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ODE with Superposition

  1. Oct 9, 2013 #1
    1. The problem statement, all variables and given/known data
    Given
    [itex]\frac{dx}{dt}[/itex] = -1.3x
    [itex]x_{1}[/itex](t)=e[itex]^{-1.3t}[/itex]
    [itex]x_{2}[/itex](t)=4e[itex]^{-1.3t}[/itex]

    Compute a solution for x(t) if x(0)=3



    2. Relevant equations
    Superposition Principle
    and some ODE related
    Anyhow I refer to this
    http://www.youtube.com/watch?v=_ECd0Jn7y68


    3. The attempt at a solution
    First Attempt
    x(t)=[itex]\alpha[/itex] (e[itex]^{-1.3t}[/itex]) + [itex]\beta[/itex] (4e[itex]^{-1.3t}[/itex])
    [itex]\frac{dx}{dt}[/itex]= (-1.3) * ([itex]\alpha[/itex] [itex]x_{1}[/itex] + [itex]\beta[/itex] [itex]x_{2}[/itex] )

    Then after this step, I've no idea how to continue, Im stuck here, what Should I do with the given initial condition? x(0)=3

    I've done some google search, some ODE with Initial condition provided

    Perhaps Related Solution:
    [itex]\frac{dy}{dx}[/itex] = -1.3x
    [itex]\int[/itex] 1 dy = [itex]\int[/itex] -1.3x dx

    To get the Constant, I've plugged in the given initial condition x(0)=3
    y = [itex]\frac{-1.3x^{2}}{2}[/itex] + c
    c = -5.58

    Re-arrange the eqn

    y= [itex]\frac{-1.3x^{2}}{2}[/itex] - 5.58
    after getting this?
    how do I proceed ?
    Imma so confusee!


    (Sorry Mods, It's quite sometime I dint visit the forum, Making such messy mistake on the previous post)
    Anyway
    Here's the (Hand written working LINK)
    http://imgur.com/VmfPJwr
    http://imgur.com/J5k5YeO
    http://imgur.com/VmfPJwr
    I'm not that familiar with the Latex Code :tongue:
     
    Last edited: Oct 9, 2013
  2. jcsd
  3. Oct 9, 2013 #2

    Mark44

    Staff: Mentor

    You're making this much more difficult than it actually is. Do you know what the general solution of your differential equation is? Assuming that you do, just use your initial condition to find the solution for which x(0) = 3.
    This is not at all related to your problem. A related problem would be dy/dx = -1.3y.
     
  4. Oct 9, 2013 #3
    Thanks Mark for replying

    I've messed up the thing eventually

    Based on the related problem dy/dx = -1.3 y
    Integrate it
    and
    apply the boundary condition x(0)=3

    [itex]\int[/itex] [itex]\frac{dy(x)}{dx}/{y(x)}[/itex] = [itex]\int[/itex] -1.3 dx

    log(y(x)) = -1.3x +c

    y(x) = [itex]e^{-1.3x+c}[/itex]
    Let C = [itex]e^{C1}[/itex]

    y(x) =C * [itex]e^{-1.3x}[/itex]

    Plug in the Given boundary
    x(0)=3

    y(x) =C * [itex]e^{-1.3x}[/itex]
    3 = C * [itex]e^{0}[/itex]
    C=3

    Re-arrange

    y(x) =3* [itex]e^{-1.3x}[/itex]

    So until this step
    what should I do with the

    [itex]x_{1}[/itex](t)=e[itex]^{-1.3t}[/itex]
    [itex]x_{2}[/itex](t)=4e[itex]^{-1.3t}[/itex]
     
    Last edited: Oct 9, 2013
  5. Oct 9, 2013 #4

    Mark44

    Staff: Mentor

    Yes, that's it, although you should write your answer as x(t) = 3e-1.3t. In your diff. equation, x was the dependent variable and t was the independent variable.

    I believe that what they wanted you to do with x1(t) and x2(t) was to recognize that all solutions are of the form Ce-1.3t.
     
  6. Oct 9, 2013 #5

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You need to decide whether your independent variable is ##t## or ##x##. What happens if you calculate ##x_2 - x_1##?
     
  7. Oct 9, 2013 #6

    Oh ya!
    I used to use dy and dx
    and now i get messed up again!:tongue:
     
  8. Oct 9, 2013 #7
    Thanks for remind that!
     
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