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First Order Linear Homogeneous Equation

  1. Dec 10, 2008 #1
    First Order Linear Non-Homogeneous Equation

    1. The problem statement, all variables and given/known data

    I need to solve for e(t)



    2. Relevant equations

    Do I use Laplace Transform for the last integral?


    3. The attempt at a solution
    [tex]
    \begin{subequations}
    \begin{eqnarray}
    \nonumber
    \dot{\hat{{\cal E}}}(t) &=& -\kappa \hat{{\cal E}}(t) + \sqrt{2\kappa}\, \hat{{\cal E}}_{in}(t), \\

    \end{eqnarray}
    \end{subequations}




    \begin{subequations}
    \begin{eqnarray}
    \dot{\hat{{\cal E}}} &=& -\kappa \hat{{\cal E}} + \sqrt{2\kappa}\, \hat{{\cal E}}_{in}, \\
    \nonumber
    I &=& e^{\int \kappa dt} =ce^{\kappa t}\\
    \nonumber
    ce^{\kappa t}\dot{\hat{{\cal E}}} + ce^{\kappa t}\kappa \hat{{\cal E}} &=& ce^{\kappa t} \sqrt{2\kappa}\, \hat{{\cal E}}_{in}, \\
    \nonumber
    \frac{d}{dt}(\hat{{\cal E}}e^{ \kappa t}) &=& ce^{\kappa t} \sqrt{2\kappa}\, \hat{{\cal E}}_{in}\\
    \nonumber
    \hat{{\cal E}}e^{ \kappa t} &=& \int^{\infty}_{0}ce^{\kappa t} \sqrt{2\kappa}\, \hat{{\cal E}}_{in}(t)dt\\
    \nonumber
    \nonumber
    \end{eqnarray}
    \end{subequations}
    [/tex]
     
    Last edited: Dec 10, 2008
  2. jcsd
  3. Dec 10, 2008 #2
    [tex]
    \dot{y}(t) = -\kappa y(t) + \sqrt{2\kappa} x_{in}(t)
    [/tex]
    If that notation helps.
     
  4. Dec 10, 2008 #3
    The method I wrote above is wrong. From what I remember, your supposed to find the complementary solution and the particular solution yc, yp.

    For the yp, your supposed to choose a form so as to solve for yp but x_{in} is some arbitrary function. What do I do?
     
  5. Dec 11, 2008 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    It would help a lot if you didn't assume people will understand what problem you are doing. Is [itex]\epsilon_{in}[/itex] a given function and not dependent on [itex]\epsilon[/itex]?

    Assuming that it is, you are essentially using "variation of parameters".
    The homogeneous equation is [itex]\epsilon'= -\kappa\epsilon[/itex] which should be easy to solve.

    Now assume a solution of the form [itex]\epsilon(t)= u(t)Y(t)[/itex] where "Y(t)" is the solution to the homogeneous equation. The [itex]\epsilon'= u'Y+ uY'[/itex] and so the equation becomes [itex]u'Y+ uY'= u(-\kappa Y)+ \sqrt{2\kappa}\epsilon_{in}[/itex].

    Since Y is a solution to the homogeneous equation, [itex]Y'= -\kappa Y[/itex], [itex]uY'= u(-\kappa Y)[/itex] so we can cancel those and have [itex]u' Y= \sqrt{2\kappa}\epsilon_{in}[/itex] remaining. Since both Y and [itex]\epsilon_{in}[/itex] are known functions, that's just an integration,
    [tex]u(t)= \int_0^t \frac{\sqrt{2\kappa}\epsilon_{in}(\tau)}{Y(\tau)} d\tau[/itex]

    That is almost what you have. But note that the integral is NOT a definite integral from 0 to infinity. It is from 0 to t (actually, you could use any lower limit- that just changes the constant of integration) and notice that I have changed the "dummy" variable inside the integral to [itex]\tau[/itex] so as not to confuse it with t.

    Instead of what you have you should have
    [tex]e^{-\kappa t}u(t)= e^{\kappa t}\int_0^t e^{\kappa\tau}\epsilon_{in}(\tau)dt[/tex]
    is a specific solution, to be added to [itex]Ce^{-\kapa t}[/itex], the general solution to the homogeneous equation. Note that changing the variable inside the integral makes it clear that you cannot cancel the two exponentials!
     
    Last edited: Dec 11, 2008
  6. Dec 11, 2008 #5
    Where is the sqrt(2k)? I think my approach is correct except the limits on the integral.
    I didn't realize I was doing variation of parameters.

    Then how do you solve that last integraL?

    [tex]
    \epsilon(t)= e^{-\kappa t}\int_0^t \sqrt{2\kappa}e^{\kappa\tau}\epsilon_{in}(\tau)dt = e^{-\kappa t} \sqrt{2\kappa}e^{\kappa\tau}\epsilon_{in}(\tau)t
    [/tex]

    ?
     
    Last edited: Dec 11, 2008
  7. Dec 12, 2008 #6
    Or is that not right?
     
  8. Dec 13, 2008 #7
    Why did you use tau?
     
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