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.
Function f(t) | L{ f(t) } |
e4t | L{e4t} = $\dfrac{1}{s-4}$ |
e-4t | L{e-4t} = $\dfrac{1}{s+4}$ |
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.
L{e4t} = 1/(s-4) whenever s>4 |
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,
L{e-4t} = 1/(s+4) |
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FAQs
Q1: What is the Laplace transform of e4t?
Answer: The Laplace transform of e4t is 1/(s-4) whenever s>4.