Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    2017 Award

    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


    User Avatar
    Science Advisor

    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


    User Avatar
    Science Advisor
    2017 Award

    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
    x = -(1/2)cos(t) + (1/2)sin(t) + c(1/2) e^{t}

    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


    User Avatar
    Science Advisor
    2017 Award

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook