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Solving ODE with Heaviside Step and Delta function

  1. Oct 30, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the solution of the equation:

    α(dy/dt) + y = f(t)

    for the following conditions:
    (a) when f(t) = H(t) where H(t) is the Heaviside step function
    (b) when f(t) = δ(t) where δ(t) is the delta function
    (c) when f(t) = β^(-1)e^(t/β)H(t) with β<α


    2. Relevant equations



    3. The attempt at a solution

    1.
    α(dy/dt) + y = f(t)
    2.
    (dy/dt) + (1/α)y = (1/α)f(t)
    3.
    finding the integrating factor
    μ(t) = e^(∫(1/α)dt) = e^(t/α)
    4.
    [e^(t/α)](dy/dt) + (1/α)[e^(t/α)]y = (1/α)[e^(t/α)]f(t)
    5.
    d/dt{[e^(t/α)]y}=[(e^(t/α))/α]f(t)
    6.
    ∫d/dt{[e^(t/α)]y}dt=∫[(e^(t/α))/α]f(t)dt
    7.
    [e^(t/α)]y = ......

    then i dont know how to continue

    please help guys...thanks
     
  2. jcsd
  3. Oct 30, 2012 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Do you know what those functions are? The 'Heaviside step function' is 0 if x< 0, 1 if [itex]x\ge 0[/itex]. So [itex]\int H(x)e^{-t/a} dx= 0[/itex] if x is negative and [itex]\int e^{-t/a} dt[/itex] if x is positive.

    The 'delta function' (strictly speaking not a function but a 'generalized function' or 'distribution') has the property that for any function, f(x), [itex]\int f(x)\delta(x)dx= f(0)[/itex].
     
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