Question

Solve the initial value problem below using the method of Laplace transforms.\ y'' - 16y = 96t - 40e^{-4t}, y(0) = 0, y'(0) = 31\ y(t) = -4e^{-4t} + 4e^{4t} - t\ y(t) = -4e^{-4t} + 4e^{4t} - 6t + 5te^{-4t}\ y(t) = 4e^{-4t} - 4e^{4t} + 5t - 4te^{-4t}\ y(t) = e^{-4t} - e^{4t} + 15t + 24te^{-4t}\ y(t) = -2e^{-4t} + 2e^{4t} - t + 16te^{-4t}

Name: Solve the initial value problem below using the method of Laplace transforms.y” - 16y = 96t - 40e^-4t, y(0) = 0, y'(0) = 31y(t) = -4e^-4t + 4e^4t - ty(t) = -4e^-4t + 4e^4t - 6t + 5te^-4ty(t) = 4e^-4t - 4e^4t + 5t - 4te^-4ty(t) = e^-4t - e^4t + 15t + 24te^-4ty(t) = -2e^-4t + 2e^4t - t + 16te^-4t
Uploaded: 2023-04-19T06:27:53-08:00
Duration: 2 min 42 s
Channel: Madhur L
Description: Solve the initial value problem below using the method of Laplace transforms.y” - 16y = 96t - 40e^-4t, y(0) = 0, y'(0) = 31y(t) = -4e^-4t + 4e^4t - ty(t) = -4e^-4t + 4e^4t - 6t + 5te^-4ty(t) = 4e^-4t - 4e^4t + 5t - 4te^-4ty(t) = e^-4t - e^4t + 15t + 24te^-4ty(t) = -2e^-4t + 2e^4t - t + 16te^-4t

          Solve the initial value problem below using the method of Laplace transforms.\
y'' - 16y = 96t - 40e^{-4t}, y(0) = 0, y'(0) = 31\
y(t) = -4e^{-4t} + 4e^{4t} - t\
y(t) = -4e^{-4t} + 4e^{4t} - 6t + 5te^{-4t}\
y(t) = 4e^{-4t} - 4e^{4t} + 5t - 4te^{-4t}\
y(t) = e^{-4t} - e^{4t} + 15t + 24te^{-4t}\
y(t) = -2e^{-4t} + 2e^{4t} - t + 16te^{-4t}

Solve the initial value problem below using the method of Laplace transforms.y” - 16y = 96t - 40e^-4t, y(0) = 0, y'(0) = 31y(t) = -4e^-4t + 4e^4t - ty(t) = -4e^-4t + 4e^4t - 6t + 5te^-4ty(t) = 4e^-4t - 4e^4t + 5t - 4te^-4ty(t) = e^-4t - e^4t + 15t + 24te^-4ty(t) = -2e^-4t + 2e^4t - t + 16te^-4t

Added by Cassandra N.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Madhur L

04/19/2023

Step 1

First, we need to rewrite the given initial value problem in a more readable format: $$y'(t) + 16y(t) = 96t + 40e^{-4t}, \quad y(0) = 0$$ Now, we will take the Laplace transform of both sides of the equation: $$\mathcal{L}\{y'(t) + 16y(t)\} = \mathcal{L}\{96t + Show more…

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Solve the initial value problem below using the method of Laplace transforms. y'' - 16y = 96t - 40e^{-4t}, y(0) = 0, y'(0) = 31 y(t) = -4e^{-4t} + 4e^{4t} - t y(t) = -4e^{-4t} + 4e^{4t} - 6t + 5te^{-4t} y(t) = 4e^{-4t} - 4e^{4t} + 5t - 4te^{-4t} y(t) = e^{-4t} - e^{4t} + 15t + 24te^{-4t} y(t) = -2e^{-4t} + 2e^{4t} - t + 16te^{-4t}