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Consider the initial value problem y'' + 4y = g(t), y(0) = 0, y'(0) = 0, where g(t) = \begin{cases} t & \text{if } 0 < t < 3 \\ 0 & \text{if } 3 \le t < \infty \end{cases} a. Take the Laplace transform of both sides of the given differential equation to create the corresponding algebraic equation. Denote the \\ Laplace transform of y(t) by Y(s). Do not move any terms from one side of the equation to the other (until you get to part (b) below). s^2Y(s) + 4Y(s) = 3 \cdot \frac{e^{-3s}}{-s} - e^{-3s} \frac{2}{s} + \frac{1}{s^2} \text{ help (formulas)} b. Solve your equation for Y(s). Y(s) = \mathcal{L}\{y(t)\} = \frac{3 \cdot (\frac{e^{-3s}}{-s} - \frac{e^{-3s}}{s^2} + \frac{1}{s^2})}{(s^2 + 4)} c. Take the inverse Laplace transform of both sides of the previous equation to solve for y(t). If necessary, use h(t) to denote the Heaviside function h(t) = \begin{cases} 0 & \text{if } t < 0 \\ 1 & \text{if } 0 \le t \end{cases} y(t) =

          Consider the initial value problem
y'' + 4y = g(t),  y(0) = 0, y'(0) = 0,
where g(t) = \begin{cases} t & \text{if } 0 < t < 3 \\ 0 & \text{if } 3 \le t < \infty \end{cases}
a. Take the Laplace transform of both sides of the given differential equation to create the corresponding algebraic equation. Denote the \\
Laplace transform of y(t) by Y(s). Do not move any terms from one side of the equation to the other (until you get to part (b) below).
s^2Y(s) + 4Y(s) = 3 \cdot \frac{e^{-3s}}{-s} - e^{-3s} \frac{2}{s} + \frac{1}{s^2} \text{ help (formulas)}
b. Solve your equation for Y(s).
Y(s) = \mathcal{L}\{y(t)\} = \frac{3 \cdot (\frac{e^{-3s}}{-s} - \frac{e^{-3s}}{s^2} + \frac{1}{s^2})}{(s^2 + 4)}
c. Take the inverse Laplace transform of both sides of the previous equation to solve for y(t).
If necessary, use h(t) to denote the Heaviside function h(t) = \begin{cases} 0 & \text{if } t < 0 \\ 1 & \text{if } 0 \le t \end{cases}
y(t) =

$Consider the initial value problem y” + 4y = g(t), y(0) = 0, y'(0) = 0, where g(t) = t if 0 < t < 3 0 if 3 ≤t < ∞ a. Take the Laplace transform of both sides of the given differential equation to create the corresponding algebraic equation. Denote the Laplace transform of y(t) by Y(s). Do not move any terms from one side of the equation to the other (until you get to part (b) below). s^2Y(s) + 4Y(s) = 3 ·(e^-3s)/(-s) - e^-3s (2)/(s) + (1)/(s^2) help (formulas) b. Solve your equation for Y(s). Y(s) = ℒ{y(t)} = (3 ·((e^-3s)/(-s) - (e^-3s)/(s^2) + (1)/(s^2)))/((s^2 + 4)) c. Take the inverse Laplace transform of both sides of the previous equation to solve for y(t). If necessary, use h(t) to denote the Heaviside function h(t) = 0 if t < 0 1 if 0 ≤t y(t) =$

Added by John W.

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