How can I use the Laplace transform to solve the initial value problem defined by the equation y” + 5y’ + e^(4t) with the initial condition y(0) = 2?

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To solve the initial value problem using the Laplace transform, we can follow these steps:

  1. Initial Problem Statement: We are given the equation:

    2. y” + 5y’ + e4t = 0

  2. Taking the Laplace Transform: Let Y(s) be the Laplace transform of y(t). The Laplace transform of derivatives is as follows:

    3. L{y’} = sY(s) – y(0)
    L{y”} = s2Y(s) – sy(0) – y'(0)

  3. Transform the given equation: First, we take the Laplace transform of each term in the equation:

    4. L{y”} + 5L{y’} + L{e4t} = 0

    Substituting the transforms gives us:

    s2Y(s) – sy(0) – y'(0) + 5(sY(s) – y(0)) + rac{1}{s – 4} = 0

  4. Substituting the initial conditions: With y(0) = 2 and y'(0) (which we need to assume or calculate), we substitute these values.

    5. Using y(0):

    s2Y(s) – 2s – y'(0) + 5(sY(s) – 2) + rac{1}{s – 4} = 0

  5. Reorganizing the equation: Collecting like terms, we can express it in terms of Y(s):

    6. (s2 + 5s)Y(s) – 2s – y'(0) – rac{10}{s – 4} + rac{1}{s – 4} = 0

    Next, isolate Y(s):

    Y(s) = rac{2s + y'(0) + rac{10}{s – 4} – rac{1}{s – 4}}{s2 + 5s}

  6. Finding the inverse Laplace Transform: Next, we apply the inverse Laplace transform to find y(t). Recognizing standard Laplace pairs can help with this step. However, we will need to simplify:
  7. Plugging back for y(t): Assuming we define y'(0) value, we can express Y(s) and convert it back using inverse transforms obtained from tables.

    8. Ultimately, the value will yield a function in the time domain y(t).

  8. Conclusion: Therefore, the overall procedure involves transforming, substituting initial conditions, rearranging, and eventually finding an inverse transform to return to the time domain.

Each of these steps firmly leverages the power of the Laplace transform, paving the way to not just solve the differential equation, but also facilitate handling various types of inputs through transform properties:

Exploring this method leads to a deeper understanding of dynamic systems, making it a valuable tool in engineering and applied mathematics.

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