Question

Consider the following initial value problem: $y'' + 64y = \begin{cases} 2, & 0 \le t \le 8 \\ 0, & t > 8 \end{cases}$ $y(0) = 6, \ y'(0) = 0$ First find the unit-step representation of the forcing function, i.e, the right side of the differential equation. The forcing function can be written as $au(t - b) + cu(t - d)$ with constants $a, b, c, d$. Enter these numbers: a = $oxed{}$ b = $oxed{}$ c = $oxed{}$ d = $oxed{}$ 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 and solve $Y(s) = oxed{}$

          Consider the following initial value problem:
$y'' + 64y = \begin{cases} 2, & 0 \le t \le 8 \\ 0, & t > 8 \end{cases}$
$y(0) = 6, \ y'(0) = 0$
First find the unit-step representation of the forcing function, i.e, the right side of the differential equation.
The forcing function can be written as $au(t - b) + cu(t - d)$ with constants $a, b, c, d$. Enter these numbers:
a = $oxed{}$ b = $oxed{}$ c = $oxed{}$ d = $oxed{}$
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 and solve
$Y(s) = oxed{}$

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Consider the following initial value problem:
y” + 64y = 2, 0 ≤ t ≤ 8
0, t > 8
y(0) = 6, y'(0) = 0
First find the unit-step representation of the forcing function, i.e, the right side of the differential equation.
The forcing function can be written as au(t - b) + cu(t - d) with constants a, b, c, d. Enter these numbers:
a = oxed b = oxed c = oxed d = oxed
Using Y for the Laplace transform of y(t), i.e., Y = ℒ{y(t)}, find the equation you get by taking the Laplace transform of the differential equation and solve
Y(s) = oxed

Added by Emma G.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Scott Stetson

07/28/2023

Step 1

The forcing function is given by +64y = 20t^8. To find the unit-step representation, we need to express the forcing function as a sum of unit-step functions. The unit-step function u(t) is defined as: u(t) = 1 for t >= 0 u(t) = 0 for t < 0 We can express the Show more…

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Consider the following initial value problem: +64y = 20t^8, t > 8 y(0) = 6, y'(0) = 0 First, find the unit-step representation of the forcing function, i.e. the right side of the differential equation. The forcing function can be written as a*u(t - b) + c*u(t - d) with constants a, b, c, d. Enter these numbers: a = b - c*d = Using Y for the Laplace transform of y(t), i.e. Y = L{y(t)}, find the equation you get by taking the Laplace transform of the differential equation and solve for Y =

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Transcript

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

00:05 So i'm going to take the laplace transform of this differential equation, which will turn it into an algebraic equation.

00:11 So for starters, i'll have s squared times y of s, and then...

00:21 Okay, so let me group all of my y of s terms together.

00:26 I'll have s squared minus 8s, and then plus 32.

00:33 And then we're told that y of 0, this is 1, so we have just a minus s term.

00:43 And then for our constants, i'll have minus 4 and then plus 8.

00:49 So this will be plus 4.

00:52 This is equal to 0.

00:54 And therefore, y of s, this is equal to...

01:00 So let me write the denominator first.

01:02 Plus 32.

01:07 And in the numerator, i have s minus 4.

01:10 Okay, so the only thing left to do is just find the inverse laplace transform of this function, which you probably will not see it on a laplace transform table.

01:22 So let's algebraically manipulate it so that it looks like something on a laplace transform table.

01:28 So i'm going to rewrite the denominator.

01:32 So this is s squared minus 8s.

01:36 I'm going to write plus 32 as 16 plus 16...

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