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

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

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

Laplace of e^2t

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 e^2t by definition is

L{e^2t} = $\int_0^\infty$ e^-st e^2t dt

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

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

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

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

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

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

L{e2t} = 1/(s-2) whenever s>2

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

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