What is the Inverse Laplace Transform of 2s+7-e^-2s/(s+1)^2?

[alexmahone]

Find the inverse Laplace transform of $\displaystyle \frac{2s+7-e^{-2s}}{(s+1)^2}$.

 

[Dustinsfl]

Find the inverse Laplace transform of $\displaystyle\frac{2s+5-e^{-2s}}{s^2+s+1}$.

$s^2 + s + 1/4 + 1 - 1/4 = (s + 1/2)^2 + 3/4$

Then break up the numerator.

 

[alexmahone]

$s^2 + s + 1/4 + 1 - 1/4 = (s + 1/2)^2 + 3/4$

Then break up the numerator.

I changed the question. (Sorry about that.)

 

[Dustinsfl]

I changed the question. (Sorry about that.)

Then look at

$$
\frac{2s+7}{(s+1)^2} - \frac{e^{-2s}}{(s+1)^2}
$$

The formula for the second piece is

$$
\frac{(t-\tau)^n}{n!}e^{-\alpha(t-\tau)}u(t-\tau) = \mathfrak{L}^{-1}\left[\frac{e^{-\tau s}}{(s+\alpha)^{n+1}}\right]
$$

The other one shouldn't be too bad. Just ask if you need help with that one.

 

[blue_raver22]

The inverse Laplace transform is a mathematical operation that allows us to find the original function from its Laplace transform. In this case, we are given the Laplace transform of a function, which is $\frac{2s+7-e^{-2s}}{(s+1)^2}$, and we need to find the original function.

To do this, we can use the inverse Laplace transform formula or table to find the inverse Laplace transform of each term in the given function. We can break down the given function into three parts: $\frac{2s}{(s+1)^2}$, $\frac{7}{(s+1)^2}$, and $\frac{-e^{-2s}}{(s+1)^2}$.

The inverse Laplace transform of $\frac{2s}{(s+1)^2}$ is $2e^{-t}(t+1)$, which can be derived from the formula $\mathcal{L}^{-1}\left\{\frac{s}{(s+a)^2}\right\}=te^{-at}$.

The inverse Laplace transform of $\frac{7}{(s+1)^2}$ is $7te^{-t}$, which can be derived from the formula $\mathcal{L}^{-1}\left\{\frac{a}{(s+a)^2}\right\}=te^{-at}$.

The inverse Laplace transform of $\frac{-e^{-2s}}{(s+1)^2}$ is $e^{-2t}(t+1)$, which can be derived from the formula $\mathcal{L}^{-1}\left\{\frac{-ae^{-as}}{(s+a)^2}\right\}=te^{-at}$.

Therefore, the inverse Laplace transform of $\frac{2s+7-e^{-2s}}{(s+1)^2}$ is the sum of these three terms, which is $2e^{-t}(t+1)+7te^{-t}+e^{-2t}(t+1)$. This can also be simplified to $e^{-t}(3t+3)$.

In conclusion, the inverse Laplace transform of $\frac{2s+7-e^{-2s}}{(s+1)^2}$ is $e^{-t}(3t+3)$. This is the original function that corresponds to the given Laplace transform.