Question

The function $x(t)$ satisfies the ODE $\ddot{x} + 6\dot{x} + 9x = \phi$ with initial conditions $x(0) = 0$, $\dot{x}(0) = 0$ and $\phi = \begin{cases} 3, & 0 \le t \le 3, \\ 0, & t > 3. \end{cases}$ (a) Find the Laplace Transform $X(s)$ of $x(t)$. $X(s) = \frac{3(1 - e^{-3s})}{s(s+3)^2}$

          The function $x(t)$ satisfies the ODE
$\ddot{x} + 6\dot{x} + 9x = \phi$
with initial conditions $x(0) = 0$, $\dot{x}(0) = 0$ and
$\phi = \begin{cases} 3, & 0 \le t \le 3, \\ 0, & t > 3. \end{cases}$
(a)
Find the Laplace Transform $X(s)$ of $x(t)$.
$X(s) = \frac{3(1 - e^{-3s})}{s(s+3)^2}$

The function x(t) satisfies the ODE
ẍ + 6ẋ + 9x = ϕ
with initial conditions x(0) = 0, ẋ(0) = 0 and
ϕ = 3, 0 ≤ t ≤ 3,
0, t > 3.
(a)
Find the Laplace Transform X(s) of x(t).
X(s) = (3(1 - e^-3s))/(s(s+3)^2)

Added by Sierra C.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Adi S

05/29/2023

Step 1

Step 1: The Laplace transform of the second derivative x^(¨) is given by s^2X(s) - sx(0) - x^(˙)(0), where X(s) is the Laplace transform of x(t). Show more…

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The function x(t) satisfies the ODE x^(¨)+6x^(˙)+9x=φ with initial conditions x(0)=0, x^(˙)(0)=0 and (H)/(phi )={(3,0<=t<=3),(0,t>3):} (a) Find the Laplace Transform x(s) of x(t). x(s)=,(3(1-e^(-3s)))/(s(s+3)^(2)) The function xtsatisfies the ODE x+6x+9x= with initial conditions 0=0,0=0 and 3,0<t<3, 0. t>3. (a) Find the Laplace Transform X(s of xt) 3 1 -e-35 Xs=31-e^-3s/ss+3^2 s(s + 3)2 Su atv MacBookAi 00 880 F4 F5 F6 >1I F7 F8 F9 % 5 & 6 7 8 9 0 R T 20 U E 2 G