Diff Eq: Inverse Laplace of the following

Question

I plugged in s-5 into the numerator and denominator. I'm not sure what they are asking for, on F(s). I figured (s-5)/(s-5)+4^2+5 would make sense.

Answer 1

First, a correction:

Then, I tried adding (s-5) to the numerator and denominator and solve from there but I can't find the answer.

You can't do that. $$ \frac{x+a}{y+a} \ne \frac x y $$ For instance $$ \frac{1+3}{2+3} = \frac 4 5 \ne \frac 1 2 $$

I plugged in s-5 into the numerator and denominator.

This also doesn't make sense, since there isn't anything resembling a variable on top for you to plug anything into.

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I'm not sure what they are asking for, on F(s).

You have the expression $$ \frac{1}{(s-5)^2 + 4} $$ They are asking the following question: "What function are you plugging in $s-5$ into, to obtain this expression?"

A simpler example would be, say

$\sin(x-2) = F\bigg|_{x-2}$; what is $F(x)$?

In this example, $F(x) = \sin(x)$, because $F(x-2) = \sin(x-2)$.

Slightly more complicated would be

$x^2 + 6x + 8 = F \bigg|_{x+3}$; what is $F(x)$?

but once you factorize the left-hand side as $(x+3)^2 - 1$, you see that $F(x) = x^2 +1$, because $F(x+3) = (x+3)^2 - 1 = x^2 + 6x + 8$.

Hopefully this clears up what you're looking for.

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Finding the inverse Laplace transform:

All you really need are these items:

• The Laplace transform (and its inverse) is linear: $\mathcal{L}\{\alpha f(t) + \beta g(t)\}(s) = \alpha \mathcal{L}\{f(t)\}(s) + \beta \mathcal{L}\{g(t)\}(s)$
• $\displaystyle \mathcal{L}\{\sin(\alpha t)\}(s) = \frac{\alpha}{s^2 + \alpha^2}$
• $\displaystyle \mathcal{L}\{e^{\gamma t} f(t)\}(s) = \mathcal{L}\{f(t)\}(s-\gamma)$

In seeking $$ \mathcal{L}^{-1} \left\{ \frac{1}{(s-5)^2 + 4^2} \right\}(s) $$ begin by noting you have something analogous to the second bullet point, but with a shift of $\gamma = 5$, and you have $\alpha=4$. To get that same $\alpha$ on top, you need to multiply the top by $4$, and then multiply by $1/4$ as well, which can be pulled out by linearity. $$ \mathcal{L}^{-1} \left\{ \frac{1}{(s-5)^2 + 4^2} \right\}(s) = \mathcal{L}^{-1} \left\{ \frac{4 \cdot \frac 1 4}{(s-5)^2 + 4^2} \right\}(s) = \frac 1 4 \mathcal{L}^{-1} \left\{ \frac{4}{(s-5)^2 + 4^2} \right\}(s) $$ Hopefully, applying the shift property and recognizing the formula to use is simple from here.