DE-BOOK Exponential Function, e^{at}

Subsection 9.2.2 Exponential Function, $e^{a t}$

The exponential function

e^{a t}

is one of the most important functions in mathematics, especially in the context of differential equations. Let’s explore how this works with a specific example.

Example 4.

Compute

L {e^{7 t}} .

Solution. Solution

We begin by applying the definition of the Laplace transform:

\begin{aligned} L {e^{7 t}} = & \int_{0}^{\infty} e^{- s t} \cdot e^{7 t} d t \\ = & lim_{b \to \infty} \int_{0}^{b} e^{(- s + 7) t} d t \end{aligned}

For this improper integral to converge, the exponent

- s + 7

must be negative, meaning:

- s + 7 < 0 \Rightarrow - s < - 7 \Rightarrow s > 7 .

Thus, we proceed under the assumption that

s > 7 .

\begin{aligned} L {e^{7 t}} = & lim_{b \to \infty} \int_{0}^{b} e^{(- s + 7) t} d t \\ = & lim_{b \to \infty} \frac{1}{7 - s} e^{(- s + 7) t} |_{0}^{b} \\ = & \frac{1}{- s + 7} [lim_{b \to \infty} (e^{(- s + 7) b} - e^{0})] \\ = & \frac{1}{- s + 7} [lim_{b \to \infty} e^{(- s + 7) b} - 1] \\ = & \frac{1}{- s + 7} [0 - 1] (e^{(negative) b} \to 0) \\ = & - \frac{1}{- s + 7} = \frac{1}{s - 7} . for s > 7 \end{aligned}

Thus, the Laplace transform of

e^{7 t}

is:

L {e^{7 t}} = \frac{1}{s - 7}, s > 7.

This result can be generalized for any constant

a,

giving us the Laplace transform of

e^{a t} .

Here are the details.

Common Laplace Transform (Exponential).

$L_{2}$: $L {e^{a t}} = \frac{1}{s - a}, s > a,$

Reading Questions Check-Point Questions

1. True/False. $L {e^{a t}} = \frac{1}{s + a}$ .

True/False.

L {e^{a t}} = \frac{1}{s + a}

True.
False. The correct formula is $\frac{1}{s - a},$ not $\frac{1}{s + a} .$
False.
False. The correct formula is $\frac{1}{s - a},$ not $\frac{1}{s + a} .$

2. What must be true about $s$ for $L {e^{137 t}}$ to exist?

What must be true about $s$ for $L {e^{137 t}}$ to exist?

$s < 137$
No, $s$ must be greater than 137 for the Laplace transform of $e^{137 t}$ to exist.
$s = 137$
No, the Laplace transform does not exist at $s = 137$ because the integral does not converge.
$s > 137$
Correct! The Laplace transform of $e^{137 t}$ exists only when $s > 137 .$
$s < 0$
No, $s$ must be greater than 7, not less than 0, for the Laplace transform to exist.

3. $L {e^{- 3 t}} =$ ?

$L {e^{- 3 t}} =$ ?

$\frac{1}{s + 3}$
Correct! The Laplace transform of $e^{- 3 t}$ is $\frac{1}{s + 3} .$
$\frac{1}{s - 3}$
No, try again.
$\frac{1}{s - 3 t}$
No, the Laplace transform should not contain the variable $t .$
$\frac{3}{s + 3}$
No, double-check the numerator.

4. $L {e^{3 t}} = \frac{1}{s -}$ .

$L {e^{3 t}} = \frac{1}{s -}$

3
Correct! The Laplace transform of $e^{3 t}$ is $\frac{1}{s - 3} .$
1
No, this is incorrect. The exponent in the denominator should match the exponent in $e^{3 t} .$
0
No, this is incorrect. The correct value is $3,$ not $0 .$
-3
No, this is incorrect. The value should be positive $3,$ not negative.

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Subsection 9.2.2 Exponential Function, eat

Example 4.

Common Laplace Transform (Exponential).

Reading Questions Check-Point Questions

1. True/False. L{eat}=1s+a.

2. What must be true about s for L{e137t} to exist?

3. L{e−3t}= ?

4. L{e3t}=1s−X.

Subsection 9.2.2 Exponential Function, $e^{a t}$

1. True/False. $L {e^{a t}} = \frac{1}{s + a}$ .

2. What must be true about $s$ for $L {e^{137 t}}$ to exist?

3. $L {e^{- 3 t}} =$ ?

4. $L {e^{3 t}} = \frac{1}{s -}$ .