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Math  /  Calculus

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

Studdy Solution

STEP 1

1. We are tasked with finding the Laplace transform of the function et+3 e^{t+3} .
2. The Laplace transform of a function f(t) f(t) is defined as L{f(t)}=∫0∞e−stf(t) dt \mathscr{L}\{f(t)\} = \int_0^{\infty} e^{-st} f(t) \, dt .
3. The Laplace transform of an exponential function eat e^{at} is 1s−a \frac{1}{s-a} .

STEP 2

1. Express et+3 e^{t+3} in a form suitable for applying the Laplace transform.
2. Apply the Laplace transform to the modified function.
3. Simplify the expression to find the correct answer.

STEP 3

Rewrite the function et+3 e^{t+3} using properties of exponents:
et+3=et⋅e3 e^{t+3} = e^t \cdot e^3

STEP 4

Apply the Laplace transform to et⋅e3 e^t \cdot e^3 :
Using the linearity of the Laplace transform, we have:
L{et+3}=L{et⋅e3}=e3⋅L{et} \mathscr{L}\{e^{t+3}\} = \mathscr{L}\{e^t \cdot e^3\} = e^3 \cdot \mathscr{L}\{e^t\}

STEP 5

Now apply the Laplace transform to et e^t :
The Laplace transform of et e^t is 1s−1 \frac{1}{s-1} .
L{et}=1s−1 \mathscr{L}\{e^t\} = \frac{1}{s-1}

STEP 6

Substitute back to find the Laplace transform of the original function:
L{et+3}=e3⋅1s−1=e3s−1 \mathscr{L}\{e^{t+3}\} = e^3 \cdot \frac{1}{s-1} = \frac{e^3}{s-1}
The correct answer is:
e3s−1 \boxed{\frac{e^3}{s-1}}

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