Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

Solve the initial value problems using Laplace transforms   y'' + y' - 2y = 4e^2t            y(0) = 1;    y'(0) = -1?

y'' + y' - 2y = 4e^2t          y(0) = 1;     y'(0) = -1

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  • 1 month ago

    Given: y(0) = 1, y'(0) = -1

    Second Order Linear Ordinary Differential Equation:

    y" + y' - 2y = 4e^(2t)

    ℒ(y") + ℒ(y') - ℒ(2y) = ℒ(4e^2t)

    Use Laplace transform rules found in the cited source:

    s²ℒ(y) - sy(0) - y'(0) + sℒ(y) - y(0) - 2ℒ(y) = 4ℒ(e^2t)

    s²ℒ(y) - s(1) - (-1) + sℒ(y) - (1) - 2ℒ(y) = 4ℒ(e^2t)

    ℒ(y)[s² + s - 2] - s = 4 / (s - 2)

    ℒ(y) = [4 / (s - 2)(s² + s - 2)] + [s / (s² + s - 2)]

    ℒ(y) = (4 + s(s - 2) / (s - 2)(s² + s - 2))

    ℒ(y) = ((4 + s² - 2s) / (s - 2)(s² + s - 2))

    y = ℒ⁻¹((s² - 2s + 4) / (s - 2)(s - 1)(s + 2))

    y = ℒ⁻¹[A / (s - 2)] + ℒ⁻¹[B / (s - 1)] + ℒ⁻¹[C / (s + 2)]

    Partial Fraction Expansion:

    (s² - 2s + 4) / (s - 2)(s - 1)(s + 2) = [A / (s - 2)] + [B / (s - 1)] + [C / (s + 2)]

    A(s - 1)(s + 2) + B(s - 2)(s + 2) + C(s - 2)(s - 1) = (s² - 2s + 4)

    A(s² + s - 2) + B(s² - 4) + C(s² - 3s + 2) = (s² - 2s + 4)

    A + B + C = 1

    A - 3C = -2

    -2A - 4B + 2C = 4

    3 Equations, 3 Unknowns:

    A = -2 + 3C

    B = 1 - C - A

    B = 1 - C - (-2 + 3C)

    B = 3 - 4C

    -2(-2 + 3C) - 4(3 - 4C) + 2C = 4

    4 - 6C - 12 + 16C + 2C = 4

    12C = 12

    C = 1

    A = -2 + 3(1) = 1

    B = 3 - 4(1) = -1

    Plug back in to the inverse Laplace equation:

    y = ℒ⁻¹[(1) / (s - 2)] + ℒ⁻¹[(-1) / (s - 1)] + ℒ⁻¹[(1) / (s + 2)]

    Use inverse Laplace transform rules found in the cited source:

    y = e^(2t) - e^t + e^(-2t)

  • 1 month ago

    y(t) = ⅓ e⁽⁻²ᵗ⁾ (-3 e⁽²ᵗ⁺²⁾ (2 t + 1) + e³ᵗ + 4 e⁽³ᵗ⁺²⁾  - e² + 2)

  • rotchm
    Lv 7
    1 month ago

    Apply the exact same procedure already given in:

    https://ca.answers.yahoo.com/question/index?qid=20...

    Any steps in there you require help in?

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