1st order non homogenous DE problem solving

  1. Aug 24, 2008 #1
    In my notes, the prof wrote x' = x + sin(t), x(0) = x0

    He did not go through the solution, only wrote that the solution is

    x(t,x0) = x0et + et/2 - 1/2(sin(t) + cos(t))

    I verified this is correct by substituting.

    So when I try to solve this DE, here's what happens:

    [tex] {\frac{dx}{dy} \ - \ x = \ sin(t) [/tex]

    integrating factor is e-t

    [tex] \int \frac{d}{dt} \ (e^{-t}x) = \int e^{-t} \ sin(t) \ dt => [/tex]

    [tex] e^{-t} \ x \ = \ \int e^{-t} \ sin(t) \ dt \ = \ -e^{-t} \ cos(t) \ - \ \int cos(t) \ e^{-t} \ dt \ = \ -e^{-t} \ cos(t) - \left( -e^{-t} \ + \int sin(t) \ e^{-t} \ dt \right) [/tex]

    ...

    [tex] x = -(1/2)cos(t) + (1/2)sin(t) [/tex]

    So I'm close, but where did I lose my e-t's ?
     
  2. jcsd
  3. Aug 24, 2008 #2

    Ygggdrasil

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    The solution to a non-homogeneous ODE is in the following form:

    x = XH + Xsp

    Where XH is the general solution to the corresponding homogeneous equation and Xsp is a specific solution to the given non-homogeneous ODE. You have found a specific solution, and you just need to solve the homogeneous equation dx/dt - x = 0 to obtain the XH.
     
  4. Aug 24, 2008 #3

    HallsofIvy

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    You have e-x on both sides of the equation and they cancel out. Specifically, you get e-tx= -(1/2)e-t(cos(t)- sin(t)) by integration, then the "e-t" on each side cancel.
     
  5. Aug 24, 2008 #4
    I understand they cancel; my question referred to the fact that the prof's answer included e to the POSITIVE t and my answer included no e^t's at all.
     
  6. Aug 24, 2008 #5
    so I would solve dx/dt - x = 0 the same way that I solved dx/dt = x + sin(t), and then I should add that result with the particular solution and that is my general solution?
     
  7. Aug 24, 2008 #6

    Ygggdrasil

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    Yes. Then you can apply your boundary conditions to get the final solution.
     
  8. Aug 24, 2008 #7
    Ok so x' = x would have to be x = et.

    That gives me x = et - 1/2(sin(t) + cos(t)). It still does not match his final solution of

    x(t,x0) = x0et + (1/2)et - 1/2(sin(t) + cos(t))

    Any thoughts? I bet that I am misunderstand how to apply x(0) = x0
     
  9. Aug 24, 2008 #8
    For one thing I've been forgetting those constants from indefinite integration. I think my specific solution should look more like
    [tex]
    x = -(1/2)cos(t) + (1/2)sin(t) + c(1/2) e^{t}
    [/tex]

    so I guess all that's left if helping me figure out how the x0et term got into the final solution:
    x(t,x0) = x0et + (1/2)et - 1/2(sin(t) + cos(t))
     
  10. Aug 24, 2008 #9

    Ygggdrasil

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    You forgot the constant in the solution to the homogeneous equation:
    XH = Aet

    Therefore the general solution to the non-homogeneous ODE is:
    x(t) = Aet + (1/2)(sin(t)-cos(t))

    To find the value for A, you should use the boundary condition x(0) = x0. (You should find that A = x0 + 1/2).


    Alternatively, you could carry the constant through your original solution and solve for C using the boundary condition.
     
  11. Aug 25, 2008 #10
    Ok, I got it. Thanks for helping.
     
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