Question

Use the Laplace transform to solve the following initial value problem: $y'' + 5y' = 0$ $y(0) = -4$, $y'(0) = 1$ First, using $Y$ for the Laplace transform of $y(t)$, i.e., $Y = \mathcal{L}\{y(t)\}$. Find the equation you get by taking the Laplace transform of the differential equation Now solve for $Y(s) = $ and write the above answer in its partial fraction decomposition, $Y(s) = \frac{A}{s+a} + \frac{B}{s+b}$ where $a < b$ $Y(s) = $ Now by inverting the transform, find $y(t) = $

          Use the Laplace transform to solve the following initial value problem:
$y'' + 5y' = 0$  $y(0) = -4$, $y'(0) = 1$

First, using $Y$ for the Laplace transform of $y(t)$, i.e., $Y = \mathcal{L}\{y(t)\}$.
Find the equation you get by taking the Laplace transform of the differential equation

Now solve for $Y(s) = $
and write the above answer in its partial fraction decomposition, $Y(s) = \frac{A}{s+a} + \frac{B}{s+b}$ where $a < b$
$Y(s) = $

Now by inverting the transform, find $y(t) = $

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use the laplace transform to solve the following initial value problem y 5y 0 y0 4 y0 1 first using y for the laplace transform of yt ie y mathcallyt find the equation you get by taking the 59982

Added by Ronnie H.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Scott Stetson

09/28/2023

Step 1

Step 1: Taking the Laplace transform of the differential equation $y'' + 5y' = 0$, we have: $\mathcal{L}\{y''\} + 5\mathcal{L}\{y'\} = 0$ Using the properties of Laplace transforms, we have: $(s^2Y(s) - sy(0) - y'(0)) + 5(sY(s) - y(0)) = 0$ Substituting the initial Show more…

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Use the Laplace transform to solve the following initial value problem: $y'' + 5y' = 0$ $y(0) = -4$, $y'(0) = 1$ First, using $Y$ for the Laplace transform of $y(t)$, i.e., $Y = \mathcal{L}\{y(t)\}$. Find the equation you get by taking the Laplace transform of the differential equation Now solve for $Y(s) = $ and write the above answer in its partial fraction decomposition, $Y(s) = \frac{A}{s+a} + \frac{B}{s+b}$ where $a < b$ $Y(s) = $ Now by inverting the transform, find $y(t) = $

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00:01 For this question, we're asked to solve this initial value problem by using the laplace transform.

00:04 So i'm going to take the laplace transform of this differential equation.

00:08 And when i do that, i get s squared times y of s minus s times y of 0 minus y prime of 0.

00:20 And then plus 2s times y of s minus 2 times y of 0.

00:28 This is equal to 0.

00:29 So let me group the y of s terms.

00:33 I have s squared plus 2s.

00:38 And we're told that y of 0 is equal to negative 1.

00:42 So i'll have plus s from this term.

00:46 And then minus 4 plus 2 for a total of negative 2.

00:51 This is equal to 0.

00:54 Therefore, y of s.

00:57 This is equal to 2 minus s all over s times the quantity s plus 2.

01:06 Now, the only thing that's left for us to do is to take the inverse laplace transform of y of s.

01:11 But you probably won't see this function on a laplace transform table.

01:15 So we need to algebraically manipulate it to make it look like something that is on a laplace transform table.

01:22 And we'll use partial fraction decomposition to do that.

01:25 So i'm going to break it up like this...

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