Question

The inverse forms of the results in Problem 49 in Exercises 7.1 are and $\mathcal{L}^{-1}\left\{\frac{s-a}{(s-a)^2 + b^2}\right\} = e^{at} \cos bt$ $\mathcal{L}^{-1}\left\{\frac{b}{(s-a)^2 + b^2}\right\} = e^{at} \sin bt$. Use the Laplace transform and these inverses to solve the given initial-value problem. y' + y = $e^{-4t} \cos 3t$, y(0) = 0 y(t) = $e^{-4t} \cos (3t)$

          The inverse forms of the results in Problem 49 in Exercises 7.1 are
and
$\mathcal{L}^{-1}\left\{\frac{s-a}{(s-a)^2 + b^2}\right\} = e^{at} \cos bt$
$\mathcal{L}^{-1}\left\{\frac{b}{(s-a)^2 + b^2}\right\} = e^{at} \sin bt$.
Use the Laplace transform and these inverses to solve the given initial-value problem.
y' + y = $e^{-4t} \cos 3t$, y(0) = 0
y(t) = $e^{-4t} \cos (3t)$

$The inverse forms of the results in Problem 49 in Exercises 7.1 are and ℒ^-1{(s-a)/((s-a)^2 + b^2)} = e^atcos bt ℒ^-1{(b)/((s-a)^2 + b^2)} = e^atsin bt. Use the Laplace transform and these inverses to solve the given initial-value problem. y' + y = e^-4tcos 3t, y(0) = 0 y(t) = e^-4tcos (3t)$

Added by Lidia G.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Uma Kumari

11/13/2021

Step 1

Step 1: Take the Laplace transform of both sides of the differential equation. Show more…

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The inverse forms of the results in Problem 49 in Exercises 7.1 are $$ \mathcal{L}^{-1} \left\{ \frac{s-a}{(s-a)^2 + b^2} \right\} = e^{at} \cos bt $$ and $$ \mathcal{L}^{-1} \left\{ \frac{b}{(s-a)^2 + b^2} \right\} = e^{at} \sin bt. $$ Use the Laplace transform and these inverses to solve the given initial-value problem. $$y' + y = e^{-4t} \cos 3t, \ y(0) = 0$$ $$y(t) = e^{-4t} \cos (3t)$$