- i and 1.
    So y(c) = Ae^x + Bsin(x) + Ccos(x)

    The annihilator for e^x is (D-1)
    and y(p) = Dxe^x = e^x[Dx]
    So y(p)' = D[xe^x + e^x] = e^x[D(x+1)]
    y(p)" = D[xe^x + 2e^x] = e^x[D(x+2)]
    y(p)"' = D[xe^x + 3e^x] = e^x[D(x+3)]

    Plugging into y"' - y" + y' -y = e^x
    and dividing by e^x yields:

    2D = 1 ; or D = (1/2)

    So now y = Ae^x + Bsin(x) + Ccos(x) + (1/2)xe^x

    Plugging into initial conditions:
    y(0) = A + C = 0 ; OR A = -C

    Now y' = Ae^x + Bcos(x) - Csin(x) + (1/2)(xe^x + e^x)
    and y'(0) = 1 = A + B + (1/2)(0 + 1)
    or A + B = (1/2)

    Now y" = Ae^x - Bsin(x) - Ccos(x) + (1/2)(xe^x + 2e^x)
    and y"(0) = 0 = A - C + (1/2)(0 + 2)
    or A - C + 1 = 0

    Solving for A, B, and C yields:
    A = (-1/2)
    B = 1
    C = (1/2)

    The solution is therefore:

    y = (-1/2)e^x + sin(x) + (1/2)cos(x) + (1/2)xe^x

    Does this look correct?

    Thanks a bunch
     
  2. jcsd
  3. Nov 28, 2009 #2
    Anyone know if I'm in the right direction?
     
  4. Nov 29, 2009 #3
    Bump.
     
  5. Nov 29, 2009 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    It looks fine. You know you can check your own work by taking the y you get and substituting back into the original ODE? That is satisfies the boundary conditions isn't hard to check either?
     
  6. Nov 29, 2009 #5
    Thanks, it makes sense but I did not think of that.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook