Inverse laplace of [(1/s^2)-(48/s^5)]

Answers

Answer 1

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I am not familiar with Laplace transforms, so my explanation probably won't help, but given that for two Laplace transform $F(s)$ and $G(s)$ , then $\mathcal{L}^(-1)\{aF(s)+bG(s)\} = a\mathcal{L}^(-1)\{F(s)\}+b\mathcal{L}^(-1)\{G(s)\}$

Given that $(1)/(s^2) = (1!)/(s^2)$ and $-(48)/(s^5) = -2\cdot(4!)/(s^5)$

So you have $\mathcal{L}^(-1)\left\{(1)/(s^2) - 2\cdot(4!)/(s^5)\right\} = \mathcal{L}^(-1)\left\{(1)/(s^2)\right\} - 2\mathcal{L}^(-1)\left\{(4!)/(s^5)\right\}$

From Table of Laplace Transform, you have $\mathcal{L}\{t^n\} = (n!)/(s^(n+1))$ and hence $\mathcal{L}^(-1)\left\{(n!)/(s^(n+1))\right\} = t^n$

So you have $\mathcal{L}^(-1)\left\{(1)/(s^2)\right\} - 2\mathcal{L}^(-1)\left\{(4!)/(s^5)\right\} = \boxed{t-2t^4}$ .

Hope this helps...

Answer 2

Answer:

Final answer:

To find the inverse Laplace transform of the given expression, use partial fraction decomposition to simplify it into individual fractions and then find their inverse transforms.

Explanation:

To find the inverse Laplace transform of the given expression, we can use partial fraction decomposition. First, we factor the denominator: s2*(s3-48). The next step is to represent the expression as a sum of simpler fractions:

1/s2 - 48/s5 = A/s + B/s2 + C/(s - 2) + D/(s + 2) + E/(s + 4) + F/(s2 - 4)

Next, we solve for A, B, C, D, E, and F by performing algebraic manipulations and equating the corresponding coefficients. Finally, we can look up the inverse Laplace transform of each individual fraction term in tables or by using known formulas.

Learn more about Inverse Laplace Transform here:

brainly.com/question/31322563

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Answers

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Answers

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Answers

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so ur solution would be 16/22 and then u would simplify by dividing both values by 2.
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You need a common denominator. You already have one. So add 7+9=16. Put 16 over 22. That's your answer without reduction.

When you reduce it, divide it by 2 since its an even number.

16/22=8/11

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Answers

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Answers

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