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 e^{at} is 1/(s-a).

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

Function f(t) | L{ f(t) } |

e^{4t} | L{e^{4t}} = $\dfrac{1}{s-4}$ |

e^{-4t} | L{e^{-4t}} = $\dfrac{1}{s+4}$ |

## Laplace of e^{4t}

Let us find the Laplace transform of e^{4t} 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 e^{4t} by definition will be

L{e^{4t}} = $\int_0^\infty$ e^{-st} e^{4t} dt

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

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

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

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

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

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

L{e^{4t}} = 1/(s-4) whenever s>4 |

## Laplace of e^{-4t}

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

L{e^{-4t}} = 1/(s+4) |

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## FAQs

### Q1: What is the Laplace transform of e^{4t}?

Answer: The Laplace transform of e^{4t} is 1/(s-4) whenever s>4.