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

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

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

Laplace of e^3t

We will find the Laplace transform of e^3t 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 e^3t by definition will be

L{e^3t} = $\int_0^\infty$ e^-st e^3t dt

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

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

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

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

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

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

L{e3t} = 1/(s-3) whenever s>3

Laplace of e^-3t

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

L{e-3t} = 1/(s+3)

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