Math / Algebra

Question $\mathscr{L}\left\{e^{t+3}\right\}=$
Select one: $e^{3 s} u(t+3)$ $\frac{e^{3}}{s-1}$ $\frac{1}{s+2}$ $\frac{1}{s-3}$

Studdy Solution

STEP 1

1. We are asked to find the Laplace transform of the function $e^{t+3}$ .
2. The Laplace transform of $e^{at}$ is $\frac{1}{s-a}$ .

STEP 2

1. Rewrite the function in a suitable form for applying the Laplace transform.
2. Apply the Laplace transform formula for exponential functions.
3. Simplify the result to match one of the given options.

STEP 3

Rewrite the function $e^{t+3}$ as $e^3 \cdot e^t$ . This separates the constant exponential factor from the variable exponential factor:
$e^{t+3} = e^3 \cdot e^t$

STEP 4

Apply the Laplace transform to the rewritten function. The Laplace transform of $e^t$ is $\frac{1}{s-1}$ , and constants can be factored out:
$\mathscr{L}\{e^{t+3}\} = e^3 \cdot \mathscr{L}\{e^t\}$ $\mathscr{L}\{e^t\} = \frac{1}{s-1}$

STEP 5

Combine the results to find the Laplace transform of the original function:
$\mathscr{L}\{e^{t+3}\} = e^3 \cdot \frac{1}{s-1}$ $\mathscr{L}\{e^{t+3}\} = \frac{e^3}{s-1}$

STEP 6

Compare the simplified result with the given options:
The correct answer is $\frac{e^3}{s-1}$ .
The correct option is:
$\boxed{\frac{e^3}{s-1}}$

Was this helpful?

Question $\mathscr{L}\left\{e^{t+3}\right\}=$
Select one: $e^{3 s} u(t+3)$ $\frac{e^{3}}{s-1}$ $\frac{1}{s+2}$ $\frac{1}{s-3}$

Studdy Solution

STEP 1

1. We are asked to find the Laplace transform of the function $e^{t+3}$ .
2. The Laplace transform of $e^{at}$ is $\frac{1}{s-a}$ .

STEP 2

1. Rewrite the function in a suitable form for applying the Laplace transform.
2. Apply the Laplace transform formula for exponential functions.
3. Simplify the result to match one of the given options.

STEP 3

Rewrite the function $e^{t+3}$ as $e^3 \cdot e^t$ . This separates the constant exponential factor from the variable exponential factor:
$e^{t+3} = e^3 \cdot e^t$

STEP 4

Apply the Laplace transform to the rewritten function. The Laplace transform of $e^t$ is $\frac{1}{s-1}$ , and constants can be factored out:
$\mathscr{L}\{e^{t+3}\} = e^3 \cdot \mathscr{L}\{e^t\}$ $\mathscr{L}\{e^t\} = \frac{1}{s-1}$

STEP 5

Combine the results to find the Laplace transform of the original function:
$\mathscr{L}\{e^{t+3}\} = e^3 \cdot \frac{1}{s-1}$ $\mathscr{L}\{e^{t+3}\} = \frac{e^3}{s-1}$

STEP 6

Compare the simplified result with the given options:
The correct answer is $\frac{e^3}{s-1}$ .
The correct option is:
$\boxed{\frac{e^3}{s-1}}$

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.

QuestionL{et+3}=\mathscr{L}\left\{e^{t+3}\right\}=L{et+3}=Select one: e3su(t+3)e^{3 s} u(t+3)e3su(t+3) e3s−1\frac{e^{3}}{s-1}s−1e3​ 1s+2\frac{1}{s+2}s+21​ 1s−3\frac{1}{s-3}s−31​

Studdy Solution

STEP 1

1. We are asked to find the Laplace transform of the function et+3 e^{t+3} et+3.2. The Laplace transform of eat e^{at} eat is 1s−a \frac{1}{s-a} s−a1​.

STEP 2

1. Rewrite the function in a suitable form for applying the Laplace transform.2. Apply the Laplace transform formula for exponential functions.3. Simplify the result to match one of the given options.

STEP 3

Rewrite the function et+3 e^{t+3} et+3 as e3⋅et e^3 \cdot e^t e3⋅et. This separates the constant exponential factor from the variable exponential factor:et+3=e3⋅et e^{t+3} = e^3 \cdot e^t et+3=e3⋅et

STEP 4

STEP 5

Combine the results to find the Laplace transform of the original function:L{et+3}=e3⋅1s−1 \mathscr{L}\{e^{t+3}\} = e^3 \cdot \frac{1}{s-1} L{et+3}=e3⋅s−11​ L{et+3}=e3s−1 \mathscr{L}\{e^{t+3}\} = \frac{e^3}{s-1} L{et+3}=s−1e3​

STEP 6

Compare the simplified result with the given options:The correct answer is e3s−1\frac{e^3}{s-1}s−1e3​. The correct option is:e3s−1 \boxed{\frac{e^3}{s-1}} s−1e3​​

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.

Question $\mathscr{L}\left\{e^{t+3}\right\}=$
Select one: $e^{3 s} u(t+3)$ $\frac{e^{3}}{s-1}$ $\frac{1}{s+2}$ $\frac{1}{s-3}$

1. We are asked to find the Laplace transform of the function $e^{t+3}$ .
2. The Laplace transform of $e^{at}$ is $\frac{1}{s-a}$ .

1. Rewrite the function in a suitable form for applying the Laplace transform.
2. Apply the Laplace transform formula for exponential functions.
3. Simplify the result to match one of the given options.

Rewrite the function $e^{t+3}$ as $e^3 \cdot e^t$ . This separates the constant exponential factor from the variable exponential factor:
$e^{t+3} = e^3 \cdot e^t$

Combine the results to find the Laplace transform of the original function:
$\mathscr{L}\{e^{t+3}\} = e^3 \cdot \frac{1}{s-1}$ $\mathscr{L}\{e^{t+3}\} = \frac{e^3}{s-1}$

Compare the simplified result with the given options:
The correct answer is $\frac{e^3}{s-1}$ .
The correct option is:
$\boxed{\frac{e^3}{s-1}}$