# Find Laplace Transform of e^4t | Laplace of e^-4t

The Laplace transform of e^4t is equal to 1/(s-4) and the Laplace of e^-4t is equal to 1/(s+4). This is because, we know that the Laplace of eat is 1/(s-a).

The Laplace transform formulae for the functions e4t and e-4t are given in the table below.

## Laplace of e4t

Let us find the Laplace transform of e4t by definition. The Laplace transform of a function f(t) by definition is given by the following integral formula

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

Thus, the Laplace of e4t by definition will be

L{e4t} = $\int_0^\infty$ e-st e4t dt

= $\int_0^\infty$ e-(s-4)t dt

= $\Big[ \dfrac{e^{-(s-4)t}}{-(s-4)}\Big]_0^\infty$

= limt→∞ $\Big[ \dfrac{e^{-(s-4)t}}{-(s-4)}\Big]$ $-\dfrac{e^{0}}{-(s-4)}$

= 0 + $\dfrac{1}{s-4}$ as we know limt→∞ e-(s-4)t = 0 when s>4.

= $\dfrac{1}{s-4}$.

So the Laplace transform of e4t is equal to 1/(s-4) if s>4, and this is obtained by the definition of Laplace transforms.

## Laplace of e-4t

Replacing 4 by -4 in the above method of finding the Laplace transform of e4t, we get the Laplace transform of e-4t which is equal to 1/(s+4) if s> -4. That is,

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 e2t and e-2t

Laplace transform of e3t and e-3t

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 e4t?

Answer: The Laplace transform of e4t is 1/(s-4) whenever s>4.