Math / Calculus

QuestionCompute the inverse Laplace transform: $\mathcal{L}^{-1}\left\{\frac{-s-6}{s^{2}+4 s+20} e^{-2 s}\right\}=$ $\square$ (Notation: write $\mathbf{u}(\mathbf{t}-\mathbf{c})$ for the Heaviside step function $u_{c}(t)$ with step at $t=c$ .)

Studdy Solution

STEP 1

What is this asking? We need to find the inverse Laplace transform of a given function involving an exponential term, which usually hints at a time shift. Watch out! Don't forget to account for the exponential term $e^{-2s}$ , which indicates a time shift, and carefully complete the square in the denominator.

STEP 2

1. Rewrite the expression
2. Partial fraction decomposition (not needed)
3. Inverse Laplace Transform

STEP 3

Let's complete the square in the denominator $s^2 + 4s + 20$ .
We take half of the coefficient of the $s$ term, which is $4/2 = 2$ , square it to get $2^2 = 4$ , and then add and subtract this value.
So, we have $s^2 + 4s + 4 - 4 + 20 = (s+2)^2 + 16$ .
This makes it look more like something we can find in a Laplace transform table!

STEP 4

Now, let's look at the numerator.
We have $-s - 6$ .
We want to rewrite this in terms of $(s+2)$ to match the completed square in the denominator.
We can rewrite $-s - 6$ as $-(s + 2) - 4$ .
This will help us break down the expression into recognizable Laplace transforms.

STEP 5

Our expression now becomes: $\frac{-(s+2) - 4}{(s+2)^2 + 16} e^{-2s}$ We can split this into two separate fractions: $\left( -\frac{s+2}{(s+2)^2 + 16} - \frac{4}{(s+2)^2 + 16} \right) e^{-2s}$ Notice how we strategically rewrote the expression to resemble standard Laplace transforms!

STEP 6

We don't need partial fraction decomposition here since we've already manipulated the expression into a form that resembles known Laplace transforms.
Sometimes you get lucky!

STEP 7

Recall that the Laplace transform of $e^{at}\cos(bt)$ is $\frac{s-a}{(s-a)^2 + b^2}$ and the Laplace transform of $e^{at}\sin(bt)$ is $\frac{b}{(s-a)^2 + b^2}$ .
Our expression looks very similar to these forms!

STEP 8

We can now apply the inverse Laplace transform to each part of our expression.
We have $a = -2$ and $b = 4$ .
The inverse Laplace transform of $-\frac{s+2}{(s+2)^2 + 16}$ is $-e^{-2t}\cos(4t)$ and the inverse Laplace transform of $-\frac{4}{(s+2)^2 + 16}$ is $-e^{-2t}\sin(4t)$ .

STEP 9

Remember the $e^{-2s}$ term?
This introduces a time shift of 2 units to the right.
We replace $t$ with $t-2$ and multiply the entire result by the Heaviside step function $u(t-2)$ .

STEP 10

Our final answer is: $\left( -e^{-2(t-2)}\cos(4(t-2)) - e^{-2(t-2)}\sin(4(t-2)) \right) u(t-2)$

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QuestionCompute the inverse Laplace transform: $\mathcal{L}^{-1}\left\{\frac{-s-6}{s^{2}+4 s+20} e^{-2 s}\right\}=$ $\square$ (Notation: write $\mathbf{u}(\mathbf{t}-\mathbf{c})$ for the Heaviside step function $u_{c}(t)$ with step at $t=c$ .)

Studdy Solution

STEP 1

STEP 2

1. Rewrite the expression
2. Partial fraction decomposition (not needed)
3. Inverse Laplace Transform

STEP 3

STEP 4

Now, let's look at the numerator.
We have $-s - 6$ .
We want to rewrite this in terms of $(s+2)$ to match the completed square in the denominator.
We can rewrite $-s - 6$ as $-(s + 2) - 4$ .
This will help us break down the expression into recognizable Laplace transforms.

STEP 5

Our expression now becomes: $\frac{-(s+2) - 4}{(s+2)^2 + 16} e^{-2s}$ We can split this into two separate fractions: $\left( -\frac{s+2}{(s+2)^2 + 16} - \frac{4}{(s+2)^2 + 16} \right) e^{-2s}$ Notice how we strategically rewrote the expression to resemble standard Laplace transforms!

STEP 6

We don't need partial fraction decomposition here since we've already manipulated the expression into a form that resembles known Laplace transforms.
Sometimes you get lucky!

STEP 7

Recall that the Laplace transform of $e^{at}\cos(bt)$ is $\frac{s-a}{(s-a)^2 + b^2}$ and the Laplace transform of $e^{at}\sin(bt)$ is $\frac{b}{(s-a)^2 + b^2}$ .
Our expression looks very similar to these forms!

STEP 8

We can now apply the inverse Laplace transform to each part of our expression.
We have $a = -2$ and $b = 4$ .
The inverse Laplace transform of $-\frac{s+2}{(s+2)^2 + 16}$ is $-e^{-2t}\cos(4t)$ and the inverse Laplace transform of $-\frac{4}{(s+2)^2 + 16}$ is $-e^{-2t}\sin(4t)$ .

STEP 9

Remember the $e^{-2s}$ term?
This introduces a time shift of 2 units to the right.
We replace $t$ with $t-2$ and multiply the entire result by the Heaviside step function $u(t-2)$ .

STEP 10

Our final answer is: $\left( -e^{-2(t-2)}\cos(4(t-2)) - e^{-2(t-2)}\sin(4(t-2)) \right) u(t-2)$

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Studdy Solution

STEP 1

STEP 2

1. Rewrite the expression2. Partial fraction decomposition (not needed)3. Inverse Laplace Transform

STEP 3

STEP 4

STEP 5

STEP 6

We don't need partial fraction decomposition here since we've already manipulated the expression into a form that resembles known Laplace transforms.Sometimes you get lucky!

STEP 7

STEP 8

STEP 9

Remember the e−2se^{-2s}e−2s term?This introduces a **time shift** of **2 units** to the right.We replace ttt with t−2t-2t−2 and multiply the entire result by the Heaviside step function u(t−2)u(t-2)u(t−2).

STEP 10

Our **final answer** is: (−e−2(t−2)cos⁡(4(t−2))−e−2(t−2)sin⁡(4(t−2)))u(t−2) \left( -e^{-2(t-2)}\cos(4(t-2)) - e^{-2(t-2)}\sin(4(t-2)) \right) u(t-2) (−e−2(t−2)cos(4(t−2))−e−2(t−2)sin(4(t−2)))u(t−2)

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. Rewrite the expression
2. Partial fraction decomposition (not needed)
3. Inverse Laplace Transform

We don't need partial fraction decomposition here since we've already manipulated the expression into a form that resembles known Laplace transforms.
Sometimes you get lucky!

Remember the $e^{-2s}$ term?
This introduces a time shift of 2 units to the right.
We replace $t$ with $t-2$ and multiply the entire result by the Heaviside step function $u(t-2)$ .

Our final answer is: $\left( -e^{-2(t-2)}\cos(4(t-2)) - e^{-2(t-2)}\sin(4(t-2)) \right) u(t-2)$