Question

Take the Laplace transform of the following initial value problem and solve for

y=f(t), where : dy/dt + y = f(t), where f(t) is a periodic function defined by f (t) = sin (t)

a) Y(s) = 1/s+1 - s/s^2+1
b) Y(s) = 1/s+1 - s^2/s^2+1
c) Y(s) = s/s+1 - s/s^2+1
d) Y(s) = 1/s^2+1 - s/s^2+1
​

Answer 1

The Laplace transform of the initial value problem is Y(s) = 1/[(s + 1)(s^2 + 1)]. None of the provided choices match this transformation, indicating a possible typo or the need for additional information to determine the correct answer.

Step-by-step explanation:

The question provided involves the Laplace transform of the given initial value problem dy/dt + y = f(t), with a piecewise periodic function f(t) = sin(t). This transform is a commonly used method in differential equations to simplify the process of solving linear differential equations with given initial conditions.

First, let's apply the Laplace transform to both sides of the differential equation, denoting the Laplace transform of y(t) by Y(s). We get:

L{dy/dt} + L{y} = L{sin(t)}

Using the property of the Laplace transform that L{dy/dt} = sY(s) - y(0) and the initial condition y(0) = 0, and knowing that L{sin(t)} = 1/(s^2 + 1), we can solve the equation for Y(s):

sY(s) + Y(s) = 1/(s^2 + 1)

Combining like terms, we get:

Y(s)(s + 1) = 1/(s^2 + 1)

Y(s) = 1/[(s + 1)(s^2 + 1)]

Which can be decomposed into partial fractions; however, none of the provided answer choices exactly match the correct transformation. Therefore, there may either be a mistake in the question's options, or additional information might be necessary to select the correct Laplace transform from the given choices.