Math / Calculus

QuestionPrevious Problem Problem List Next Problem
Standard 18: Problem 1 (1 point)
Find the inverse Laplace transform $f(t)=\mathcal{L}^{-1}\{F(s)\}$ of the function $F(s)=\frac{7}{s^{2}}+\frac{1}{s-5}$ . $f(t)=\mathcal{L}^{-1}\left\{\frac{7}{s^{2}}+\frac{1}{s-5}\right\}=$ $\square$ help (formulas)
Preview My Answers Submit Answers You have attempted this problem 0 times. You have unlimited attempts remaining.
Email Instructor

Studdy Solution

STEP 1

What is this asking? We need to find the inverse Laplace transform of $F(s) = \frac{7}{s^2} + \frac{1}{s - 5}$ , which means finding a function $f(t)$ whose Laplace transform is $F(s)$ . Watch out! Don't mix up the formulas for Laplace transforms and inverse Laplace transforms!
Also, remember the properties of linearity!

STEP 2

1. Separate the terms
2. Inverse transform of the first term
3. Inverse transform of the second term
4. Combine the results

STEP 3

We're given $F(s) = \frac{7}{s^2} + \frac{1}{s - 5}$ .
Because the inverse Laplace transform is linear, we can take the inverse Laplace transform of each term separately!
This makes our lives much easier.

STEP 4

We want to find the inverse Laplace transform of $\frac{7}{s^2}$ .
Remember the formula $\mathcal{L}^{-1}\left\{\frac{n!}{s^{n+1}}\right\} = t^n$ .
Let's make our term look like this formula!

STEP 5

If we set $n = 1$ , we get $\frac{1!}{s^{1+1}} = \frac{1}{s^2}$ .
Awesome!

STEP 6

So, $\mathcal{L}^{-1}\left\{\frac{7}{s^2}\right\} = 7 \cdot \mathcal{L}^{-1}\left\{\frac{1}{s^2}\right\} = 7 \cdot t^1 = 7t$ .
We just multiplied by $\textbf{7}$ to match our original term!

STEP 7

Now, let's find the inverse Laplace transform of $\frac{1}{s - 5}$ .
This time, we'll use the formula $\mathcal{L}^{-1}\left\{\frac{1}{s - a}\right\} = e^{at}$ .

STEP 8

In our case, $a = \textbf{5}$ , so $\mathcal{L}^{-1}\left\{\frac{1}{s - 5}\right\} = e^{5t}$ .
Perfect!

STEP 9

Since we took the inverse Laplace transform of each term individually, we can now add them together to get the final answer: $f(t) = \mathcal{L}^{-1}\left\{\frac{7}{s^2}\right\} + \mathcal{L}^{-1}\left\{\frac{1}{s - 5}\right\} = 7t + e^{5t}.$

STEP 10

The inverse Laplace transform of $F(s)$ is $f(t) = 7t + e^{5t}$ .

Was this helpful?

Studdy Solution

STEP 1

STEP 2

1. Separate the terms
2. Inverse transform of the first term
3. Inverse transform of the second term
4. Combine the results

STEP 3

We're given $F(s) = \frac{7}{s^2} + \frac{1}{s - 5}$ .
Because the inverse Laplace transform is linear, we can take the inverse Laplace transform of each term separately!
This makes our lives much easier.

STEP 4

We want to find the inverse Laplace transform of $\frac{7}{s^2}$ .
Remember the formula $\mathcal{L}^{-1}\left\{\frac{n!}{s^{n+1}}\right\} = t^n$ .
Let's make our term look like this formula!

STEP 5

If we set $n = 1$ , we get $\frac{1!}{s^{1+1}} = \frac{1}{s^2}$ .
Awesome!

STEP 6

So, $\mathcal{L}^{-1}\left\{\frac{7}{s^2}\right\} = 7 \cdot \mathcal{L}^{-1}\left\{\frac{1}{s^2}\right\} = 7 \cdot t^1 = 7t$ .
We just multiplied by $\textbf{7}$ to match our original term!

STEP 7

Now, let's find the inverse Laplace transform of $\frac{1}{s - 5}$ .
This time, we'll use the formula $\mathcal{L}^{-1}\left\{\frac{1}{s - a}\right\} = e^{at}$ .

