Question

Determine $\mathcal{L}^{-1}\{F\}$. $$F(s) = \frac{11s^3 - 20s^2 - 5s + 20}{s^3(s-4)}$$ Click here to view the table of Laplace transforms. Click here to view the table of properties of Laplace transforms. $\mathcal{L}^{-1}\{F\} = \square$

Name: determine mathcall 1f fs frac11s3 20s2 5s 20s3s 4 click here to view the table of laplace transforms click here to view the table of properties of laplace transforms mathcall 1f square 19188
Uploaded: 2024-03-29T01:36:03-08:00
Duration: 4 min 17 s
Channel: Melissa Munoz
Description: determine mathcall 1f fs frac11s3 20s2 5s 20s3s 4 click here to view the table of laplace transforms click here to view the table of properties of laplace transforms mathcall 1f square 19188

          Determine $\mathcal{L}^{-1}\{F\}$.
$$F(s) = \frac{11s^3 - 20s^2 - 5s + 20}{s^3(s-4)}$$
Click here to view the table of Laplace transforms.
Click here to view the table of properties of Laplace transforms.
$\mathcal{L}^{-1}\{F\} = \square$

$determine mathcall 1f fs frac11s3 20s2 5s 20s3s 4 click here to view the table of laplace transforms click here to view the table of properties of laplace transforms mathcall 1f square 19188$

Added by Gloria J.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Melissa Munoz

03/29/2024

Step 1

Step 1: Perform partial fraction decomposition on $F(s)$: $$ \frac{11s^3 - 20s^2 - 5s + 20}{s^3(s-4)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s^3} + \frac{D}{s-4} $$ Multiplying both sides by $s^3(s-4)$, we get: $$ 11s^3 - 20s^2 - 5s + 20 = A s^2(s-4) + B s(s-4) + Show more…

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Determine $\mathcal{L}^{-1}\{F\}$. $$F(s) = \frac{11s^3 - 20s^2 - 5s + 20}{s^3(s-4)}$$ Click here to view the table of Laplace transforms. Click here to view the table of properties of Laplace transforms. $\mathcal{L}^{-1}\{F\} = \square$