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Solving differential equation using Laplace Transform

  1. Jan 1, 2015 #1
    1. The problem statement, all variables and given/known data
    solve the following differential equation using Laplace transforms:
    [tex] y'' + 4y' + 4y = t^2 e^{-2t}, y_0 = 0, y'_0 = 0 [/tex]

    y_0 and y'_0 are initial conditions.

    2. Relevant equations
    Using L to represent the Laplace transform, we have that

    [tex]
    L(y) = Y
    [/tex]
    [tex]
    L(y') = pY - y_0
    [/tex]
    [tex]
    L(y'') = p^2 Y - py_0 - y'_0
    [/tex]

    3. The attempt at a solution
    Taking the Laplace transform of the entire DE gives

    [tex]
    p^2 Y - py_0 - y'_0 + 4pY - 4y_0 + 4Y = L(t^2 e^{-2t})
    [/tex]

    From Laplace transform tables (using M. Boas, page 469) we have

    [tex]
    L(t^2 e^{-2t}) = 2/(p+2)^3
    [/tex]
    [tex]
    ∴ p^2 Y - 0 -0 +4pY - 0 +4Y = 2/(p+3)^3
    [/tex]
    [tex]
    ∴ (p^2 + 4p + 4)Y = 2/(p+2)^3
    [/tex]
    [tex]
    ∴Y = 2/(p+2)^5
    [/tex]

    [tex]
    ∴ y = L^{-1}(Y) = L^{-1}(2/(p+2)^5)
    [/tex]

    Then from tables (pg 469, M Boas, mathematical methods for the physics sciences), we use the following relation:
    [tex]
    L(t^k e^{-at}, k>-1
    [/tex]

    becomes

    [tex]
    k!/(p+a)^{k+1}
    [/tex]

    This all makes sense and I'm confident that this is correct so far

    But somehow this means that the inverse Laplace transform of [tex] 2/(p+2)^5 [/tex] is [tex] 2t^4 e^{-2t}/4! [/tex] (which is also the solution to the DE - since it's [tex] L^{-1}(Y) = y [/tex]). I don't see how to make this jump and I don't see how that is correct.
     
  2. jcsd
  3. Jan 1, 2015 #2

    LCKurtz

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    Science Advisor
    Homework Helper
    Gold Member

    ##s## is more commonly used for the transform variable. Use $$
    \mathcal L^{-1}F(p+a) = e^{-at}\mathcal L^{-1}F(p)$$where in your problem ##F(p) = \frac 2 {p^5}##. You need to multiply numerator and denominator by ##4!## to make the formula fit.
     
    Last edited: Jan 1, 2015
  4. Jan 1, 2015 #3

    lurflurf

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    Homework Helper

    that is just a particular case of
    $$L\left(C\dfrac{t^k}{k!}e^{-a \, t}\right)=C(p+a)^{-(k+1)}$$
    which we can get by combining
    $$L\left(C\dfrac{t^k}{k!}\mathrm{f}(t)\right)=\dfrac{C}{k!}\left(-\dfrac{d}{dp}\right)^k L(\mathrm{f}(t))\\
    L\left(e^{-a \, t}\right)=(p+a)^{-1}$$
    into
    $$L\left(C\dfrac{t^k}{k!}e^{-a \, t}\right)=\dfrac{C}{k!}\left(-\dfrac{d}{dp}\right)^k L(e^{-a \, t})=\dfrac{C}{k!}\left(-\dfrac{d}{dp}\right)^k (p+a)^{-1}=C(p+a)^{-(k+1)}$$
     
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