Find Laplace Transform of e^3t | Laplace of e^-3t

The Laplace transform of e^3t is equal to 1/(s-3) and the Laplace of e^-3t is equal to 1/(s+3). This is because, we know that the Laplace of eat is 1/(s-a).

The Laplace transform formula for the functions e3t and e-3t are given as follows.

• L{e3t} = $\dfrac{1}{s-3}$.
• L{e-3t} = $\dfrac{1}{s+3}$.

Laplace of e3t

We will find the Laplace transform of e3t by definition. The definition says that the Laplace of a function f(t) is given by the integral

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

Thus, the Laplace of e3t by definition will be

L{e3t} = $\int_0^\infty$ e-st e3t dt

= $\int_0^\infty$ e-(s-3)t dt

= $\Big[ \dfrac{e^{-(s-3)t}}{-(s-3)}\Big]_0^\infty$

= limt→∞ $\Big[ \dfrac{e^{-(s-3)t}}{-(s-3)}\Big]$ $-\dfrac{e^{0}}{-(s-3)}$

= 0 + $\dfrac{1}{s-3}$ as we know limt→∞ e-(s-3)t = 0 if s>3.

= $\dfrac{1}{s-3}$.

So the Laplace transform of e3t is equal to 1/(s-3) when s>3, and this is proved by the definition of Laplace transforms.

Laplace of e-3t

Replacing 3 by -3 in the above method of finding the Laplace transform of e3t, we get that the Laplace transform of e-3t is equal to 1/(s+3) if s> -3. That is,

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FAQs

Q1: What is the Laplace transform of e3t?

Answer: The Laplace transform of e3t is equal to 1/(s-3) whenever s>3.