The Laplace transform of e^2t is equal to 1/(s-2) and the Laplace of e^-2t is equal to 1/(s+2). More generally, the Laplace of e^{at} is 1/(s-a).

The Laplace formula for the functions e^{2t} and e^{-2t} are given below:

- L{e
^{2t}} = $\dfrac{1}{s-2}$. - L{e
^{-2t}} = $\dfrac{1}{s+2}$.

## Laplace of e^{2t}

By definition, the Laplace of f(t) is given by the integral

L{f(t)} = $\int_0^\infty$ e^{-st} f(t) dt.

So the Laplace of e^{2t} by definition is

L{e^{2t}} = $\int_0^\infty$ e^{-st} e^{2t} dt

= $\int_0^\infty$ e^{-(s-2)t} dt

= $\Big[ \dfrac{e^{-(s-2)t}}{-(s-2)}\Big]_0^\infty$

= lim_{t→∞} $\Big[ \dfrac{e^{-(s-2)t}}{-(s-2)}\Big]$ $-\dfrac{e^{0}}{-(s-2)}$

= 0 + $\dfrac{1}{s-2}$ as the limit lim_{t→∞} e^{-(s-2)t} = 0 when s>2.

= $\dfrac{1}{s-2}$.

So the Laplace transform of e^{2t} is 1/(s-2) provided that s>2, and this is obtained by the definition of Laplace transforms. In other words,

L{e^{2t}} = 1/(s-2) whenever s>2 |

**Remark:** Using the same way, the Laplace transform of e^{-2t} is equal to L{e^{-2t}} = 1/(s+2) whenever s> -2.

**You can read:**

Laplace transform of sint by definition

Laplace transform of sin2t and cos2t

Laplace transform of unit step function

Laplace transform of Dirac delta function

## FAQs

### Q1: What is the Laplace transform of e^{2t}?

Answer: The Laplace transform of e^{2t} is 1/(s-2) whenever s>2.