# Laplace Transform of e^2t | Laplace of e^-2t

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 eat is 1/(s-a).

The Laplace formula for the functions e2t and e-2t are given below:

• L{e2t} = $\dfrac{1}{s-2}$.
• L{e-2t} = $\dfrac{1}{s+2}$.

## Laplace of e2t

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 e2t by definition is

L{e2t} = $\int_0^\infty$ e-st e2t dt

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

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

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

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

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

So the Laplace transform of e2t is 1/(s-2) provided that s>2, and this is obtained by the definition of Laplace transforms. In other words,

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

Laplace transform of sint by definition

Laplace transform of t

Laplace transform of t2

Laplace transform of 2

Laplace transform of et

Laplace transform of 0

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 e2t?

Answer: The Laplace transform of e2t is 1/(s-2) whenever s>2.