Math / Calculus

QuestionPrevious Problem Problem List Next Problem
Standard 17: Problem 3 (1 point)
Find the inverse Laplace transform $f(t)=\mathcal{L}^{-1}\{F(s)\}$ of the function $F(s)=-\left(\frac{1}{s^{2}+36}+\frac{8 s}{s^{2}+81}\right)$ . $f(t)=\mathcal{L}^{-1}\left\{-\left(\frac{1}{s^{2}+36}+\frac{8 s}{s^{2}+81}\right)\right\}=\square$ help (formulas)
Preview My Answers Submit Answers You have attempted this problem 0 times. You have unlimited attempts remainina.

Studdy Solution

STEP 1

What is this asking? We need to find the inverse Laplace transform of $-\left(\frac{1}{s^{2}+36}+\frac{8 s}{s^{2}+81}\right)$ , which means finding the original function $f(t)$ that would give us this expression $F(s)$ after a Laplace transform. Watch out! Don't forget to distribute that negative sign and keep track of your constants!
Also, remember those handy Laplace transform formulas!

STEP 2

1. Distribute the negative sign.
2. Apply the inverse Laplace transform.

STEP 3

We're given $F(s) = -\left(\frac{1}{s^{2}+36}+\frac{8 s}{s^{2}+81}\right)$ .
Let's distribute that negative sign to both terms inside the parentheses.
This makes it easier to see what we're working with! $F(s) = -\frac{1}{s^{2}+36} - \frac{8s}{s^{2}+81}$

STEP 4

We'll need these formulas: $\mathcal{L}^{-1}\left\{\frac{1}{s^{2}+a^{2}}\right\} = \frac{1}{a}\sin(at)$ $\mathcal{L}^{-1}\left\{\frac{s}{s^{2}+a^{2}}\right\} = \cos(at)$

STEP 5

Let's look at $-\frac{1}{s^{2}+36}$ .
We can rewrite $36$ as $6^{2}$ , so we have $-\frac{1}{s^{2}+6^{2}}$ .
Using the first formula with $a=6$ , we get: $\mathcal{L}^{-1}\left\{-\frac{1}{s^{2}+6^{2}}\right\} = -\frac{1}{6}\sin(6t)$

STEP 6

Now, let's tackle $-\frac{8s}{s^{2}+81}$ .
We can rewrite $81$ as $9^{2}$ , so we have $-8 \cdot \frac{s}{s^{2}+9^{2}}$ .
Using the second formula with $a=9$ , we get: $\mathcal{L}^{-1}\left\{-\frac{8s}{s^{2}+9^{2}}\right\} = -8\cos(9t)$

STEP 7

Putting it all together, we have: $f(t) = -\frac{1}{6}\sin(6t) - 8\cos(9t)$

STEP 8

The inverse Laplace transform of $F(s)$ is $f(t) = -\frac{1}{6}\sin(6t) - 8\cos(9t)$ .

Was this helpful?

Studdy Solution

STEP 1

STEP 2

1. Distribute the negative sign.
2. Apply the inverse Laplace transform.

STEP 3

We're given $F(s) = -\left(\frac{1}{s^{2}+36}+\frac{8 s}{s^{2}+81}\right)$ .
Let's distribute that negative sign to both terms inside the parentheses.
This makes it easier to see what we're working with! $F(s) = -\frac{1}{s^{2}+36} - \frac{8s}{s^{2}+81}$

STEP 4

We'll need these formulas: $\mathcal{L}^{-1}\left\{\frac{1}{s^{2}+a^{2}}\right\} = \frac{1}{a}\sin(at)$ $\mathcal{L}^{-1}\left\{\frac{s}{s^{2}+a^{2}}\right\} = \cos(at)$

STEP 5

Let's look at $-\frac{1}{s^{2}+36}$ .
We can rewrite $36$ as $6^{2}$ , so we have $-\frac{1}{s^{2}+6^{2}}$ .
Using the first formula with $a=6$ , we get: $\mathcal{L}^{-1}\left\{-\frac{1}{s^{2}+6^{2}}\right\} = -\frac{1}{6}\sin(6t)$

STEP 6

Now, let's tackle $-\frac{8s}{s^{2}+81}$ .
We can rewrite $81$ as $9^{2}$ , so we have $-8 \cdot \frac{s}{s^{2}+9^{2}}$ .
Using the second formula with $a=9$ , we get: $\mathcal{L}^{-1}\left\{-\frac{8s}{s^{2}+9^{2}}\right\} = -8\cos(9t)$

STEP 7

Putting it all together, we have: $f(t) = -\frac{1}{6}\sin(6t) - 8\cos(9t)$

STEP 8

The inverse Laplace transform of $F(s)$ is $f(t) = -\frac{1}{6}\sin(6t) - 8\cos(9t)$ .

Get Studdy

Studdy solves anything!

Start learning now

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

Studdy Solution

STEP 1

STEP 2

1. Distribute the negative sign.2. Apply the inverse Laplace transform.

STEP 3

STEP 4

We'll need these formulas: L−1{1s2+a2}=1asin⁡(at)\mathcal{L}^{-1}\left\{\frac{1}{s^{2}+a^{2}}\right\} = \frac{1}{a}\sin(at)L−1{s2+a21​}=a1​sin(at) L−1{ss2+a2}=cos⁡(at)\mathcal{L}^{-1}\left\{\frac{s}{s^{2}+a^{2}}\right\} = \cos(at)L−1{s2+a2s​}=cos(at)

STEP 5

STEP 6

STEP 7

Putting it all together, we have: f(t)=−16sin⁡(6t)−8cos⁡(9t)f(t) = -\frac{1}{6}\sin(6t) - 8\cos(9t)f(t)=−61​sin(6t)−8cos(9t)

STEP 8

The inverse Laplace transform of F(s)F(s)F(s) is f(t)=−16sin⁡(6t)−8cos⁡(9t)f(t) = -\frac{1}{6}\sin(6t) - 8\cos(9t)f(t)=−61​sin(6t)−8cos(9t).

Get Studdy

Studdy solves anything!

Start learning now

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

1. Distribute the negative sign.
2. Apply the inverse Laplace transform.

We'll need these formulas: $\mathcal{L}^{-1}\left\{\frac{1}{s^{2}+a^{2}}\right\} = \frac{1}{a}\sin(at)$ $\mathcal{L}^{-1}\left\{\frac{s}{s^{2}+a^{2}}\right\} = \cos(at)$

Putting it all together, we have: $f(t) = -\frac{1}{6}\sin(6t) - 8\cos(9t)$

The inverse Laplace transform of $F(s)$ is $f(t) = -\frac{1}{6}\sin(6t) - 8\cos(9t)$ .