Question

(1 point) Consider the following initial value problem: $x'' + 36x = \begin{cases} 2, & 0 \le t \le 5 \\ 0, & t > 5 \end{cases}$, $x(0) = 4$, $x'(0) = 0$. Solve for the Laplace transform of $x(t)$. $X(s) = \mathcal{L}\{x(t)\} = \frac{2(1-e^{-2s}) + 4s^2}{s(s^2 + 36)}$

          (1 point) Consider the following initial value problem:
$x'' + 36x = \begin{cases} 2, & 0 \le t \le 5 \\ 0, & t > 5 \end{cases}$, $x(0) = 4$, $x'(0) = 0$.
Solve for the Laplace transform of $x(t)$.
$X(s) = \mathcal{L}\{x(t)\} = \frac{2(1-e^{-2s}) + 4s^2}{s(s^2 + 36)}$

$(1 point) Consider the following initial value problem: x” + 36x = 2, 0 ≤ t ≤ 5 0, t > 5, x(0) = 4, x'(0) = 0. Solve for the Laplace transform of x(t). X(s) = ℒ{x(t)} = (2(1-e^-2s) + 4s^2)/(s(s^2 + 36))$

Added by Adam G.

Elementary and Intermediate Algebra

Alan S. Tussy, R. David Gustafson 5th Edition

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Please help with homework. 1. Consider the following initial value problem: J2, 0 < t < 5: 9 + x(0) = 4x(0) = 0, t > 5 Solve for the Laplace transform of x(t): X(s) = {x(t)} = (2(1 - e^(-2s)) + 4s^2) / (s(s^2 + 36)) Help with formulas.