STEP 8

In our case, $a = \textbf{5}$ , so $\mathcal{L}^{-1}\left\{\frac{1}{s - 5}\right\} = e^{5t}$ .
Perfect!

STEP 9

Since we took the inverse Laplace transform of each term individually, we can now add them together to get the final answer: $f(t) = \mathcal{L}^{-1}\left\{\frac{7}{s^2}\right\} + \mathcal{L}^{-1}\left\{\frac{1}{s - 5}\right\} = 7t + e^{5t}.$

STEP 10

The inverse Laplace transform of $F(s)$ is $f(t) = 7t + e^{5t}$ .

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. Separate the terms2. Inverse transform of the first term3. Inverse transform of the second term4. Combine the results

STEP 3

We're given F(s)=7s2+1s−5F(s) = \frac{7}{s^2} + \frac{1}{s - 5}F(s)=s27​+s−51​.Because the inverse Laplace transform is *linear*, we can **take the inverse Laplace transform of each term separately**!This makes our lives much easier.

STEP 4

We want to find the inverse Laplace transform of 7s2\frac{7}{s^2}s27​.Remember the formula L−1{n!sn+1}=tn\mathcal{L}^{-1}\left\{\frac{n!}{s^{n+1}}\right\} = t^nL−1{sn+1n!​}=tn.Let's **make our term look like this formula**!

STEP 5

If we set n=1n = 1n=1, we get 1!s1+1=1s2\frac{1!}{s^{1+1}} = \frac{1}{s^2}s1+11!​=s21​.Awesome!

STEP 6

So, L−1{7s2}=7⋅L−1{1s2}=7⋅t1=7t\mathcal{L}^{-1}\left\{\frac{7}{s^2}\right\} = 7 \cdot \mathcal{L}^{-1}\left\{\frac{1}{s^2}\right\} = 7 \cdot t^1 = 7tL−1{s27​}=7⋅L−1{s21​}=7⋅t1=7t.We just **multiplied by** 7\textbf{7}7 **to match our original term**!

STEP 7

Now, let's find the inverse Laplace transform of 1s−5\frac{1}{s - 5}s−51​.This time, we'll **use the formula** L−1{1s−a}=eat\mathcal{L}^{-1}\left\{\frac{1}{s - a}\right\} = e^{at}L−1{s−a1​}=eat.

STEP 8

In our case, a=5a = \textbf{5}a=5, so L−1{1s−5}=e5t\mathcal{L}^{-1}\left\{\frac{1}{s - 5}\right\} = e^{5t}L−1{s−51​}=e5t.Perfect!

STEP 9

STEP 10

The inverse Laplace transform of F(s)F(s)F(s) is f(t)=7t+e5tf(t) = 7t + e^{5t}f(t)=7t+e5t.

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. Separate the terms
2. Inverse transform of the first term
3. Inverse transform of the second term
4. Combine the results

We're given $F(s) = \frac{7}{s^2} + \frac{1}{s - 5}$ .
Because the inverse Laplace transform is linear, we can take the inverse Laplace transform of each term separately!
This makes our lives much easier.

We want to find the inverse Laplace transform of $\frac{7}{s^2}$ .
Remember the formula $\mathcal{L}^{-1}\left\{\frac{n!}{s^{n+1}}\right\} = t^n$ .
Let's make our term look like this formula!

If we set $n = 1$ , we get $\frac{1!}{s^{1+1}} = \frac{1}{s^2}$ .
Awesome!

So, $\mathcal{L}^{-1}\left\{\frac{7}{s^2}\right\} = 7 \cdot \mathcal{L}^{-1}\left\{\frac{1}{s^2}\right\} = 7 \cdot t^1 = 7t$ .
We just multiplied by $\textbf{7}$ to match our original term!

Now, let's find the inverse Laplace transform of $\frac{1}{s - 5}$ .
This time, we'll use the formula $\mathcal{L}^{-1}\left\{\frac{1}{s - a}\right\} = e^{at}$ .

In our case, $a = \textbf{5}$ , so $\mathcal{L}^{-1}\left\{\frac{1}{s - 5}\right\} = e^{5t}$ .
Perfect!

The inverse Laplace transform of $F(s)$ is $f(t) = 7t + e^{5t}$ .