Question

Find the Laplace transform \( Y(s)=\mathcal{L}\{y\} \) of the solution of the given initial value problem. \[ y^{\prime \prime}+16 y=\left\{\begin{array}{l} 1,0 \leq t

A) 1-e⁻⁸ˢcos(4t)/s(s²+16)
B) 1-e⁻⁴ˢcos(4t)/s(s²+16)
C) 1-e⁻⁸ˢsin(4t)/s(s²+16)
D) 1-e⁻⁴ˢsin(4t)/s(s²+16)

Answer 1

The Laplace transform
`\( Y(s) = \mathcal{L}\{y\} \)` of the solution to the given initial value problem is Option A) 1-e^(-8s)cos(4t)/(s(s^2+16)).

Step-by-step explanation:

In solving this initial value problem, we start with the differential equation
`\(y^(\prime\prime) + 16y = u(t)\)`, where
`\(u(t)\)` is the unit step function. Taking the Laplace transform of both sides yields
`\(s^2Y(s) - sy(0) - y^\prime(0) + 16Y(s) = 1/s\)`, where
`\(y(0)\) and \(y^\prime(0)\)` are the initial conditions.

Applying the initial conditions and solving for
`\(Y(s)\)`, we get
`\(Y(s) = (1 + sy^\prime(0) + 8y(0))/(s(s^2 + 16))\)`. The unit step function
`\(u(t)\)` introduces an exponential term in the time domain, leading to the final Laplace transform
`\(Y(s) = (1 - e^(-8s)cos(4t) + se^(-8s)sin(4t))/(s(s^2 + 16))\)`. Simplifying further, we arrive at the final answer: 1-e^(-8s)cos(4t)/(s(s^2+16)).

This solution aligns with Option A) in the choices provided. The exponential term
`\(e^(^-^8^s^)\)` represents the effect of the unit step function on the solution in the Laplace domain. The presence of cos(4t) in the numerator reflects the sinusoidal nature of the forcing function in the time domain. The denominator
`\(s(s^2 + 16)\)` corresponds to the Laplace transform of the homogeneous part of the differential equation.